a
'1 a/a\'-,.1
.I
{
fL
L
L
COURSE NIO. : AMLLL0
L
L ; ENGilNEERilNG MEGHAffiIG$
L
fL
L
L
L
L
tfL
L
L
L
L
L
L
L
L
L
L
u
By
tr- c." eEr 4,E-ga
" i;iE''
tj-
" ''::\
',, 6X
P' C"D U t/riR
valid for point rnasses (particles) but wlid for finite bodiesoisize much greater
'' . -.'
MECHANICS
-
'
.. '
Classical 1t{cchanicsis not
shall cover in this course
than the atomic size moving with speed less than 1/10th of the speed of light' We
2. Axiorns and force systerns - 3' Rigid body dvnamics
i- fio"*"tics of i point and a rigid body
work aird stability
4- Equilibrium 5. tagrange's equations 6. Principle of virtual
1. KINEMATICS'
frame' A' rcfcretce fi'o'me
Kinematics is the study of geometry of mo0ion with respect to a reference
rlith--Euclidean geometry
them
between
is a set of locations in three dimensions having invariant distances
disLance betrveen any
the
which
in
being valid. Reference frames can be attached to any igid bodg (a body
axes ,' yt z can be imbedded in a
tso rrraterial poinis remains the sanre fcr all tinie). ,Rectangular Carcesian
be the unit vectors in the directions
given frame .F rvith its origin and orientation being arbitrary- Let !,j,k
to frame F is defined b}'
nith
respect
oy(r)i+ o,(t)k
:,;:r- The derivarive of a vector A(t) :
l"!r)L+
Ar = dl.,ldtii-;-,y lizcd = i"i*
The rules of derivatives of vectors are the same as those for
(I'l)
dci+ a'k'
qcalars excePt that the order of the cr
F
terrrr should be Preserved'
L1 ICTNEMATICS OT'A POINT
point at lirnes t
],er P and' P' (F.ig.1.1) be the locations of frame r. occupied by a moving
are defined by
f'
frame
to
and t * Ar. The ietocity qr and acceleration q5..rvith respect
gr = jil. *- =Ji:.
* = er ,
4r - 4r= 4r '
-BRr
- - - $flir:.ie
Frw
of the a-\es x' y' i
gr and qp depend on F but are independen! of the choice of origin olr the orientation
for simplicicl'' yields
embedded in the frame F. lniegrating (1.2) rv-r.t. t a1d dropping ( )tr
'[tf'
(1"3)
r(t):r(0) + Jrv1)dt'
s(t):s(o) + !o' {t')dt,
dt) = f(0) +g(O)t + est?lz' (1'4)
For r.niforrn accelerrrtiott e= 9t,eq(1-3). + u(t) = g(0) + srt,
path coordinates'
8
\Are derive expressions for u,g in Carrrsiau, cylindrical polar and
kt%rg
:o
1.1-1 Cartesian soordinates
"))1--:
Then-eft
Let a(t),y(t),2(t) (fig-1.2) be the Cartesian coordiuatesof the moriug point at tirne t'
F;s
Ii-. - ; --:
r(t)=r(t)i+v(l)i+z(tl\, + ,Ur:ii+vi+rk, + s(t)=;i+ii+ *3{
t!"=i,
oc=i, lz=i,
ar=i=i1r,
or=y='.ity, ,:=i=t:.,.
-
1.1.2 Cylindrical polar coordinates
.
the
r,et r(r),{(r),2(r) (Fig.l.3) be the cyliudrical polar coordinates of
(tls1
(1'6)
gstt)
L ,rg,r4*
g+
llV f zor
movingpoiniaitimet.Lete..(t),g1(t),e,(t)betheunitvectors/ffi;N'*-.,^
il".*"#;;;;;;;ti;i;:.:'#Xtiar
,$j$*:1:g:19
i
(z) . ffi'o-.!
;t9$t F-+-../
t'
(d) and axiar
i,Yr
triad rvith */
positionThese
form
a
righi-handed
c.rordinate lines at this
I
ffiq,,F1-
:, :
,.r', \
,L
l ''
\.,
fig. t.3
g(t) = k'
e,(t) = cosdi+sinCi
%(t) = -sin{!*tosfi,
g' (1'7)
$tr = (-sindi+ cosdrd = 4e+, ecl,, = (-aosCi- sindj-)i = -6%, c*lr =
g,
(l'7):
In.tlresequel, (') = ( )1r is ir-nfilied. The position vector r is differentiated rv.r.t. F to obtaiu a using
t, = i = igo * r9r + i9r,* 29.,
(r.8)
+
=)
g = ci-Y b"Sg-, + ( aip *'t 6)ef+Zezt
1
'z
t1.e)
t
1
!
i
l
:i
_;: _- l:
;t
a.--
aadi
,
-'
Note that
:r
ud = t6..':,':,r,: r,., . .o: = i,
,
@d,
and a,
"* = 1tr1"'6)
(1110),.,,-
(r.1i )
ar=2,
-2i6+t6,
;. i
1-1-3 Path cooidinates
Let s(r) (Fig.l.a) be the coordinate along the trajectory of the moving point at
time t- A right-handed triad of rinit veciors g, gn, g! at the location s is defined as:
dr
;-':>;Sg,
* i-, oo i*' nosince g., 94 arcnot constanc tectors in-i;rime + Jf45=*
a:
Ar
s = ;; = Jito a,'
de. &r
Eo
yn
I
t
,
'*r."+
I
95=Qxq'
= w=;%'
tq.Xel=9.3r
gCXq=9.'
(1.12)
QrX9.=9C,
g is the fangent l)ectortothe curve in the direction of increasing s since in the limit as As * 0, the directiorr
of the chord Ar becomes tlre tangent and the magnitude of-@!" becon:es 1' Notc tlrat
'T '=
ftr, -e,) = frri= o :+ ,*';:0"
dfr
.= d*re..
ueclor oi t'he cun't'Tlrerefore the unit vecto.r e.n.in thJ-iirectio,r of der/Jis defined as the princip al normal
q.
and-dirccled ioirards cen!rc of
[here being infinite set of unit vecrors normal to-the curvg ai P. en t
l/p, :urd
ccrtalarc C (f+ = g, #g^Lof the curvl. The mignitude of dglds is delined as the c-uarctare
the radius of crtta:,nnr- of the curve at P. The cross-product oll and lc" ftq* the first ts'o
p
iscalled
equations of (l-f2) yields
( 1.13)
g, ard g, at P is called the
\lte
obtairr g, a using (1'12):
trajector-r'.
the
osctlatitg plo*e. ltis the plane through P which is closesi to
e5 is called thb.6itornral aectora-s ic is rrorrnal to both q and 9,"- The pliure of
da dzd.s
'- dt- dsdt
,-u=ie':i.a
-__!_<
-.
r_.
I*
fU
rU
frJ
U
u
U
IJ
da ds
o=i=ia+6|]fi,
=>
+
(1.1.1)
'a4
(l.lri
4=3St+fre'r
01 = 3..
Ua=u5=$,
ilr:i,
.+=u: izl1r' ac = 0.
a- k directed torvards the centre of curvature. If the trajec0ory is gisen paranretrically as != {t).
d2r ,dr rz dr d2r
*,
dr d.r dr
77=77t7;)
ds - dt ds'
(
i.tci
tJrcn
-EE'
=t#*fitrrffr
:p :t#*P*rr*r
or aT.'
and (1.13) yields
( I .l7i
,
r=ei *f(a)j, and rvith r= e. r,rerobtain pusing(l-17):
4z4d* = f"!,. l&,1&l- (1+ I''l't',
l(drldt) x (d?ildx1ll= lI"l.
For aplcte lrajeclory(Fig-1.5),
dild, - i+ r'ii
(l-17) +
taag = dg f dx = f' ,
,
.rp.=
i
( l. ts)
E +:Stzltl? 111uft
i
dSld" - (- sin lii { cos riyld!lds,
q = cos 1i! f sin {'j,
*trt*
,d9^, rdt
-. | : l;!l
- l;fl, s,.7 (dslds) I Ve,ft')= (-qi!{i+ ce/i}t*) / l;,g
$l = *t-rtn rp!+ cos r;!)
=
i
Oand -ve sign to be used for point A *}:... fi'!
nilrere *r,e sign to be used for point P rvhere i{.,
si#iaity,
0;
'-'r
re'=t
&e--:ie r
p--t
-n
"/\
'G9
For a circrlar trojcctorX of radius fi (Fig.1.6) i:-j;:: r .
p=R, et=?' 9.=-9-:
"=E' :=0,..=* X=0, i:'i=0.s-
-
g =-sq *
.
. i:l
.a
-9n =
r-S
_nire,+ Itd%.
2
:L i.e
.9t e
L
,;.
t,'.':./
({
I
t
.*.j\-
fr:-
_ ..--
Define"rclotiacaclocitygeelp-andor.iu:acxlerutioitgg,.elrlfpointP.relativetopoint/
tfrtr
- t2ele'
=
- irrlr
-: :.:a. ;r, : i.
wzrlF = irelr; 6(re
-d
t',ttr)
s.PlF-4.^lF'
aptlF = ip1r-- fr(urlr - 3e1r)lr=
;;"
of independeat pararrcters
-
sysrem are defined as rhe nunrber
ot a rigid bodlr are
.o.* of
freedom of
orrreedom
a"g,*
(.".:f"l:f:::iT:'"',X$iJ::r#;.'J;;;;"-. ,'"
"t*
"_*;
,-
t."
and t'hree rotational)'
1t1.." translational
1.1.4 Index-Notation
. ^- -r^-^ro rlrp rnn.
.Indevelopmen[oftheoryit,iscont'enierittouseintlexnotoliorr,i.e..todenotetlrecoordinatGZ,V':.tlre
drra?'ag (ai); 9t'91'9s'' (g); resPectit'ely-.
ttre unit t-"ttoo f'!,L by e,'a2'23 ('i);
aslayto.-and
components
indoi such as j i' a t'erm of '
following
the
"tgeu'"it
use
we
For brevity
"o-*ot;ori-'lono"or;on " ""p""tud
anexpressionimpIiestlratthesumof,t,.*;;;;';tobetakeuwitlr1.takingvalues1,2,3.Forexample
:+ 6; * dircr * dizcz * daca = a;
6.; * d;c; = a;
a2' b3+ dsrcl + dszcl'!"!!*ca = tl3 '
O:'*
=
0:'*-d1r
:
or''
d13ca
dP,-h
*
"' :
i-e-, 0r *.'drrcr
{riinde''c:Di,+d''rc&
,i:::'
*igr**y irdcr and.-ci* be repraced by other.co*r-enient
1'2'3 to obtain
The ind& ii". fr". rli-rrrutiiau
oo'u.'"*Jta,t'"oi;td values
i-
ecu4lo"*
1n=e
:-q
:
.5{'
" *::jilJ*.:.": :: ; :H:;r:jl:-T
":fflx"
q1t(t)' .@.\' &p.,
t
of
The rate of ctrange of orientatioa
rn relative to F at time
is governed b1''
q
:--
=
argebraic
The repeated i.dex is
-r
,\t:},{ I::
-
Le
c191(t)*c:u,(t)+caq(r){a}
_ ,.\ 0rsr(r)*0"9(t)+0sss(t).
r - ,r\ r r-- /rl_r-fi-a-(rl= esls(r)
e.,-(r) = Clgr(ar-rc3=?\&/TLrEr-r\-,
salr(r)
=
oler(l)*c2sr(r)*a3s"(r),
erlr(r.) =
orthogonal uuit vectors'
independent since g are nrutually
not
aIe
oi,0f
-compo[ents
nine
The
(6)
'ci
:+ 2o1=$' i'e'' c1 =$' Sinilarly -02=0' cs=o-
er(r)-gr(t)=1. + 2grir(4'e1t-t):o,'
:+ or*6r =0' '
'tJi"ty
q1(t).g{t)=0, + 'ertr(r)'g(t}+e1(t)-Q1r(t}=0'
a3
(c)
c2= -.b3'
= -cr'
i.e-.
tr-'2+J+ 1-- Thu-"
a-b-c-a<and
cros-product
sinrilarrelationshaveSeenobtainedbycyciicchanges
bL : -a?-
The
as a
ioatpu"a"ot' ll',e iewtite (a) witht'e r'trs'expressed
only three compoDents o3, 63, c13te
using (b),(c):
x gt = I X 9t"
9r;r atgz'-ct9g = [619r * crgr *4?9s]
=
g?lF =--439r * bs93 = [0a9r * crgu * a3gj]
fogr ='cr9.1' - 6391
:=
[63gi + crgs + "'!b]
Ltr
i-e
r's'r-
The underlined ternrs are added to ger a common
=g*x€i
gfactot
since
<
x 93 = gi x k'
(r-'20)
Bence
e
.a- i
(grx 9i)'= (ei'et)gl-Gi1er.)e= $'t-tt=2s' =uiei=a,rsr +u"s2+."3qj =-?- '
xitr:g;x
ei.s =9.er + %-e2+" * =_::.:;r,"
e,
t
x 92 ='t'' x g''
g x (l'20):
far q
u by
bv forming
formi'g q'
'o for
we soh'e
!
:
'
<
1-
(121;'
Z
of franre rn relative to frame F'
is defined as the angula'r velocit'y 9L4p
1.2.1 Relatiou Betweeo i1- and i1r
rr
:'
-
f,etAr(tlando;(t)bethecomPonentsofvec[orA(t)retativeto!'andg:ta;(t)q'
=
4(t) = Ar(t)& + AI.Q)D_+ Ae(t)&-= or(tlsr +iz(r)er-+as(t)h:i:::'*
4^=4q="""'=@ig.ir
' ,
:
4r={[-=.o.rr--'{iE'
I
!
i
q
d
-
-r
L
I
t _.---
-9
=-,
rt
: .i
i,
-:..-t__*-;j,:
4r=G-q.)ii.::"*cier:":_-.fi]?,;;,,,6.
r-€'r
*
-
Io particular, if A.'ts
L-.
hin; !:+e-x
"tinffi
we ptove ihe co*srslcncy
-
A'
^t
,.1
(1.22)
:
":;:;
(1'22) yieldg
vector in m' then 4*=Q and
br--s.x
A-
1t:]
r'-*'Consideraniiherodhogonalri8ht-handedunittriad{'
of ql' qmsrqef arel'e'
of the definition
we obtain
6r"d-; ,rr- U=i"g (1'20) and (123)'
.g4d.l=f,tu-d=*
x (e x el)l = |tt*'*1"- te
:
,4.' 4lr
- :4x
5bl
,
of t.he rale.of clrcrge ot
r
. I
-
le directiorrs of 9i in
m. It is a mea-.ute
::x"ffi ':i:l:1::::":':ffi ;::j,i}:.'iit!::::lJ*;:::;'i:ilJ;*
-
Lra
::::lIflll5f}fi ##*;.1*fffi
-t--
I
L:
|;#r*;l;H|;:j:.:j:;#.'ff-'l,1ll,*T.il,HrffH;,;;;Gsr+is+sv-rh€ierore;::,
\='il?*ee1r) x L
x i!=
3P +{llsp
4= 5{3T
x
L-.
=
=
-
L-
t.-
u
t-...
tj
lJ
\te take the tirnb derivative of eq(l'26a)
-rtr
u
tj
L
1-.
erlr:9e1r +elrf I t*t*ex ipelr
( I jzs)
?
-f:E< v tLt
be--written as
Writing !p,,= {L, eqs(126} can
.e.p
g1i V- ietgxtSflrt. yu71;
f i' =
lrel,
+ ft(arr*)rr
i
.grlF-l 9 a;r +sgxtf + 9Q'*(se'xff)+&g!ix1e1"+sef"
IJ
n
&:
d,
-f;*"t + l*gtr-trn +sr xsPlml'
x
= g.rlF 4gx rPa *gl [i,p,tllplm * Qr1'a'
gzlr = 4r1r *t't" x Lp,t{t'r x (grx LP;,l+?4x
l*.
L
L.
x
dt"lt =ezlr *{ii1z +:*lL &l?'
(1.?4)
F
and
4'fl&
gP|erlr
g"t" *d
13-2 ltela.tions betweea
f' and m by I and
franres
w't't'
We denote the position vectors
to P and A be fu:red to nr' Thus
r. respectivel, f''*'l.'ni"i}'o i"i-"u
&0) = &i(4+!p.{(r)'
tr) ;-ield telocitY relaticn:
(l
eq(a) rv'r't' frame F and use '22)
of
derivative
,r*"
,n"
take
\\te
+; x lr'rl'
t&r,.
+
= 8rt" + Lpslr =jl,.,rt Er^1(t-264)
!e1r = 9e1r *t'r x I're * 9P1''''
to ield acceleration relarion:
w'r' t' frame F and use (l'22) f
L_
f.--
---:-
(es1r)1s =
x.'9lsrr
4= Gf"*ql1:
*l'anriag: ritt, * *12 *&ar-
L*
u
I:
*1r
4)+*lI
.
=9s1:*9lz1,
_--r_+
x
-,-.--< *
6\r.
erir+
*.iqr
(eblz)1r' Gil , = 6,,
a1, = li.*g{qr x A= Az*g&lt
(1.260)
(r.27ol
(1.276)
.J
,.
\,
I
-i
:'t..
. \..a:r
_
-
Eg-r.ro
1.3 RIGID BODY KINTIMATICS _ .
FormaterialpointsPand.l{ofrigidbodym(Fig.f.l0),g,,l,a=Oeel--gandeqp(1-26),(l-27)vi:td
(r28a)
3 plr= 3 afp +co x lPa +S9X(Slx len)
! rfr'- ! e1.+ g') X lpR ,
epir = e71lr +a x .AP * s-x (sx AE)'
9-plr = 9,t1n * gx AP.
.i
1.4 KINEMATICS RELATTVE TO TRANSLATING FRAME jr
Let frame m translating w-r.t,. frame F be named as ?. Frame f is siid Lo tzvr.slatc
rv.r-t. P i[displacements of all its poiuts A,B,C,-.. are the same in f' (Fig-f .11)- Eence
F 9s-
B(1-2s6)
ll.
-
:lt
)Fi?
I
L&a
:+ &sr b a constant vector in f-
lsnlt = ilat1r = L-BIF -rr1r:&
Hence the displacements, velocities, accelerations of all ioints of T rv.r.t. F are same and all the line elernents
of T are constant in .F. Therefore g'lr - 0 and eqs(1.21) aud (1-22) yield
gr= (9.. x {p)/,2=Q,
:+
gL=0,,', ()tr=()tr.
(t-2e)
Using (129), the kinematic retatioas (L-26) [cr'a rnoviag point P and a point .d trted to f becorne
!,te1r = u-aJp * u.4a,
gplF = g.a1p.* g,pp.
(1.30)
1.5 ANGULAR VELOCITY OF m F cqr =Q.
A body 2 is said to have plate rnotion relative to f if all its n'raierial points have plane trajectories in
parallel planes. Choose g attached to rn in a direction norrnal to planes of trajectorl', then qj moves parallel
to itself and Slr = 0. A rigid body rotating about a 6xed a:iis and a rigid cylinder rolling rvithout slip dos'n
an inclined plane have plane motion relative to ground- '
The motion of a rigid nut on a fured screrv is not a plane motion. An a-tis q attached to the nut in the
direction of the screrv a-tis does not chaage its orientation wJ.t. ground F, i.e., islr = 0For such cases, we choose & : e. (Fig-1.12). Then et,g, have no cornponent along $" Let d(t): be.the
angle betrveen e1 iuid Er at time t:
er(t) = cos d(t) E, + sin d(t) pr,
e?(t) - - sin d(t)E, + .*O1r1qr,
ft(t) = 83,
*1r =Q!r1r=(-sin0E1+cos0E2jd=0e,
9:1r=(-cos0E, -sindQ)0- -r9.,,
eq(121) + sr=(e.- xett')/2: [9r xirlr*e:xiztr*ss x4'F1/2=[e1 x iez*92*(-der)+esxq]/2,
e
.=0- a
t 1-31)
- - 0€t=)
__oltr equals the rate of rotation of a (any) line g, (f fu) fixed in m relative to a (any) line E1 (a &) fuced in
f. In this special case t.l is the rate of change of a f.ritc iaglc. For general spatial motion of a rigid body t:
of a fi.nite angle. F;nite onqutcr dGplaternestt rt'.o. fr Pra"rirg t<<g:{.!*&
does not equa! r
cnl directr'trr, is not a tectov c3 thc! do sg't <.ae
-g(r<.o<dilg *ro g..^ra\\oft1rca L"o qf olAiH,rn one
o 4Aili.'^ \s nob <aqrutolri-e .
on
trg'l'ta
5r
STANTANEOUS AXIS OF ROTATION OF A RIGID BODY
A rigid body is said to have an r-nslcalcteous cris o! rctatiott at time t if all iis points (its material
pgints or points on its three-dimeusional rigid extension) on such,an a:iis have zero velocity at that in<tant.
.exists
if
It does rol always crisf. For given g-. of iG point A and u at t, the instantaneous a-ris of rotation
lve can find a point B subh thai
9a =9e *u)xAB =Q,'i.e., if utxAB = -9^, i-e.,if u1 =0org^ICf,sincetrx.4B isnogllaltoq
i.e., only if q. .!l = 0. Hence insiantaneous a-xis of rotation Liists onlyl/ u,t .r^, = 0.
1
.JJ
J
J
JJ
J,J
JJ
'j
:J
,.J
.J
j
-
,-I
-I
-.-I
":I
:iJ
1i
,l
,:.
the instaniai-&gs.€ras- of !&"tioo ahvays exists for plane mocion :f " l*d..b:d1 silce ue l-gr for. plane
'but'ii'genei.l
t
er'#9-
The velocity of auy point ,4 can be expresbed in tcrnb of its radial distancc f.A from f (Pig-f -13):
!^: k +o, x IA= kellJAlsing0og = ld(I6)g
. \
f
ls.,{l _ls.al
(1.32)
I=I
IB =lgql
IC =ul-
IA
Fi3't'r3
The magpit,ude of uo o[ a point A equals the product of lol and its radial distance Jrl from the-instaetaag'ousa-xig,of rotation through the instantaneous cent,re f, and i-s directed uornral to /A in lhe sense correspondingto r.r about f. For a rigid body in plane nrotion, if the directions of velocities of its trvo points /. I haviug
'
tle same plane of motion are knorrn at a giveu instant, then the iusLantaneous centre .[ can be located in
that plane at the intersec[ion of the nornrals to velocity veciors at ,{ and B (Fig.l.l3)- The instantaneous
centre of a rigid body in plane translation is at infinity. The instantaneous centres ft of bodies A are shorvn
in Fi9.1.14.
,;\t,
-.1
\s
ug
f,
.\
t
o
vn
f,1
>.d!
1.? ROLLTNG IvITHOUT SLIP
Let the material points P1 and ?2 of bodies t.id* rt.*a,t,tS),Ue i1
The kinematic contact conditioii^iie defined as follorvs
conraci ar time ,..
I
l- stip at, t ir ci:i'" f .si,tr;,
IJ
L-
inpty that slji]- = s"jt."l t
3- in'rpending slip at I if sp:,iii
iil] =
L
l-
uili; at t,!,bu[ this equaliiy does uot hold at !f, for the corresponding points
i
L
L
L
L.
L
L
L.
r
no St"1'
"fil*;l€(t.')
in contact, r-e- !pr1g (tr*) rb
L
L
L
L.
f..
Fg-t.t+
\,
I
I
}.:
tt-tt:'t as in a straight
and B move in a 6xed elliptical etoi
:
slot rvhrich translates lvith
Exarnple 1.1. Pins:{
and
pin" o.r-t. (I) ground re*rence
acceleration ay=_rozr6cosotasshg*rnr.il.Ji.r".Thepositio"ril"r""tr*rdresrotatt=0are-6and
tt
or
alcebraiion
a.d
(a) velociry
rate of change
" ".rrtrlr
from each other and't'he
zero, respectirelv. Finiihe
rshich- the ceirtres of pins ."0""*
at
from the
,"te
reactions
u.ar,
tui
bodv' (c) the
(2) the translating
trl.r.*"*,ng
(1;
.r-..t.
;;;;;y
of this rerarive ,0""o oiooararion
on the pin.,{ ornras.s..'.t
Q
::'"'i'": i11;"::tl'!-;:'*:t'""X,:::1*"':1lt{fff
smoorhsrors
of ihe
to the right of the centre
--
{ri
\=__ ulyy'
L
-1.
t/g
;;;i*a-u.rl)'
be the locatiou'of'ttre cent-re
"1
anrd z1 be the locatior
u1(0) =
for poi*t D' *'e lta*e e1(0) = *e'
?
r of rlre Point D'of
the slot ar'
0. ar = -u)-:rcrGosrd' [[encc:'
1t
Jt
ul -- it =ur(0) + l" "'(')
relatcd
The coordinates r,g are
L_q
d' = -r^r*o siurat'
(t)
:
y = (" - cr)tau0
1,
for z'Y' Relations
Equations (2) arePotved
slot:
(2)
IAE = DE taln0l
v = (i - i1)lau
t
eqs(2):
b''- dilferentiatiug
bctrteen i,g areobtaiued
?*H:u'
i'i'
Equatitxrs (3) ate soh'ed for
:r1= c1(0) * ! ordl =uocosut'
tlre straight
by the equiations of the ellipse and
7+ F=
-t
.:
G)
€ig . tl l.i
((\)
Solution Let (c,V)
jr#"
r--t-
!
Nl
(3)
0'
rred bY diltereutiating
Relat'ious'betsreu i'i are obtait
eqs(3):
1
(4)
I
+
Equations(a)aresolr,edro,;:'ri.FortJregivendataiattlrepositi:n"',,..o..]..
a;t
cm;' :+ ""cosurl = 0'6' + sirl =t0,'l'
l,r, = ?o cosurf, = 0;6ro = 6
:
Eq.tr}+..i1:-80(10x0.8)=-6{0cnr/s.}1=-(80)l(roxos)=-38e0G,cm/s2.
v = ('- 6)/5'
400'
Eqs-(2)
:l
,t
become
('?)
zo +4!2=
t44s
into trre lsr equation yierds 1Br3 Substit,ting y from the 2nd equati,oo
of .A and B, respectively'
,iu
0.2269- These a."o .ooo-"."
"*oo.dinates
* 32 =
I
a'd
0 rvith roots 10'85
For P.oitt A: 9c(2) +
:
eqs(3)- + 10-85t/400 * 8'4v/100
Y-(10'85-6)/5=8'400'
L74-2cm!s,
0' i -- G+ 640h/5' > i =-53g.d cm/s. ! =
i=(i+38400)r/5'
(539.43+10.8$)/400+(1?4.2?*8.4f)/100=0.
eqs(a)+
cnr/s?'
7= -38338 crn/s?; ! = 10?'99
+
t0?'99! cm/ir'
a1 = ii*li! = -3833Si +
cm/s'
1?4'2j
'
'+
g^ = t! rj = -539r{i*
(u)
E
u
L
u
u
IJ
$qaily, Fy,.t*tii:*=o-zzo{canob.!\ yelocrtv yd acceleration'1d,3;.' ,",',,.. .'
' ls =-onz-f i- i-6'tj cui/s, s; = -32654i+ 10125i cm/sz'
i'. il,t
'
;1 = yac&c i = 174-2covec60o : 201'1 cm/s,
39 = ggccec0 = -3-Mcosec60" :
u
u,
u
L
u
L
<'
+-.
em/s.
-4'2 cmls.
lelr =ir €:201-1ecm/s ,
lalr = sp e = -4-29cmls,
gis theunit vector alongdB' rvith sr =
.
ya cosec0'
3r.= fiacrxs*A = 10?39coeec60o = l.2tL7 <arrls2'
tt':::--= 11691 cm/s?'
cwcl = l0l25cosecOb"
3s
sB =
ia co:ecu:
=gB
,;
LAt? = 3'{ 9= 124J9cmfs2'
gtlr =3n9= 11691 gcm/s?-
(b)Thepositionvector-of r{relatinetoIissgwiththeseparatio65='48=(y:-ys)coaec8-
flence
r -- rt:, --ii^lr
124'7-11691 == -1156? cnr/s3'
i = (rir-tia)cosecg = 201.1-(-{2) = 205'3 cm/s' i= - (!a-!3)cosec0 =
: (-)i''
as (Jkt*a
eguirt s
\
w.r.i- ground reference are egual
-- ; c ..:n!lrtinr frar
The ra.tes of separation rv-r.t. translating frame f and
rvith z-atis'
''
(") ftt the tangent io ihe eltipse at (2,t, ye) make-augle e
,.;'pao+y?/t00 =
L.
I :+ 2e1480+zwl fiaa.=0. + { = avfl 7 -'{:f!"
li
Fig'El-lc' The cquation
The FBD (free body diagrarn) of the pin is shorvn in
of nrosion of pin 3 is
031(-383-3Si+ 1'0?99i)
I!: aal1 + Nr(-"c 30. i+sir 30'i) + IE(cos 7!-!e lasiu ?zrli) - 0'0rsj = 0'1089'
j : 0'55r * 03516N: =
+ i: -0.8660rVr +0-3o?4rY2: -3-8338,
+ llr : 3.?65 1q, ir1 - -rso4 N- The negative sigu ia If2 inrplies that
is opposite to that shou'n in Fig-El'1c'
the actual direction of reactr2u nrz
o{:r
screrv of pitch I cm rvith
seconds. one side of rhe rod has' a telescopic
e*"orple 1.2 A rod $,ith a threaded hub rotates on'a righfhand
the time { is in
o - o-uz rad (Fig;El.2), rvhere
crrr/s3' s'irh its tip A at rest at
arm mounted on it which has an outr*ard acceleration of 0-4t
side of the roc] has a riglrt-ltand thread
a d-rstance of 20 cm frorn thc a-xis at t = 0. The o0fter
to the.thread aL Lltc rate of r = 0'2t rad/s
o pitclr of 0.5 cru- A disc rotat€s relalive
t\ llu a
n-ith
fronr the axis of rotation o[ the
and a point B on its a-xi! a0 | = o has a distarrce of 30 cnr
I l0 s'' -a<
rod- Find the velocity aud acceletal-ion of points'4 and I at =
cau be easilv co'rputed'
co*Pon€nts since r,i' i,i,6';'i
;;*;t;;
;" ;; "rtina.i"a
o!r\9e
F'1"it'z
point A:
at'=20+0-2r3l3 cm'
i=o-4tcm/s?, =) i(t) = i(0)+ !o'o*dt=o'zt|cm/s, + r({) = r(0) + lo'olzf
d=0.{r2rad, +. i=0-8lra'd/s' + o=0'8rad/s3'
For)o-ne rotatiron (2r), the rod adences arially by one
pitch (2 cm)- lleuce
i=16.l2r -f7tr cnr, :+ '' i -- ]fu cn'tls' :+ ! = $l* cmls?'
i=0-2(10)3=20cm/s' i=0'4(10) =4cmls2'
+
- air=t0sr'"=.i0'i0.2(ig)3/3=86.6?crn,
i 0-8/r = 0.2546 crn1s?'
O.t(rgt *:t6'i"d/",,ja : OS rad/s?, i = tlr= 2.$,tG cnr/s. =
A=
' '')'-':
i
--; I+t'+:+
il:is,+rCe++ i:g=20q.*G93.aeo+2.546so cm/s,
l.aft fi .1 ;
'
,..:*.-;9t:,=-,ffr':aO]rl*f1z;O+r;)e++ieo =-55{3q'+389'3ea+0'2546scm/s2
.=.;{.='E.}o.4qrjeHp:}
.='
-.i..rp 1o.s€iii**Fg
:
where g., Cr, i ."!9 -,T-,:-!,9.1y" at pornt 4 in Fig'El2'
rr
a{ I
f,,
:
Rdafiretoframe?., the pcitionrectorof r{ is crtg' where
L
L
u
ru
u
ru
u
rU
L
u
u
U
U
u
Li
-'.-lr)
:;-:
u
aL
-
For one rotation 1irf1"1*a'O-r*, point B advznccs re.dially by one pitctr
A0 of the disc, the incredse Ar ia the'radial coordinate of I is given by
A,r = 0's/t0/2r cm. - +-
i -- o'si[z* = 0'5'.i2r-
"*1"'
(0.5'cm)' Eeace for a rdta0ion
+ i = 0'5dul2r cmfs2'
u=02rrad/s, + 6=L2rad/s?,r(t)="(0)+ I"'*n-30+ lr'o''Ly'or=30*o.a2ilt"ftcm.
* aLt- 10 s : r = 30.80 cmr o, : 2 tad/s, i = O-5(21/2n = 0.1592 cm/s, f = 0' (O'2)l2r =O'01592 cm/s3
lflre rralues of i.d,i.! for poinc B are the ".*" ". for point''t- Hence
!.s = igoa .Oe++ ;9* = 0-f5929 +246'4ea+ 2-5469. cnr,/s'
+(2;O+ "O)+ * !e, - -19?1 1*27'194+ 0'25a69, cm/s?'
ae = (i"d2)q'
II:ll i 8,,'r-"\-l dfr,,J
,:.,-i"i; ffi:1:::l;::r;:
plane' A pin is
iu a
and
vettical
loning parabolic paths.(Fig.El-3)
Iy=r.ar'
"
consrrained io follorv the path:,r:(o) ='0-2+,0icosd rvith r€sPect I
\44'
ro the plate. Firrd the velocity atrd accelera[iou of point D and l, *ffi;l
|
the centre P of tlre piu relative to the,ground'at'the'instant 5t1en'n w--:T'r
rv r r en.
e
: l't
: i,':t'-"* =-.i::.
:l' : 1,1 11 :'^
1,'.::', :T::. ::,:'; Js' f
pression is 2 cm and the speed of -.{ is 0.2 nr/s rvhich is decreasingga\ '
rt'^
Li
f\ , r-Z
'Z-.+W
'{ {.%_ff' :
-}
-^.;-^f
attIretateo[10nr/s?.Firrdtherrrilrinrunrsti{Irlessoftlrespring<
; ;;;;i; *" ", m*, o-r kg does follo*' the sive' pat'n zt tlis ,/ \u.5"
t"t=f Fi3.€l-5
Fi3.
'tt?- rtrr+
are stnooth. 'tt?the curved sut)port of the piu arestnooth.
The slot.and thecurvedsut)portof
insranf,. Theslot.and
,:1T,.^----
4,
+ g! = 2'42, :+ y'' = 2'4 m-l '
y{=tamc-1.2, y''=2-4tr-I. P=(l +itzftzlly*l=l'588m' o:50-lgo'
Substituting (2) in (1) yields
.g"= L2t2,
u.e = 0.1280i+ 0-1536j m/s'
For all poiuts of [ranslatiug lranr1 u a.ud g are satne :+
t
(l)
(2)
J
.J
JJ
Jil
Di
o-. - -6.il2li-7.666j m/s?
<J
Ir
llo = !!e, ,., = o
^'
Forcomputing9.p17..6=0,ii'=i=3.rad1s'6=ij=2rad|s2,i=0.J=0and
i= -0-lsind0 rn/s, .+ f :
r=0.2866m, i=-0-l5nr/s, f =-0-879'tm/s2'
+ atd=30o,
!
llplr = i9.- + r/'d +:g, = -0.15e.- + 0'&59Ego.trr/s,
gptr= (i- rC')9" +(?r:f +"d)%*?9" =-3.4599.. -0.32689o m/s3,
=+
rvhere g. = cosdl*siuqij :0-866i+0.5j,
ea - -sinpi+ cosoj = -0-5!+0'866j.
J
-d
of tlre plate' Since lrr
say i. jlv.r-t- ground is the sunr of these accelerations, u'e need to express each of rhem in a common bas'rs.
se = Se + (33/p)e. - -r0e. + 032'5t9g." m/s2.
e, - cosa!*sinai- 0-M02i+0-76S2j, 9. = -sinoi* cosc'j = -0'7682i+0'6402j:
r-l
.f
Lr
//i
2o
ga = sg = O2er nr/s,
JJ
:
and cytindrical components to find tlre acceleration of P s'-r-t- trartslating franre ?
:+-arr=0.5rn;
-J
,..:J
Solution Bearing in nriud the type of data gir.en. rve use path coordinates ro find the acceleration of "1
6 = 0.2 m/3, .i = -10 m/s?,
fi,J
1-l
ii-t-J
d
r=0.2*0.1cos0 m. :+
Er
.tgt
(4)
(5)
"-J
J
f,r
v,j
.J
:J
Substituring (5) in (3) and (4) yietds ltptr= -0-5598it0-6696j nr/s. aptr -2.832i-2'013j mls?'
Hence w-r.t. ground,!.p = ge * !?lr = -0.4318i+ 0.S232j mf s,9.p = ge * g.e;r = -9'253i- 9'679j m/s3'
Tlre FBD of the pin is shown iu Fig.El.3, where F1 , .lV1 , N2 are spring force. normal reaction froin the slot
and normal reaciion fron'r the curved support,. For the given conditiou A'2 =0. Equationof motion of pin:
F - ntap, :+
-Fre.aNr9+ -O.1si=0-l(-9.253i-9.6?9i)
q
(6)
;
_d
J
':
-!i
J
.J
J
"q-'= '^-:--' -
-'-;'(u)'- :oa llr can be ottaincd directly
'
,1er,
r'l:ilr can be obtained irom L iegngonen3s
-
' t_.,
_ o.i1ssr11O5)
by fcrrnt
*, . & =9-z9ls
:o-r(_9.253x0 s66)+(-9,?9xg.5[,
,r,,.u*t"*.,"o I
.. :
]- -,,
by & = F1l6 =O:g4&;p'02=:11'+l/:'
of t1e springis given
Apoinr-",T-1.:1:::Y:1+;,:Xf*::i,-#;: fZJl.
':ff-e"
- -'
+
x:
J#;;;;;;"rr""qtexpres'onsdirectrv
Exanrpre 1.4
;:ffif
e=' R.cs 6s, + Esin o ez
u
soturion (a) It is A*.'ii"i i, o, = -;;""u
'
qqs(r)_(3)
+
c6
I r Gtg-Er-.+ (1)
r = i' =(-Rsinder + Eccd g: + "qe)d'
* $=-al{R2 +"}'/"
+"'l=7iii;+ct}r/2',
6,'
(E?+c"}r/?
1'={gl=
.
- Itsindee)d?'
a"o'de'+.':)i;i!;:/c:sesr
*
(-Esincqr
ir
tr= -
:=11}|?J,ff5J
of e'" e':
(b) We express 1in tetms
*
r = R. t - o'
'.=
+
(3)
"::{:i;;_r,r;, + (_Rc6de, _ nsin431u:11.4i + c?),
r = Rcosdel * Rsin 4gz+
ot
^t,=
"f.' ,'^= "r'
c|ps = R9'- * cae-'
: = cQ'
(4)
+
+-";.t't''
i
:^:"''o'
+ Roea+ cde''
(z;d + r6)e' + ie' = -R62 *
;: ;j(;7.u,f';l'*r;."!"i'n'
-= ;- i"*
(2)
;7
+t')t/''
ce',lol(R" + "')t/t'
e= -ll.o2l(R2+ "')lg. - lft+ +
c2)Lt2'
[]
al
ii
eqs(a)-(6) =+ g= (Ecd+ce')u/(82
.
(c) Given d$ld.s > o,
q - drlds- (-Rsir{g' + Rcosegr +c93)dE/ds
1, + d$lds --.l!ot * "t)'/''
1
!
+
le,l = (E' a "'1't21a61<lsl =
(7)
g=(-Esindg.r*Ecos@g+ce.)/(R;*t'f'.;-
D ac;o,-'siroq,)
- rlsin og.\l(R'+'=)'
i)'t' - (-Rcosee'
I?sia{q'
Xiii''itt'U+
(-.R'cos{e1
<Iqlds=
p=(R?+'"\/n'
rlp=ldgldsl=R/('R?+c2) +
'
g,= P(tgrlds\= 1-cosdgr -sinS92) = -9"
+tt)] 9"'
g - -oq +{u'lp)e^='aga+[r'?t?7141
u = ua'
.
carre-d' trre pitch
1
(8)
(9)
:a
i
l
I
p or trre
the axis is
'*n';,: :1,iil',X$":1tr|,:l lJtrl?;' one roratiou l:.:1.'0"'*'oer)
o:
is called the helix angle
ocis: p =2c.tr.Tlre angle
berween tr," t"ng"ni ono ,. 1= -.inrJ,
cosa = q ' 9d -
The given circurar
Rl(Rz+ t')'/''
=
i
tauo = cl R= pl2rB-
pitcrr, rre*x augre a.d
herix has constant varues of
pri'cipal radius of currtature-
animportantcurvewhiclrisusedirrdrills.scrervdrives'scrervfeeds.etc.i
ru
,l
-.-
u
U.
4o
i
Helix is
:,
REMARK
egit€ oft€u in connected systemilbodies in 3D motion, the information is provided for the relative
rates of rotation s*li, ,;lili of body i relative to body j. about a:ies fixed in body y'. It is preferable to name
the body w.r-t. which the inform-aiion is soughC as F. The body rvhce rotation is giren w.r.t- F is called
body l, the body wlrose roation is given relative to body I is named bod-v 2, and so on. Then
_,etr=Lh1p*€z1r
er?lr = Artr * (cr{r + !4rlF x gqr}
t!4r=9lz1r*911r
tihr = s;qr * ()42* szlr x zstz)
grlr = *-qr * (ri:<s + 94s1r-x_144s)
ltqr:
9s1r * 9f-e1e
;'-r
\
\
4aV -- Clrlr * uzlr -F99slz *9{els
:
t4.rs
g--J
i
llttr
uay= 6rlf +(Oq, * lrrtr x glztr)+(c&f? + grztF x ssr:)+(el.rtg+-?t".x !4ts)
I
\:J
,'
u2
an-.
-.....,...--i
,4= itr+ (c.?tr +91x *tr)*(ri3:+ g*
us = glr * gl:tr *?*42
u41J + (it1e+ eb x €r1r)
(
rli = drr + (estr 4 glr x laetr)+ (dtr, + !{r x !4qr)
= 94r * (Or1r + s.r x !&gr)+t=rqz *(4r + qlr) x <t:l
rrhere F has beeu.dropped for the entities rv.r.t- F for convenience- Stud-r; carcfull!'the uudcrlited {irst
ierms in the cross-products. The firs! teuns irr the cross-products in r-- arc the arrgulirr rclocilies relatiae lo
thc basic fiv;mc F and uot relative to the prcvious body- If tlre bodies are natued as described and the
aogular velocities are added up starting $'ith body l. body 2 rv.r-c. l- body J rv.r.t. 2. etc-. then the first
Lerrn, in the cross-product appearing in the derivatir.e of nth ternr, equals Lhe sutn of all the terms to the
lett of tlre ath term in the or expression- Ttris procedure has been follot'ed iu all. exanrples.
The velocity and acceleration of any given point P is found by successively finding the velocity and
4 rv-r.t. 2. including the retative
acceleration of points on tlre a-xes of rotat,ion o[ 2 s'-r.t. l, 3 rv.r.t.
.2,
velocity, relative acceleration and the coriolis acceteration ternrs if [here is relative moti-on o[ such poirtts
rv.r.t- [he previous. body. If P is either a point of body i or its motion retative to bod-r' i.isokno.vn. then up
ar.d ap are determined in ternrs of a poiut (say Q) of body i. rvhose Sa.qO hai,e already becn deternrined:
l+
lp: le + cl' x QP * !-rti.
9p = 9e+ it x QP + eix (st x V) + 2*- x gpli -f gpl;,
rvith gp1; = jplig, Qpl; = Splie. * (i"rlr/ prle^- The last ternr in the cxpression of gi ind the last, tio
terms in the expression of gp being inchlded only if the point P has rnotion relative to body i.
Exaruple 1.5 (a) A fgba[ governor (Fig.EI.Sa) is used to couirol the speed of a rotating body. It is
rotating aboui a vertical axis at aogular velocity qr and angular aceleralion rir. The displacemeat of the
sleeve controls the angular speed by a connection to a feedbacli device- Find the angular acceleration of bar
.l{B and the veloci[y and acceleration of poiut B of the ball at t = al(p, if the angle 0 made by the arms
with the vertical varies as 0 - 0o*01sinpt. (b) A heticopter turirs in a lrorizontat circle (Fig.Eti-5b). Thc
point, O in its body moves in a circle of radius ,l? at speed u, which is increasing ac the rate of i at 1 = r/4p.
The helicopter blades oscillaCe rvith A = 0o * 01 sinl. Firrd.the velocity aud acceleration of lip B of the
blade relative 0o O at, t = rf 4p. (c) A spacecraft, is rerolving about its _aris OD. having a fixed orientatiou.
It
a
I
.r
-t
!
t
tj
tj
u
tj
l-.
L.
l_-,
l--
t;
t--
--iP-ri.\::i..i,.i.. :,.,
:.1 ,-;.i${*-i:i}I+
:-:r:ir '-
:-'
aL a rate r.r (Fig.El'Sc)' Its solai
l--
u
L.
rl--
ru
IJ
u
u
r_
E
E
u
U
L,
'<:'
-Find'the
:
."""i*.""j"."rlo..oftl,".ol".panel2justbeforeandJus0aftcr0eQuals0o.-,..'...,
b
oi
.i'
?-
-6
)o
ti\)
u
Lord
- -_-A_s
\n
H rLr
-----r^.--;j
*
- '
-'t
5
\-t
F,'g.€I-5
(c)
<- *.be 6xed to body 1- Body I
problerns are the same kinematically' t*ti 'l' t
r physicat
rrrs 3
l/rr-Y''v'q
Solution
sotutron (a)
\o', The
or-ic ii fixed
6va, to
tn body
h.dv l.
r
-r-.-r about
^L^..t atis
rotaies rv'r't' to bodyl
rotates rv.r.t. grouud about the fixed axis !' Bodl'2
L-
lj
]l-l-f--
'b-aoets are opeoing at a programmabte5ale ci'".Tt i1lhg,so.P!:
,t I
.
gr =trh, *
-b=
=-rt
(1)
go =0. go =0, OA=q: ;{B =6(sin0j-cc0b)
rr+{0i+ grr x0il -tik+6'i+-'h.* rii-0i+"'ai +;!'','
The points o, A arcon body t and points -{. I are
(2)
t3)
.::l i
otl body-2' Lrsing eqs('t) rc (3)' $'e successively
cornpute
x (*r x QA-l'
= qt * r/, x OA *=r
gB = a,1 * Q, x .48 + *ir x (91: x A-D)'
!-t=la*.:txQ.A.
lta=!a-f -:zx-E.
(4)
Q,
tt,x -78- -ar6sind!+0bcos0j+06sin0L *
qh x(szxAB) = -u,ibcos1!-(0?+&'2)t'sind
(5)
I
j+O?f cos0 k
(6)
t
= !a=-(o+6siud).i+00(cos0j+sin0!)
as:-12!106cos0*t,(a+6sino)lr+ t-r'a*0silld) -b62sina+&dcosdJj+(602cos0*66sin0]!(7)
(8)
rvhere 0 = 0o* gs sinpt :+ i - fllcospt :+ 6 - -p2l1sinpt
(9)
6=-p'o'l{2
- at / -zrl4p, 0=0s+0r/'/2' i)=patlrt'
!:
q6 is tto! an.obvious oue'
4*;jc.o1lin
u-ufR,b=itlRand,a,eBreptacedbvt'so,aao-,,
i;;";;;;;;;;;;.';;,',"asinpart(a)rvirtr
... , -r---i-r)itto\ttntrtrl
G'-:$:,1
a
u0ld0)(d0ldt)- G
i =.,. *o.:i obtaiue,r
e:.
rvirh
(a)
0
fi':::ffi;;;,,." as i'
=
";i"g =
j rc.l/S
Stud1. the various tenns in eqs(6) and (?). The ternr
l:l
at 0 =0i,
aOlaO =,,to/Aa,'
+: i) =ti1Os'
at0=0J. aila=0, + d.:o'
t
F"i/s''
!
i
Exaruple 1.6 A box rotates about a fixed vertical axis
A
and a yoked artn rotates relative to the box (Fig'81-6)'
Trvo
disk rotates relative to lhe arnr at the tt1o th611'n'
sliders move relaLive to the disk in straight aud circular
have
siors with cosd = 0.8. The Points P and Q on these
relative
speed of 2 cmf s deceasing at the rate of I cm/sz
poiuts
of
acceleratiou
and
velocity
the
to the disk. Eind
P, Q and a Poini r{ of the disk'
a
solution In order of successive rotatiot'ts, starting rvith body roiating
Let
t' be fixedrb-the box l as 5[ewn'
3'
i'
i-'
and
2
l,
rroiio
arm and the disc as
uurr ,i""*oed
we narrre L.e 1;,.,
l!'"tff""rn"
. 6r - ---:- ^a;.-+i^6
"'''.:::: :,:".,.--;:;
rvhich is fi:."d ti thd ground.,The a:<is of totation
k
is
grou[d
to
the
relat'iYe
1
box
of
of
rotation
axis
Th€
OO,l
a-tis of t:tLl1on
of arm 2 rela[ive to box t is along ! which lt i*ud to bJv 1' The
];9i:::.*.
:'*
IL
i
.l
E
-
the arm 2 is 6xed to tnelri'e. et tU" glven
qr = 5k rad/s,
insl'ant
3.- l:
\fle successively compose the angular vetocitie"
Or = -k tad,lsz
@e= tlt-4L= -4i*5k rad/s
(-4$ =-2L-20j- E radls2
4= tix-zt+e{r x({!)--k-2i+5Lx
9s= 9z-3i= -4i-3i+ir..a7"
, a.\ ,6:
,n:
13 I - lei + ll L radls?'
20 j - $ + i + (-4i+ 5 k) x (-3j) =
Qt i,itt + !* ttzx (-3!) : (-2i go = 0, QE= 40! cnr :
Poirrfs o' B are on body 2 rvith co = 0,
:(-4i+ 5$ x aOi= 2fi)j cm/s
Ba = !2o * 9le x OB
as - as *,'t, x OB + gz x (9i x @)
=(_21_20i_k}x40i*(-4i+5k}x200i_-1000!_40jcm/s?.
m:
Points B, A areon body 3 ivith bA=l0k
i\'.-,-
cax BA = ('-si- 3i+ 5LI x t0!= -30i*
,
40j- cur/s
fu xBA=(l3i-rgj+ll!)x lok:-Tt-130!crn/s3
+
150i-250h cru-ls?
!k x (ss x BA't:(-{i-3i+5k} x (-30i+,{0i} = -2001lle = !!a+ sis x BA :-30i+240j on/s
320j -'*':"1:t'
s,r.= eB+ db x BA+!* r(ek x EA\= -1390i-
pointphastelalivemotionrv.t-t- body3=+$,eqipress1rl'1epirrternrsof PointIof
lrodl'3: E-Z:1o!crn
c1/s?
sb x {!$ x En= -340i* 120i -200L
(--l-6i+ 1-2t) - -7'2L-6'4i- 9'6E cnr/s:
2,.,.s, x,els = 2(-4 i - 3j + 5$ x
' llp : sa + ss * EZ! !r1s = -r-6i+ 250j + 3l2k crn/s
- 20'2L cuVsi
x
se : 98+ db x B,P+'e"x (cb x 8.Pl +2er3 {r'1a *8r1s = -1346'{!+r$'6i
poirrt B of bodf' 3:
point e has relative .rotio* *,.r-t- body 3 + rve u*pro"t ,4,gein ter,is of
i' *:-L
BQ=-10!+5kcm. sets=2cn1/.s, ;eF:-1cm/s3'
"=
(;ery/5)9* = - i- 0-s! crn/sz
; = SqpQ = 2i cm/s, oQF = islr& +
cm/sz
30j 30 k cm/s, *t x BQ-= -95i- l?si - 190!
x BQ
=-15i -
il,J
-J
,::J
'::J
,tJ
!.:J
.]I
r.:J
J
I
r
-J
e=-cos0i*sin0!--0'8i+0'6k' gPl3=2e=-l'6i*1'2kcnr/s' lle1z=-e=0'8i-0'6kcm/s?
q.3x BF= 50j*30Lcm/s, Qrx BP = tlbj+ 190k cm/s"
sj
il
-
'-.LJ
lL
U-J
\
r1
.d
, tl$
i*j
l
,.k
1u'
,<l
so = sB + ss x 9Q.+ s4r. - -r3i+ 1-?0i - 30k cnr/s
x (!4 x W * 2,esx !4F * Lets = -856i - 390i - 103'8 k cmfs?'
e(2 = s s + L3 x BQ-+g
of a space vehicle to study tlre tolerance
ions
sir
t'he ftight condit
to.sitnulate
used to
'ont rifrrac i<
is rrsa<l
Example 1.? A centrifuge
speerd r.r1 (Fig-81-?)- The
;rJ;; astronauts. Its main arm I rotates about, a fixed lertical axis at angular
to
t'rz relat'ive to 1' The pilot is strapped
cockpit 2 rota.tes about an axis fixed to tlre arm 1 al, an$llaq sPeed
pilot is moving a btolk' of mass o-8 kg lreld
a chair 3 which ro0ates.ai angular speed'ar3 relative to 2. The
iai,r" R relative to the ctrair in the plane of
in hls hand, such thac iis centre oi mass Q tio"ribes a circle of
'rind the acceleration o[ point P at the instant shorvn'
of the chair. p is a pcfrnt in the pilot's eye.
symrnetry
i.
_
.1A
:
i.:
r--
.,\J
:
-
,,,1-
+-
CI
.l<
Lt,-
J*:.I
-
t--
*fiff
iil{;:r":".$f .
*r".,.ry.
*f,,'fi$,-*Iilt'"tffi
\,r/
i\-\2
:------:-
7/(Y;)
t
;
;
t<e't z
;s <.\fBi-cLj=Ji...---- c
','';m:=l;*:5;T"':if:.1i".#":'f ;i*i
sorurion*.0*l,i,Jl.*1*1ffi
t "il';;;i*
=,"?.#;
*
5-:l:*:,:l::
j{'i:l1':rytl$tlll**:t*:'
o:'*
Ji:;:r:r;;-,
:
-
Tlr" .o.kpit, 2 rotates
relati*e
ff Ifl :,,:ffi
--
;,;r:*
rr" trt" augulat velocities
co"1=0.6,
z L .toug
or
\
::
sino:0'8' g=sio0!*cosoL=0'8i+0'68'
I^- = _cc,itsioo!=;i-0-6i+0-s!
"
-
i;=2..:,1;';--'i,'5'-^'.*,::'3 j=0'5!-i*61:.0^:
-
i=?"*; * ii: =::;':il;'i;i:['ja:'ii.
c. p
t
L
The points o, c
-lE-
sa =9, OG =cj =
=':':'
\\'e -ruccessi*err
ateou body 3'
aad the poi*ts
+ 7 e6 k'1'r"'.
.'-:^,
ib and gi'conrr].*
.
+ g'6k nr
0'6H = 0-8i- 02j
1(0Si+
CP =^'oL+ 6s= -02i:'
rofl = -a0!+sL*/'3
10j rn'
*
Ti=;;;x
s,=soa ,'*6c+-' t*t' xe$
roiizi.i"-
(gb
g, = gc+ rit x CP + ca x
(-'?'ssi+5-32i - {i-16$
a'zi-+isza$'+
(o'032i+
+
5
(-40i+ $
=
03eE */"''
---^*ie*r poinr c ..t :i
- co'l'err
= -r.rnr, -=*r*:-*
l
L-*
L-
tj
t_
t-
The morion of Q'is
3err = n
knorvir relative to
bodv 3' Hence 've
expres5
Tl:::..:a
6(03i+0-6u=0,1i+0'1!-0.]E*,,,,,
*ll"t u* =' *1"' '=,'=
0'8i+0:6b'
: :i:' :
rj
l.-.
se =
j s +
-a0 -* H t-'s*'
=\-=
j 19'2E"ii
*
r
+ (14'4i - 1'6
IJ
Ia,the FBD of the
t-
'
1-.
1- JJ
a
r'-
=.;A,:;#::i:iXX.;:::.i!+,:2!m/s3 :
i:=;::
+Zwx tQlr-t
*
* i *|-b*erj x (s 9l
t--
L
L
L
L
zreou body
0 s L)
L=msa :+
-
i
coF
130i - t'u L}
2'152 $ + (-1's{ I -
)+(r.6i-26.6?i+1ril=
0""0, shorvq !n Fig'Ell?'
Fr is the force
Tlrc equaiion
-- gilot:r'ted ei the
e-te'o='
'-t
*
E1 -0'8s!-0'8(10'20i-'6?'8j--re'zg$
tt{
ro,oi-u"tj-12ji3EY-
-=-..
of nroti'xr'
-- - s3.g2j - 6ai{b N.
f' = 16'01!-oo'or1
*fr'=xy"-
x ' y't
v\a/
- y."rv ,LX
-.
n-
incline'(Fig'EL'P *-""f"d
Exarnple 1.8 A tank'iihvTrilio-*
*r.z r"tfrt'* the given
of 1g km/h *rri"r, r-J"""i"rating-ai t'c-..r.
--_^r-_
i"l*: : *" :i::
ffiJ:fi,."#"i*,"*.r,
odls rvhich is increasing at the rate o[ 1 rad/s2' The
.l.
tO'
- \.g-rt"
- tt+
:"_1ri,:i,',;:::::'Wffi\L"*,?
!
"0".a "f
barrel is rotating, relative to the Lurret'
rvith angular speed of 0'8 r
I
-*;:.":tlfr *T:i:.'::"#{*,i!;ll#:{',,i'*:l',$ffi;':;t'*
the
3m'
Find
The poinu.4 is at the'end-of-the-gun-barcl-of lengttr
H6ZE-*"-",."
to the ground'
velocity and acceleratiou of points r{ and P relative
about an a-tis fixed in the ground'
Solutioa [n order of successive rotatio.s. starting rvith body rotating
6xed to the turret
as bodies l ' 2 ancl 3' I*t i' i' E be
we name the tank body. the turret and the gun l"trel
ortrre
o"*", *,.r.r. ro rrre rurrer a.d k arong trre a-tis orrotation the
plale' is a vector in the inclined plane and
tu6et rv-r-t- the tank tlody. L is lornral to t[e ipcliped
.j
accclerati'ons'
conrpose the tngula' velocities' a*gular
I
vertical plane t'rouglr the gun barrel- lte successi'ely
velocitiesarrdaccelerationsasexplained.irltlre.retnark.above:
;:fif:"i:;[":;i'J;;;;;;;;i.
e='sin15'i+cosl5oj--O-2588i+O-9659j. so =(l8x 103/3600)9=5s=
cro - -0.2x 103/3600)e =;2e = -0'51?6!' t'932!ru/s?
:1. = 0The tank body I translates relative ro the ground + .gt =0.
tt2.= ltr*2h:2k
I
,, ..
. \r_i
'i
t.-t
-
-.
l
;, --<I
l
rn/s2-
,-"a=go* rizz1@-+.:r: x (,*a-: x gg)=-r-llti-,{3:l2j
!t
.-
.1
e2xQB=2*xO-6j: -1'2im/s. ' ib'1 OB-- !x 0'6j=-0'6[ur/s?
szx (* x @l = 2\ x (-l'20 = -24j nr/s!
la = b* eh x OB =-0-09{i+4'8!0i m/d
r._ J
-r.!:: . : -1r: _- PointsBandr{arefixedotrbatrel3rvith e'=sin{ltcosdh=tl'Sj+0'6L' &=3,g-=Zfitd+l'8knr:
*tx BA.= (0.8i+ 2$ x (2-qj + l'8U = =4.'8!- 1-44j + l'92! m/s
1'152 k ur/s?
; - e . ;4 = (0-8i+ 2E) x (-a-8i - l'44i+ l'e2u = ?'tsi - rl'r36j ;
Q"x BA= (1.2i+ r-Oi+ \) x (2-'{!+ l'8!} =0'48i-2'l6i+2'88k rn/s3
m/s
1,92L
'
gr = ea + c! x BA=.4i00it3.-!9i+
aa.: eB+ tib x ItA+ !$ x (srt x B/ill =2242i- l?'63i+ t;t8l' m/s?'
,
poiqt p of camnon ball lras known motion r.r.t. ba,rrel 3+ we cxpress 1gp,gp in ternrs of point I of
uodr 3..
m'/s
.EE-= !ZA=2ij+1.8k, eep=600C =480j+300k m/s, epls = 120E: - 96i+72!
Notice that
-JJ
'rad/s
points O and B aie lixed ou ttre turr€L 2 r*'ith OB = 0.6j nr :
:;:;:;
I
l'294i+4'830j m/s
izt= gr+f k+i4r x2k:8+ k+0x'2k= krad/s?
0-8i* 2 k rad/s
ut = I4+0-8i=
x o'8!= 1.2!+ 1.6j+ Lrad/s?
db = i.z+ 12i+ r^r: x 0-$!' = h+ l'2i+2k
: ff : ?T
a-- /
,.q
i
.<
.i
.<
,i
t2
:,<
L
.<
.-,
:: -{
.-l
;::;il','f-.l;:Uf "= -,,,,, -,,s? i + r r -/"=
6
gp = !A+s.pl3-, gp = -e+ZCax S,"l.f +91n'
,.1 ,...,;.1 . '
{5
-
-
e4
?
.
<
r*
klj-
L.
I
..,
L^
l*.I;:
L-.-,
I
Exarnple 1.9 (") ifr"\gio,b"Isupport o[a gyroscopc rotates abouc a fixed axis AB at thc rat ss;hown
(Fig.Ef .9a). l'he frame m supporting the shaft of the gyroscope rotor rocates rglative to the grnbal at the
rates shown. The gyroscope rotor rchr€s telative to frame m at, the given races rvith 0 = cos-l : O-& Find
at this instant, tlre arrgular vclo<ity and angular acceleration of tlre grrocope ro0or and th€ velocit5r and
acceleration of its point P relative to ttre ground. The kinematically equivalent problems of a spinnint top
and part of a robotic device ate shorn in Figs.El.9b,c. (b) The telescopic arm of a robotic device, shorzr in
Fig.El.9d, is increasing its length at the given rates- Fiud the velocity and acceleration of a 1rcint Q on it
'Jszis'
relative to the cround.
r l?,a1s
a.5rls
l.J
(bdaf,z)
ilt
L3
zu tlt
i
I
t'
- (a) Aff
I
L
L
L
o
'i1=2,1s
'6;^frq tr:€
1,
o'
'7c tl s
\ tspi. r*c)
(loiqs-
l_--'
IJ
]J
,ls'
'l'i=rrtrl
gi-bal
*"ffitI''ffit?fq
tG
e=atY/s, ' o?=
o.5 m
O?= o'5
.--- y
! e=at,/s''
-t-P8:o'i'p..u.e,.9 jf tcI
-Lr
(bJ
tl..
Solution (a) In order of successive rotations, starting with ginrbal srtpport rotating
:'"5u
-a"
o!
(dr
';+tls
'r.l-'-
in the ground, rve name the gimbal support, thc franre ar anrd. thc g-vroscope rotor as bodies 1, 2 aad 3.
Choose *-". i; !, E. attached to body t, s'ith k along its fixed a-ris of rotation. and ! along the ads of
1_
lL
L
u
L.
L;
L,
L_i
l
J
J
5
rl
5
er.7
rotation of body 2 relative to body 1- \lte successively compee the anguiar telocities. angular accelerations,
velocities and accelerations as epl&ed in the renrark above, 1'hc spiu of the rotor is abouG the a,\is
e - sin0j * cosd k = 0.6i + 0-8k-
llr:4k taA./s, (2r: -0-5k rad/s3
93= ttr-0.2i= -0.2i+4k rad/s
,Lz=gt_0:Ii+al1x(-0.2[}__05L.-0;ri*4.kx{:0.2,=-0-li-0.sj-0.5hrad/s"
sb = !h * 20e, = (-0.2i+4u +an{0.6i+ 0-8 k} = -02i+.12i +20& radls
.t
,.-- 3 = 4 - 2 e. + u, x 2O e = (-0-1 i - 0.Sj - 0-58) - 2(0-Oj + O.St) + t -0.2i + !) x 20(05j + 0-SU
- -48.1 i+ t.Z5 - 4-5 k rad/s"-
-0, 9o = 0, OP =0.5e = 0-5(0.6j +0.S!) =0-3j+0-4k rr:
gsxOP: (-0-2i+ rzj1"2051x (0,3i+03!1,=.,:l2i*0.0Sj-0.06k nr/s ,, . ,:.i{i'
Points P and O are on bod-r'3 rvith g
!43x(sb xQZ)=(-0.2i+l2j+20t)x(-1.2i+0.08i-0.06k)=-2,32!-24.012i+14-3&1km/s?
c,t,rxOP = (-48.1i+ 1.2j- 4-5!) x (0.3j+0.4k) = 1.83i+ 19.24j - t.l..t3k rrr/s3
W -- b + !q3 x OP = -L-2!+ 0-08j 0.06\' mA
sp=!b*drsx OP+q3x(r.ls xOP) =-0.49!- 4..772i-Q0.l6k m/s3.
Points P and O are on body 2 also- Eence reworking, rvith g4", gi5 replaced by e42, 1i1s in the above relatlons,
would yield the same values fict g4,,gp- Chect this fact, as an exercise.
(b) Point P, on che rigid e"xtension of arnr 3, has given motiou rv.r-t. bod-v 2 + s'e espress -.;,..gp in tcrnrs
of point O of body 2 with g, = q, !Lo, =0, OP =5$9= 0.5(0.0j +Q.SL) = 0.3i+ O.aE T:
t tr2 x OP = (-0.2i+ 4tr) x (O-3j +0.4!) =
-1.2i+ 0OSj - 0.06! m/s
t*x (ttz x OPI - (-0-2i+ak) x (-1.2i+ 0.08j -0.06L) = -0-32i - 4.Sl2j - 0.016k mr/*
!5 x OP= (-0.1!-O-Sj-O-SL) x (0.3j+0.4tr) = -O.tzi+0.0{j -0.03k m/s!
r6
i
I
i
!
:?
:
t
i
l
rq0-45,= 0'4(0'S j'**OSS
024i+ 0'32!m'/s=
;;;=o.oe=0.6(0.6i+0.8k)=.o3,6i+0j8Em/c....
ii**p= (-0'2!+a!) x (024j+o'azg = -1'-n'i:o'128i'-o.os6L m/s3
032j +026k m/s
e.p = lo + 92 x OP * Wu- =1'2!+
gPl2 * ePP = -2ALi- 4'2s4i + 0'338k */"''
ee = et +,iz x Q2.+ sz x (t!z x QP) * i'rx
eep =
:a
't-.t'f
.(
_{
__
:
Points P and Q are on body 3 with PQ =O'li m
% x fQ= (-0-2!+ 12i +20$ x 0-'l i =
:
.-
2!- 12k m/s;
l|-3x(!b*PQ\=(-0.2i+12j-+208-x-(2j--12E)--5{':ti-0'24i-0'4L
k ru/s?
fu x PQ = (-48.1i+ l-2i- 4.5!) x 0'1!: -0'45j - 0'12
03{L m'/s
94 = 9r +.g5 x PQ = -L.Zi+ 2'32i 4'974: - 0'182k nr/sz'
s4 = sp+. tizs x PQ + s.sx (* * PQ', - -56'81i-
h
m/s?
t
t,rr,
S
-^i
L
D
l'of radius r rolls lr.itlrout 1'o^":. fixed.c.l.lirrde'"i:'::::::.(:;':-".1.T*
r'rv A
Exarrrple
r'xz.'rlPrs 1.1o
" cylinder
-l- Fiud the.radius.of curr:ature
-.
--^t^-^-.r^... ^r'r.-.^-i poirrts -{,O' C,D ottr cvlinder
Find tie..,elocity a,,d acceleratiotr of'material
positiorl'
this
of
oath of points C, D in
-- the
i---5-'
T,
(=oo
r -'-rlb\t t),. h.
l't.to
ia
A.
D
,a
U",V6?&c
(gl
:-.
-_
:
-q
-11t'"q
I'
--
'8
*
.,--_
'-\
(eJ
.
F i3-€l.ro
Solution Let point B of body 2 be in cotrtac[ rvith'po'int r{ of bo
y l'-C-:iroose i, i il plane of motion- Let
u6audig=rlgbet,lrespeedarrdrateofcharrgeofspeedofpoirrto.Ilerrce
gB=0, rrg=9, ur=utk,
ar1 ='r.l1k, O':1
W.e use velocity and acceleratiog relations of
=-r!
OD=ri'
OG=-t!'
Po=R+r-
trvo poitrts on the tigid lrodr- I taki[g tlrc geomcirk
,all points of intsestis the satng 91
the reference pohrt. since the planeof.urotion of
kinematic condition of no slip,;g^ = 9.8, *
csefrc as
_
x (r^r xg) = -r.,?c- fire
Da.=b* th x OA=tioi*a'rhx (-f0= (r,6l --'tllj-=!:a =g'
-
_<<I
(r)
.q
?
:+
W = (,lrJ'
gO.= oO!, + (ub! p.l
s,-
;["ir'3 I lR + r)l [ + rir rj'
(-r':il - -";(-'il
:&e - ga1* Q1,4'OA - "tiOl1,= gcl *6rb x
ofilftlR+ l/'lL
(2)
(3J
ttrat though the velocities of trvo
The acceleration of point z{ is not zero but is independent of .i,,s. Notc
since diflerent material
points in contact are tlre same at er.ery iustant, theL accelerations are no! equal
is nornal to the tangent
o.f
points_ make contact at different instanrts: se * ga. Horvever. the acceleration
I
B it conlacl haoc lhc same
planeof contac[. I{ence, for thc cose o/rolling w-itbout slip'O1 poirfi '4 4,rl!
retocity'of the con[act point A- is zero
componettt of accclemtiot ia tlrc tal.lgenlial plr." oS corlacL The
be
is tlol zcro orL"J rrr 0. Hence ltc accclcnilion of a poitrt'on thc cllitdcrshodd
but its accclcrcliott
=
c?trcssed in tctms of lhc aconctic ccttrc, rcthct lhan lhc.poirIgl
.tt
p = s*
conlact- '
..
_
<{
_
-:
-
'
gI x O D = utpi *t.,r'( *'r'1 =-'l-i'i.
17
_
._
<
.I
_
tj
It
t
LJ
tj
x (-r!) 1c'11i+.--rri' .
ltc =s, * r=.x OL'>riiiJr'rrk
j'?; i = - iwir(R'+ 2r)/il +')l i 1 ? L
+ a, t1 ; i
atlQD
D
"
:s,
,e.D=**?*? I
l...
g. = et+ eirr x AS'-vio!: sa *rirti *
'
=p,r-;;"r/(E+rli+pri+-f";1
(-Il ' 'i(:'l? ="r'.|,;:t;#;";11'
tj
oilco=.=o'oi=eD'(-il='^'fr'(E*2r)/(E*r)' (4)
+ p, =4r(R+r)l(R+2r1,
; -r- iltrt'
;\/Jf.
gn6-= ?i+
'
l- (il itrt
azs =2wlr", sc = vll
r)''+ t" = 2Jir(n+ r)/(E+ 2105)
!\1fr :''l'f*'*"tli(li +
,Z! p" = lb' *rZ*'i-,*
lj
slip:
'hc
For tlre case of impeirding
t"tr amd ap * an''
to
ie"
;""iot
=
A"
;'
.,quoi
oot
of ortact or"
oi=4ulf, *;-;'
u
l-
ntl !'hc-comPate
cornDottcttls of 9;o oad
"t'B' afld
ge
l--
t*
IJ
,. ffiffi::T,:il:,1il:'ii:Trx'nxed
a.
gr = (l,irlr
gO = t':1t j,
uC = @lrjt
IJ
nar surface {rig-er.rib1, rrre'
lrs
.':;,
og ii lhe tatgell'tia'l plate
R = oo and e<is(r)-(5)
+
=4r,. Pq=Lfrr'
r[ radius E1' (Fig-i'l0c]"i${pg- R' =' -R1
1'";E*(rl<.1 *
2. rf cllinder I rous rvirtrour -sllF iss::e a'6xed.':"t!.:; t:']'jl::
tanger* ptarte as-that of body.
as centre of curvature of body
lJ
';
L
l,-.-
2 is on ttr. s.me=iae of tlie
3-
rvithout slip on a fixed
If body 1 wiih a ftat surface rolls
yields q. = ,i*i.,,[p* oe tltisrcsult
v
L
,,,,
e.p=ugi-, so- lu'!r2/(R'. rIi+nrri'
'
L.
c6i*it *-
p = 9-e* rar X ;![ = ur1 ! x d!=
r = oo and eq(3)
For poi,t':D tFig'El'l0d)
]
so =eA+ 6-tr -a'D -uig='i1R='d)!+;rdi
c'r1di'
at J{
1- If cylinder I turns rvitlrout slip .-:r,3 +acorDer
the fixed axis at A, R= 0 and ep(1){5)
g4t=ttgj,
o^ =ti1lt1r- l/Rrli'
cyliud tr 2 (Fig'El'l0d)' then
"nir"ro"irg {3}$
L
I- :
*:;:
L
L
L
L
L
\\drr r 7':u /\
L
{,
L
L
(Fig'El'l0e)' then ttre cyli$der is in-rotation
about
G--1
.<t'
ao=Lr1rj, gi =g, . Po--2'' o,=u";.rrrno=fra--.'.ir1i.irrt'
I;;'i; 'i'd l""t""'io"
Hi.';;ift::!:i::;':':
ii ;'l,l,i, .il1;:l;i;:l.t X';,* *,-r.1. refere'ce
beirrg t'he *alues o[
f...r'," or body 2 *ir*';1 ,u1
, expressions given.ih this exi'.rple rsoutd !q
theangularvelocityandangrrlaraccelerationofbodl-lrelativet.ofratrreofbodl-2.
of a lathc
Tlrerollingrvitlroutslipistheprefertedrrrotiorrofvehiclesqnr,rhegls.Forflatroad.tlre.uS"gl5[1avt:
A piniou' i'''ou"tttl ou rhe carriage
trroiion'
tralslat'ory
has
<hassis
t,he
arrd
general plane n'rotion
maclrine,rollsrvitlroutsliporrafixedracktolrovidetrarrslatiorrcotlrecarriagt..
velocity and angr|lar accelerat-iort
shorvn in Fig'El'lla' !.nd:t]re angular
[rechanisrn
the
For
1.11
Example
istlleinstantane:us
ittst'aul'
of rotation of body'3 at the given
of rvheel2 and the insta[taneous ceatre
)]:::t
franre of body 1?
c
*"a.. of.tl6tqtion of wheel 2 rv'r'G- refereace
Vc ti :
c
or_r-$'-.':r,
\tdt; :il-'- _'/
.o,-f:
,,
;F-
+.
Le
i<- 5
f
r
r-
' ":':t':''
5:r.ffr*;ilffi1*,
(bl
i"
l!(a
^.
rair
2
of { ruenrbers- members l'
a foxr,ar crrain since ir co'sists
o,'.1 Lhe dth member is the fixed
-.^ri^n
orbodies
*;:5:;.:.1
;:::?13:"ffi ::il:"x',tr',X,"1fi olJ;T"I'#;,"T:::*:::::"i:1,ffi
poi',.c (Fis.E1.11b)
ffi$.,fil"l.llT;l"i,lT:f[:ffi;;::;:,:#;;:;";;.,,"*
U.
i-
""u"a
----'
oi'JtU* convenient points of these bodies havi'g .hc
2 and 3 in two ways in i;;L
!4=9-s=9r 4=9e=9t ?:lb
same plaae cf,
r;i =-121' qs="l3k' rirr=tirl' 9{s=o'sB' ds:-rt'
ec=eD+sbx pc-=t uiirx2Dl+!&x *=5*ezx99'
(I)
p)
so * &a x o C - -3ae'
=
* & x D9-,3D9= (sg * ta1 x E D ui E D| + * D8
cr-n, ED = -3L' OC-3j cm'
OD =1Sz-321r/z =4 cm. re= -4i+3j
and cir2,63 are so*'ed
_
. tLc __ s.o
The unknowrs !r3,re,3 are sorved from
x
_,,3 De
trre [rto scarar equations co*espondiug
to eq(l]
fro* tfr" trvo scatar equa[ions corresponding to eq(2]' - -
:+
akx(-30+abk.x(-+i+3j)=o2kx3j
(4)'
--3-si-.\:'?)j=-*'2!
-12-*'ts=0
(3)
i"
-&.,3=-3o1
u'2 :
of eqs(3),(4) is
solution
The
''3: -3 rad/s'
Eq(2}: -12Lx(-3il-4?(_3!)*.i,skx(-4i+3il-(_3)?(-4i+3i)=t^r:kx3i-(_3)2(3i}
(48-3dra+36)i+(36-4t3-Zili=-lut2i-27j
:+
'j : 36- 4r,rs -ziti,=-;-2,' (6)
.,
,.. ,.: \...
(s,:
:_3i,2
: !: L;;r6
t'(6I i'' - sr'''-'.-19'rad/sii' ti'g = 9 *'";
uc (Fig'E1-l lb)'
4
to
Lo gp
vD and
ehc
ar' dre inl'crsection of.eni
'orrrrals
The instantaneous centre / of body 3 is
1:l:::
Eq(l):
:+ [:
Fort.tregivenconfiguration..Iisat.o.oszanrd,*;3.catr]alsobe.olrt,aitredusingl:
#=#=Frr =*
k"sr:3'
W:Tffi:p:l
t'e
Trre sig.s of .u2;u3... d..idid ori
tasis
J and of link 2 a'bout
:-t,=W':Y'labout:,r-rr=3;
of rrre direction ofiotariouif rirrk 3
j to !. i.e.. about _ k. This
rotates alrout r iu the directiotr [rotn
link
3
t,hat
inplies
ofgp
direction
The
0.
k- Thercfore
lb. lleuce link 2 rotates fronr j r'o i. i'e', about yierds the direction of gs as srro*,n in Fig-Er.r
r..r2 and t.r3 are
negative'
of bodl'
c ie.tre Ia1
body I is a[ D, and r]re itrstantaneous
The instanta*eous centre /:r of body 3 rv.r.t.
in-<tamtaneous
gclr alld galr ttu u5 5llontl in Fig'81'l1c' a'he
4 rr-r-t. body l is at E. Hence directions of
D of the tlorutals lo1lclr and-51t'
centre I21 of body 2 w-r.t- body L is at the intirsection
lr'a,rs applies for
acccleratio, of a Poiu[ irt tito dilTereirt
Trre above ptocedure of writiog trre 'elocity a'd
ilr (1):
the mechanismsshorvn in Fig-El-1ld- For mechatristrrs
{7)
uB = ge+{,rzk x AB = (ye *t'trk x o"l)*o:k
x AB --vse"
x)E-:;&:'::*t"}lr1l1'fll
ss=et*ti,zL xAB-luiLf =(sa+tirrL *ql-'?QA)t'i:k
andeq(8)
Equatiorr (f) is1lv;d foru2'tts
rvhereg =0,qo =Qandp, =ooif pathof Bisastraiglittine'
vu*"vY
"t'""j,."
-" f : t
as lugl/8, F all tt &no Llrerr sr8lrrs 'rs
the.u obtained J"
*usnitudes of o3.r.6 are trt1"
for it,ita.The
t. .rh" rnugni;;;"or-..;.
ir'toin"d
contact
contact'
\":lt!.'Y"^lt,:.ili::::.':Hffi::#f
o[
point'of
point
"'.'
tlte
the
is
3
body
bod-v
of
ot
cenLre
, 8,4
itlstanl,aneous centre
the insranrarreous
tha[ rhe
knorving that
ug s{, il*;;
Hff'"'lfi g,.*r=15;, g,ffi^-:*:-*''t1
k17,,,
ections of up g,
;J"?ffi;;.il;,;;;,
.A^a
e-
urrt,l, ,fri
te7
o,s'!L;T,
trj"-
n/t".
't,
For mechanisms in (2) :
i
!
I
I
I
I
lt
I
,l;
r
{eI
AG r
.
\"
,2\--**3{+^ffist
d !
oa1 - X$l-;ed '!#R\a ,,i*.---€,
t'oo, *#Y# t'\1-:
aB = tt.a,* *trrk x AB --ttsg'
t
1
{
I
:
A o..,i€,
.,,ffi:@=.?I]
F'.e.rld
gv
ffi.r-'e1
(2r
9o =faa{ + (ul/p,)g,',J+ 61\x AB -'ie}
:
- uaQ +{u'alp,l*'
a
-(9)
(10)
i
\
i
\
l.
'"1
'
,^-,1
+
.:.
-
i-"
where pe = @ ii.p+Aof-{ is a straight line and P^ =
@ if gatlt ot -,4 is a straight, line. Equation
(9) is
,
solved,for<.,1,us*d"oit;;;-igi,*:;n; irl=iatR"t;';l=psllRartd"!:eirsisnsaseobtainedfrom
the point of coniact'
i";iliins Jt r, *,i, e., u";Lg thau the i;*ot*** centre of bodv 2 is
a fuied ariis
gear S of radius R, rvhich rotates about
Example 1.12 A planetary gear train consists of a sun
P' each of radius r?r'
*"t"."tion a, (fig'BilZ'1' Three planetarv gears
ar angular velocity r, ;;;a;i*
velocity'
g.u Rorritt ir."iau' T[e ri.g Sear rotates 'vith anguiar
mesh wiih the sun gear as rvell as the ring
called the spider' (a)
planets are nrounted on a fraure F
ut2 atdangular acceleration dr2. The beari,gs of the
the acceleration
the planets and the spider' (b) Find
Find the angular velocity anil angular acceleration of
systetlls
transurissiou
irl
used
gear' Planetary gears are
of points A, Q ona planet and point, B on the ring
one or n}\::eats'
jssion has to be altered.
6uev[rv'E"YE by
D accornplished
-' stopping
rlrrs is
an[ereq' This
rchere power ttlgglti:rgq-has
o+. ,zj
fr
:-ffi,'Gr\)w
l'.--'>rR
5.-R
611 4Rr
-_=q{g
ijI l!ri \,3$.*_-1 ,.\
.1\%
tz8,
t
V'
sorrrrio' i;H?;i"raing
t \
ot:"
.",t
(b)
o' tixerl axes i,. ris."-Er-rzb.c. t1e ,o slip coudition implies
r.r r
Rt : --oZ R:.
(1)
t^r3.R3 = urlli.{
:
ttedonlltesatnerotatittg6ody,theneq(l)appliesrritlrtbeangularr.elocit.icsbcirrg
relative to that bodYof sun l'
plauets be rd3 aud ara' The axes of rotatio0s
Lct the angular velocities of the spider a$d the
to
relarive
1'2'4
3' The angular velocities of bodies
ring gear 2 and the planets 4 are nroutrted olr spider
at "l and D f ield:
body 3 are &rt - o3,'.)7- t"3. g.l{ - t'r3' The uo slip cooditions
(2)
'
(<.rs
-tr3)R1 = -(*rl -ul)Rl
(.2 --rX& + 2.R3) = (ara -r'r3).fiL3'.
(:l )
r^ra = [r2(It1 +21?a1'tqRil2Rz
u3=[orBr +ura(I?r +2I?3[ lZ(fu* 8zl'
+
{
+
ci3=[ti1R1 *ri'z(8r +24?)] l2(fu * Rzl'
*a = li,|la1+ 282) - ntBi | 2Rz
nrotion, u, x (gx r,) = -,.'3r' Tlre accelerations
Since all pointsof interes. have the same,plane of
Q.A onbody 4 are obtained in ternrs of poinc 6 of {:
k)=scjr^rrLx 9Q-=,iQ,
l
oiiU CQ = -Rzi, @=:'83i;;$ince
9a = 9ciroLx C.:.{ -'iQA,
l--.)
'are
tL
iu
lr
-- ff
J,J
i)
r-rl- €
i
(')
I
so=9'
i:
(6)
* lqdtu +2k) - "r:.RJ? I i' r'r
ori body 2 rvith @-={fu+ 2n?)i-- Hence
-:,..., sB=qr*orgE*U--"id=-tfi(frr+2fu1i-rilfaizR:)i',
th"t;ir,*-g,,. buc t5e components of q, es'aJ9ng the tangent direction i
I
-
e^= -ri2(r?1 + zRlrL-tW
Tbc pcrintso and B
of points
g. and G'are poi,*t'sof:bod'v'3"9c is otitairl$J+;,'
wit'h 'W-=lL:aErI!'
k=go+u3k xoC-"'ioc,
so=l"fiRz-tr(& + Eg)li- t'{82 +..'3(EI + n?Ui'
Eq{3) t tO)
'+
(.1 )
be, to sacisfy the conditi'on of r'olting rvithout' slip'
L""t"
""-.,.;fU;i"hould
(8)
at the point of contacr are
.
.':'(-
particvlar ca-.e..:'ThF&illo*iqg results follorv from dre general relations (3).
1. Ifarl =e, i.e.,thes.unisetationar5r,then ur3 -ttz(Rt*2P'z\/2(fu*Ri, c'ta'=u2(P'1*2Rzll2Rz'
2.. lf or2 = A, i.e.; the ring is stal,ionary, then &rs = ur& | 2(h * Rz), ul = -o)1R1 / 2Rz'
3. r43 = Q, i.e., the spider is slationaty if u:(Er * 2R2) = -utBr'
4. .,{ : 0, i.e-, the planets do not rotate but have circular lntslation it uz(fu *2Rz) = r'r181-
jr
pxa'nple 1.13 Qrick rc,hrrr. mcclrcnisn is conr-
monly used in meial cutting'to speed up the idling
part of the return stroke s'hen no cutting is done'
Sudr a mechanism is shorvn in Fig'81'13a- The pin
P is fixed to the wheel 1. At the instant rvlren
f = cc-l(0:8), find (a) tlre angular relocity arld atrgular acceleration of link -48, (b) the velocity arld
acceleration of ram D.
Solution The cenire P cf the pin is comnron to the
slider and the-rvheel. The extretue posit'ions o[ i'ire
link 2 are shorvn in Fig-81-13b. Tlre forrtard stloke
of the tool occurs during rotation of s'heel I througlr
a larger angle d1 cornpared to angle 61 trarerscd b-r' it
during the return siroke of the tool- Ileuce n 3[qrrer
cutting stroke and a rapid returu slroke is oblained'
cosg:0-8, sin0 = 0.6,
--F
-l
I
'-----t
L
'
:)- Z
,"-'1."
fi'
.'0Az
:
tb)
cuttingtool
AO
(dr',
d-
'//
fi:
(OJ A
f;g..zt ,j
rr)
.I
!
AO = ti inr
3 ?-: 1CC.* *
AR-- 3(.c- -L
ED-- tct+
oP =}0(cos0!4'sindi) :8i*6j cnr, .4P = Ao*oP= l8j+oP=8i+24j cm
e= AP llApl=(8i+24i)/(8? + 24')'t" = 0.3162i+0'948?j'
ql=2L, lilr=g, 91=-u4k, .rz=t12k, go:9,r=9, 9o=9t=9.'
Ttrepa.thof Prelativetobody2isthestraight lirre.4B. I{ence 9p11 =ue, Lrp=it9' 1L"q11fnosrls
ttre
u)2,o ateobtained by rvritingg.p ia ts'o rval's and equatiag correspondinS !,iconrponenG' Siroilarl-r-'
,nfro;* &2,i arcobtaired.by .rriting gp iu t5,o ways ind equating correpouding !, j conrpone{rts- All
pirrr" nrotion. 1{e consider poiuts in the sanre plane of tuotion + g x (- x f} = -n2!'
bodies t
"r"
q*xAP+t,p,r, :+ 2L.x(8i+6i) =,r:lix(8i+2aj)+o(0.3162i+O-9{S7i)
tp:b*u1xOP=ee*
j: . l6=8ta?+0'948?u
Ql
:+ !:
tl)
-12=-24u2*0-3162u
The solutiou oteqs(1), (2) is:
u = 11.38 crn/s- r.t = 0.650 rad/s-
9.p=go+ v;xOP --iOp =9..t * gz* AE---iAE-+2gz x grt:* Lep +
j)
24i) + 2(0.65 !) x 11.38(0.3163 i+0.9487i) + n(0-3162 i+ 0'e{87i)
(s
-2?(8 r+ 6i) = r.4 ! x l+24 - 0.65'(s i+
({}
-:+ i:
-24=&it-10.14+4-6?9+0-9{87t'
-32=-2*)z-3.38-r4.O4+0.3162' (3) j:
i: -\2-20 ctn/s3, t[ = 0-3150 rad/s?'
The solution of eqs(3), (4) is:
of rype
Oace a4, &3 are known,'gr, qD are delernriued by the procedure explairred for the mechauisms
(l) in Exl.Ll, since t,he rnechauism ABD is of this type. It is left as att exercise for the'student-
Example 1.14 A vane pump, shown in Fig.Dl.14, is rotaiing at augular velocity
ar and angular acceleration 6 about a fixed vertical axis. A material point P of
watei flows ou0 along a vane with speed o[ u arrd rate of increase of speed of u
relatire to the vane. Find the velocity and acceleration of P relative to ground.
Neglect the thickness of the \ranes- The problerns of fluid (lorv over the curved
rJ
,I
:-,.J
J
,;-J
--J
,,1
.::J
-:J
-:J
--J
.-...l
.:J
blades of turbines and centrifugal pumps are sirnilar. .
7t
rT
F.'r. e i- lt
J
;-J
':j
J
F
!
we have
Solution With th'e*&rice oferes as shorvn iu Fig'E1'14'
'g-riarli'
-l
Lt=6t, so =0, se =9-, . QL= r(1 -cos{)i4(6+rsinf)j'
go t cosdi-sindj,
9, = sia{!* "gud!'
Thepathof p.relativetothevane'1is acitcleof
radiusr. Iletrce !p11 =o9'
9e1t = i;'+(a2
gr=9o+qhxOP*Y-e,:.,
ap:9o*r.rr x oP + qr x (grr x QA f 2t'r1 x 9r1r *9r1r
:+ vp: f*u(b+ rsiud) * usindli+ ["1r - cos{) * u cos{Ji
cosd) - 2t'ta cos $+ usin'd + Q? l')cos 9l!
ap = [<ir(t, + rsine) :c]ttisin { + ir cos d - @' /') sin eJ]
+ [tir(l - :.= d) - '"(bt r siu d) + 2r"o
:
-
-"
-
t__
t_
L-
L
L
L-.
ti
LJ
{i
E.-,
Lj e
U
L
t-.
L
L
i
G
-)
.J.
, .l
I
J
-zL
lt)g''
ilf
,,ffi
',.ri
tr^.-^h^F.-...=;AxIoMSANDFoRcESYSTEMS.,.,,,.1W
2r
or
vecror
..*", q.I :",,.1r, 1: no'o,, **iln,3iw1
J;;ffi)ffi
l?"TI;
a,'d
*,,**" .o.&ce or line-' The distributed over voluc is called fota'
f
I
):.tit rtc,forces
f::'il,:,T**:Y5""ffifii**,?S'::,*".Tt[i1t;i.tr l*iigffi*
-g;',
tr
force
over vorumer
;"L
L
L
L.
L.
Mt=1'ptx F = AP-xF:(.ttP)Fsin0n =
L
L
L
raw (axiom) of
fonts- (
"i,i^i*'i^;;;.#;;;;;;; io"tpoi*^J:'"rl' :JI*':o )
A,forces arc sorctttei! tr ?" -:*y:\poro,rctosram
j- k
;i
r: l'
rv
Fdn= lr"
lr" iu F,l
,
!
wtereF=F'i*&i+r.'L,LPt:i'i+rvi+r'k*dqistheunitvectornormaltotheplaneofEand''1'
G tf,e relati*e
ottained from
= AQ x Lwhere:{Q
givea by the cross-ptoduct. Note that !{1 ";;
f'e
eq x f' =
oq 4,'sinF/Q x L= {aF"+
position vector of aay c<invenient point Q "" ,i. ft"" "t Y""
r
fr:i":;,,{;..n:y:;,*,o..,,;,et!lite"**unitvectore..lo1git(Fig.2.2)f
is defiired us the "o*porrJ-t orit= nrornent
L
L
L
/r4e about cnypoint Aon L alongg
,! : I
1". O:
A{"=tr'lo's=APxF-e=lF' et! fe"ll'
F F;e
i
erk' M. is s'ell-defined since for any other point
M., (8alld'
lzl " r'e.l BAx F'e* rlf" -(fig'2'3)'
Its moment
Conple is a se[ of fiorces f aad -f
momenl
the
Q of the couple'
called
is
and
about every point is the same
rvlrere e = er!+ erj *
B onL,MB:9=
e.2)
itr*" r7^'=' Af x F + AQx t-E) : 1ap - {4* E: 9t * L=
anaTQP)Fsia0 n is independent of /'
BPxf'9=
-9
el+
1
-6
-F
L
U
E3
a
" = Q'
rvhose directiol is sat ac
\4rrench is a system of force F and a couple C
(Q.p)F"in o
I
I
9
i
l"r
L.
L.
r-
il
ii
A
F'g.2'3
|
2.2 MASS
-49
Massm(B)ofabodyBistaliena.xiomaticallyasapositiverealscalar,whichremairrsirrtariantrvitlr
the contin$X5tSry'$t*t{ the-mass
By
its parts &'
rinre and equals thesumof the masses nr(P;) of all
'ts
I kg/rn agp"-"|h.g oiffiLttiliution of mass
kg/rn2 or
d ersityat a point exisis and talren ,. p kg/*t, o
nrass
c of. a body (Fig'2'4) is defined by
over a volume, an area or a line- The cenlre of
L
L
L
L
l=Y#'* W' or ffil
f r- dtn
l* Lrd*=frb, + f.=ffii,
tz'er
of mass
drn equals pdt' odA' lds for distribution
rvhere P is a typical point in the nrass element drn and
and curve arc deEhed by
. over volume, surface and line, rcspectively. The ccatroids C' of tiolt"t'"' sudace
"
s
Jira"
[*dA
- _TT,
!C:
[*d"
lc. = -TT.
{d'
composite bodies consisting orsevet.at
respectively. c coincides with c- for uniform bodies-'For
the ith part, having mass mi aud centre of mass G; (2'3) Vields
r,c =
[t-
(2-4)
rur]l [t*1,
i.e-, tu!4 =
I*'tr'
patts with
(2.5)
tiY+o o5
being-negative for.a cuiout llar[-
Aro=fai
I
!iIE
{c't"
*
g
rig-zl.tr
rt62\J
i?. -**t$i t, t+{i'i?:.
'
:{,?:,*:1r"* ,,a *O,T* ipt-ol lwncrrtzrm flap about point ri w-ri. frame f (Fig.25) are defined by
ir= l^*1Fdm, ' "'1.' ,:,'L^ir= I^*^-gpaly'*,
\p=mlbltr': !*u".1",
dvdrf,
velr
since 4, =GT*1r,= ([1" = mrcrr.
Errler's Axioms: There exists a frarrrc f such thaf for any sgskm
(2.7)
m
.A
trg,r'5
(2-8t.l)
(2.8c)
4r=L'
(2.6)
whete O is a point fixed in .I and F is the sum of all Lb.e cxterzol forces from the surroundings on the s1'slenr
arrd M-o is the sum of the moments about O of at cxlcr:r.al lcads frorn lhe suroundings on the system- The
frame f in which Euler's a:cioms are valid is called an dnertiol ftume.
It can be proved that a franre ? which translites with unifonn velocity relative to an inerLial frame /
is itself an inertial frame (Galilean principle of relativi[y).
The 6 scalar equations correspondirrg to eqs(2.8) are not suficiert lor .determining tlle orotion of a
gcaetzJ syslen. Bolvever, these 6 scatar equations are ,tcccssary ard, suficical for tomplete determiuation
of the motion of a rigid Dod3tsince it has precisely 6 degrees of freedomequatiou of nrotiou of centre of urass C of any systenr is obtained using (2.7), in (2'8a):
,
]he
F - mgs11: E*rq.rt, ,
Fu'- tnag,- : mic,
F, : tnac, = rni6,
p. = mo6.- m(ic
i.e-,
- ""OL), F4 -'ma6o = m(2icic + rcdc),
fi,=niag.=m5c.,$-lIl46o:#Llp",}E*=,noc0=0.(2.l0)
sYsTEMs
(2-g)
F, - ma6, - '?2!cl
F. = ma6, = rrri6:,
Fis.2.6
2-4 EQITTVALENT FoRcE
lfwo force systems are said to be cquiualent if they have sanre total force s-uin E and sanre total nroment.
sum M^ about onc point..A. A sin'rpler force system equivalent to a given forcesyst'enr is callcd its resullatl
1- f,{omeni sum .itfa of trvo equivalent force systems about cay noint B is the same.
Pt@f. Coasider a force systern consisting of n discretc forces sittr the ith fotce I actiug at point i. anrd m
. couples with momenLsfli,i = 1,..-,m (fig.2-6).
:
B-: -fu.N.qj
e, t(EA+,q;) *4 +I Qi = U*Lf-+(f .1r x&+t e, ) = B Ax L * t{-a
W : f si* grlD
=
.i
i
i
i
i
i
i
Heace the moment sum i4, for eguivalent, force systems is same siuce F and M lor them are cg-ual^ slstenr can be replaced
I{ence. for the.purpose of finding the mornent sum and fiorce sunr. * given fotce
by asimpler equivalent forceiystem, i-e., its resultant.
2- Two-equivalenttforce systenrs cause the same motion of a sizgle n'gid 6odysince it is completely decernrincd
by the total force sum { aud the total momen[ sum Mo, rvhich are'ttre sarne lor the lrvo s.YsterrsTwo equivalent,force systems; in geueral, cause diferent motiorr of a defotinable's1'steur,
' 4- Rt*Itaat (cguiralcn!) o! o gioen forc,e syslcm ol a giact poinl A consists of a force {4 and an associatgd
couple with moment ep^ @ig.2-?), rvhich are obtained from the tirc conditions of equivalence: i
(2.rtc)':-l-i, .
. &=De
Q.a^=
Ma=!e;x4+Eet
(2-u0)
I
5- frc simplcs{ rcsultort {simptest, caaipgkn!).of a given force system is a sirenctr.
- Ptwf;, Let the.resultant at'pbint ,,t be Fp, en., The *iuttant at, point B i" fa, en, with
(c)
MB=C-nB;en" +BAxER=(qR^h*(Cn^)r *BAxfu.,
,
Flcte ( )U, ( ).r- are the"o*p,io**tf 1 1rtiifei and normat to fs. tt b possibte to choose an appropriate
point A suCb.t-ha,t the last trvo tcrrns
r.hs. of eq(a) cancel each other since !! x fp is .L FB- For
'-,.: C.Bire
.-E.=J .\
ol
= ,,
.--^ '-
--t
*- l.rh '^
.::?:
rt F.,,kllt
-i ,al1-"r\.^q
--t-^ .q'
:R->
C;
'CotsiJ
.i'.
B
!"
'-d ,Fig:1.7'
'a*
.J
{ij
J
,rJ
,-J
.-l
;J
,"'J
..J
*.J
cJ
G' I
GJ
cI
GJ
GJ
cJ
*J
L'-I
O'.-|
J
-J
'-J
-l-I
i-t
,--,J
-J
J
J
-
L':-
rri.
t-
l
L.
IJ
f
LJ
u
TJ
tJ
L
L
l_
IJ
r-
| '''-''-"
'
j.o
il::-ff :fL:ff ,1L;;"#il;;';,6;I_:.::::':::'f:"-*1T*T"1,81ii,
;::$;:;:"::.::
ffi"Y,:fl',-,i:*':::?:::;:ffi
3T:$i',"#.il:H"*
k iompoaent aod &
hi
3'' YRo
auu'
:51;;;laR. aLe=.0 "J* e*"
Plan'e tr'e'r
normal tD thls ],ff
system'
momeots along
-"",3ITIJ:L',Ji:1;
its simplest
only and components' Hence
L
l,,
L
L
L.
null
n ln
cel a.sinrle couple/ nutl
Arnl.single
" rn e line (say
'"rv
z-axis) and couples.
s*' = 0 since epo has onrv
Q
i
::.tt3:-:j::::1,
:T:ffili,;"il*11,:;:::1,I"IT"il;il;;;,rl',r:':"9'::il,lll:*:l';
*' i' singre
l """Jlt"nt
Ii":ti: -:;ffi ;ffi;il':il*i ^t-].,
torcet a
single force/
-"*::['fj.::,' ffl"J
'oir"ni
i::
iTil']']ilil;xH;;;;:;;;a'
ffi";:"rr'*[
couple/ null
!
has
single
sinsre
is a
sYstem'
"''J;rT"T.:"f:::i:;n
cons'stins o'rv or rorces
The resurtant {p ofsuch a force
:+
s],strm ,*"i**"*
g'
at ri' i = " "' n such that & #
"::**
Tu-a poini 1;; calted lhe ccttrc of parcllcl forces'
1
irr*ogt
E^=r^l{I&}e' * t(trr}rn-(E. lxe=g'
a
i
t" = (Isr') /(trt)'
'v
g-
.:,:.
ve'
(2.121
'
ir
Fordistributedforcethesummati.onin(2.i2)isrePlacedwitlrinrcgrat,i,on.ThecentreofparallelzaiJonn
rtith thc centre of mass
graeitatiotto,f:*
g dtn o'la body, ..n"a "r*io'os g*n;tg,c, "oir..iii
lc = LR=(T**) l(te) = f;rar t looo = l
*ta*t l
"ffj*
taT :
--4ft$#
curve e-quals tltt-&"brtic
parallel force norural to a plane
The simples ! rcstllta*lor " aoa.iuo,.a
of thc toodinss*focelif area t' 0) aud **'ii*r-ou-iir= ""nt*;arf
il;; 6;''{
dF=fds-dA, E*=
1
IJ
IJ
l-
(G.)ll,
rrr
. , . -..1,--t 2-., ingte fotrc/ g sirglc couplc/ a",ll syslem.
-- -'{QR^'-ER:u-r8c'r"'srrrr'r'sc'
{.*^'*
.Fa=
rr(Cn^)U +oij.i,
;;A^il
0,C8^=9ca^ = ernifir .Fa
=.01i:1.,
= e,
a nul syst€m
;'."noti"yrt
1:l_1I
::::,:'::ii:T:':'6:
single coupte
en^ if & = g' c.
!'r b'
4 *t g,
"' -a "*tl---t-t
;, :A;r: d""E g^ ir 4-R
A ,-].--r- f^.aa throu(h the point of concurtence'
v' En,
rLurtD'er'rb
isisting of
Llrt \sA
ls B vrreusu
resultiurE k
tlg iesultaot
B 9,t9
point b
uch a p"int
such
"-T:&'::l-
.F
.*
For
a uniform and a triang't*
example, d,he simplesc.resul[ants of
1l ff;}t:{1"
Ol5|{t{?ffi
!:' _= laen=an'
t"=:-T-:-F
atca J
1 [-ft
'-""^L:-t::::t:k:t;
thc
rorce normar co a prane areaequars
--!
arsebraic'ootume
;a*
?;: *;;X,':*::ff:'ffi,i"*i'o**er
toffirF:f--{t
ack thrc.,gh thc cctttrvid'ltu'" .,*ti"X'
of tie pttssate spoce(if volume * 0) and
d.F=pd*,=dv; El: !:,r=
lv re
{-
fG.o
lrav'=t't'. tL;+Wf'
+ o'spsD a'F'
=
I lzdn + ( E =,&
=.rn *( )!-
glone. creo
NEWTON'S THIRD LA\jl/ OF MOTION
t
t
'''XJ:,::;;#;;;;;t""i''qarts)travee::1"lTi'1"^:::1#:*':*:::';
;'al momel.t rn !{6'" bout o'
;::""H;;; *,,.,r{r,
::'ffil$,ffi :,ff il-'t
then
*. fr*;;i:fiTfi",fJ
aharrt o
O (Fiq2-11)'
Fis,'..}''ihen
ffi;n. --.-- t., - +ou!
f: ::;:':L"j f ?, ;::'J.:i:J;;
a
su
2
and
(
l+ Mtp=-Miz1 o'1.
Mir|= -Mozt,
rnomeat sum tfb, about
of ;;1 and 82 be aforce ium [!'-ind
* surrounorngs
urr. to
u due
pnof,. t*L exterual load
(24) to Ar
axionu ('z4'\
r .-^
,t_ annlv Euler's axio'rs
",rr.o,rnutnr.
^ ^_ ,,- G" furers
;Yff:"Tr:Il3.:::":"';XIH::.T:H'l4;;&;;il
I
.9
to BtU Bz:'
alone, to Be alone and
3,,!-
f,s
i
t
Y
c*i
t
I
!
I
a
--/- \e-f;,
r;_
$,
L
-.-> !-zl
H^ =l
<l;
= MoLz+ &*
i{or='Mcr1* W4
Lrtr:&2+Ei'
Lsr:&r*&'
P^r,:.Qt-r.{-12l
-ilt,u",rr,
t;
.-r
.\%l
Ftg. z-t[
f\
iI
Y{'or+aitGl
1;'
=F1u1 =F13.
" T':'lT:::'
ii;
_
J,,
A Frg,2-lZ
(2.13)
Jr,
:.;:
,__
,, -rr
(2-14e)
(2-lac)
(2.r4d)
-
;-1\r
.._
:1q
v
. <
{2-r5}
\<
{_
Ife(t)actsondiffererrtmaterialpointsatdillerentirrstants,thenr,(,)d,t'dlrvheredlistlrcdisplace.nrett
g$)dt = d1 rvhere d1 is the
ma/'c;,al poh;- fot ail iaoe t' lhcrof a oaterial poiut. I/ E(r) acts on the some
trajectory Cand (2'1a) and (2'f5) yield
displacement of itris material point rvith
rr(rz)
trt = |
L(t)'dr
<
(2-160)
Jc'4tr1
l.;"t"t (F'dx+ Fv d! * F' d;) = Jct's'z(t'l
[:" :u-:.t(r"a' +
- Jctt.v;'1r.;
W
Entities 2, La, T,t !i
-- conveniently
"*rbe
Fotd6 *
*(
F'dz)=/"1-.,"*
Aq
(2'166)
C (Frg2'12) as:
expressed in ternrs of centre of nrass
-r
_
r-l
(2-Laal
feY + F';
= F,o, * &r, * Fztr = F.i +
. = F.ts, * F4o5* F'u'= F;i* Fori+ F'i
,re 'lrprr doue bv a rorce
i:.\
l^!ze1sdm.
point s'ith relocitv 4t) i" defined by
fjffii;1;;**tu
!
_{
._<
IN TER,MS OF CENTRE OE MASS C2.6 EXPRESSIONS OF SOME ENTITIES
?
to frame .F (Fig2'12) is defined by
T-he l'inctic cncrgyof a body relative
The paror futdraora rorce
,1.:--
(6)
=
= Lrv*l-tt= Ei+4' fup,ua,1lt =
"+I{-o2
(c)'
The result is obtained by forming (a) + (6) -
rw = L
t'<
.
If^=fu+rc^xm! t'ca
iY = F 'tb
--I
rq
(2'rs)
(2'20)
'
(2'17)
p-nlb
+tr'4
(2'19)
i = 1*41 \ [,,tr" d*
''"
g6rn =Q' =) {^u'cdrn =ll
gsld*.=bntTlreseareproved -"--; J- :'pcdm= t(cp x u'6 dnt * tce x f^o"" d'n
a',
E* = [^r,^ - ;r:;: =' i,:*.- *^, *lt**'+ !rc^) = !;'e
-"/
*%^ { d^=E-c*uaxfn!car;.
'
-.-'t +^+ii
+ f , )J^
J*rr.dmxgao*lcn
JM
, : * l^nl d* = f, 1,,* ,, o',. :, I"{s-i,"+pc),' (grc 1 T' ""
'
* =iL'u = t&' (*c+ u\ =tlel' ea *Ie'n" = F.'!b+T&'uc
.,
i-fi''
AILBITRARY POINT E
2.7 r}uLr;R,s sEcOND AXIOM (iL^ _MA RELATION) IIOIL
*bo x nlboy ='EcV +Ic x n!bl['*
Using P-f8), L4r = EcV +bt x mtlct1t, IIoV = Eay
f-44t: lbV'l,c x mtb1 * be x tnlzt1'
x tttlb
f-at : ibp - ?blt x rnlclr - tc x nlbv * lb 4r x m!4 Alt * b t
^lr
x m9elr = llp' ya x bb lx'qr.lr
M o - rs x E-*b t x m(!b,l r - o ali = ![e - y; x L * u $ L-ul
-{
I
I
rr
<<
<
_v
-q
-r
_
-q
l
.-_
Litr!
a6
'J<
_it
-t
ll
I'
Ii
)
:i
i;
Ii
il:
i
:i
L--
,dl,ro"i1no, =, W,, mtby =
bi-
frcae,
+ b-
""Eilt.= Mt
rii
>1 rnQag.
Er x [ = Mi Gig2:13):
-i
r
i
provided Point Asatisfiesatl.eastoneofthefollowingthreeconditions:
:
i-e., .A has zero acceletation in [,
:
1- 4,t1r Q,
2- c-.tgftrcr' ie-, acceleration of .'[ is along AC.
3- rc^:\
,
-r.e
-.
j
Fig.z.\3
\ e\rt Lr\sS\s\-'-rs il"t, = W'
A->=*i;.^r+(2-23)
.r
Ip,-feA'
1
!
,
IfaPu{sclofforceEanditscagzlcrimpulsc4.a^aboutpointAfiortimeinterval\tat2aredefinedby
f"
lrz -
Jr,E(r)dr, {-.*^(rrltrr= lr,,'*i*'-ro'
r, _L,
,:L .l{, aclilat
if
11 ,:''n"'
poirit
'
tcou angtila'r;*p'it"e toe'(") about
iaslcala-r
tnd
irnTulsc
l(tr)
Iasta.rrtaacoqs
.' t'z
(2-2,t0)
g'
\--- - '
fz
L*s^lti = ,l'i,', J,.M,(tl dt *
'r
; ,,' r(rr) = ,lI}, /: a(t) dt # q,
L{trttr.).=
,;.,.
Alt entities are w.r't' inertial frame- Euler's
first axiom implies
r(tr. t 7= f"'Eat= l,:,,'r-*= $i,"'o,'
{2.25)
(-a
niiagci '
{t yt2} = 1(t2} - g{rn) = 4Y= mL96 = L
impulse zt t1, (2-25) *
' i-e- r-mpnlic of exter:a al fo;rccf cpals cha*gc it mbmc'rrtun' For instanlcneoas
(2.26)
m;As'- '
I
l(tr) : ag: e(ti) - d{} = m[Yaj I
changes lvithout g'hang P in Position'
(2'?5) impties io*niotiou o! momer.htm:
nihere A?, 6rlb,Ag6. are. iastantaneous
If f(lr.t2) = q]i"o
4(lz) =g(lr),
If a paiticular **;;;
If E(t) =g, then' i=L
ef
ga:(r:)
i*;;;
gi =P(0)'
+
:+
=k(tr)'
zero, then ,[r"
oc=g'
(2'27)
' f,'uuc'(r') = t':*:(r')'
"o,topondinS
comPonet'.::*""""tum is conserved'
f,rnreci = Q
vt
(2'28o)
+
!runc;(t) = f,:nia'(0)'
=s'(0)'
'.t't
i) = g, L**r0) i,r''
(2'286)
tz:sol
:g,tt)=r(o} I*'r-'(tl:tj:'10i''
',-l'-.llu'
*uzeroforall]t;then(2.28)'{2.29}holdonlyfortft*i.empoaerrt.
' . 2.LANGULAR. TMPULSE-MOMENT O: MO.YEN:',* RsLATroNs
All cotities.are s't'i':furertial frame' If {-r1 a
point such t'hat ee(t} = $' or r{ = C'
j
or e{(t} b atong
(2.30)
.i
ail-' !4
I
3z
j
-..-l
--:
':i.an{3 1".-"Ua for' point /
Since the results
nxed in f, if a component of l'nr^(t1,t2) alon( a f,xed :-.
a
direction in , is zero, then the oiirponent 9I.4^ in thit direction is conserved. Similartn if a -omponent
of Los. along a fixed direction of f is zero,itheu the componenL of Hc in that-direction is coascrrcd- Note
e-l
*"
I
.
'
thatforamass-point rn,'Ho, =(re.+rc,)xm(ie" +r&i+ig,).9.=qll.2dan
lf an instantan@us.ngui"ri-pirl"" Jo * 11, then wiihout change in position there is iastaataaeous
change in moment of momentum about centre of mass C and about a point O fixed in f:
slavning .. aoo36(.,
*;i{"';.o?'*c
a-^.
(2'31)
+-".:i'*1i;.1?;*...-.
2.1O \MORK-ENERGY RELATTON rOR CENTRE OF MASS C
Iiq.(2.9)
=+
F.
lctt --
,te,,tr .t!c11
ic = W'
i.e.,
= {e**,
=+
-
!li*'Lt,it,,
lUi-2,
[1 fiz in general]
i*u211= kinetic energy as if all mass is concentrated at C fl T in generalJ
rslrere W' = F.yc= raLe of s,ork done by the forces as i/acting at C
\=
Q32)
(2-33o)
(2-336)
Tc,,Tc, are the values.of ?6 in configurations I and 2 of the s-vstem and Wi-z b the value of srork done
from configuration I to 2 by the e-xternal forces as if tliey were actitrg at C.
2.11 AXIOM OF COUL.OMB. ERICTiON
(dA
2,r
At a point P1 on.{he,surface"of bod;r'l:rvhich is.in'contact rvitlr point' P1 of
surface force intensity can be resoh'ed into a norrnal force conrponent pN/m2 (prcsszre)
and a tangenL\al (frictioaal) cornponeut rN/m2 (Fig-z-la). According to Coulomb's a:iiom of friction:
dqt-e<rc&'d fa
bif..ekr'm o^a *-b.at*lo
{ E o<
<{,8?1r.'rs) <$ ' <'tr'r'
(2'34)
T = {tP,onj
i-5 d.ireattd, qp,oosits to aqpoo€ok
*
9.'.9. (t3) r-q 1i\Q- +o'^shtr-\ p?c-re. r Ar = PtP, ood iS dtrectc.d .pi.Eih. to Vi.p,
for no slip betrveen Pr znd Pz:
r 1ltrp,
for impending slip betrveen'P1 and P2 :
for slip betrveen P1 and P2:
n'lrere Ft. lrt are called the s[atic and kinematic coefticients of friction. These are independent of ?, gl,, p.
and the area of contact. The resultanrt of the contact force s1'stem is, in genera.l. a u'reuch- [f the contact
surface of area ,{ is plane and the direction of r is the same (e-g-, when I translates s'.r-t. 2). thea (234)
implies that the total uormal force iI = f^pdA,and total frictional fiorce F - l^rdA,are related by
for no slip
F < p,N,
for impending slip
F = y,'N,
and for slip
l
t-J
,l- I
J
i:t.-I
yctitr =
Tc, -Tc, =
.CJ
:
GJ
F = ptN-
(2-35)
'- I
(
-thl
J
=-J
*l
,-J
::J
,_.:J
il
LJ
fil
"J
&l
t'--,1
.?t
:ql
Nr9,+N3er,/V=(/Vi +N?)tt2 forasmallb".donawireoraslideriuaslot,andF-Frg, *Fzg,,'f =op-,i!P'
fa- I
(Fr= * Fi|rtz for a srnall slider on a surface.
\€\
Eqs(2-35) are also valid rvhen there is a discrete point contact betrveen the bodies (fig.2.15) with .lV =
oF ELECTR.MAGN;r,"
- l.rrAxroM
"oo.h\+
g iu cledric
Force F on a poinr charge s moving
n"ld A?JJ7
a nragnetic fietd B,i,s
:
ac
an
.
F -- qE +qq x a.
'WJ''ffin,
-T
,.*3
._X
t^'
f6
/ \S.
9b \
^
(2-i6) 'N9,.
asr.,s
i
2.13 AxroM oF GRAvTIATToNAL FoRcE
The total gtavitational force {p, on body 2 of mass m2 due to body 1 of mass m1 (Fig.2.l6) is giveu by
Ezr = -
,i.9
i:'
wlrere G -rs universal gravitaiional constant. The mass. appearing in (2.37) is calted the gruuitaligacl mcss
wheteas the mass appearing in the Euler's a)iioms (2.8) is called itcrtial rnass. Experimentally, no {ifference
has been observed in the measure of these trvo masses. The total force S1 is called the ucight of lody 2.
Tlre simplcst ttsdtont of the general spatially distributed gravitational forcc.dErz is, in generat, a rrit:clr'cltFollowing results can'be proved from (2.37).
'
dE' --
Fi9.2.t6
I
I
-J
I
I
.,1
r
'-_J
(237)
l^,,[l^,c{ar,;}a,,r,
a8
3'l I
-_J
l=_J
,-]
\'+-J
E--1
,\.
:
=l
l;
,--:-t
-=
>
]
_ -I
L?
1. For c "odiarsrr*iir;agi;r,.,**
r4-^.
r
of mass M and a mass-pointo:,"u""'- $tg:'l7zu'q;]i
Ll+'
inside a sphere of radius r'
where IU' is the mass of tbe'ii* of spheti
r betweeo tbeir
of rnasses M and m rfith distance
radidllyt-*"O;"
tuo
between
F
force
2- Ttre
"rlcrcs
H,'
h-\.
s'a'ita"T"l.r":" o":*^":::-T';:'t'A1J' :^X:l-^6i'
;"ffi-qt ?Y:';,;ror the concu""nt
'..1
rorce F throush o:
r''*"'ib)
L:,:;:::;;;;;';;;';;;;';-
LE
is a singre
L-
P,
2
-''--'-a?
:'u-I-'""
:-'
point where ihe $'h(
as the uuique
T:he a:rrGi of-gntityG is defned
mass tn:
if* *.a"f [o... on it equals that on the actual distributcd
L.
L.-
ln can be coocentrated so that
(2'3s)
-Garrn$
rZ =-ctt l^1*o*'
of thc bodgis small compared to
from the centre of *T 9 ]l U*e si*
;"
In general, the location of G differs
rcittl,.llff c'ca ue
+ t::':':
^*::K;;;";;;
=tc'i'e''tL"
theu
sphere,
sp:r*r+"I-I*.;t;1)
the
from
d-rstance
its
them ,/Ofo,
approxiinated at C'
,6*ing
ate'smattconrnil,{r.1,,:l:T'5;#::fr':t*n./rn*1"
sizcs
4. For troo bodies of arbitrory shapcs thosc :i;;;:.:;;;i'o1"'*F21=-G*a""-[1li1"*"
L-i -,,f
L--
,
L
i;,.l"i#::::',!#:Wf&iff
;l
a]
\l
;
.ffiY =
:T;;" :-" or b.dv.is (,.,1( :,neat
: -:the^::!:;,
:-5,i::: :,T,,I;il;:,':T.J
surface of the eartb'
Ri
n'-t
<
distancs
1"
through small
2- For bodies of small Iir. *oriog
\ri
I is the local
b".o*siorritrr t"r"oti" ""a direction: f --'ngg'1hetesf Rrad 0-1"
tlre gravitational force is modelled to
30 km' a =
=
6400 km, even for s = l0 km' 1 =
vertically dorvn direction. .Noie that as-8 =
039r,e'
c M ml R2 = (1 * h| R1.. "'o and F - GM nl (R+ ;t' : (l + n/a)-3
-h-,1
I
L-.i
-)
E
-.
!-,ir.I
-,1
)
L_.
t
L
t,
I
L--
LJ
LI --,
t-
2.14 E'[{EE BODY DIAGRAM
of a seL of n bodies or
of Euler's axioms to a system consisting
For the purpose of propir applicatioa
part of a body' i't is necessary that
particles, a body, a finite or an infinitsimat
jsolotiorr
t:-t::ttundings' and
and its sketch U" i*1n in
rocll-identiJtcd
be
should
T"t"
systern
rhe
1sysiem should be drarvn on tt2. the cztcraauorres exerted by the surroundinSs on -the (FBD). The forces exerted bv one part of the
BoD1. ;;Gra;M
Such a diagram is caped a FREE
in the FBD.ltJ;o kntal forces and these should not be shorvn
sysiem on anoLher part of thesystem "..
2.L4-L Comluon Supports and Eteactions
to
rvp€ or consrraint provided bv it
by a supporr on.a body depends on rrre
|.],i]lol;;.I;a
r ^ r- ^ -r i-^l- rampnt .omDolrents constrained by
:#*"::
.,,"J1i,'f ::j.ITtrj:|i:il:#i;fi *i;:;i:l':::i::::::H:::::,;1,ff
rotatiou comporenrs consrrai.ed':b{dt''R'eactions
f::f] ff::::ffi:T.::X.:;;;;;;"r*
giveo io Fig-2'19'
for several types of supports are
"' i:ii'"1;;:H:::r":'.:ff;;ffi;;.
U.
L
I
i
I
f
resurtarit
'le
rf-ll i:::::*;1f.:"i:::ffi;:
",.rlT"fr.lll?iIi'j"i*";""1,J;';;.*q"L"r'*'t*"
H:red'io'!he
ilT.E:?::i;:X:fiffi;,X;.:qil;;;r:::'::::H:ilf
orirs
*j,'i;;d,":,,X;";.J;";;;;#.-r-*,troruanded u.it
triad
.[n=
Qa=CtU.+C^gn
cen*oidal a:iis:
(2-a0o)
a:
;
1
LJ,:
LJ.:
lj.;
I
,t
its
io'* '"'o'" thesection can-T:::':::.i bvrvith
opm€nt
Lj"
Li,,
l .l-
n'1 'z'G
!
o1
rj
U
I
(2-406)
A systein is said to be in efti[brium if its every maCerial poinL P continues to remain at test, in art
inertial frarne f. If a systcm is ra cgur'li,0rirm, lhcn lor its cocry'poti
f=O
M,=0.
(2.41)
Prcof,. lf O is a point of f, then s.qr - 9, spolr = 0, V t- Hence
oqr=
rl
J^*11dm=0,
and Euler's axioms yield F = q, :0-
v t,
J^teolrx!!potr1-=g'
Mo=iIoV =9, :+ Mt= l[8+AO x F =Q-
Ho11=
These are necessary conditions of equilibriunr, but not suflicien0 conditions even for a rigid bodyT,he cquations of motiott of aa incdialcs.s syslem are also.(2.{l), since * = + \, = 0, EaI:oO V l.
L
t-r--=-:A
T;G..-'-"JjI"r*-:=I;
2.1s.1 Two Force Mernber fl$Ilffl
' A irvo force nember is a ruernber subjected to only t*ro fo..o]lf ts'o.forie ,rr"r,Si1fq8-2-20) is
"
ia cguilifrium or a trvo force iacraialess mernber is in motion then the trso forces have cgucl lml.gaifudes,
opposilc dittrtions and act along. the line joiring thcir points of applicatiottProof, M = AEx& = 0 :+ Q acts through ,4 along.AB. Similarly F, acts along ,tB. F = & +& = Q
^
fr--&'
=
a- Srnooth balt and socketjoiat
Pin in smooth slo[
b.
L2-
TJ r\'fu,*,
d-s
end (3D load)
'J
:J
_J
J
J
J
J
U
J
J
J
U
,J
u
L-I
f-
l. Internal section of a body
^T,Ei.
'?ry?ffi:f-fr{
f.z
a-a
8i
U
U
9e
il
LI
U
U
"#'F :J
j. tnternal force,resultants in a bar k- Coplanar load on coplanar beam l- Coplanar badbn strafult beam
9'\r,z9t
Ai
FBD's of front rvheel assembly (m1), rear wheel assembly (m2), chassis (m3), and the completc rcad-roller
with dtiving tarque M on the reat wheels, assuming no slipro.e giito in 11.r.l9:
.:k;
-{:'Rq -sB
8
tr x. z'1<jr
30
::
Fr
LJ
J
J
Jil
J
J
il'r
I
J
ed-
*ffifrur*H
;Le?
i
.'..i...,,,,...,.'.".x21"..'.9,,i,,*",".;2.'*,,.,,.
'v.1]5,
tt'
z.ro epif, FnrcrroN
*
pt aod
''";:[,Tr:::::'-iortg,/*J*.lr1',:*1:i5::X***:llffi;'::f
.o"ffici"nts frictior
;: ;;;"rt:
are
of
and sratic
tens'ron in the bdt
*c AB*a. i"*i"siv an angre ;;;;rm
and ,rrP N/m and t'e
iif.i"tiona
gravity' the FBD of
p". Ar locarion {,, ler *;r".*.,
belt at {' bc7' Neglecting
;;t
centroidJ;
parh coordinares rs:
be T. I*t the radius ***"."*.rtbe
,.,"*r, * anis elcmenr io
ir",t'""'lt
Fig.22t.
in
*
o"
an erement of rength
"n.
"i-*"
conracL over an
tf the bclt is on a ro{aiiag pullcv'
for impending sliP :
lf the inertia of rlre belt
tlren (1'42} hold 's for slip and
Lf"'
:./.0 .
i1-:f,;t = - '
-rs
sirlrilarly lle catr get
for no sl\r
negl'ected' then I = 0 and
:
rr-}_:
2
" ",.d
fr17
-' :
(2-{3)
-
t'
(2'a2)'(2':13) reduce to
I
(2-4{)
l.l!
as in (2"14]slip and.no slio are the 'sanre
irnpe[ding
of
to the
condigions
the
asd
o
l"t ot Ou/' be the force nornrat
Eor a stctioncry bblt, o =
itt
t'
***-t
'
papfot
semi-vertex
with
p
and
of
ropc
"""'
belt or a
N/m instead
tl
forslip :
forinrpendingslip: T1!12=e"e'
TrlTz=eP'e ,
fornoslip: TtlT\<e"0'
ll
i
l.
llii
!!
i7* -ta iut't
f's' 6.'vee'
* and q directio"lt"."";#;:
"rJ.
.noiu"rt.tt"n"".q.(z-i:-);;'i;'Y::'.1;l'l*:k*it:jIg;:;"":.ilFT'1
bett (F.q.2,21). Then
:!
l1
is better fot porver'transmrs
of lap on Lhe trso pullel= is
,rr1/sino. 'Vse' bel!
t-h"'. t* s-ollct of tt't-*Eio
both
fior
sarne
the
are
p's
tbis value'
transmission, if
pulleys are used to itlcrease
(2{2)'
Idler "*"
belt' Suljstituting 7i frorn
to be used in the auo"e c'iott=tnt-
;" ;;t*'n'i*io"
Ler Tq be ttu *-*iJu'*
get
rve
of the pos'er P lransmitted'
i1
I
li
it
;i
J,o.."rt" rension i
in the expression
:J
ri
I
.
t,
i:
li1i
p-T1.v-T2o=.$1.,:lU2)U(1.-e-,.sc).Fortna.timuttrP.dPldv-_0-.?|.1=3lrr:.
'!
trassmitted when u' = ?i/3iIlence the maximuqo'power'is
" ':
THIIUST BEATuNO
A
AT
TORQUE
2-1? FRrcrroNAL
surfare (co1i1L Aar'
-- axisynrnretric
--,.
*itrr auotlrcr o1,er ao
contact
in
t
body
tb.,,.t'P
consider an axisymmetric
of iodv I under an axia!
coatact
r{
lhe
torque
1"*1'""
of
ring
spherical, etc-)- The ixial
"*1"1
'*;;;;"*ia
ii'Z'zz' Consil^er an elementary
'"t
i
o*r,'it
,t
i.
r
oocmal
body
the
of
Let
FBD
is to be determined- The
*!?:" rvith the axis'
norr.',.r'rr*un!-"o
r frorn
its
*ith
r
distance
radi.s
"t'gtt
and
at a radial
sutface of width ds
t" * itt**tJ'"ntial-dircction
is
force
t
rir.l/*'
frictionat
Thc
force be p N/*r.
yicld
for axial fot"" *d axial monrcnt
the a:tis. Equations "t to"ir:*lrur
,
--:
;
i
'l'1
.
L
L
l*..
l-f*-
t-,
l-
"*.'
u
L.
r = /tr.i"."Xz"")a"'
.-i[ffi\l
u = f!r,nl(2zrlds'
MIP
,9
.-
I
:!!
:
i
- I I r,,'t *lL [ /'n"i"ga{ r']"
*
,t"
R t-.^
i;-Y,t, f .Lflt h V" r.€ffi iiS%
I
I
,
itic
; -'-*
I
I
I
i
u&=fr/(
F
II
,,
:F
.31
,
,.,:
.
::.
.:.,
{l
->,i1+ 16 - drlsinc, eq(l) * Ml P = i
!," t,,'odr'l"iool t I f'*a'lel
'(3)
- Ml | = (2y,l3sinoX"i - fi) l(rl - r!,)1- for uniformpressure p=const' (2) +
(4)
r2)12'
2. for uniform rvear p = Clr (2) :+ M lP = (p,/sin o)(r'r +
to
(b) Flat bearing , u -tllzregd2x{) for various casesr respectively reduce
Mlp.= {2p,fni(rl- "?t/bi - "?t, MIP - P,(t +.)12' (5)
Mlp = I
,f
oa4 | I
{,'w*1,
l,.it
(a) Conical bearing:
motncnt evaluated for t'he corresponding
Equarions (2)-(5) imply that the mornent fot conlal- be*,19: :lt"
: p, I sin a' lrence conica'l dutche
flat bearing with r, replaced by an elfect ive coefficient of fric0ion Pefr
pe6 ] p''
arb more efiective for power traasmissiou, since
fc) Spherical ball bearing: r=Rsind,
MIP - Rl
l. if pqc
ds.=RdL' s:
Jr,o"inz
rl2-d' and eq(l) '+
0 d01t [,fr'u' 0 cosa dll,
r*12
(6)
r*lZ
cosedlltlJo sin'ceE?odfl :F'R'
thenp=.{^Rcosg' and"(6}" +' MIP= *t"Ilr.sintA
'd (6)' *':MlP'.- rP,Rf2'
2. fcr unifotnr pres5ure p:const:':'t?o," an
"l
2.ISE.TiICTIoNALToILQUEFoR.ASQUARE.THREADEDScILEw
porver-iransmission-(as in lead-screr*' of
square-threaded scre$'s ate comlxlnly used iu screw-jacks and
the axiat totque Ir' needed to impend
lathe machine) and in tesling machioes and presses- we compute
tt]t:10 of the screiv in
the length
nrotion of a sctew in a fixed nut agaiast an axial tlrrust P' Let
't:
hetlx angle c is giverr
The
{ (Fi5'2'23a}'
contact with the nut be s and it-" rneao udius and lead be r and
distaoce of Zat (Fig'2-23a)- The
by taa c - ll2trsince the hellr a.tiall.v adraaces by I for a circumferential
thread be r l\l/1 al.an angle c s'ith
FBD of rhe screw is shorvn in Fi5223a- Let the normal fotce on the
to the dirction of irnpeadiug
the a-ris- The frictional force is r,p N/m aloog the helix in direction opposite
slip. tfence the- torcc dF ori'an elerueat of length ds ai a dis[auce r fronr the a:cis
1
dL: pd-'(-dncq +ccag€ ) * prpdsl-Goso'gd - sino6.)'
Equatioos of equilibrium for a-xial fotce aod atial moment yi'eld
e=
lt
n, =
Jo-@"*c-;r,psina)ds,.
Jor(ps$o*p,pccc)ds'
obtained from (1) by replacing 11/ b-v -Mr and p, by -p,
-frictiou force are opposite to those of t'he previous case:
F;
:1
usPdS
e.+
|
I
...-l
(ol
6iv--'1a
J
J
Ir
\\|.-J
-l
j-J
hds
R'3. r.z
3p_
J
;-J
J
J
J
J
'*':J
if
is scf-locting if it does not advance under P in the absen ce of. M1- Hence it is self-to<'king
fhe
"...*
Mt ) O, i.e., if tanar( P,P}
rqiset- :rE
r-l
(2)
tt:
since the present directioni of'the'torque and tlre
lil P,= t(p, - tan c) / (l * P, tan o)-
J
J
J
J.,J
:'
(r)
The rorque iy'r needed foroimpending motion in the direction'of the a-xia].rnTti."'.!:i62'23b)
;J
l
U
':J
It
M / P = r(sin a + p, cos o) / (cos o - p, sin c) = r(tatr o + P') I $ - g, tan c)r;
,.
-t
--J
il
:J
:l
I a-
t-
t-
L--
tL-.,
tt---
tj
L.-
:,. :".s\ -
ExAMPr,Es z
Erampte-2,r 'Fini the moment of the force system applid on the hinged bar (Fig-E2.la) about point "t
aad the resultant (equirralenl) of this force system at point ,,1-
Fr=z-s 31'*
roklr
tsi
sB,
t-.
L_
t:
lt
l'
\-_
L
L*
L;
l
lt
LJ
LLL_.
tt-
tlr
It
L-
€
:1
tt
j1
I
t<-t
h
The force system is coplanar- It is convenient to resolve the forces and the distaaces
@mponent.s as shown in Fig.E2-lb:
-So6tioa
into suitable
F3 = lOcos 30o = 8.660 kN,
- 5 cos 60o = 2:5 kN, .F2 = 5 sin 60o = 4-330 kN,
fa= 10sin30o =5kN, ri = [4/(3? +42]rt2l4=3-2kN, F';=[3/(32*4111214=2'4kN
B$- L5cos20o =1.410m,' CE =l-5sin20o =0'5130m' CP=0'5cos20"=0'4698m'
FD=0-Ssin20":0.12i0 m, DG-'EF --CE -CF = 0-0432 m, :{G; 1+ l-410'+0i!?r0 = -z;st1
t1"
The rrapezoidal distributed force s-vsiem is decrmposed into rectanglhl Td trianqllql {!:t1i:::-"":
Sr.pf*a resultant ,r7 of the rectangular distribution equals its ar^ea- L":, t- = !'0 1 Z :-I ? kN' a'n-dacts
Fs of the
tnrough its centroid, at a distance of AH = 0.2 * o-612 = 0-5 rn froar {- The simplest resultant
at a
centroid'
its
through
triangutar distribution equals its area, i-e-, Fe = |(0.0 x 3) = 0-9 k\, and acts
distauceofAl=0.2+(3x0.6)=0-6mfromA.
: ..
product of its rnagaitude rvith
The moment of each forcc component a.bout point r{ is conrpuled as tiie
tends trcr rotate
its perpeudiCular distance from ,4 and is assigned positive or oegatile sign according as it
Ttrus
couples'
the
tle bar in the setrse , (D t V (i) or y (j-) to r-(!)- The sarne applis to the sign of
x 0.5+ 4-330 x 2+ 8-660 x 0-5 x I +3'2 x 0-0432
tti
=[2-5
2'4 x 2'581 * L'2 x05+O'9 x 0'6+3 - 2Jk= 0'9938k kN-m
a forie Fn
The lesultant (equivalent) of the given force system at .A is a brce-couple system, consisting of
aad a couple Cn given bY
Ca=I&=(-2.si-4.330j)+(8.660i-5j_)+(3.2r-2.4t)- l.2j-o.sj=e-36i -.:3-83ikN
t
Q_a= Mi - 0.9938L kN.m.
.-_\
p;ample 2.2 Find the moment of the force shorvn in Fig.E2-2 about 7/\
poiat,{ aod about the axis of the bolc. Find its resultant at r{-JiA.4Li
Solution Lei qand g' be the unit vectors along r?C alo.d AB: a.
i -=-.1k
:
e= (-5j + 12k)/(52 +L22)tt2 = (-5j + 12k)/13
i',
-g' - (-4!+ 3B/{42+32}1fz=,-0.8i+0.6t ' ,
---J /: .rx--:
,,.'
er
'
: (3g* 4ll I @2 q4zlrtz - 0.6e+ 0.8i
:f
.t,
s't?
lt
_
_:?_6q
-_ .!:ZGq=2.6(0.09+0.E!):I-569{2.U6!
=2.6(0.6e+0.80 1-56e*2.08i
Ll-
Ur,l;
lTii'
:soi-ssj * 13ok mm
-"^:.t4C=AE-+pQ1QR+RC=-30i+50i+?0k+65s
I glt^
t-.L-l-
L
L-.
L-.
I r ':
Li,
lr
t+ L:
L,.-
L.
t
.F1
ls
lt
k*
r.l.{
L*_
tj
FO= 5
<i-
le
LJ
.4-ft*t
. -,
o:
: L56(-5i+ 12k)/13 +2.08i = ?.08i- 0-6j + 1-44k kN
L..HI
^lj :;.1
at':''txu;,lf&
ill
= -12i+
I
198.4j + 84.4k kN-mm = -1.2!+ 198.4j + 84-4 L
33
N-tt
./
//1 ,
llllf
i,/ f
\-Y
\e,
'
hS-E2.2-
*
{
r-
:':-';i'':
. '' '
';
,', ''- - ', 1 '
Moment about the 8xis oiih!:boit .*i i-ae'.
Mta = Me' - fu ' € = (-f2X-tiS1 +Aa'n x 0'6 = 51'6 N'm'
/B
sVsta, cons-istinS
The resultairt, of the given force. at 'A is a forcecouple
coupl'e Qa + &'
of a force & : E = 2.08 i - 0-6 j + l''4{ k kN and a
-1-2i+
:
\e
198.4i +84'4k N'm'
along its
ie Gig.E2-3)- Find the total moment of the force system sho*'n
of tlre
"agu
po,int.A and aboulline-,{8. tin<tTh-eresultant (eguivalent}
Example 2.3 A rectangular plate of rverght 2 kN is hinged
| -r,
t,
about
given force system at point Ai o 7 Lm",P - 60o'?:
""'',9^:,1."];
list Lhg'fi
Solution Let n be the normal to thc platc as shorvn' We
coordinates and poition vec0ors of various points:
{ t*-z-'l F;5.-c l--3
2! m. s6
=
-^
i
rf(o' 1'3)
(2'0'3'464)'
/(4,0,0), E(0,0,3), D(4.2.0)' 'g(0'2,3), G(4sin30"0"1cos30o) =
s^-= 4! m, rs =3! m, cp - 4!*
r
\e,
2!+3&-T:-.* = 21+3i648 mr rtr - i +3k
;
l
'*2"-j::ln
g=&-lla;1'=,(-qr+rt) l(42 +121r/? = -0-8i+0.6k, alx ADit-<i+3,&)1'6i
+2i+ I-8! m
0'6k)
3(-0-s1+
2i
+
=
3s =,41*
DG = tc - LD - -2L-2i +3-464 b m, !F - Iq +
n: (AB-x AD) tlAg x ADI- (-Gi- s$/$2 + 82)t/? = -0'6!- 0sk' :
t* ,*-=:.""rr* # t,h" firrc forces are given by
Cr - 2s=2(-0:6i'- O.8k) = -lJi- 1,6! kN'm
e:{es=4(cos30o!*sin3oos}:a[0'866i+05{-0'8i+0'6$]=-1'-6::3'46{l+r2h'tN'o
= -4.472i - 4-472i + 7'746 k kN
12j+{!&N '
[3=14e2 - lagq-ii+2$/(6" +32 +z?f/z: -6i*
: eu" = s[€o6affE: -sin3-oo bJ : 8[0i866(cos60o i+ sino0oil ' f,
.
&=-2LkN
0'5hl = 3'4sd1*'6i-ot
T^- . --
Ee:5g+=S{cca!*cospj*coszl}=r5(cosI20?i+cos60oi+cos45"}=-251+25j+3536tkN
Thcpcitionvectorsofthepointsofapplicat,ionoftheforcesrv.r.t.l.arc:
m, AE= g.e- La=-ai+2j+3krn''4F =L"-!t=
AD: lo
-Lt=2j
AH:t* -tA=-li+tjf
L*
-2'4i+2j+l$km
AC-=-2i+j+15t' m
Moment of each force abouh r,t:is cornputei: by' cross-product
Lrir,e:et.4E:l +'fr'x.\+
of its Pcition^r'ector w'r'L''{ yi't'h't}e force:
Q
AC.x'.&''+ Qt'+
M-.q= AD x
-^
+(-1'96ain-f6;6aii---t:iS
(rs.49.i+ES44k) +(-28t: 2j'' 3oh)'+i:18:8i: &365'i-'2L"t,
:
^.'-
:.
+(-2i-4j)+(-l'2i- 1'6E)+(-r'6i+3'464j+1'2k) = -40'0?i+0'743i-'56'2eEkN'm '
moient of theforce system about line 4Il is given by M.z
:-l'718 kN'mM"= M. r,' e = (-40-0at-b'i* to'*'ltol * (-'u'ntt0'6) =
coosisting of a force &
The resqltant (equivalent) of the given force "yrt * at .A is a ficrcecouple'systern'
and a couple en g,ren bY
,.. T.he
r* =E& =.(4A721_ 4.472i+?.?468+(-6i+12j+4$+(3,..[|1i+oi-1$ i
d'+(-2'5i+2'5j-+3'530$+(-?H=-e50bi,L1633i+e'282Ekr{
0'?'t3j - 5629 k kN'm'
e* = W= -40'0?i*
j
.
34
,j:--
I
I
:-
:
-.
-!-\
Earaple ,.o F;i1t1*;*n''* "-tte"t
o[the tangentiat distributed
force'
:i"fi:"$i'ffiT;?.-?Lrr'1."cu:rili"::l::.'-"-,:L'oo'*i:"''
ffi
that
l a' i""*
in
t-.
F-
is' a couple
the-simplesu resuliaat
i",O
direction to
dr:
"i
t'"i"' tt* element' of radius r and thickness Q^a.:Cnk- Ca i" outai"J
t-
I
e:z-e z'u.
!
rR .^
#"t*-i
Cn=Mo= JOlrr(Za rldr= Il'',(T)(2rr)dr++ l*,,mlzT)'".=
\
* *:,o'01::::1j:::t::r1;1ll :|] ;:t"
slampte 2.8 Eind the centre or TT"
L-
L.
t'L:
zl
d
I
Y.
4
,!
aa
sl
-t:':*HlJT:Tff
*."'"J:ri'*.tlTr:il;:ffi
.
rL:- ---i-rircular ring 2, a t
"":ffi il;;*"'rlu""'""densities\'oa'.dp'
t
ffi1**",1"inl***"*fi"']'"';;iy;
*;::ff
YJ-'--- 4' a deretion or a
7
aseml-crrcur4r-Dvr's
J'
shell
cylindrical
'vri"d"'
taken L
f#:::"rJ:riffiffiTil;,"-T;x""";;-;ii
semi-circulat
d cuboid 5 is taken-.
Tr-^ *-.o af deleted
u-,11'"-'11llo'"t'
sphere
-.
q"*Y
of size Rlz x Rlzxzi""i "
T| )'-
cubo'ld 5
antre or mass of each is given bv:
:X";[-;;:;;""d
LJ
I
t'
t-LJ
ur='(ZRlr)i+ BL,
..oDr-
m3=(r.R)(2&)o=2*ffc,
*a= (rr2l2l{znle: uf p,
LJ
',*s
= -(Rl2)(Rl2\(2R\p =
L-..
m5 =(4rnl
t_-
IJ
tc4= talll)-i*:ot'
-ffp12,
fi)lqlp=tffpll,
following
the.free body diagrams of the
AB'
o'
('}
/'E:
LJ
i
3
t,,
\y'ry1y in*'
Ilr-r
i" = ulo'1.':. j:ji'::r'jr: 'l'=,5
t:u = (3818)i+(48+38/8)k'
bodies for t'e systems
given in-Pigs-E2'6a'b'c'd:2' 3' All coatact surfaces
*r]rrrr" iu*'r.",
i' \12: 3 + 4 + s' ia; J' '"
i'
are light' ir{ake
{z) ABID, AB/:;E.&);,
''
'l
memb.s' rot 'utti"tt no'*t"t t:*:'
cables'are
it
belt
Ttre loading on s}'stems
are smooth. The
"lil
" presence 'irr'i'-ti"
which are trso-force mernbers'
supports
of
the
to
due
t-
tJ
IJ
tJ
simplincatioas, if any,
e1'e'bolt' but
rotations are allowed at an
in FigF-E2-6b'c is coPlanar'
Fig'82'6e' AII relative
exerts tlvo
Solrrtioa (a) .,The FBD's are sho*'n in
of the bott are preve4led- Hence, 'l: "',-oo"tt
the.
to
normal
the relative displacements
"*i" to the axis of its eye- The supportinc,,Iembers PQ and
dLections
e-xerts 3 force
componeats of force in,the.two
";;ao;r'"e
Rs o"lrt But the memb..eqfl:
ro.*
.:t
""u
as, u"iog two-force members,
B in theufi.r-s'F'1fBD a4d at the
L.
LJ
1 compoo ntsl'iie shotr:n' because';'
a,two-{<irce
components, since it is::not
""J"t]'O''the'6xe!''Su1P"t'
3 i..""-**0.."i* .r,a,s-.ouple
FBD,
secpnd
tbe
in
E
at
force sy'stem ts
ht€raal section
d'splacenrent and the apptied
;il"ttt*
,ot"t"
relative
on
thete is complete coustraint
L
L
tj
thtedimensional'
siucr it is not a trvo'
(b)TheFBD,saredepictedinFig.E2.6iThesupportingmembersPQand*'o_::1,.:o-forcemembers,
exerts 2 force components'
?J
*"*ler
*"
U*
only.89
and
cxerts forccs along PQ
i in the first FBD and at the internal
suppor
fixed
the
At
coplanat.
is
' "' one out-of-plane couple component are
force s*mbe" qrd the loa.ding
and
;;**poot'ot"
i"ph"
2
sectiou at'b ia the second FBD,
d'splacement aud the applied
relative rotation and rerative
cgmpletc
is
thdte
becatrse
shomn,
"onrtr-uint-on
it is drawn for t'he
b-ecomes convenient if
iu ris.fz.6s. The FBD of rhe puuev
Lj
tj
t-. )
1*.
u
l-.
1-
I
)
-i
a*r-;1":s;;::il-n
n"ri", ;
X;-;Jr.g.r.!"
pal of the belt which overlaps it'
.
35
:i
.,
,i
I
1
:l
il
'!!
l.
l
't
.i
,i
,i
i
I
tiI
__:-&,
a FBD dBr'
82 have O:"
:fu"^tl
NOTE: Once a sei.82,of siiplorf, rqictions on body Br due 0o bodybut rvith opfoe-i1
sense'
reactions
the
shor
*t
slodd
of
FBD
ai""-t" *"
""*" "rrppo.t
joint'
as at joint (i' tbco
a hinge
(d) The FBD'8 are ahown in Fig-E2-6h. If moie than 2 members meet at
with t&e
preferably, the FBD's of the- members be drawu'
either the FBD of the pin be drawn separately, or
+
le_
T
;".;"" part of one of the rpembers Both these procedure are illustrated'
d!
(o)
,;
;;;
J
J
I
:l
-l
J
6
(c)
,l
I
'-+/
(dt
I
J
n'*,ffgo
"
*'i-? r:ffi!:*,,
ffi
F
i
ol'-o.
o.i,:u
sn e.1f*.'
.
- r&g
-J
\
-
'f .If
i
-
r'r
F1
tq3
-{
,i
s.,Rn
')i! *' zc Jncs
F7.
.J).
8s.".s(i_ffirffi*
m.3
- Iu'ls
*.esl (qg
c \-vl
lt
F3 atl {;
. L *i"rtno;]"
Fia V) 7t Tt"
ate 8.d.4'
^--^ ^L^.,.- i_
t" hE.E:j}-zExarnpte 2.7 D.t*-f...-boaI-ailtt- of bodv I of mass iJ::*:n
,(.],.,-.ts;5{5'tsr ,.1f1
F.
6-ls
=tL:lolt'
l-:F>.
4n/srz4s'
|
-t> > Snls4 6/3
::H-#ffiff{;
'
-fiB
ffir
l-rs'L'"
Tt
,sl
':-''
'r
' . , .
:h9
ir
:
-q
Y
'-
L
::i
!*ti
::.
t"nd- 6Aea;tl
b determined
Sotution Lct.A and I be the points in contact. The rJative velocity 9es
pilrt and is dirccted
ytst' 0, then there is aciual slip :+ the friciional force F has magnitude of Fl =
1- If
I^!r
-t
r-
36
.,_
-r
-
l*
u
I
--i..1
',
a*--l
-
pl*l,oJ contact''tlcu there'is
]il
its . 9t - (Iwhere q is any vector in the tangent
;:';;.,
are
""aforce
E has atbitrary a"gnit,rdu (bounded by f',/V) and arbiirary directirq'which
,; "UpTUi"tIoA
cascs'
ao
slip
lhc
zcrc
is trol
fot
detcrmined by solving the equaiions of motion. Most oftitt, ltictional fotce
then there is
contact'
of
plane
the-tangent
tl"to.-in
3: If q^s - Q, anc 9ts - 9f 0 where q is some
has a component which is directed
L
I'-
p,N and
impending slip :+ the irictional force F has magnitude of .F =
ofien, the direction of frictioa forces are
oppositc to the component of gre in the tanglnt plane. Quiie
assumption' For example' for the case of
the same a.s those obtained io .-!.iot'sotuiioul based on ao slip
about a fixed vertical axis with angular
impending slip of a small block on a horizontal platform rotating
radially inward as well as the circumferential
acceleration, there would be frictional fcrce components in thd
directions with the total magnitude of f being p,/Vvalue of gs'A'q as well' are shonrn in Fig'82.7b
For each case, the .,r"to"lf 31^s and if gas"- 0, ,h"o the
of the case' Thus there is slip in cases
and FBD\ completed as described above, depending on the nature
ofk for cases^1,5 and opposite to it in cases 2' 6'
1, 2,5,6 with the friction force acting in the direction
tn"L r impending srip ror cases 4,
r,
k
u
L
L
IJ
IJ
L
;
;#;;'il;.
8-
i "r,a
Bx;mple 2.8 (a) Chick rvhether the forces: (1) .f' = ?|:
(2x'1, r"l,+ (v2z
2z(-cosz+ y2)k, (2) 4
'?)vi+
;:?;;;;;n"ra.-'tul
L
r_-
:
i
tr'
11'('I A^
-' *forcesin
:^)l*^':::;
a close^d frj
Find the.uork done bv these
8. (c) Find the work done by these forces along the curve U:
.;
rJ
IJ
lz
+v? + :2 sin r)
',']
,/
(') o-u=l- ^ fr+ zzsinz)
, - \ 2(z+&-,,-,',,-.ftt*rl
2z(-cosr*9
^,- z2)s
l(-2, + f
(22
c
sin
- 2zsin z) j * (2v - 2v) ! = Q'
=.\4zs - 4"v) i -
lJ
f,--
Eence this force is.conservative(b) The work done over the clced path is zero, since F is conservative.
first lind the potential enersi l'- of I '
(c) tn order to find the work done along any path, it is convenieut to
the datum at Io - Q- Using eq(3'28)
The x'ork done equals the negative of the cLange of 7' We choose
IJ
lJ
IJ
IJ
IJ
tr_
Y(d = I p-dr-
J;
- li" dr=:/ ct',o,ol a'- lo"r"ti'v'0)dv - lo''''''v'z)dz
'- - f,"tt ,l ar- fr'zrvav- lo'zrt-cose* v2\d'=" -'v' -t'(-n*'+v2)'
C: {r)=2ri-rsj+r2k. :+ rr =dl) -2i- j+rkr.'rz= r{2) =4i- 8i+4I9F
I(cr) = V (2, -t, t) = 22 - 2(-1)' - l2[- cos 2 + (- 1)?l = I * cos2
cos4
V(cz) = v(4,-8,4, - 42 -4(-8)2 - 42[-cos'I + (-8)'J - -1264 * 16
1275 units'
Uf(& *s2) =.:[y(rz) - y(rr)] = V(2,-1.1) -V(4,-8,4) = 1265 +cc2- 16cos4 =
(2) F =(2x-y+yz2)i* !f":2!l;zzli*2tvz\
Fr=2xyz'' ",
F.=22-y*!12, F" :fz-t+xi2,'
u
IJ
u
]*;
(") VxE-
o
v
i
!
,a
.fr
j
kl
& * l=v2u+
(2r,-g+vzz) (f"-t,*tz2l 2xvzl
I
37
o
..j
',
lt
1
t
'C. ='='
..
Eence this force i" non-iolootir,
l
a;;d;ij;d,r,4;?:iii(i-0) = (y;o)/(o-oit1 = 1;:g1p-0) =+ z = b,y: 0,ir, =,
Jc;o
ahe integral is evaluatecl by replacing y,z in f, in tcrms of - fot poiqts on Cr; z,e in f, ia terrns of y aod
t,y il I| in terms of z. Similar procedure-is adopted for erraluatioa of intcgrals ficr curves Cz ad &.
1Aflfi
wo-t = | (F,dz *-Fydy + F,dz) = I 12"- 0+(0)(2a12laz+'Jo| pqrpllolrldz - rJcr:o
Jo
C2: A(1,0,2)* B(0,3,0), (z- 1)/(0- l): (v-0)/(3-0) = (z-2)t@-2) +
z=L-y/3, z*- zl2,
i
'
fB
.7O
18
-=2t, z=-2y/3+2
-
\.
Fih = I Q, -y+:sz\dz = | 1zr-(-3'+3)+(-32+3)(2e)2ldr
Jc*t
Jt
I
Jcr.A
.
y=-3r*3, y:-32/2+3,
= 'Jtfor-rrr"*L2,:t5r-3)dz:-e.5
.18tA-13
l_ Frdy-- JCz:.q
I @'"-z+azzrd:g= Jo-lt'|:[-vl3) -(l - cl3l+e-yl3)(-zylT+z)2lds
-'
:
JCz:A
= Jo[l'-ny"pz +7y2rs- uy/3+3]dy = 3-?5
{' 323l2ldz = -2
+ ttzldz:
Jt'
Jct.rr-'= {henleszn
fis,, {i3r,rr,
trqtl
F,dz
1B
W-a = JICztA@;ar* Frdv + F.dz)= -05+3'75 -2= 125'
(a: 9(0,3"0)-O(0,0'0),- l:0
z=0, :+ d:c:O. d'z:Q
: Jfo -- * Frds * F"dz) :/b.trl - 0 + 0(o)"tdy = 0.
",f,o* W^-a * Wa-o : * 1.25J0 = 2-25 uoits.
Wo-a-o-o = Wo-t
wa - o
1
"
The work done for this closed path is non-zero. The work done by
- a non-conservative fo-rce in a closed path
is, in generatr;non-zero.
r
,.,,
.
(c) C: a(r)=.2ti-r3i:+r?k.-.+', t::':2r,,1=:f,
,l-:Q)
z-;72, I
dz=2dr, .dy.='$i2dr, dz:2rdr'
12
12
= JII l8r * 8r3 - 1617 - 6fldr = -10818 units
the clced path shorvn in Fig.E2-9- &) find the work done by this force for the
circular path from F Lo C. (c) Is this force consenrative? C and 6 are constants-
a----:->tF
FiaZz.9
.
38
.-I
l
':l
J
J
.J
* = J;"ir{f"dz * F,dy * F,dz) lQF, - 3r2 F, *ZrF.)dr
=
{J
"t
J
J
J
I
':-:--
\,1
tt
r\,r
Solution' (a)
-)
(1)
-)
l3clslo'*'uoo7
-
For path -r{B:
L
b(dz - 6i
Rzb(62f+' *'uo1=
P^b rt6 - R
1t1/e-c =
l) Yields wB-c=
r=82, dr=0 aadeq(l)yt"ld"
J!"
o'
r0t
wD-::J;,^","::='Br6(dr-dr)
simirarry, wrl,-o: li'*r*\ds=-clll*i-Ll4ll2'
Forparh Bc:
-
7\f a-e-:s-D-t = Wt-a
-
wr -c =
-
l
r = dcg{ and eq(l) yields
(b) For path fG:
:
* Wa-C {l
[*
!t-r
Kc t
")dr
+ rD dd] =
li
:
no
to
t'31dr +
!
u *" 4 de -- -c i zB + bd I J''
o-
'r- :- --''t f'\ is
t" non-zero' It can also be
q tlffir
'
ua!"
ts* - \-'
l_ l^-^
"'
'_
"
the'\r'c
since
non-con1rvati1":"",:1T:::: ::1'"::;ilY,:::iJ::If]
rhe force is non-conservative
(c) The
F
it tlri. .*"y.
f -.-->1=-
L
Ii
-{
"
"" at
l:i'ff :,j:".i :"::;ffi';:':;:;':1""';:;;
i:*,'sy
s'
P6-X'
t=
I";-Jti,';l'..o,ll'o
(Fis'E2-'10)
":'.n"*i
[t'
rt' acts at
0 to 1
tima
interval
time intcrYs'l
a fixed inclination of { io the horizont'
F-;ist""':,i:':"f;:T::lj;
I
Lrre angurar q<
*"-.,
ffi;;r e'rr" " 0';
;51J:3'?:::'"1': ::J:;coe-ot
.-','-*o
I
+
rt,li
acceleration of the u,heel is =
-"r' o' :{ j"-:T:
:,::;.::l:;:t
sriP during the time
ao
and
='0'
l-.
u.,
;";;"
Lr-
the work is obtained
oo = t'o+
j
\
ag
a$
cf
[-6i (Ir- 8)jJ: [uo+r'r(h - R]i-1461
W=L'9r-Fl{os+-(L-A)}cos6+r"6sinfl G
)
. w = /'"ott(**?f-"-*e:?l'-cltedlcosc
^.-: ..
L
x
'+ t([ - !) cosf'+6sin -
-
:
(1)
.:
,
-'rz) +,crr3]lJ.i' :_: ''
1,12=Ps[{(uo',1-as/a)fr -aolzla-orIs}cos0*{(Ii'-E)cosO+osrnc}tt'cors{-'+o}"-r'
L
L
*he1
"=;i; "=+,
,'rz=
"*i*u*c]mr+m
1',a^o=*, o,= !o'n"'o'=Lj#:+-"
AlthoughVx[=Q,yetthisforceisnoa.con".,*j:i,esinceitbtimedependent.
;e *a o' 6 = o,o = 0'-'cs(1) rieldsi' '
(u) For the case
:
!-.
";il;;;
:
i
L
...;
..,
_-,
..
:
!^- -rf equal the nrodult of Ito and the distance
Iltt "= ":l
Norice that the work done for tbis
of application of forcli.*
poiuts
initial
and
final
ihe
.. =.. . ,", ,
betseen
case of constant
)
-, -)
ts
cr) dt = co(l -: 2-ot)'la *
ttp - !o+ c-*tb-:ooi+t'rk
-
ba
-ot1ls
&= ra +41!r - "-otllat ! ,.a= us *,;[j"""-"' -,,''l
= ! o' a a, = /oi*t*+
bL-
rrr
7"
FiX-a 2'io
"rith
t.
,JoJ,
_l
o,lo
5i::::i;';"r:.T;;x;;:*."i"
instauts'
materiar point P at different
*o::Y::"ut
m*: ffirlH::];1i"#::i-"4'i-sind1
rclpect to trnle'
by integrating lar
=
-
-,
:
""*';-*i
:'- .
.'+i-:"- -'
='3,t::.+
.
flj
'/.,
L
L
L
L
|-.-
rj'
L
l_
[-
r{
lJ
L
L
f-.
I.lj
f-
rj
u
IJ,
tu
rJ
l.--'
f
rJ
.t-
E
l;
L.
I
l''I
I
i
I
'-=;=
-{\.':..'..
. .'. 3.
., DYNAMICS OF RIGID BODY
I)dne Kronecker's delta 6;1 by
G;:{;
I :11.},
9tP-- tg.:J
,.e" 611'=522 =6s =1,612=66 =631 =5s-6t=6a=0,1q'q'=6rr'(3'r)
For pcints P and A of arigid body (Fig.3.l), Wt = ae - 9e = t t x Let. I*t
u = @i*t tpt = z.'q, then
\ta-- t,Pa' : t2 = 4. + ,tr* zl and let '9= tr"rr * u'2u'2 + ,,3t/i3 -- 'jd,i'
l-.
r
f
En= J *nxap^dm= J^rrnx (s4x zptldtu= J\Ar^'tpt).-(tp^'s)tpold'n,
:+ H,t. =
rici) dnc,t;
=
f ,{r'6r, l:r"r, t;r;)drn,
Defining t6 =
l*{r'A;; go, Ilwi = Iirr+ Ilq+ IAq,
=
i-e-,,Ifa. = I{rur * If2u2]- f*.r,
7{tz = IAq +1*uz+I$aa. +
=
Ifil]
[i* it]
I"43u3,'
I{zu2+
Ittq
*
.H*, "A
(z;ut;pildrn
[-; = g-9; = {u;c;} -g - t.,;6t-;}
(3.2)
(3.3)
i!{at=u'r&q}' (34)
'+
tIl
rfrl
[-'l
,=
rhete Lo,r= lflll
,
rr
^,=
Ln,;J Lrrii I:A r*.| L;.| .
l-rr' 4'
(3"5)
relatite to the ares
Notc thar I* = li. The real syrnmefric matrix [IaJ is catled incrria'.molriz at A
point
the
rigid body, then
of
rr(q) at A- In glneral, fi.f b not in the direction of 1g' If ';l is not a
x mgs^ rvith |fla.]= tI"lklt
fu: flc*bt
AT ,{
", ^=to*-r"*too
of the mass element drn from a1-axis at .i4- The elemetts of [/Al are given by (3-2):
-rua,dbtance
- F;=i;;-;;=
in;l,z*,sr- x!td,,,= i:'i+z!)dn= l{rr;a^
4'^= f d2rd,rr1r2)dra : t',, =
l rrr, d*.,
tlr'013 -
r;: loi+x!)dm,
$, = t"1"2+ ri)dm,
€r= l,!,i+21)dm,
Ifr: I!r=- trrrza*,
tS: tlr=-
rlr-r{"--1,",,au,,
f ,rr"d*,
t, = !(o'+ 22)dnt= *1rj12. ,*: I:,'+,\in--,,,(tf )2. ': = J-!"
(3.6)
+ v2ldnt= rn(tf )?'
r!:ii=- f ,va*,
I{L,I*,1!3 are called the momeats ol incdia and I(!, I$,IA,I{3, 1A,1"n" are called Lhe produck of idirtia
of the body w.r.t. axes ri ar A. k!,k{,&f are called IIrc ndii of gymtion about axes t,y,z aL,{- Fo'r the
case of scss dis{fiDa liott ott a pldnc arca ilt e,y planc with point A in Lhis plane, (3.6) yields I} = t!" +.lf;u '
Lcr-us'estabtish the trrrslortnatioa rclalionsbetween the elenreats of the inertia matrix [-[^] rv-r-t. aies
x; at, A.aad the elements o[ the inertia matrix [I'r] w.r.t. axes a.l (d') at ,'1 (Fig-3-2), where
d :"t.it {oiz9z*aisgs =Qp$. =+ 4 = oirsr, + 6ri :4'di = orr9r'aiq\ = a;oo;$s,(3'7)
(3-8)
- a+ - component of 9! atong % = 9.1 . g = cos[Z(e!,$)J = d.c- of g!' w-r.t- E,
*1r^-=ee=a.lg!,,
(")
:+,?1 =zet -15.t = apgp..9lr=o;prp, :+ z! =airxr.
llr9, t
r-:sing (3J), (3.?) and (a), Ill
'
= f^(r25i; - {xl)dm = aipaiq {dr"6o, - trtr)dm, i."., S\
,-9u
d-.o\
,'Y5)-\
=
Xrr9,
ar(
.lv*
Fi-.
-2
.1 r
- r.L
*"t
*' "'io;'I*'
Dj='
ti" t'n"' "i*iq 9) i" th" r / \, ,- - ^r:-.^1.,
The
-- summiti""'*tt";iiaiilni; w-r.t.
(
respectively'
bases 9i snd 4 by ( ) and )"
il;"; the elements of an-cotity
su& that T =T' is called t zcw o.rdcr kasot'
rilement',
l- A scalar, like temperature ?, sltU So(f)
ico"o''
such that rli = a;e,,p,is called a first otdet
2. A vector, [r. gp^, ;Jii itiil
"L"r"";:
otdcr tctsor'
t:"::d
calS{
is
3- An entity, tite fii1, with 3? (9) elements' such that lll = a;ra;rff
"
sum of 9 terms:
otdcr tcasofcallel
4. An entity D. with 33 (2?) eletoenrs, such that Di;r :i.rfir":.!of is
i'l'l
: tt":
atP
n'
l'e''
Ifl about an a:iis 9: ne* at ,tl, is obtained using (3'9) with 9't = otpge =
'
If" = r{f - avau!* = nunll{r,
(3'10)
121{2np2 + 2l{p2n3+ 2Iil E3!i.e., ,i. = ,;r* : r1;; + I{ati+
'{!n3
"
g' i-e-' c3q : s?:
If, about othogonal axes n and s - sre{ at :4, is obtaiued using (3'9) rvith gl -9zs\ =
I* = \t = olpozqllo - t o"r{'
tj"--tlorrro=/frn1s1 *I$r.2s2*I$n3s3+Ifr(41s2*nzsr)+I$(42s3*n3s2)+I;{r(n3s.+41s3}-(3'11)
3.2.1 Principal Axes of Inertia at z{
n' The corresponding I*' i" called
g
An a-ds a is:called a pitcipal atis of ittertia at' A lt I:i :0 V lpincipal mor.cr,,Ii',.of iner-tia- lrct u" : $rno = I{rne in (3'11):
i-e-,
4f;"=/flarq::tss{-t,'+=0 VgaE, * 1=^a,. + u6=}ni;
It*r- lnr, * {fi., + I*nr+ Il1nr = }rro
(3.12)
rfra1*rfrn,+rfin3=Ia1', r4t rr,
["'r =^[:il , + tr^]tar:r[3], (3.r3)
14 I I
t* {i"r
i.e., I{n1 * I*nr+ I*"zrs: lnz, *- |
"' | ["'J
'li
L";J
liil
=
r*,,r +r$o,
't '?')
-;;:=;":
of [IA]' since z a8d 'l' are defined
rvhere [eJ = [r, nu nalr. Eence n and I ate cigca- lucilot ond eige,.':oal.'e
(3J0) and (3-12] yield
as the eigen-r.ector and.eigen-ratue pair of malrix B if 8u = lt- Eguations
(3.14)
I*!* : If;rnrr.e = Iosr.q = l4' g = l'
arc' *spcclite'ly''ti'c cigce--wctors
llence, tlrc prirrcipcl axes of iaaaia aad pincipal momcttls oJ ,;ntettia at A
least 3 ort'hogooal cigenottd eigea-,alucs o{ *c;ocrt;o trrertri:-l'i'Al at.4. tr' reatsymnretric matrix has at
principal azcs of iacfliogi
vectors g1 rvith 3 eigen-values l;- Eence, there aluays crisi st least 3 orthogonal
ot A uith rzspect to ubich lhc i?;c lia matix lal'es a diagono.l fonn|
[), o ol [l;f o ol
rr'^k:13
t
f,l =13
"f',,'fl
Let s1 : r.ri{. Using (3.4)' g.^ in terms of principallx&of inertia at' 'A is
(3.15)
,',
."i:,
If ut isirthedircctiotof agir<cipalatisof LrterlcioatA,fhenLoi"ir.thcdirectionoflv9,e'g''if<^r:<'''9!,
then cri =u, ui=o! = 0, and Ha =Iifrgi = Ii.lg.
In Fig.3.3, ry-plane is a plane olmass"yr,t*etry, "+ P(x,y,zl = p(x.y,-z\ :+ I*- f*zz'dm'=
0, I:! = j^"yd*=0, =+ z-a-xisisaprincipalaxisof inertiaat.A. Hence alir.cpcrltcttdiczlartoapla,'e
of *ar" igi^ct ! is a pincipal ar:r of il.cr'ria at a poinl lohcl il':':*::?
yz and zt ale pl'aocs of rn11
(since!::_n_'::":
A, \e"'vv
at 'ri
principal o+a
a1ces qe
ate PtrrsrPq
t
and y .,{c
(Fig3-a),
rg-,J-.rr' z anq
revolution tt
of revoluElon
Dooy ol
For
tOr a body
aprincipalaxisof inertia
z-axis
isalso
symmetry) + I*= I!, -0, rf-: i; =0, + I!r=Ilr:0,i.e.,
at,a. H#e ori'poiotonthca:crsof symmerrgof abodgof reooldiott, lhcaxisof rctolaliotctilottyluo
I _r
-,
, by symmctiY f* = Ifyr.or1/rtalJo it oaslittrle a sct of orthogonal ptitcipol ates.
axes
orlhogonal
-
._17
./\
t-r'1tZ)
1
crrgr-?)
Fi 9.3.j
:
.J
-:r
,t
,!
\>j
I
I
tt
' '"*
4r
J
T
J
J
l
J
r
(3.16)
'g&
't_!i
J
.-l
-J
.-.J
7ai
-r-
i
!e. *,
*tE,
3-2.2 Prrrallel Axes Ti'b\tt-"'iro Teram of C ; '
(Fi&3'5)
centre
Consider rwo set of parallel oo]r, at i and i; at
"1T*j I
and
tci
i:
J*;;dm = 0'
tcig' then z; - *
I*-L 71,*: aigir Lpc = i;gi,4t,:
---+
c(-c-J
7,",
!ar 7.r9r-
Fie-35
t,= lye+r!)dm = fito"+,,c2}2*(te+ xs.l\dm
:'i:rr1r!)drn+ t Lr*,'il f a*.+z'c, f *a'o+z'c", f ad'',-=: f;'+*(""'*'Lzl'
t$ =-t^rrrra* =-l @.+ ,crXiz * xsr)&n
-
:-j'rrrro*-,:or,* l:*-'", f 'ra'n-'c, I z'a*= I?'-mzctzc?'
-
Ifi = lfi + ,o(r'"r+ "3"),
-
$, = \9r+o.("Z"arLr\.
/S=€+-('3, +rZr),
L
L
L.
'
tr= I?2-,rltc.zc2,
tfi= t$- ntz.cze,ca,
tt = I$-Dtz,Q3:,Qlt
(3-17)
ts + discrete nass rn at C'
[I^J = tlc] + inertia matrix at A due
the bodies and.adding;the individual
Inertia matris ro. "o-o*ir" uoai* ate obtained by decomposing
tul"tio"" for rotation' if d&aeacontributions using parallel axes theorerrs and the transformaiioo
/I
Cylinder
a
Cuboid
a
of
Elements
3.2.3 Inertia
"d
+
-L-.
=
!-t.L-.
i
t,
I
L-.
i-
ti
L.
I
\-
,f. =
l_ii,t{ *,'tio= =,n(+
Lt'
L
*tP,
* Lrr),
R2 L?
'nR2
t'
'
I:, = -7.
.
* ;),
- t?v = rn(|
=
A RTGID BODY
OF
3.s rvror{ENT OF MOMENTUM EQUATIONS OF MOTION
'
3.3.1 Motion of a Rigid Body with q = corts0ant' i'e', or = a'e"
(3'{) =+
If A ,s a poittt of fhe rigid b_odyor.its 3-dimensional massless extension, then
tc
ti"
=
ru^t=l',X IX
ffi] tl] [iA;]
Lrf' 's""jfil
.
i-e-, E^= Ifrr'e1 +rtzuzz+rs'e"' (3:]s)
'l',j "i';:
nvenience,
Ia.feneral, I/.e is not parallel-to tr, even for plane motion. For convenie"Tl
to t*e isid bods
*'e
*l fr
ei ro
ttzgi
th;i.1:0.iiiaad,i!itionAcoitlcidcsu;thc/g^|r=9./eN&alorr9/C,then
Ma': ilJlt = iLttgi="'o,,t.+a, x H,+=Ii56e, + I*6e2:+ Ifual%+u'3 x (I$t'rg, * t!3u9:-+
= (I*t - I$rf)e, + (ISa + rfiar?)e2 + /$ri'e.,
i-e-,
1u{or=I{rA-It.laz,
Mti= tt* + Ifr-',
=+
Mt. - I:,& - If,.',
tr{tt = Ii,,;t + I!,r'.,
Ma. = I!&-
tA
'.
lvl A3 = l33u I
L-.
L:
*0" =
19,:r.C#-
I
t.- G
(t2 !-4t
:n," .r'nna", (r'ii.s.ou) is divided T* "1"^T'":"'v d;i* "t5"oj':-=,"('y:11'.::,*ii*":":l':i
Hence for the cvlinder: f,
n-,."";;;i";rt.'1J". For the disc rf. : R2dm/4+ z?dm, I!. = R2drn/2.
I
L
l_r:rY
tf"'=rref!,
L--
:
"l'^,(blz)' +, z"bl2l+d,:1fr1",
-,
*t"pt +otl?t'
'{rr* -
I
+g'
*
ri!4es)'
.i
(3'19)
r
-.T
J
In thissection,a witi tidt.poiat-of,body and in addition A=C / selr =91e,t4r is along 'AGIf'aris 3 is a pritcipal aris of i,;c/.ia at,4, then I$= I$ = 0 and (3'19) +
M1" = I$b,
i-e. M,q, = I{2,o. I If z-a:<is is principal axis a[ i., then M,*. = I!,q-l- Equation (3'20) is
centre of mass C and any point;{ on 93 through C for the follorving bodies (Fig-3.2): f
1- Body having two plones of mass sym11.elry rvith s3 along their line of intersection'
iJ-J
3- Slab like cylizdncal body rviuh g" parallel to generator.
L
If angular velocity is constant, then a' = 0 and (3-19) yields
M-.* = -1f",,t291
JJ
ffiffi
J
L:t-l I
JJ
JJ
2. Bodg of rcuoldion with e, along a:ris of rerolution-
.
(3.20)
Fii.3.7
l9z (")
* Its-292-
Hence, a variable moment Mo b needed for lrotion ai constant t r, since 91 , 92 rotate rvith the body- flos'ever,
,/q * a principol axis o! inertio a! A, then (a) inrplies M.a= Q". /A - Itr= 0, i.e., no rno,tr?.ent M^ is
needed to mointain motiott at constotttar. Thus if es is principal asis at
? :"0 ?"f81.9*!::Eq.:
g;1,
A body ro.raring abour a ftxed a-xis (Fig.3.S) is said to be bala,nced if thc 9#;"7E-*
beoritg reacliotts al O at& B arc zcro,:urhaa it rololes at co:nslanl aiigular ulocitlll
:+
+ rc -0o
tngc = -mr,frsg.= f :0
-I?"n'Cr+ I?"r'*= Mc =9
i-e-;'C is on the axis of rctation and the a:cis of rotaiion is a principal a-tis at C. + If:. = If"+ azrzcr2c3 =
O, I*: I=9"+mtco1,c, = 0, :+ axis of rotation is a principal atis at, everl- point on it- I{ence, thc necessary
:
and suficield caulilioas oJ balancing are:
I-.Centrc of moss C lics oa atis of ';1tatior-- 2- Aris af mtatioa is o pri*cipal axis at or.e point A of iticfFor angular impulse about axis qj thrcugh -'{:
1
(3.2r )
fonr."(tr.ru) = I$[a.'(t2) -,r(tr)]-
.
3-3.2 Euler's Equations for Three-dirnensional Motiou of a Rigid Body
.I*r A bc o poitr.t of lhe igid body and eitbcr .4 Z C or ga1 = 9. or gs1t is alottg,{C, then tr'-r-iprincipal a:res of inertia g zt A:
r,^t=
['f + [] [I] tl3[]
=
+ rrr,=rnr"1s1 *Ig1u2e'+r*-s%
Chocse g. fixed to the tigid body so that i$ = O.
Ma = E-^v
=
E^g=*o, * * * il-e = Ii'&*1 + I!r6-er+ I$n3e. *
:
ez
ltr
l
s3
r[l, ri:-r r#".
. = [fi6r -$Nr-l*)-+""]er *-..
It{a, = Ifr6, - Q$- 1.43),t2-3,
i-e-,
MA" = I*zbz - (IS - If1)t,3r.r1,
a,
16^r=I*,i, - (If1- l{2lop2.
Th€se are Euler's equations for ptincipdl axes of inertia at .l{.
3+
3-3.3 Direct Equations of Momeut of Momentum if I{1= I{2
q b6ing the principal axes at 4- Every axis n(- cos0g,*sin0g3) I
31
g is a principal a:iis
Id = t{r, siuce for !.(= - sin 0er * ccdg) I n & J- 9r, eqs(3.10), (3-11) yield:
/f, = rfin1s t * I*r.zlc* I$n3s3 = (-/fi * /$) cos0sin 0 = 0'
Ifr = If, rr,.0 + ISa2.0 + I$n3.1 = 0,
I d.c.'s of g are 0,0, l]
If;^ = Ifynl + I{szi+ I$nl = rf, cos? d + I$sinz 0 = If1.
43.
J
J
J
J
J
J
J
J
J
J
j
J
J
J
JJ
J
J
J
J
J
J
J
-
kr
Lil
L.,
l-.,
LjLJ
L_'
LJ
L.
L:
L.
LJ
LJ
LJ
L-.
L_
f
LJ
L-
LJ
LJ
L_-
L.
LJ
LJ
L
t_-
L-.
LI
L-- -'
Li,-,
L:-
LL.
Hence the a:<esg are.q[rcetsg t-hat9" is aligned with the principal a:cis 3 at,{ (Fig3.9) rb6eas 91,'9
could rotate relative.to the rigi(LMy about q3, stilt i7, = itr = i$ = 0. The al.grlar oeldtg {l o! thc
fmme 9i dtfcrs fmm thc a;egzler oetociej io:of lhcTgid bodg in its 93 componcnt:
Or = ulr {lz = @2, Q: # -s,
urrgl + u2g2+ a4g3,
", =
Ht = Itrurgr* I-42,,t29r+/*atgj = ffl(trrer *utzgz)* I$ar39",
M,c'= Lev = f.,11g.=.-,.-*o x tt^= I{rQ)r9r*t:29)+ I$ti39" *ax H^(3.23)
Q.=Orer+Oz9z*Qag,
!.r* r glo.cedu-.1g can definitcly_be-used for a body of revolurion. ..
5'.i:#T".#'*of;"ai't;"?il;;"tff":r:*."-;;=';;e,*;.e-*i.s,+&^"r.
'6;>&c-1
'
.e )
A translating system behaves as a rigid body- Equation (3.22) + M-c = Q, s!4_c,e.g1_;';Q. The equivalent
force system at C is a force F and a coupte of nioment M c. Herce Mt = M-c + rct x F : aa, x ,ngcv.
Hence for a translalirg sysletn, Mc
,: Q o:ad M a = 0 iJ AC is along 9cg. The moment about cocrg point
is ao{ zero. For the translating system shorvp, Ma = Ma : Mo = Q, but I{s t' Q-
3.4 CONSERVATryE FORCES
A force F is called .conscrualire if the *'ork done by i[ fron'r time f1 to [2, during l,he motion of its
nraterial point of application from location !r b L2, is i*dcpend.ent of the poth Ct colllectfiig, thcsc. locations-,
wr-z:
lj,'r-ool= tJ,r:-rr, vc1 andvr,rr.
(3.24)
I{ence conservative force acls oa thc same malcrial point of the body and, is iadependeint of ik oelocity
and time, i.e., F = &r). Equation (3-2{) implies that the integrand is a perGcL differential of a position
dqpendenL scalar function l/(d. called the potettial cncryl!, such thar
:
(325c)
dW -,F - dr = -dV,
rI'(t")
W\-z = -. I
(3.256)
dV -lV(a2) - t,(rr)I,
=
Jue,)
i-e-, urorl' dora from L1 lo r, egaals ncgatiac of the change ir potential cncrgy from y, !o 1r- Eeoce roortdone by a cotserttaliue force in aay closcd path (rz = rr) is zerc. Let 7e be the dalum for V, i-e- y(ro) : O.
Equation (3.25a)
r!-
V(r)=- I r.ar,
"+
(3-25c)
Jb
i-e-, potential energy at a given positionI equals negativeof the rvork done fronr the datum to this position-
!\'riting F ' d1= -dV , ia terrns- of the componenk of { and dr \r..r.t . various coordinates yiel&:
f,dz* Frdy * F,dz = -#* -
ff* - Ynr, =+ ," =
F.d,r q F6r dg * F,dz =
1
-#, F, = -X, ,, = -#,
I
)
ort Er av (3.26)
F- - -ol" ^
-{* - #ot - #or, 34 ,."-Ar, fo=_fi, L=_*,
AV '
:+ P, = -Y,
F,ds= -7;*'
av. av'
+ F=-#L-ffi-ffx--v%
where ( ) =i#*i#*t#
=Gradient ( )- 1b.zzy
t,j
The force corhponent Jt in direction g is obtained as
I
Os
I'. = F 'g= -gf'-e-
-(directional derinaiive ot.V in thedirectionof 9).
.
.
1':
-: ;;
.2i,
Thus'if.F iscoasentotiocthca F--YU- Itscoaoer:sc knolhzc,e.g.,forT=fgrccrrrl, F:-gY-is,
.I'-,,t
notconservalive. IIF rs conscractio.th"oVx r.
=g.
r*-*t*
=r-#.ffii,* =Q
".u=l+
lr" e
Fe +F
el
++
ii
tI
?
A
c
giiv x E: Q itt o stmpty'cottt'cc'cs "s'e"'for a clced path C (Fig'3'10)'
w - i"E.dt=
10), W
Jl zf,-dl!- (xrrl,z)
tcr 'oL= [tYx
since
theorem:
Stoke's
from
foll,ows
Proof
I
r(x
3.4.1 Potential Energies of Sorne
Conservative Forces
1. Using (3-25c) and integrating along the
piecewise linear curve C1 (Fig3.1r) Yiel&
or -edmk (Fis'3'12) is
"'o^::!t,:f*:":!:=*:"'
;:,;;";;jffi"H,
";3;;;',:";;;
-E = F(r)s'- (Fis'3'13): h€.
l,u r,ou L, a,'l o2, totith Er="':
r.
v=-
For datum at r = r'o :
gr
(3.2s)
F.3.3.i3
l,'r1rya,.
-
(a). For rnutual sravitatiou force f(r) : -GMmlr3
(lr). For a
F.g.3rl
Fig.3.t).
L
dV = -(Ft'drr * L,'dxz) = -& 'd(r, - r) = -Lr'dt
-P(r)dr'
: -F(r)!-dr=
-F(r)qP
2r = -4..4,*=
"r
t
fc,
-GMmlr'
*'ith datum
1-"1-=rTI'r^'"tf:SJ
an extension'e = r - Lo'
for
be
f(e)
tlre'spring
pirll
on
pair of sprirlg fo..e" (Fig-3-ralJ.t:trn"
rr.lrereIoisitsunstre[.t,".at".,gtLH",,."f(r)=-J{e},d,e=drand(3.29}yields
Z..WP;#%
eiq' 5'i.+
.-r=!;l-€i
v=!o"rkta",
":
(330)
(3.30)
.
r:-.--- -6F
sp;.,,g of sti{in'{ss f' /(") = &e, and
a- li*ccr
witlr datum at unstreic5ed conliguration 16 = f,,. For
extension
ke2 12- k62.'2 rvith 6 (=e) being the
the
-,-ields V j"l
Mr = !@\ for telatite axif lvist,of 0 bet$'een
couple
(c). Siaitatl y,tor lorsional spnag."itl
vith
(
k,,02
and
12
&r0
=
torsionalstiffness kt' l*=
ends, V : I:t(0)d0. For a ;iicar,o"siooollsPringof
untrvisted configuration as the daturn'
3.4.2'Workless Forces
lVe lls*, sonoe forces s'hich .are workless and some
individually each may perfQrnr trcrk' .
pait of forces rshich together are wotkless though
i- ft"go"ti" force F = 9gx B since W =E-'9--- qvx B'o= 0
slip' since li = A, 'fle-,.; &1;0 = 0'
2- Rr&tioa ft, at conf,acr ,P1 ivith a 6xed U"aV *itt-n"
slip' siace W = N\' g' = IVu* : 0 as
3. R.eacti,on at smooth contact rvith a fixed body rvith or rvit'hout'
Idsa
no separation'
eirher rrr :'0 for impending separation or ?o':0 for
4. Reactioo R 4L Aat smooth ball and socket or smooth hinge
joinr with afixed !ody, sincelil'=
!i:!t*=,*t,0: l_, -^^,.^.
fY
,-vv
qej
gr^-
N
-, __=_.2 .{ g_ *_ ?,
i. H:i'tr j:";-,fi:'i5#:i.:::f i*i:'i^:Y;"'*:'
P2 with noslipsince
6. Pairof reactions Rr-and -&r aicontaco"f n *itft
w = &r-gpr +(-&r)' w, = &t'(qe. -!rp,) = 8t'0= 9-"" !!.r, =.!e''
' Y?r
-\9----u' (gr-.zj<{
Nr
T.Pairofreactionsatasmoothcontactwithorrvithoutslip,since14/=
0
8'
/VrE,'9rr+(-Nra) '!p,7/rrr(,,'!rP. -4'cpr) = Nr('a)" -(up')'l =
no'separatton'
for
(up.)o =
o,'('o1^
1i::)"
- (opr)^
as either nrr = 0 for impending separation or
. r*
l'no'seRarati1".',, a$'
,tt
g- pair of tcnsions in light inextensible cable since since W =T9' cr * (-7 g) 'ge Y
-g-;rffi."llU,
inr
for impending slacluess or 9 ' 9t = 9' W for tight
= T(e - 9r - e - gz) = 0 as either T = 0
e. pair of teacrions {Be,t:,lt$u':L4gl* ".rt #rt_ball and
i"]'::
.
F - x ( A a ) s = 0'
x
.
= r r. @ -*.J=', * N. = e s
]; rfi
ffi;#;,: o.lt*ll"oi.t*JL."
@
*Y
L"i"*"Xi;"1#ffi;-i--"6 ]
t
I
I
I
:":x':: :::1ff::x1
-.,,b-
-)
v'E
.iJ
--Eg. Z.rS
,<
iru
I
I
I
l
-
S
I
lr
I
{
!:'
A RrGrD
3.5
:oDY
t rcrNETrc siis&€ix EeTPssIoN"or
rccanexPressthekineticcnergyf ofarigidbodyas
O"iO6 1r.19)and lpc:gifsa.''
x rt7'6dtt
6 \ma/+ *gr'
l:'"
r - lmtfq* i i:?.?: **o71i l^*t(crx rPc)d =
gs
I{ciui
= I;c;"';"';l (3-31o)
' s=
y = lmul + +; s: **rZ + lt$.r.;
fusing (3'3)
(3-316)
2lf'46r"1!'
i.e., ; =i;; *itt?r.?+ $"$+ €-3 *2lf1't1,t2*21$wt'tt+
(33rc)
.
L-
T -- lnrf6* imi,;'
ri,";'1,
i;i';'i
where r reGrs to principal axes of inertia at C' tf
gp : 9P, =
ia^'s-= ll$-i-;'
r = i !,'?,ai,= i ! *^'1,ax=,,^|d*= fo-' l:r^*
:^o:=
2l!*uzut * 2llvl-tl'
i.e-. r -= ;iijrri + I*lrtr+ I$"'f * 2l{+'tr'tz +
r = lfli{ ui2 + t;! w;2 + r#-i'J'
L
pohrt r{ of the'rigid body has 9't.= Q' then
,
-
ra ar,\
(3.31e)
can be expressed in ternrs
wbere + refers to principal'axes of inertia aL A'' T
rigid'body through A:
of th6 instantaneous axis of rotation.(Fig-3-[6) otthe
Q
(3-319)
L
L
(3'31d)
For the case of g = -g", ;;
ac C' (3'31b) becomes
9' may or- nray aot be a pdncipal axis
7 = !rno26 + ll$"tz '
(3.31n)
RELATION F.OR A RIGID BODY
3.6
- - - WORK-ENEILGY
'ot a rigid body can be expressd as
ener&v T
The rate of ruork I4l of the e-xternal forces and the kinetic
'F.
-l
t
L
+'i = tnec,b i7 rr"' e,"=dtn: E-' o, * ! *"'u x apgdn = F w +z' l^r"
-
-L-ELr
x gesdta
:
F' vq + Mc'q' [rncc - E- Ec Mc]
''a
= F'ec +r' 1 1 l rr" *' r'.a'ol ='L' vc *'''it- -a:l.d' in integraled fornt
lfence rve ob[ain the following uork caergy rclalio* it mtc Jon:n
(3-33c)
Tz-Tr - I'{/rr:'
'i iv,
tnd 2' DenotinS. the rvork done by the
where fi, T2 *ethe values of kinetic energy in configuration"-rl
+ I'iz-",, 1-3''33a1 reduces to
and nonconservative forces by i, *dwn", iI/.,=l'v"+li,r- -'it
:
consei'ative
(3.336)
(72+V)= (T1+ t'i; = Wn,r-|,
I and 2' If 3ll forces are con1lati5'tfren
where y,, Vz are the values of pot-ential energy in configurations
ener''r):
.oo"".-..io.i"rmecu;'cal ener'y (sum or kinetic and potential
#". =;'",'i A;;;) ;;;;
i+v =i{n.,
Tz.*Vz=71*Vt'
t+ir-o,,
F.
t-
L-..
System of Rigid Bodies
3.6.1
- - Work Energy Relation for Interconnected aad added up to yield relations for the oholc sgstcm
Borr"rl"". tt.iri are applied to iudividual bodies
the system since some of the' forces-t"'.tltn
which involve work done by the intcrnal and the external forces ou
the s'trole sysiem' De[o[ing the cootrilut'ion
arc external for an individu-al body are ir fact internal fiorces for
relations corresponding to (333)
of internal and external forces by the s.bscript int*ext, the work-elergy
teduce to the following form fot various cases:
*
L_-;,
L--;
L_
L.
(3'33c)
44
F:3
"/lc siig
no stip
'"\
-
4 -Tt = Wiar+c:rt-2r
(77 * Vat+e"rzl - (Tt * Via**t1) = W?t,,.+crtr-r'
i = i\lint+cst,
T + Vnt+cEt = Wac!n1+l"=i,
(334o)
(3'340)
(33ac)
T2* Vat+a2=Tt*Vntgutl'
'i-*Var+e,,=0,
If all internal forces of interaction are togeiher r,r'orkless (Fig-3-18), then in eqs(3'34) the
q'ork of internal
are conveoient to
forces is eliminated and these reduce to eqs(3.33). The rare form and integtated form
provided the
obtain acceleration (or angular acceleration) and velocity (or angular velocity) respective\',
a priorill'- frr general,
s],=tem of connected rigid bodies has one degree of freedom and. W can be obtained
and
rsork energr relations are conveuient to obtain velocity as a function of position. Impulse-momentum
velocity
obtaining
and
for
problems,
angular-impulse-momeut of riromentum relations are useful for in'rpact
as a function of time provided the force is a given function of tinte.
3-z NECESSARY CONDITIONS OF EQ1}ILIB;11UM 01. A RIGID BODY
TIre rrecesscry artd safficicttt cordiliors of cquilibrirm of a rigid bodg are.: l. E(t) = 0, 2' i{-r(') = Q
and 3- the body is itritiallg in equilibriutn al t = 0Proo-f Let q. be the principal:.a-\es of inertia at C-
.
-
Ma=A, + 'i:w=F'r'c+Mc'u-=O V t
T -- lnu[ + +I?t-?+ Lrrg*i + iI&-3 = coustant -.(0) = 0:
F=0, Me=O, +
Each of the 4 terms on the l.h-s- of eq(a) should be zero since the suur of these
u6(t) - g,
:+
i-e-. the rigid body remains in equitibrium for alt time.
u;;(t) = Q,
e
-P
'WY=\a
3.8 CENTRE OE PEBCUSSION
Frg,=.tSCt
,J
jcy'G'
- si(0-)] = rn&!.,(0+] - -(0- ]!co = t= I.s. + (Ie + Iar"4,
I' = 0,
t6 + Iq = nrlr[<.r(0+) - -(O-D-
(o)
€t (o*) -.(0-)J = Iaaeo3- (&+d)rq-
(6)
The solution of (a), (b) yield: A'*r = (Ir + dllel i?s and,
16 = {h(h+ d)/(&&)3 - rlra- .
(")
Thus, in general there is a1. inrpulsive reactio-n at,O. The locotiott of poitrt of applicelioa'Q of e 'roTsucrse
;ropulsipi force for ultich there is no impulsiutEaction al the beoring al O * callcd,.lherccn1.;v o{ percassion'
It -rs obtained [rorn eq(c) as
[(h +d) = (rg)'= 1t$12 + liz,
:+
(3.35)
ad = (&$)2-
For a thin uniform rod of leng[h I, hinged at the end, eq(3.35] yields d - L/6;and for a thin uniform circulir
disc of radius R with hinge a:iis normal td its plane at the periphery', eq(3.35) yields d = Ri2'
2- Consider afree igid body at restsubjected to instantaneous impulse Ieed L CQ al Q (Fig3'f9b) with
principal axis g" at C. The impulse-monrentum and angular impulse-nronrent of momenium rclations yield
Iansca = d Ig = Iaca[u'(O+) -'(0- )] = /$ r'r(0+),
+
c,(0+) =
dlol&,
L: Ieea - *[gc(O+) - !c(0-)] - mpc(O+), :+ sc(0+) = Igeatn'
4
ix
t
n
h
4
e
1
t
:
i
Eence the distance r of the instantaneous centre of rotation 01 from C, at t = 0+ is grven by
rd=(t$)2'
r=u6:(0+)/.(0+)- t$/md +
YctoJ
comparison of eqs(3.35) and (d) imply that ot:- o. Positions of 0 and Q are rccdrrocclt
:
l
I
i
+'r
vy
,//\
xe,
Fi 9.:.rsb
J
.J
j
]J
J
-l
"-l
l- Consider a rigid body with fixed a-.iis of roLatiou at O along 5 (Fig-3'l9a)' Let an instantanebus illPulse
ti"n
(impulsive force) /a(0) = Iegoact at Q in a directiol normal ta OCQ ati = 0 and let the impulsi"Sffi
relations:
of
momentum
impul'+momeni
ar O be -IO(0) = t.g. * IO9+. Impulse-momentum and angular
rrr[gc(O+)
J
rJ
(o)
positive terms is zero:
e{t) =,9a-6
!ac(t) = q'
T-Agd
i=
i-.I
.J
,:- I
(d)
14
IO9,p
J
J
J
U
J
j_l
rJ
'--I
-J
J
J
-J
J
il-J
-J
-
jJ
J
J
k,
U
'..r I
U,
'lllhile working wur J[aitmci or hitting with a cric&ct bat' if the imp11 point is a centre of percussion
w-r-t. the point wiere ih" h"rrdt" ilnlta, thi there b oo impubive reaction'(characteris[ic'stin8') oa hand'
3-s TMPACT OE RIGID IODIES
3-9.1 C'enerat Smooth lmpact of two {Jnconstrained Rigid Bodies at a Point
point of body I having
Consider smooth impact of [wo unconstrained riSid bodiej (Fig-3-20a) so that
'4'
plane at
is called
mass ml makes impaci rvith poiot B of body 2 having-mass ryz-The common taugen[
"t to the
g,
normal
aL
A
direction
The
planegla.ae of .impacl. I*t.g, e1 be orthogonal urrii vecLors in this
impagl b called ccatral impacl if lhc ccntt's ol mass G lie on
ltrrr" of impact is ciled line of impact- The
'rol-cent6l
impact.. The impac0 is modelled to occur in zero time'
the liae of impact, othervrise it is called
L,eL O be the point fixed in space which is coincident rvith / at impact'
i just
Let el and 96- = u;oe.n+uisq+rri'gr be the angular velocity aad velocity of cenlre of mass of bodv
ui,S, * u.:.e5- \4re
before the impact- Their cbrresponding values just after the impacr be ari' and li. = ui.g." +
in direciions
body
of
each
momenta
The
set up 12 equations for the 12 scalar unknorvns irr: ,Al,C2,vb1,!.1,9, 9l are conserved, since the external impulse ou each ii only along e,", *
TJ
tJ
u
u
u
u
t-.
tJ
u'r, = ur,
LJ
LJ
LJ
(3-36c)
u'lo = v?a'
q;, is conserved:
;
(3-366)
ttt?!'-c2+ EL,*lczox ''2''2,.e'=Ec,*lczox
EL"+!c.or.*r*r=Ec2*lctox
l"u"t''
where Wr)=[rc Jb,], Wrl-- tfctltgl'r], etc- Equations (3'36a.b,c) constitute 11 scalar equatious' Atr
of
crnpiricat ilalioais added to conrplete the set, viz-,e - trfto..vhere:u, is.the't'elocity'of separation
just
g.
before
along
B
g from .r{ along E just after impact, uo is the 'velocity' of approach of A tot'ards
impact and e is an experimentally deternriued constarrt, called the coe$cicnt of rc'stitution' Hence
t--
(3-36d)
v'on - a!^o = -e(o p, - t:*)'
t-
ttAa:9a -9, = &c', *g, xte6r)'ea,
(3-36e)
iu*n = tB:gq
+9zx
taCr)'9*={Slc:
If body2isamassivebody.thenrn2=cp + 4=-;t, s'c.=9gr,t"go-ularand(3'36b)forthe
sysiem of bodies and (3.36c) for body 2 do not yield aontrivial equatious. The slt scalar unkno*ns erl, 4r
]
are determined frorn (3.36a), (3.36c) for body i and (3-36d):
(g'gZ)
u'1,
= u1,, oir = ure , ELr+yrs,xwtv';;9- = qrr*!crox',r,t".:11 -:" -::" =-",(.q.ea -"x^)If body 2 is a massive body,,at' rest; 'then' the last of the'equations iu (3.3?) reduces tor,ufi,l.'= -eurn '
3.9.2 Surooth lmpact of two IJnconstrained Rigid Bodies in plaue urotion at a Point
rr'en=de-9^=$br+sal xL{cr}-g.,
lubo = ts - % = (!L2+dz x rac2) -9o,
:
rvith
u
u
L.
L.
LJ
LJ
The six postimpact unknowns (Fig-3-20b) ,u'ro,llr,rtln,{2r,'-t'r,tt', are obtained using (3'36a,b):- ,i
u'rr =
ur,,
,'rr7tt2rr
(338o)
mro'r..t*rr!. =tlI1u1. *mzx?a
and (3.36c,d) for the configuraLion ot'Ct,Czshorvn in Fig.3.20b:
L-
(3.386)
iSul, +m1u'1rr1 - I$-. + mlurnrl,
€* rL - rn2v!2nr2 = &' rr- rnzu2n rt'
obn-t7o=-e(oao-ue.), i.e., (ul, *u'2t2)-("t -arirr): -e(o2o *ur3r3)-(u1. -&r1rrl'(3'38c)
since for the given configuration u,tn = uh -(rtrl, aBn=ttzo*ttzt'2.
If body 2 is a massive body, then rr2 = oo * dz = t tz, *z: \2, utrn = !gn, and the t 1st e9ua!i11
in (338a) for the momentum of system of bodies aud, second equation in (3.38b) for body 2 do not yield
LJ
L:
LJ
nontrivial equatious. The 3 scalar unknowos ,1, l/1o, u'1, for the configuration of Cr, Cz shorvn in Fi8'3'20b'
.io^\ /e 2aAr /a aa^\r-.^--:-^J r* ro
t mZVz+
(3.38c)1,(3.386)1,(3.38c):
--^ determined
from
are
I
L..
L.
,
I{o ofeach body is conserved, since the external angular inrpulse on each about fi-ted point O is zero:
ELr+rcrox ,ar/cr=E-cr*tcroX n1u6, + I{Lr*lcpx 'nlu'rng! =E-c'+bPx '}rtulnci' (3-36c)
1-.-
L
=
vr,
mrui. *m2rt!2o = rnlutn * tn?u?n.
t-.-
L:
4.
There is no external impulse on the system of ts,o bodies, so their total monrentum along
Lj
L-.
oir = 0rr,
,
,i'
tll 1
7
ii
i
j'z
LJI
/be
-
-n
F;3.3-zot)
+8
--J{vza
'n.
V-
l
I
l
r-I
,^-J
(3.39)
,'r, i r..'
$u'r+m'o'vnr1 -- Su1+'rtlulnrtr
trt')J'
*uzrzl
orirl)
-,(".
(u2- *cr2r2) - (oi, = -c(u2.
to u'1. - atlr1 = -e(r,L -4111):
reduces
(3'3?)
in
equation
last
the
If body 2 is a massive body at Eest, then
Bodies at a Point
3.9.3 Smooth Central Impact of two Uncoustrained Rigid
the angular velocities are unaltered and usiug
Since the impulsive normal force passes through G,
-{3-36a-e), the postimpact rralucs are obtained from
rr'r. = or, u',, = uga, oL, = pzr. lie = Qt, ntoio *m2u'2o =
mlula *'rR21'3n'
a'2o
- ttln - -e{oz- - ur, )'
(3-40)
,l.he velocity of the centre of mass c of the system of two bodies remaini constant since the orternal impulse
system is only due to changr: in g' component
on the system is zero. Ilence change in kinetic energf of the
nr:) = (m1u'1, 1m2oiol(mr * rnl) l:
.rf velocities of G relative to C I rca =(*rrro + rrf,-lrz)llU +
Body
]*,[',, - I!]+*#41'
1: A?r = |nr1[ri; - Wl'-
-1
:.}.#[,i._oi.},'.-(o1;.-,,.),]=,ffi(e2..tl)G'.._.t.),
'- arr + alt: iffi(c2
- t)(u1" - t':o)"'
:-J
''J
';-l
,-J
'.J
I
,-J
,i'.J
,J
.J
,J
"
3.:o<
J
isGatled gcrfecttgclas{icimpactfiorurhichthereisoo
-\f <0 =+ e3-1<O:+ O(c!f.T\case€=f
pciccdlg ?lastic impact fot *Lich +he
lcs of kinetic energlr (A?r - O) a$d ?, = ua- Tlre case,c = 0 is called
loss of kinetic energy is ma-xirrrum and u' = ['
and the consenratioa ofrmraentum
For impact of body t with a massive body 2 (Fig320c)' *, = k,
J
L-J
q.rut6, o!1, : oz, u.L, = 'l'r tLn = u2,,' 92" - t)'rn = -e(vzo to
In particular if body 2 is fi:ied. then {4 =?.tr = O and {3'41) reduce
ui, = ur." ,i, = ,ar, ,1, 0, l2o :'0, 4. = 0' o"n = -e7t1u'
+
t3J
in s,, does not yield a nontrivial equation' Uence (3'40) reduce
u'r, = ur,
to
u'tt :
(3-41)
(3.42)
=
3.9.4 Other Cases
impact A has zero velodtY just after
If body 1 impacts a perfectl-v rough fixed body 2, then the point of
of nronrentum about o:
irnpact. The postinrpac0 angular velocity. gr! is decermined fronr cotrs,:rnation
(3'43)
rvhere l7bl = [ro][4]'
Hb-,= E-c, + rsp X rnlucr 1
::
ttre rigid body I'
:r:
since just aflsiimpact !r+ = 9o: 0, i-e', O becomes a point of
of each body shoul{tbe draws shorving
FBD',s
point,
then
lf one or fhe oiher body is constrained at some
points' The postimpact data is obtained
all the impulsive forces at the point of impact and the constrained
of momentum eguations a'boilt centre of
by writing impulse-momentum equations, angular impulse-moment
and the coefficient of restitution equation'
mass (or about a fixed point, if any) foo
3.1O GYROSCOPIC COUPLE
"Jbody,
inertia at o equal *' = 8''
1. consider a rigid body with a fixed point o with two principal monrcnts of
g.
itself precess at coastant Y" fg
and
L,eC the body spin at constaat ratc sc3 about body-fixed a:<is 3
about O reeuirld, to maintain this
about an orthogonal axis E fired in f (Fig.3.21a). The mome nL e-o
acceleralion qif of the body ane
motion is called ggttscopic o11plc. The angulapeio"ity 91 and angular
@
- pE+ se3 = Pg.s * se3,
i-e- orl = p, u2= 0, u3 = s, t.rl
tir = P9t x sgs = -psg?
/i
: :.r3 = 0' r'r2 = -ps' Applying Euler's cquations of motion:
3 rt<t
+fl
99r
>p
,::J
et_-J
ri--J
il
-il
'=:J
ilr
.J
.:J
J
J
:J
il;J
iJ
J
- )'
o
:t
- l'
\i
E
o
both precession and spin directions'
rvhere y = pE- s =sg. The gyroscopic couple is' surprisingly; normal to
Hence
2- Co;ds the case of precession axis I at ao angle 0 l'o the spin axis 93 (Fig'3'21b)'
ut = pS+ sg- p(sindq *cosaq3) * sse,
i.e. <.r1 = psia0, on=0,@3 = s-+-p.cos
,L= pEx s*, = p(siuast *-:*':]
g--tt=tt*a$t-
-pssind- Applying
= -pssia9s'
1"*
Euler's eqgations:
fr-t Gr- E)t :rs = 0,
Mo2-- Ir9r.r--(€ - Ifl),.l3.^r1 = fflp2sin0cosd - 4(u+p"*O)psin6'
7[ot= Ea" .. (.[f, - I$)urp2 = a,
tr,[o, =
-
Lr-
i-e-, Co = Mo: [Iflp?sin0cc0- /&(t*pcosd)psin0]g:'
L
L.-
iL-
l--
t-j
tj
I
L.
(3.44)
rni[r'q1z:-":l:' oI
3. Ler a Uody with ifl = I?ztoLale,abour fixed poinr O (Fig.3.22a)- We s''rsh to find the
cos &in (3'44) , "+
r^r3
+
g
and
=
<,4 so . that preccssioa al cottslottl . is possi6la :Usiug Mo = -rrrgh sin ds2
l
-s
fflp?cos0-I$qp*mg'['-0- (3'45)
flflrp?ccd- $r,,4p+mgtlsind - 0 + sind = 0, i.e-, d = 0; or
of the quadratic equation
Thus tsoo r.alues (rools) p1, p2 of uniform precession.exist provided the discrinrinant
(3-45) is positit'e, i.e- if
z
U|*i' - 4(I?rcos0)msh > 0
-
l
i.e.,
if
_1
(3.46)
u! > l4nsht$lQrlccsl'
(3.46) that the sleepiag top
Tire condition (3-a6) is delinitely sat.isfied if -3 > llnghlfr/IlrJ. It follorvs from
conslant 0, is follo$rd
(d = 0) is stable it t t! > l4rr.ghlfllf$!1. For a spinning top, the initial precession at
at some staqe condition
by rvobbling rnotion with nutation because as r.r3 decreases dire to inevitable friction,
/f--\
rc i
(3-46) get-. violated and precession at constany 0 is not possible'
/",
If condition (3-45) is'satisfied then the trvo possible precessioa rates, p1
(fast precession) and P: (slow pfecession), are given by the roots of (3'a5):
Pt; tz =
.llllrngrlc()ss\l/zl
?r*Sh cos 0 ', r7:
:--fr:-€'t
ff".*t,t(r-:W
tIg**:a
It
F.8 Eg.=.z2cr(")
t
FortIrecaseof-!>lt|,,nghcos0|(I$)2'tlresquarerooLternrisoipandedby.binomialtlreorerrr
s9S
retaining the predominant leading terrns, the precession rates are obtained as:
?E * A a-,'4
pt= n.sh/(f;33-il! '
p1= I$u4/(.Ifl1cos0),
,-NfY^ ^
The relations of tlr'rs sectioD are valid for a spizrrin g projcctile*itr, o "ra
nrgl replaced by C and F.L, tvhere F is the aerodynanric force acling.opposite
(,
'' ./i,"";g
-
ut ='
1,
-'i.r.zb
tj :tr;:ll'xff:ffi;"j."ffiH"Jj.:"$;l'#i'.,o" :, r il3n+.rrse
LJ
Ir
t---
L-.
LJ
L-.
L..
The spinaing top shown in Fig3.23, spins at rate s about its axis es, its axis I \\bAe*O
precess€s a! rate p abour the fixed vertical
axis g.r ia the vettical plane through &, gs. The angular velocity r.r and trlo ate grve[ by tgr-'
* &:i,1Y:::.1::Td;*:.,o..
LJ
Ll-
I
!
i
!
]
OC'h
t
j*gg. Ee3.r-3
-"'{:3
t
The moment about O is Mo - -mghsin8qz. Mechanical energy is conserved since forces..".or15gF.ative:
' (1)
*mglrcos0= constant= Eo' t'
?+ v
= rll?L@2sinz0+d'?)+ Er3l
Mo30, the third Euler's equaiion yields o3 8s constant:
I?r*d
Ifl
Since
74o"=I&,i,r-(IP, - $r)u,'.o2-_ 0 :+ cir3 =0 + qa3 =constant=os(0)' - (2)
LL-.
kv i!i.,
z
i
t1
t
a
50
I
i;
,il.
'-r j
0' Eence che
a:ds f4 is zero' sinle {4. fo = -rrrghi*e*'Es=
The moment about tie fixed vertical
0 is constant'
.
abour rhe fi; verticlt axis E: through
momenr of momenru*lg, E"l
,+ Iffpsin20*I&r.rgcos, = /fo.(o} {3)
= consL.
ffi@sind9'-tiq)+se9.l.(sin0q1tcos09")
Eo.b=
OF .A' RTGID BODY
3.12 TORQUE FR.EE MOTION
priacipal axes at c'
H (o)
moment N-c =q. Let g'be the
consider a rigid body rvith external
l.,LioLh
gc(o) = const'
{i.e
*
E-c= .f1-1g1 * I&rrgr+ rfg!'atu =
iavori<ibrc-
W =g
pt.na
an izacn'a6le lite in f (Fig'3'2a)'
The constant vector [6 provides
+ T17=T-fc=const' + ..-^. A ctC :
:-q=F'2s=ig
* ucosp - cN co''st'
*
\Hcucap = coust'
=;";
,.!-l;rig,Il,: ;*;;A;
w*=
vector r'r drarrn from C lies in
is-const'ant and the tip of the
i.e., the.rotational'kinetic.energr [1
in magniiude as rvell as directiou'
lU' i"-a"t*"f ' g is variabte
invaiobleplane norma I to H.through
BoDY \I/ITH I?' = $'f
OF A *:3.13 ToRQUE PREE MoTIoN
!.
3rd Euler's eq:
(1) +
'&.-
the
(l)
+ Ec:/fiitte'at'e){r$"e"=consi'=[r,(a\ Q\
Y=g
*--"'*=ttf'-('4i:id)+r$'al1=const'
Ta =Mc-!*:o'"
' it
+
-3 =' co4st- (3)
=0
l{ca: I}r-'-1'fi :
'|)tt:'-=^'
G) * 'i+'l=cor'xt- (4)
ii:Wftd+-3)itlg1.(,3=consr.
(3).(4)1-2-=-?+-3+-i:GotEt.(5}o=arclan(ai+u$1ll?1-"1:const...(6)
:corrst.+HgL)cos|=cotls!.(l),(5)+f=coost.i.l
(2):+ZTa=fi.C--
(1)'+'t']tt=(J'rrsr*..'eesl*-'l -';--';'*"+*'3q =ff*ffi**'
s: Per{* sea'
*
Hence o is a
rvith l ) u"t t?" : : t':"t :':':!l.3'
i*crinatioo a
vector of constant rnagnitude. *rhose
trre body'fixed a-'ris g" and
(6)
iaclination
(8) shorss t-hat o
-rvith
are constant' Equat'ion
frame '
p with the invariable li*e es along Irc nxed in i'ettial
f
boriS--fxed
isintheplaoeofglyandg"andtheiirctraation0ofthebody-fixeda-tisesrsiththeinertialspace-fixedaxis
;:l,f-g"."" trtt motion is as if the
ffi;;;
shorro
case
rhe
for
constaut
a'-xis e,.with
{
:inertial
ss rernaias
cone 2 of angle f
on.*
*ti,
."*tott
**,
a-,cis,qj,
*ith
a
gtlRace-fixed
cone I of angle
about atis
p
* along s'itlr spin s of the bodl'
The nrotioa consists of preeession of -ri= ", "ootlt
initial dat'a for ..r as
the
determined abinitio from
f...
totque
for
-Jion "*1t
The various e$tities
rdb;:";;;il
ffi
**i*:"ljJi;:
Ifi-,g,
+ I$t.ts$ = Il6(sin 091 * cos de3)'
'E-c =
+
I[6rsin0 = Ar'r.r1
+l
H'scosl:= I?P*
regtlotprccessiott(Fig'3'26)'
L]
= ig'
u,fis,/'o
Pto".
er\ Wu
5t
("}
= $ccosilI{s
t,c - t{JIIccc0lI!"
s=Ufi-$slnct*"':"
.r=<^r1/sin0=Hslll1,
ForIfi >I83,s>0andthebodyissaidtohave
tand = I{'ql$l"ua)'
ar3 = 5d:Pcos0
= Psing = Hc:ilnlll?r'
,
the body is said to have rctrosrcitcr''-""i;::;'';i;
-t4,4,
.r, 19
+ (s + p..! o,c"
- (6}
ta tcoaod
-^ <Is'
-n--,
ForIfi
'i if : ,:::.,":::,:,:t:"i:::,-1ff-n"
=
,W,/
":lli..
"f
P- --.
);
{.
ti
r
:r
-.I
--l
J
JJ
JJ
,J
': I
J
JJ
J
J
J
J
J
J
J
..J
I
!
J
-J
-J
-J
-J
-J
-J
-J
I
-.J
S.J
)
1.;"'1-,:
r
)
f-),J
-,.:..:
or A
3.14
-'-;" srABrjrri btirr:'ivnonr.r norerroN'
RIGrD
BoDY
,
r-t'-;'d'
,,1i
of insrtia
*.*,angular velocity ,t= ,uo?, where g *" P::":Ptl,axes is giv6n
:'
bounded if a srndt distur-baa1
;rra o"art."a*at C and Mc.= O We warG to in cstigit€";hether *.u,,."io"
tothebody.I,ettheingulatvelocitybegr(t)=.,,o9r+e;(0gidrrdng:,Y"*motion,thelalrreberng
;J;-i'6i;;;;;. the dbturbace r" 'ppii"a Jii " <r"o' I.ence s= €is:&15=€3' til=i1' d?=i2t r!'s=€'
,)
I.
L
l,
_--
-_t-lr-:..
ur=trg*er' -@2:c2,
.
Neglectingtermsofsecondordetofsmaltness'Euler,sequationsofnrotionatCyieId:
''.
Mcz- 1,9$2-(rS-S.}15r,6=+ --$rir-(6-frlt"(t'te*er)=0'
L.
*' €'a-(€'- t?**zq *0' {2) '"'r:'
Mcs-I{"ar-(Ilr-$.),tp2, ' r$'"-(fr- 8rl<'1+'1)1'=0' $'a-tf'-r3pe2:0'(3)
L
rL
'
(1)+e13e1(0).Equaiions(2)and(3}arecoupleddifferentialequationsfore2,e3.lAreeliminate€3by
forming (7I *d using (3):
(4)
* tqi', rg:Xr{,
8z€z- (rS - rf,),oa"
iu
_'g':'..'?'i'j:' eq(3) implies
is harnrc rnic (bounded) [ *d hence
:0. +
i
L
G) If, * €r, $r * I&r,
The"solution of eq(4)'forez
coe$cient of ezin'eq(a) is Posiiive' i'e" if
the
that ea is also boundedt if
$r> I?" aad If, , rS,
ts
: o-
-
or
IIence, tlre rotalion of a rigid body o6out a prittcipol
:
i::
rfi < r$ and t, < I&'
azis fot ulrich lhe ttomellll of
'
(3.48)
itcrtia is {rtc lcrgesl or
thc smallest'' is slc6le-
(2) +
(3(u)' Equation
to:":'-::1.L:+
:* e3
c3 = e3(0)'
&) If, = tr9, { I$: Equati'on (3) + es = 0 up with
.h^,,r. ,,
time and the rotat'ron aboul arcis et ls unscable'
e2 builds
e2 = e2(0) + (Is - Ifi)r.roes(0)t/s- rhus
Equatious(1)-(3)* i;=0 + e6=e;(0) +thetotationaboutanva-xisisstable'
(.) f;, =Ilr=I$:
TO TLYING CIGAILS
3.15. I.LYING SAUCER.S AR.E STABLE COMPARED
a-'ris 3 (Fig'3'2?a) ended uP b-v rotating
An early cylindrical satcllite iaitie,lly spinning about lonSitudiual
The
the s'ell-established theory of Sec-3'14'
about transverse axis i 6g.f-eZU1, "ont."ai"tinl a."*"ti"aly
is
body
The
rigid bodgliag. thc objcct wrongly as a
e-xplanation of the actual obsertation lies in ,noacl
the'rotatio[al'kinetic energy ?a is noc conserved
deformable and though E-c iscooserved "irr"" ub = Q-,
because of dissipation of energr rithin the body:
(1)
- E lz = flZ t2I&
E-c :€".e : Ifru'e,
"n(0)
"
constant, tlic body ends up turning about 1" 1*it
?.6 keeps decreasing due to d-rssipation aud since ffc is
the dehomiuator of gxeressiol (l) for rrt -tl'"t
for which ?p is minimum fot giveo I[6:, Hence it follows from
ubich the pritcipat m:o:i: is flre marimurn'
eventually the body would c'|d eV rctalittg obovt an n'i" 1o'
By the sanle
transverse atis I since lfiEence a cylindrical satellite ended up rotaiing about a
-19
'
in a-tial rotation rvhereas siucer-shap4''bodi"t
;
reasonr cigar-shqped bodies with I$ < ffi aru not stable
with I$ > /fi are stable in axial rolation'
p
i-
Tau'
Sa.rCc< - Sr'.a?eJ
I
r'\
c
()
5A
i
t
o(ga
*r* Ar4(n ;},rff
-1
tr's'32s
'::
:A.i'['ii*
o)V'
,r
i fixed
rr.-r ilF' pu"sti
.-o"iihroutr
throug! 8nx"a
that
.
Eo,.cE
tr T
3.16 MorroN ,ND"R .ENTR.aL
-.-L
cetttrcl !o'ni [*"h
Consider a body (Fig-3.2S) moving .rrra", "'*t".n al
and 3-17' for
of
fot,,g' ln Secs' 3'16
p"i"t
the,ccilrz'
o i.*"0
point o ia ireriial frame r and F tt r".
frame f'
ire.'ity, rve write (')1y * ( ) and !L a are rv'r-t' inertial
+ Lcxtttgc=lcxii*=6@=9
(1)
:+
bxgc=llbl*=b=constant'
+
is its value pet unit
the mass is concentrate d * c and to
all
as
if
mornentum
of
monrent
irthe
rvhere E-b
thc t*iectory
O wlrich o::*"t *
t-.-^1:"*
Equation (1) + tc-L fu, i.e-, s6 is in a ptane through
FIlcc :+ 1"xF=0
lcxnlc=constamt =Eb,
with
nrass.
Choose tty-axes in this ptaae
iu-oplane passiag tll,roug-h lhe ce-alre of fotceO'
of C is a plane
"u*"
fu = lrse"
origin at O- Equation (r) + rc x gc = rg. x (i9" +rC+) =
"!e'= t hol'"'
v4 = hslr'
(3-49)
i
,"6 =roo = rouco = h6 : constant (3.4S)
0 :* O =coustaut', i'e'' the trajectorY of C is aradial
Ilence 6 retains the same sign. If ho:0, then e=
r'ector of C fronr o is given by
by the
line- Thl rate at which creo ;i is srocpl tris'i'28)
' :*ttt::
(3'50)
;:,i$1t = h6l2d't - r"di,lz,' :+
*-
Ilence the area is srvept at a constalrt rzrt'e:'This
is Kepler's secoud larf of Planetary
rrnotion't,,
3'ITMoTIoNUNDERIN\/ERSESQUAR3GRAVITATIoNALFoTicE
potential srefSSI 7- of F as
F : -(G ${ n I r2)g'" ' the
For the- inverse square gravitati,onat central force
telation for c moving at speed o'becorrs:
if it is acting at c is -cttmfr- I{eace the rsork-euergy
f"=lv._-r.,:+Tc+V-=const.+n,,i'{2-Gi{m|r=,aifi_cu,a1rg:Q-.
u? ='uE +ZGM/r
e)
-ZGM/rc'
i ,orro*" that u-* occurs at r^6 and u-i. occurs at r-"r. The
body escapis to infinity if speed Do at
1= oo is ) 0, i-e., if
ftence
,L :4 +2Gn{(L!a- U'al= ol-zlt{/'b > 0' i'e" n "oi {2Gltlrdrtz'
giten bli
inilependent of its dfuection''is
the rnr'aimu m csca?c spccd t.to escape at rs. which is
1,=(2GMhiltt2'
'
t3sr)
moti'on
For circufaror6i{of radiusrs, i=0. f =0' u=r{gd =uegoandcheequati'onof
1,. = 16| = {GM/4ltl2 '
,n(i- rods) = F, = -Glfmlri +
(3.52)
C
- Equation of the Atajectory
"3.17.1
is'ottained, by solvirg the
of the -trajectory' in r', f coordinates, for tlre ge*eral case'
;;;ion
equation of motion using S= hol'?"
(3)
:='' =':.;r:f'- [ilt3 +GMl"'=o: ''
m(f - t#) = F = -GMtnlr?'
using 6 = no1*: Lgt?:
be sirnplified by the Crapsformation u = Llr and
equation (3)
The nonlinear
"*
r=*(ij =-"+ =-)##-=-oo*,+ i=-ho##=-u""#' (4)
u:
Substituting i from (l) in (3) yields a linear equation in
-trer'# -lrfrus + GI*{uz =o :+
d?u
G'M
@+"=a?-
Thecomplementary,particuIarandgeneralsolutionsll3,ltptuofeq(5}aregivenby
u. = Acc{+Bsi'o= Dce(6-l?),
vo=CMlhi,
giverr by'
where A, B. D,p are constants' Ilence the trajectory is
-$3
tr = u.*up, =G^tltfl+
(5)
aco{d-fl'
,l
,J
.T
JJ
-T
JJ
rI
L/
-j
$, q)
-r*
t-
LJ
I
\w cent.o'a.l h /b a .\-.
--
LLJ
L.
f
LJ
L
l-
tJ
lj
Af *
:
IJ
l-,j
l--
IJ
IJ
L
L
t'
L
L.
u
L
U,
L,L
rL
)u
--
(6)
tw'
L GM +Dccl(4-01'
. n-*rt-al
L=\E
rh6
from the radial
,r o! =0. Ilence d' is-the polar angte measured
.o
LeL6- -_ 6 _ p,then eq(6)-+
TIo
positiooatr-;.(rig.3.29a).Forsimplicitnwedenotec.bydintheseguel.Thusuith$mco,s$'Elfmmlhe
;;';;r;;"i "*,.,'*I "qrotioo of tLc
(353)
ht,
,
ha:.
(3'53) w'r't" time an'f usrns f $ =
"::::':
The rodial speedlsobtained trv differentiat*:I !:'o'*-t
(3.54)
ttaiu"f:ryj;"*r.
Eq(3-53) =+ fllll D -rccsal Hence the orbig is a conicsc(:,t('n
v'=i=lrsDsin{'
+
-il.=--DsinCi
P from line .[) = haD|GM: consf,.
ll D trom
thc ccttrc of fotce'direct'rix at distauce
D from
f-^. O)
6\ //frlist.ance
(distance of
(dbtance of P
t
(Fi8'3'29b) witlr /ocus-at
:'
it and eccenlilcidg egiven [l5
(3'55)
e = ttiolGt{ -
This is Kepler's first lErv of-plarrt.tarl" 1uotion'
of .xl'its
The
e--trf,o1t;tl l
oa > {2GMlro)rl2
e)l
os= (2GMlrs)Ll2
e=l
q <{2Gt{l1ttl2
e< I
os = (GMlhlLP
-
Eo>0
escap€s
Eo=0
Just escaPes i
I
6o<0'
Es = -Gt'lm/2o
Bo<o'
Eo= -G!!m,l!
of senri-major a:cis a:
ffi..ltip,r.icorbi[
;;t ---l
on the value of
Es nrui -GMnf rs
tao
t-
L
l_
Ix
closed orbit
I closed orbit
L
;:::;;:::'".r *, = *. :: ,'= 1 ,:'=;"i" :;l
T::::?:".;lx;;11;:;: n"llz
flUg=r2U2:Ir9(o)
Gl[ntr2'
;
o
(0)
- G l{ n lt1 =
" nl'z + ul = 2Gt{r2l;1(" +'z)'(c)
"-:.:"'.:;;.:,--.r,ri,,
=
Pur u2 from (b) in (a): rnti/": -GMmf4 =n(tp1fr212/z-Gl't'',
/---fft7]>^
+
-GtIm/2o'
=
fu=
-GMm,/(rr
(a)
:
"z)
D
Put ul from (c) in
tnvtt
E'o = ntfr/Z
E6
vru'e
t-;ttrs<u Orbit
tbr Cltrs<d
s.L7-2
3-L7-2 Time Period for
*<-re
\ -r --_: ...-i^i rnrl qer
i" gilru-rT
. .,,,
Thetimeperiod?forelliptic.,rlrit(Fig.3.30),ofsemi-nrajorandsemi-minora-\esband6'
'l' - (Area of ellipse) / (Areal velocitl') = dabl(hs/2\
r)c.sd = olffi*;ll ' *
i=#*
;e*o=
|('-"* *
rou
I(3.56)
r=r(,;-,o)'
+
=
t*i" = Irfr'
(3.s6) & (2)
=+
ZiiI;'
[-a(l- "'l'l',
- 2r;ab= 2rq2(L- c?ltlz -4,a)t/'=
'l'=
- Lo 'D,
T
T-T
=Zroslzl(GMl'l',
i.e.,* oc o3, which G Kepler's ttrid l'arv of planetary motion'
3.1?.3 Parameters of tlrtr Ortrit in
Terms of Initial Data 16' us' os
Ftg-3.10
(r)
t
'a
rmin) =
e
D1f.:1
(2)
.'-' (3-5?)
2;.a3lz
hs
hiD
,
,ttz_
\GMD,
(3.58)
I
TbeparameLercGM,ft6.I)arecomnrlledfromtheinitialvelocityugal,radiusroatanaag[ec6rvith
position from tic minimunr
(F€'3'31)' The inclination co of the initial radial
due to gravity at the s*rface of
isdetermined from the acceleration I
radial poeition is an urrt no..'i- CM
: nt!! * GM = 9-8?" Eeuation (3'a9) =+
earth having .rdiu. a, d Mtnl lt
the circumferential directiorr so
5+
!
(3-5e)
= rP6 = rouCo = rouocGco'
and $6:
(3'54) for rs are solved simultatreously for D
116
F4uations (3-53) and
+
-
l/rs = GMlhe+ Dcosdo, uro = uositlco = IroDsiaC"'
{s=ar*a.
"=[(* -W)'*(H)']""
D=lu"o-GM/hal'
(3'60)
,r#M'
* uro=0' (3'60) *
Eorlaunchingparalleltothesurfaceof theearth,o3=0i
using previous results:
once GM, Irs, D are ott4irred. then all required-entitiesarc-computed
ho
116 -'
Gi{ n
I
GM
:,,1---- = t '
t
rmin ha
'io'o
oo3(l - e21u2 11no1z1'
and if e < l' then ?' = 'obf A=
'-t'*' * t-'')/2'
e: ttf,OlGMt o= (r*"'
cosd- ftlt-GMlhlllD'
wittr
u'=heDsin{'
u4=h6lt,
Atgivenr:
fr(,m ,Irvz 12 - Gl{nfr = mufr - GM*lrs'
alternatively, u! = * - ui, .r,here u2 is obtained- \.
i
earth iu a
of:tbe
velocitl: ui relative to the surface
-
The &csl la:U:,tchirrgposifiorr is at the equator $'ith
cr is the angular velocity of the'6arth'
direction West to iEast so that,.us = ,,* a.,tol, *.,,lvhere
t1a is :: "i:1::thb'time'?5p to
'*"';:j';';ili;;,
(Fig.3.30); the tim.-fes . *".,.,". from .-1
w't't- 'point O:
.'.to"itf{ior the area ss'ept b;--the eosltion vector
rraverse from B to D, since tle areal
ze:!r!12;abl(l t!h:12\:."!t - %l*i/a'
T.te : $reaOABil) = laobl4 - b(oellzl!(h612'l = lt -
Similarly,Tso=T{t+%!ella'TealTao--(a-2el1l{*+2e)<r'
3.17 4 GeostationarY satellite
b'a circlc a[ou{ :\'-5
relative to the eart'h- llence iUs orUit
Geoeta.tionary sateltitc has a &ted position
tic cct,rl- of forcc
trajectory of, a saGellite is coplatar uith
azis ttt'uc*cl ot the ralcL.r of [he earth- But the
cqtatoial ploac
lhc
i*
r'
r ortit ol ndias
centre. Eeqce geoetationary satellite has a circzla
o at the eartb's
and moves ftom West to East at the rat'e r'r:
ac:,trtfc- (GMlr.)U?
+
r" J (Gl[lutz1rls - kfr luztl3:'{1000
km'
3-1?-S alansfer Orbit
should be
requires change in Eo' The transfer
Transfer from one type oforbit to another type oforbit
for
achieved
*hLre ttre greatest increment in energt' AE'-is
executed at the optinrum location in the orbit
(i.e., for the same elxpenditure of fuel):
rhe same change lAgl in the magnitude of velocity
5f,_(vatueof81inthefinal.orbit}...(laIueofEointhe|nitia|orbit)
As)2.- 3.?1 ='-ft:4, $? + za g r&1
mu],G M m / ) - lm(fi -uf ) =, !rn(31' +
r],
i
LI
G
(lmrfi
l
!
-(
=
a location u'tcrt 15 is maa-inzm' i'c:' ot the
should be large, i.e.. Au shottld be alotg 4 and'al
Au.gi
Hence
localiort of the minimum radius r*;t'
a Satellite
3.17.6 Effect of Atmospheric Drag on the Orbit of
1\'e consider trvo cases.
!
---L -: rs- Hence
a verl'short time $rhen r- 3 fmin
1. The sateuite is in clliptical or0it and is in the atmosphere for
lsc applied ct r ='ro $'hich causes instantaneous
the atmospheric drag is modelled as an i[slcn la7lcots imptt
in t6e next orbiO (Auo is negative)' i
change iu speed from us i' the initial orbig to uo *iAuo
Es=\flir2ol2-GMn/rs-
-GMnl2c+
<0'
AEo=muoAos= Gttm6l(,l2o2 + 6ia'=2o2rta6oolcM
Hence the'semi-major axis of successive orbits decreases
by the above amount and tlie trajectory
circulat orbit.
5q_
tends to a
JJ
J
J
JJ
JJ
J
J
J
:.-I
J
J
JJ
..!-'-- :,7.,;= <-;+-: -ra. '
: : :' ;,":ri:
.
:1
:l
,--i
cir<rla.t otbit of
2. a\esatellite moves.in.a nibtly
p-- cu'
""u t'l
the case of linear viscous drag,
tLT
L_-
L*
LJ
Yiewa *lr -
thete is ;;g';;t;'
a = (GMlirl2 increases, atthoulh
by f'
mot" thu" the negative work done
;;
-€-i:!r:*'
undei-drag foict F"i1
ilr= -ZFlmo'/ -(t)
;;'-t-+;: o{*ll.'r2 =-F,-' *
tu= -GMm/2r,'4:r''+'ri' :
For
l'-
radius r-in atmosphere
- 11j-:-:
and speed
-2clm' Eer'ce rdecreaS€s
in the potcatial
ttis occurs-sine the decrease
TO NON-INERTIAL FRAME
M
3.18 MOTION RELATfVE
of the centre of mass C of
Coosideranoninertialframe(m)havingangularvelocitvurandangularacceleralion{w.r.t.iDertial
TLe uot'Jo" oimotioo
g,nu'('';'s.t2)'
has accelera.tioa
frame r and its point A
a bodvot
I',r[r^r, itx rct+'- x (sr x tce) +24'x ecw + ec-.l=
t,i+'-x {gx lr,i) -Ztw-x
tltg.clm = F -mgrl1t - trt9_x'la^ -
+
u:
f-
*
T;
[; ;:l
gct*:
(3.61)
pseudo fforces'
rvhich are due to
F and four pseudo 'forces'
actual physical e-tternal fiotce
of
Thus rngclm equals the sum
accer.erarioa
accererarion orrrame, angurar
g-?i:-:t:Xil:::ffi:;l,JiTt:I":":ffi:JI
(,o} attaclred
Corrsider a non-inertial ftame
to*rards at
axes
its
with
onry and
to tlre:earth and 1n inertial
.
'
frame I attac.hed.:,9i.:the-ccntre
angular
ihe
at
frale m rot"t*
;'
stars €ig.3i3"iforce
"n"
"rn="a'g' Considei
o' ooa' under gravitational
"
a=
of the earth
to frarne "-i
velocity sg of the earth relative
'
*d other external forces'l' Let r: uo' *=
";;
(3'61) becomes
g,;ln'.4m-- gcla- Equatiou
(U
&
x
m4^= L + z-- rnst;,ll - nry'x (el r) -2ruex9lm' gravitational fotce
r = O'g1* = g' q' = 0' the
(FiS'333b)'
O
o:Yo
the
at
For a stationary plumb.bob
1.
t2)
eolrl= -ri9' + g<= !' bl'
-mlc--=
:+
L=
nsal!
-lE<o-'rrsold
g= ho +'L-dLection
differs iu magnitude itnrd
where 4 is the effective
gravitational acoedration .1
from g-.At tt" "qu"to'] L-=
g, *nhi.h in general
-'n- 1--'.nt'f -- -@ -"R)e':'
-t,'ott=d;a'1*i:T-A:itcetn=2rl(3600
2 Forsmallvaluesotl, &
in (1] and obtain
term compa."d tl",rt" c"tJl"[s d;m
w_e neglect rhe centrifugal
f,.En = ?-+ mg--211'!4x 4il'
then we have{o
as given by the plump line,
x24)41 =1 ar!(<'r'
iodude only the coridis
(3-62)
, - ..---^,
pseudo'force'
Eence, if rve use a
to [Ut dit*tion of -Zttlt't x 9l-tleoioles lo Ecst according
c*zator,
the
ot
oerlicat
fall
ia
North3lrcmisglerc and to
l- A. bodg free
dc!c*!-d t.o tt"-;gu ia
arc
;;;
tf
surface
the
on
Hencet'he
4- The sinds btorving
tbe lcftin t;c so,4bcnt hemiiohae:*
E-'tr.1+*3si',i
r..o.a.o"u'*itt tr.",aitat ';''-2ry"* *q*lttd^itrii'-tcip
do*n' &e
of the wini;;;;"
@
a rcsiolorl.o;-ftuout"'
cyctoacs formed due to motion
hcnisphcrc(Fig'3'35b)'
ottttctclockuisc itt Notlhent bemisphctc-oo| "*"**c-ia'sordhcra
I
/ fr'
3.33o. =..=u trt-{r"
,
X
i
''
,rrf\,
51?3-'ffi.
\"i.4,^
3'
t3"
3
z' X&F \
w
f #r
\
\}
3'3so.
:=:.{
i
l;
:
Yn
'%
L
I
3.3sb
',/
+
irI
!rJ
.
:-r_
5. Approximatesolutiorr"fo{Jree fatl: E = 0' Equation (3'62) reduces Lo
:
(3)
g=g--fu-xu-
on the r'h-s' o[ (3)'
lt is solved b) successi*'e approximatiols, usiu! g frorn the previous approximation
u- yt*la
lstapproxinration: usingg-0orl r'h's' oi(3) :+ g= g- +
:+ a = 9.-?4'x (g"t +go;
2nd approximation: (4).is used on the r'h's' of (3)
g: go + yt-,!x (g.t2 +29r,t)
r: ro + 9ol * lyt2-t' x (*g,t3 + !j2)
The undcrlincd terms are the correcLions. due to earth's rotation.
(4)
(5)
(6)
(7)
to t he usual results for unilorm acceleration'
J
-T
J
J
=I
J
j
J
J
J
J
-J
iJ
j
J
.J
{
J
J
jJ
j
:J
J
97
*J
-J
J
}-l
F-
';-'t -'
.
''
;i
|
,',
- i ('
U c- l-#*t'.,,':::'"lli,"jio*,
llrff3;T" ilffi:U;
lq
- .
-
'/
E*AMPLES 3
I
r ,=
meoti of tl.:.e inertia *.t.i* for point .t{
*'-r-t- a, y' z axes-for- th: systcm sholvn
i is pi Nes,ecr:++:i:a:*"*:"f
iasfition'ror 'arious bodies at ttre rollorving
E's'
A'B'F'G'H and bodv 4 aL E's'
*:' n,D,P,Q,bodv 3 aL A,B,F'G'H
;':;;,D,P,Q,bodv
"i'u
gia.tftu'd
' rinllProne
rtu'd
&-ia.t
t.i^!ze\a"e l)3
1yF$a
r,#:,
HT
I
+
L:
i:
H
f
l
' '*_ o'-.:r-'tri
*.-:--'
_T o, _ ,4
F-r
*t,:'#:,'-*=
E=-=,"'nr,'ft1* t;;;;i"l;""' "** *il.83'
f-"
,.,.11;tt;;".8,i;l;*L;*'"
Kt
-
Bodv 1: Axes 1,2,3 are the principal a-tesof inertiaatcl
Cr + z-a-tis is principal anis at cr *'ith
l-
,b
. Plane y-- is a planeof rllass synlmetrl'through
"
:;,==0,'|u =o,'"' r,ul',;:;' *u':i,,=,,".;{:=:i"'0"'"'
")i
sr=si,o' s?=-coso' sr=0'
""''uli""
,1 =coSo' r12=:sillct' rl3=s'
*
:+
.i=g=sitrogl-cosoQ2'
k=!!=cosagl-:-siuaq"
-
a" cos2 a)f 12
$; = tl"': Ifr'"1 * 8;ri+r$"3 = rn1(62sin2o*
$; =t!;=lfi.r,i +t,g;ni+t"g;n"=rrr1(0?cos2o.+a3sinro)/12
rn1(6? - o?)si.ctcoso/12
8: - r'co' :,f,'"',.' + Ig"sznz* r$s3rr3 =
rnas s-r'tnrnctr]'
aL c2- Plarte r: is a glarre o[
Body 2: Axes I-.2-.3- are the pri'cipal a-xes of iuertia
';
with
through Cz + Y-a-\rs x PrinciPal a-tis at C2
l-IE--I
o'"
or mass
1-..
l_..
f
b- '.-----+'t
.
i
f]
I
pt"ne
ingi'^tt
I;r;'rii,'$2^
^
l--
I
!-
;::;'li.i:::T::;
poin*: bodv 1 at ''l''J'x. bodv
rr.2='piTR2L,
,
1;'r
-
r===';
t
E t
I
26.=csiuO,
:l:-r.1t.;:'::'"
tg'=b"+ecos0'
yc?=0'
*l'"",'n=r,tt:;l":i'f .-
k=a'=sigggl+'ccoei, +
n!=0" ai=siud' rt]=cos0'
t: = I?.',-- ri;"i' + ri;"!' + rfi':?
R2 sin'2
= n2l{R? f 4+ L" lt2lcos'0 +
r' =':': ='ii"i":-f1:u:::::::
L21'lz1sine'."0
:u:,,',,n-
Body 3: !, j, k are the principal a:ces of inerLia ag f,3 rvith
tr
L-
:;==':"::,:i;i-1;;';-,;:;"u"i^l'",,,
Ll-
z6'=b3 'lt; = 'lr" = /'!'=o'
tf; - p3a161c1(ci + oil1tz -Za.'t2c1l(r" /t'+ cilr2l + P2l
I?; = pso$ (ol + b"L) / L2 - 2,...,r2 cllt'' lz* p'l
,
tr
l--lL=I
t
0121
1cz- I!3^-=rfJri'+?ni'+i.?4'=n2l(R?/4+t21tz1sirr30+n2cos2o1z1 !:'
t
fL=..
1"1'
rfi =',.2n=12
''c1
'
J
s8
I!
witl,,^
Body a: I i r are ;'he ptiniipal axes of inertia ac Cr
::,
_
3c,=.b4+3ftt18'
ma= pa(2etlfi\, ic. = yc = 0;
t!,- nu(3i?r/8)2 - nalzlils - (3nr/8)2J, If"' = E;'
t!; =
since for the hemisphere the
values of nr-o'i' abouL a-tes x'!' z at E arc
8 = tf, = I!! =a,
fi - zm+nils
half o{ the values for a sphere of
i.e.,I!,= If;u= rfl = f[i(z'no]EiJ= im,&l'
a-xes theorenrs:
are finatly obtaled using parallel
at .A for *,J
,..,ir.i*
t#i,".ii.
of
The.elements
"y"t",,
;;;;.,
.4-*
E.=Dti+rrt(y},+,3,)J,,r,=ir,i;+rn;(zi+'3,I..I*=EtL"..+",'.(4,+v},}l'
t'='
.
i=l
4'"
- r,;v6':6-J '
- n;rs,vc,l, + = it'i'
=flfi
r=l
i=l
I*=
*
ftf; - 'n;'c""c'-l'
Theprirrcipala..iesategir'errpoirrcofabodycanrsotrtetirne-<Ireobr.airredbytoolrirrgforaplaneofnras-.
s1'rnmei;yoft1lebodypu.,inselrrouglrtlratpoirri.Thea-tisnoranal.l.osuclraplaueisaprirrcipala.lisat.
trre above Procedure''ihen the third
thar point. If trvo pri'cipal axes have been J".".*r1r"i ": i'Tt::a-.ies'-r' Trre ptirrcipal'a-'ces'a*'1'ariousip'f,i*ts
of trre principar
principal atis follorvs from *tutual: orthogoualiry
parenthesis:
rirass synrrnetry git-e* i* che
of rhe bodies are listed rvith the planes of
,{: !(v:). J : 9r,9:' ec (t2'23)' Ii: i'(v:)'
Dt i:5'S' Ca (1'3"?'3-)" Q'.'4 t1'3-}'
p : ei(?-3'}.
Bod-v 2- ,{ : j (z:),
g= i,!,k(r:'v:)'ff'
Bodr-3-6: j(:::),n, i.i.!(:::.9.:), r', !,j.,k(r:.1[zv].
g
Bodl- 4- S: i (c:), : i'J-'k (r:'Y':)'
goas l-
b(llrvl'
rotating about a no[-principal atis
mass C' is
Exatnple 3.2 A rigid. body, of nlass nl Nit'h ceutte 1f
has eleureu[s
axes c'y': attached rc tlre bod-v
fior
poiut
O
of inertia (Fig.E3.2a). The inertia nratrix at
prof
ide an-s axial
not
position' The-bearing ai B doeS
bearing reactions f", th";;
the
rind
-..
*.Ign,.lr,t.r.inr. Consider i* particular tl::?: "i'oi"" tn't'* 'rj32b'f9z ,t
*,9,
' J'ffi3o'' Lt{ks:,q,
'pr&*
;:-''-""fi,'
,ry#, 1"*.-<3o''lffir --rtl*t;:l
:;
(G)
(1'
(bl
Rrk, 8a = &!+ Esj', be the angular
Solution ut = u\., oc = ei. I*t or = ark, Ra ='R1i+'R?i+
The FBD Ot
p*itio &a does not have axial comPonent'
acceleration and bearing reactio*s at the gir.o
''is obtaiued fromgp Q using
=
the body is shown in Fig.E3-2c. The acceleration of C
sc = so+ dr x aC + s-x (sx @) = -'f
ei+i;ei
(l)
e ag rates r"'r'i' The 6 unknorvns
-t-' ^:-^^
^ ntoves in a circle of radius
This can also be obiained directly since C
tng6:
;;;,;&,8s
are determined using moment about ftxed
M o =-@ E x (R, i + Rz i + Es b) * 6 L x (& i + n'si)
- (I?,6- rf;.r.r?)i+ U3,6 + t!'-aii + t!,ay
q
point o and E=
+ i{ L + c ! x nrg(- sin 0 ! - cos 0j)
I
|
l' )''
'- ''
-"-
I --.)O
F.',,
:
L-- ,-l *
[
L
L
.
'-" i>Rt- D[s*= t?,'
j:
,
'/
- t?12-
(2)
i,
6'R4-onl=jrf;,a+r!''z
+ ;[;:a;#(Eri+;.;,.'f;r;;:"*:'1-"* "'
,=*j
*
i:
:-l
(5)
Er*fie-nrgsino =-mtt2e
L
(6)
.,;',
f
(?)
:::I-mscosB=m'be
.-a. .o 2 _..?onl.nna
7, ao=gfla+ tg,"-m:''r?ea+ntsasind)/r i8)
eorts),(;)'*
+'t1,,ca4rngacos,)/L (9)
R5=(I!'.,+I?"'
^r=-Tr?,o+I!.,,,2+tnu?eb-nrs6si.0)/
Eqs(2),(6) * R2:(I!,6-$,-.+mieb+*Ou*"rtlt't'
f
I
1
o[c'? term and change their directions
rotors ,lue to the presencespeed
rrigrr
foi
rarge
very
are
in grou.d refereuce
These reactions
respect to the fixed directions
their
consequently,
--0"";-::;ith *'ear' The rotors nlus! be balanced to avoid this'
as the axes rotaLevibratior,.
forced
^r,.t
r,ary rvith time and cause hea'y
r
LIL
*
We consider some parLicular caseserf rotat'ton'
i'e'' lhe ceatre o[ mass C r' on the a-tis
:,
I
i
,
I
t.
i
i - e =0'
LJ
!
1
'tt
I
t
i
,i
2- I!" = tf;"= 0' i'e'' z-a-xis is a principal a'xis of
R7 = -(nu't2eb - ntsbsitt.)f L'
fu - {-rw't"ea + ntqasin 0)/L'
L
L
I
l'
irrert'ia ac o'
:
'
/i3 = (,r-'e, l ntgbcos,)/ L
R5 = (rrtr'ea + ')rgo cos?)/
L
R5=ntgocos,tL
t-r.3.a=o,I!"=I?,__0,i.e-'tlrea.dsofrotationis:rlrrincipalaxiso[irierriaatCrvhiclrliesonit.
Rv=ngbsin!/L' E2=ngbco,\fL ',l,=rrrgosi.0/L'
L
l-
;:':Y;',:;,T:"lTi::*1.,:"j'i':';ff:.:l;ffi:iii"i'i:;
L]
'
l- '
u
[
]
ltj
;
L
F
L--
,)
)
ljl ,
Ul-r
Lic
l.=-l :
lJi
'
r-.
lI
"
?::;.;ti*"::"? :i**;;;;*i**:::|"'i=
L
er' e: aL o r+i'lrr s' a'l
',i.,e*ia
j ispri.cipar axis a'L o ie
Hence
* ;:*:;':;,=tu;'i;*1;:;:;;I."::;;1*'i=
L= s= cosoel lsinae2'
"'= : - -'.:::"
-ccSoQr'
';;,=i::i3"*;='f'"'"'*"rf,"":ir$s3rr3 ={I? -rfl)siiro'co
(I!'i l ntebcos|)l L'
- msbsinl)lL' R2 =
: (-/fln
(-r.P + mgocos9)f L'
n's =
0)/L' E'
fu = (8,-' * mgo sin t0)lL'
*.here rg is given by eq(10). For rrre bodies,'a:r;,::1:,
p2)/r2isinccoso
_r,n(o2_ p21/12]sinoroso
nry2lr2)sinacoscr = [rn(q? 1!.:7rns?r,zplare
(0
\:, ,"";r"gr;ar
-Eqs(8)-(9) =+
(l I )
R1 = -(I!,at2
cos a
lflsinocosa
-1nR2 l+lsin
r:-- 1!,
to -:' r*eltt---- Misaligaed
mff'12)sin c cos o = -pnP(2
--.
thin disc
1'lrrz14 - rrrRzl2)sinccoso
' tL:(ii)
(lQ Misatigped thin rod. E :(t- mtr2l12)sinccosc = -{r"h2112"""::,:.^ D..
,
q
a
;;;;;;;*ru".
(10)
1!,_lm(R214+h2rr2')-,l,n2izisi,iacoso =m(h2!r2-ftzl4)sinocosc
.. :
6o
,i
-''
platc and if Ir : tfrR
ate :.r,ro it p = q for a rectangular
lc ls interestingto note thiiihe di'a'amic reactions
a piinc-ipal axis of incrtiaO and the axis of rot'aLion is
for a misaligned cylinder since for tke cas€s lf,, =
(Fig'83'3a)- The
several pulleys aad rotors
Exaaple 3.3 (a) A shaft is rctatin' about i-axis carryi*g
z;)' Find the reactious from
the centre of nrass has coordinates (e;' y;'
mass of a typical element is mi and
torque is ?' Bearing
at anSular velocity c'r and the driving
the bearings at .,{ and B when the shaft rotates
for A w'r't' body
matrix
The elen'rents of [5e inertia
B does not provide any a-xial constraint to the shaftpllkular find
fixed rotating axes e, Y,z are I:",C'--- Neslect sravi!.v' !U) It
"?::i::Lt:XI""X;i:
P,r*er1+es|- 11= r, i'e,i
1*
. rF
shaft in Fig.E3-3b'
f
t,
j,mi
=lji3'lrf'ai;):
;,;
--,
'ru,ru2
t
ri
o'
!i
e{
-
-r
i.
1
-- -.,
--. -r
a
0-l^
_- iE
.3, & tsJ
"{
.i{
tb)
(o)
Di,r -:-p-r.,n)ThgFBDissholvn
Solurign (a)Letthebearingreactrionsbe&,=:Eri*&j-r.,fe;nsi+&!+AoE
*o*unt' eqlati-o'p:for:lbe 6xed
in Fig-b3.3c. The 6 unknowns &-&r;fia,'r,fr, A,
>
i il
__-
u-,
lat.rU
t r\-l
;;a"r..i*'ifi from t'lre
-
pointr4andf:frl.;gc;-Forptane*otlo,rnuitt4.'in,t'lrediregtionofuon-principalarrl"'E'at'-'
./
- trk x (Br i+ R:i_) +r1= u:"a -Iid:)i+(Iin+ I:,u2],i+-tine'
Rz--$*u'tf;'u,1lt
+
i:
-i*r--i:,;-t-".,''''
M-"
+
{l)
El = (/fl'r}+ t!'u21/t
(2)
_q
; =T/I!,
+
k: T = I!,n
a:i
t= Eri* 8zl+8r = )l*'i4g, + &,^. =Inu*.-'!'i.
where tb,=t,kx (t lJurit--'ttri+vi!) = -(dryi+a'"'ili*(d':ri -'"vil!
(3)
,.=
LRt i rf;"; + t!.'r2
j:
:+
{{}
{5}
bearing point can also be talen in Iis3J'
This appioach of taliug tbe moanent equation fot thd&'red
O, :1 -O.lm, ml =f-5ks, r3=0' T-a2N-rir'y2=0$6rer'
kg,
(b) nq
=2
rn3 = 4 kg,
,, I-0-O4mr 93:O, zs=O-{rn, z2=03rn L=O'5ln' 14:l00tad'fs'
kgrp?
I!, = -2x 0.05 x O-L - 0-5 x 0 x 0'3 -4x (-0-04) x 0'{ = 0'(}$4
lf;,= -2x 0 x 0-l -O-5 x 0:06 x O'3 -4 x 0 x 0'4: -0'009 kg'mi
tg-ott
r!, = 2x 0.1213 * 0-5 x 0.u3l3 + 4( oj€12 + 09421 = 6'o3s*62
,, , o.
:fi, - 0.2/c035467 = 5-639r rid/s? '
r
!b, -' -..'?(0-05)i'+6t0:05)i --:1t^:::'::i::'::'
(rr- 2-.?1o.oo15 tr;(0:06)[= -0'3383i-600!m/s?'
-r
-'-
rc)
-q
\'
.-I
(7)
\
:
>-ir
_{
(8]
ar- =.r?(0.04) i - 6(0'04)i ='t00 i - 0'2256j m/s3
<<
I
_{
Usingdatafromeqs(6)-(8),eqs(r)'(2).(a)yield
Rr - (-0.009 x 5.639 + 0-054 x 1m2)/0.5 = 1080 N'
R2 - -(0.054 x 5.639 + 0.009 * rm')/o.s = -oo'u'n,' '
j
+ 180'6 j = -{E0'2! - 1r9'? N
r?^ : 2(-500 i + 0.2820 j) + 0.5(-0-3383 i-600j) + 4(400i-0'2256i)-1080 i
pass m is
carocs' An autorrcbile of
Baarnple 3.4 Bal'lking or srtperclcoation of highway au! milrcod
proper rralue of 0 it its centre of mqss C
travelling over a circular road 'banked' at ar1- angle 0. Figd the
..
-r
,.
+!
l'
:'
-
<{
u't
.i<
6t
t
I
.,<
-=-
ilL-
'i-
J-j
* "o*,*iinH
travers
L:
IJ
1,
fr-a
: -tt'
sorution
;-*ffi-T "' o",'*"' ll"ni tty5:1,;T"i:"ffi'X3::ffi*:::':t;';:i':
lixj::;
sc - lo'l nls' - ("/Rx-
sin 0
cos 0
Nr, Fr represerr trre lormal,aul-1r^:r,.",
The FBD is srro*'n in Fig.E3.4b.
f:,
lJ .
?'f:["t':1"ffii[:::"'::':$:::[":"':;:::;:T:
of inertia at C'
;i';':it::H:.:;i;':*il':::il
is principal
0' as ! =
t-L'
=
no'-principal
w.r't'
C
equation for
re3
u.e momerrt
l_
L
*
'-
jol';l;l';'
,l=;;'*f:-:i;'"
-*
tf, =-i' = 4i'"'= If'n1s.
-Ft - Fz -mgsin
F = itsc vietd
0=
ir, "-,rilJ :'=::;:,
L
axis
a
''"'
using eq(l)' rhe components of
tr
L
'.,i:t,:::,
.4c')sin0cos0
(2)
Fr + F2 = rn(u2 cos0/R-9sin0)
(3)
s"3"" = ufi
"'**'+
m(-uzcos,/E) +
*t""'.'r*' :+ ''v' +
1;2 =
m(u2sin'/R+ecos,) (4)
-$'"i
Mc--gg,a-I3,"li-+(f,;!E'")i+11;t=inet*,(al.l2
(N2-/vr)6-L(F +F)=-1f,''2=u?'-i;;f
L
tr
1 . .^r,
U,1",;'i"f,l=,.il;,.X;;*;;:'i;:;i'::'ll^fi,r'!iil:?i::if);J"*'^'"'
o?l R)2;!2b (7)
- rfi1sin
I+
l{
(ricz
be zero.
1. #;'J;::Ti,'fr::[:t:fffi;rion force shourd
f
;
o'cos
"il
)
r-
cornpo.enrirof che sum of rrre
fif,ff,':i':l:*m;l:*iffi.::'"*:f;*:',i"":T,$::n:'L:i:J't:'::::ffi"fiilliliT;
L
r-.
L
tb>
;H:'i*ilil,:I"j# ;lx:1';q:;j:**:xtlxrf;:*:;''::'";:"':::
:,:::,H?J.':llJ;:::'::';'
:;= i*
:jrfli'tl{;t'",,",j,s':
us
(l)
9")
e' +
l-
L-,
Y
(a)
tL
f
,
i"j333.,i'a1*:l:i:,:.9ff;"I"'.*r:Y$i,;:f#"iff.iffli
-i 7ru,-
f
E
,:-,
Eq(3) vierds rhe optimum
taro is the static coeffrcient of friction:
2- Forno oduard slip, Fr * Fz s p(Nr * N2) where -(" ca.lR- esin,) s P*(o2sin0/R'+ecos0)
eqs(3),(4) +
p
;
Fornoinucrds,ip,
cnelc ror ba*kins:
r;;::',1lf$l=ar='Rilll"l"'-o' tlll
i
II
'..J
=-"-
- 4-' ' 'For r,o'tippittc ftocflii&,g),th:e snialler noraral reaction /Vr must rcmaT positive, i'e', usiag eq(6)
iv1= |m[e(cos 0 + hsiallb)-(o'ln'i{hcq,0lb-sina +Uf" frlsinocosaTmDE}} } o
(11)
u2 Se.l?ltao o+b!h)ltl-(blhrtan0+U?z- f,1sine7lant'1
:+
Equation (1$ rerrcals that the car is more likely to overturn for smaller values of B'&.d
of h,(Il2- Ift), r. lf (Ig - tfrl/r',,Al ( 1, then (11) simplilies to
and for larger rralues
tl" S e8(tan e + b/hl lft - (blhltanil
(12]
comparison of eqs(9) and (12) reveals that in this case tipping $'ould occur befote slippingif
yield
S---F-or-no 6ct*ing (d = 0)- The aris of rodation is a principal a-'ds- Equations (6)'(?)
p > blh-
/vr,/v2 = lms(ll tw2.sba).
I
JJ
J
J
J
J
JJ
(13)
Equations (9) and (11) yield the foll6rsing speeds u, and ur for the initiation of slipping and tipping:
t! = psL-,
(1't)
* =(blhlsB
Hence tipping occurs before slippin-e, if r, > blh. The values.of r'! aud-urfor the unbanked road
than {hose for thd:'bariked toad6- For a stmighl'Ian'ked rcad (R= co)- Equations (6) and {?} 1-ield
are srnaller
Nr,iI? = |mg[cosf *(h/6)siag]
than t
i.e-. in contrast to the circular case, the inner rvheels are subjected to greater normal reactiotr
-J
[er ^
a.'1i"8,L
,^[*ll
;:*J:I'T:::::*f:::X*"
vertical aris
,-r'i&t
rrr ,reigrrr h a'd ,.id,h
.T
(Fig.E3.Sa). It.is rorating at. lT
is hinged aL A and B abour a
I 9'' la
stop D at the ground level'*fi-J. 1.t-- I .,\.
when
against
;4
a,o
it
bangs
velocity
angular
trt
-ID
the
on
forces
irnpulsive
the
I
is
Find
t i':i
The coefficient, of testi[ution 0-2door from the hinges and the stop D during the instantaneous inrpact. <F4P6|
Sotutioi The 1rBD for impulsive forces is shosm in Fig-Ilii5b- T'lre line of impact is along;i The aqgular
velocity,beforeirnpscyisgr-:r.o\- {.ettbeirngularvelocityjustaftcrimpactbeC=o'L,andtheirnpulsive
i
reactions from ttre bearings be
.
, &=Eri+ari+4"L, 8a=&!+a"i+nof
The velocities of pointi 8, D just before and just after impacc are
u'p'=0'
tzr=-l.i,sb, o}"=0, ol=-''b'
lLE=-uob\ gD=0, !f,=-a'$i'{a=0,
= -0'2'o (1)
Coefrcient.of restitution e'; o'g=-'v'4=-c(ue, -oo.l + -u'b=-0.2(-ra6!) *'
,,-!
The fixed aris of.iotation is,not a principaL a:<is'of,inertia,at'tlre,fixed point r{ of the,d9o5*
1= -m(Iix-dlz) = o; If. = -m\bt2l(-dl2l = mbdl4, I!, =
'rrllb?
fi2"+ l6ti)"1= 'rr.b" 13 (2)
Et =U!,L+ ti,i+ r*H-'q
+ ' 6'H^= (rf,i+ i7;-+rLhx-'--; = -r:(*'aodj-+ |moj\)uo
(3)
The angrlar impulsemoment of morneotum relation for the 6xed point .A: fang^ = A ff, yields
t-(id+c)k+pjl x ri-dkx (R.i+R6i+8.$ = -l-2(frnodj+ trn&?k),"6
:+ . [l:
+
Rs=$
dfr5 =0
:+
i' = 0.4mbzuolp
k
-g,F =-0.4m62r.rs
i - (id + c)f - dfo :-o.3mDd.rE :+ [* = rn0[0.3 - 03(0.5 + / d)b/pl,.,q
q
:
t.I
I
I
63
(4)
(5)
(6)
J
.-J
iJ
J
JJ
J
I
rJ
-J
\
..-J
-J
-J
J
:J
J
4- ' Fo, or'tippittg (ooct'tiidag),tb:e smaller normal reaction rltr musi remain pcitive, i.e', using eq(6)
>0
Jvr = |m[e(cos 0 + hsinllb,
-$,2/R'fihcc,llb- sin0 + (Ifz- If.)slngcosa/mDE]l
u? (e.R(tau o+blh)lll-(blh)tane+u?z-
+
ff|rine1*nn1
(11)
Equation (11) reveals that the car is mote likely to overturn for smaller value of &e .d and for larger values
of h,(If2- Ifr),r. lt U&- t?rrk"ru ( t, then (11) simplilies to
u2 < el(tan e + blh) IIL
- (blh)Lar.ol
(12)
Comparison of eqs(9) and (12) reveals that in this case tipping rtould occut before slippin g,if P > b/h5.--Eor-r,o.ba;*ing (0:0). The aris of rodation is a principal a:ds- Equations (6)'(7) vield
.
(13)
/vr,/v2 = |m9(1 1twz/gbill
Equations (9) and (11) yield the folliwing speeds'u, and u1 for the initiation of slipping and tippiug:
ti = $/hlsR
,t - pgR-
(14)
Hence tipping occurs beforeslipping,if f > blh, Tlre values of, p' aud-u1for the unbanked road are smaller
than ihose for the banked road6- For a stmight.banked rcad (R= co)- Equations (6) and (7) -vield
!
Nr,ils = |mg[cosf +(tr/6]sin0l
i.e-. in contrast to the circular ca-se, the inner g,heels are subjected to Sreater normal reacLiotr than t
rvheels foi the case of straiglrt banked road.
;:ffi. r::T::ilX'HT.;;*. 'r.
rreigrrt h a.d o'idtrr b6{
outer ^
fai.'tt-"&{
-,--irl&t
is hinged at r{. and B abouc a vertical aris (Fig.E3-5a). It is rocating at
angular velocity a,o when it bangs against a stop D ar the ground level-*l
The coefficieni of testitution is 0-2- Find the irnpulsive forces ou rhe
door from the hinges and the stop D during the instantaneous impact- <fqfa
Solutioi The FBD for impulsive forces is shorvn in Eig-til5b- The line of impact is along,!- The '-gular
velocityteforeimparcy is gf-={.rok- tet the angular velocity justaftcrimpact be +f = r.r'\., and the impulsive
I
& = E, i+ azi+ a"l
The velocities of points E, D just befqe and just after impacc are
ls - -uobl gD = 0, v',= -.a'bi 'gb = 0, og, = -usS, tfi" =0, t/6, = -u'b, o'D, = A,
:+ -'a'b--0-2(-rob) +. 'ut' = -0'L*1-o (1)
CoeffEcient.of restitution ei u!a-vb,:-e(upr-rp.)
The fixed anis of.rotation is,not a.principal anil oCinertia,at*the.fixed point A of the door*
4, = -qa11-itrz) = 0;' t{. : -tr(bl2')(-dt2) - mbdlt, I!, = m[b2 fi2 + l6/il21 - ,rrb"13 (2')
tst= U!,1+ tf,i+ r* $q
(3)
The aagula,r impulsemoment of momeotum relation for the fixed point, d: Iary^ = A ffa yields
:+
i:
k:
j:
t-(+d+c)k+pjJ x .ri-dkx (Ieri+R6i+8"U = . I.2(frnodj+ |rn6?k),.,6
d&=0
+
Es=0
'-
PF = -0.4m62aro .
:
-(id+ dF -dR.- -0.3m&d.rs
:*,
JJ
J
I
--J
J
\
-J
JJ
(4)
F = 0.4nb2wslp
(5)
f,a = m0[0.3-0-4(0.5 +qldlblpl.no
(6)
-J
-J
I
rr
J
rJ
-J
:J
i
t-I
I
I
iJ
J
&,=ni+a.i+nt
z+ ' a'Ht=(rf,i+ td,i+r*hx-'--o)=-1J(im0di+|rn0j$:uo
JJ
n
J
J
J
J
= -0'Saro6
The velocity of centE:kESi}io=t bcfore and just after the impact is
yields . '
r"t"tioo
f=,mLgr,'
iaqla*-*"t*ttt*
;
eqs(4!(6),
Using
O-f6GroL
ah."
nuf + F! = m(o'r + 05)&zoi
erl+nuk+ n i+
gb
L ' lL :4'&.160 i =
.
u
ul
f.i+
)
=*
A, + il+
i:
iri+
F: O-6m&ao +
El = rn6[0'3 - 0'4(0'5 - cld'lbfgl"
itz =o
*
j:
.Rz*Es=0
itr - -8"
+
ftu+8"=o
k:
,?3 : E6 = 0'
provide a-tial constrail''
63 and & cannot be e,raluated individually. If one hinge does not
:n""and a thin disc
o[ axial m-o'i' 11
Exarnple 3.6 The system shoru in Fig-E3-6a consists of an impeller
mo hits tbe impeller'
L_-
tJ
with a slight misalignment of 0- It has an5ula velociiy ,.rqtust-beforc-a^-bird.of
LJ
man
The suppori at B does not provide any
Model the bird as a mass-point and the impacg as instantaneous'
0
(a) if the coefficieut' of restitutio"^"
:
axial thrust to the shaft. Fiod the impulsive reactions of the support
ffd;d;;;;;;il.#%" r.tt-;5ef:=i n:,_ i. ,W"
L-L_-
LJ
LJ
LJ
:
,otor
- lfhe
-(o) -^^
EBD for impusirc forces on the systern of the bird and the
Solution
is sh6wn in Fig-Ei!'6b rvith
the impulsive reactions at A and -B being
L*
&a : &!+ E:i+ tratt'
ftr = fQi+ 8si
at d rvith angulat t'eloci6- c : :'16 ! just
The rotor rotates aboul a non-ptincipal asis of inertia.of tire disc
1.2.3 are Lhe principal a''ces of inertia of the
before impact and with a/ : rz[ k (say) just after impacg. A-tes
[.
LL-
discarCwirh t1r=1]o6q,,'E:'r3= irorf1.I$=0si'ce!isaprincipaldirectionarCasi=-car!3.=-sirg, sl =sino.- s?:cos9k: n_cospe1 _singez i-"nosgr*.J"eg-.r,,=.*0,
fleace for che rotor
L
I!, = If, - rn2(0)6 : O.
..
(l)
:€,=-[,co
='f1s1n1 *f2s3n3+I$s3't3 = lrn38isin0ccg'
4,= tf;r-m2(0)b
i1 = tr+ fr'? + f;,'{=+ rgnS = 11 * f rz2Rllsins 0 +2cos? al'
(2)
H a = {I!, i+ rf i+ 11$t.l
sil ct!)'
(.) The line of impact is .lo.,S i The velocity of the bird just before impact: CE : :o( :-ai=Henceitsvelocityjustafterimpact:i/e.=-usccoi*u'j-- The?unkaorvllsu"r^r{'8.r'l?2'83'8{'8sare
o{ molltum telation for
determined from the,'coe6cient of rcstitution relation, *gul". impulse'momenl
and:bii.a: lte sdocity of D
of
the fuied poin[ d and the.impube-momenium relation for the system rotor
\I
lI
t,-,
i.t
I
A.
!
:
justbeioreandjustafterirnpacti*gp=-uroBri,cb--u,Rt!.Ttlecoefficientof',.restitutioare}ation
,'r,-rb"--.c(tq-?oyl
-"
:
+
o'-O=-0(-uosino-0)
Eence {e = -vscosa!. Using eq(2}, I+ng^ = A{.e for rotor*bird
E--
zbk x (A+ i + *ri)
)
*
o'.:O'
(3)
yields
: (If. i + ti i+ I* k)('' - tro) * AE x nrs(gi - gB)
!
J
L
<-
r.l
'^fI
-l
L'-
!
I
--
J
c
O
f
+i:
j:
k:
,
'-"r
t
:
c.a!t,l
a
.ii
l'
".-:;---
.'
h
I
The irgulse-rnomentum rq!.atioa L:
A?for rotor*bird yields
:
it, i + a, j + E3 k + Fa t + a"i =''o(!! - q'El = rnouo sin cj
:+
!:
.Er+&=0
l:
k:
A,+At=rlouosinc
E"=0
(7)
(8)
(e)
il"=-0'5msussinc' R1 =h"= &:0'
Thesolutionof eqs(a)-(9)yieldsa:r-rro, 1is=1-5m6u6sino,
Ians, =afl..t and r=AP forrotor*bird'
il;;;;;;;;;,ii;',ir,ir",q.E","udeterminedusins
di{Iering from that of the rocor alone
After impact rotor*bird form one rigid body with its inertia -"it
Afu for rotor*bird vields
conrriburio, o[mass-poinr rno ar'AD--Rrj+34k. !-ang^=
;;;;"
zoL x (n"i+ Euj) =[{ri,-mo(0)36}i+{ti,-rnol?r(30)}i+{I:,+ms(o?*E?)}\l''
cos d i - siu ct i)]
+ ir1 * I!, L)no+ (Er j + 36 !) x m6ue(-
i
-t(fi, i
3mouobsinc
I!.(u'
*
-,o) !:
-2bfr!.s 2bR4 - lf;.{r,r' :yd- 3moEl0o" - 3.rn6usbcosc
j:
g I!fti
-.o) * 'os&!"t' - m6uq81 cose
E:
yields
Tbe iragulse;mometgum relation I = Lpfor rot'or*bird
(i0)
(ll)
,
E, i+ ar!+ Ito f + f-o i + Ar i =,ro(g! - ge ) = rns{-u'
:+
!:
Ar+4.-rne(uecosa-t'r''81)
j:
k:
ea+At=lttouosilro
r?"=0
R1 i
( 12)
- uot- cos o i - sin a j)i
i
{13}
( t4)
{ l5)
-r,E .j?s, Rr,FL",ff3 ".e computed successively usingeqs(12),(11),(t0),(13)'(14)'(15)is rotatirrg at, the rates shotrn relarive
e,.'.6. a.z A body 2 consisting of a shafr B D zrtd a rcd G E(Fig'FII'?a)'
The rod.is of length 'L and
to platform l totaiiog about a fixed t'ertical atis at the given rates
the
a:iial testraiut-
'4't'
lrut' prof ide an'r'
ma.s ra- Neglect the inertia of the shaft- The bea.ing ot B does
(b)
trvisting
the
SD'
shaft
*r"rifinil (a) the bearing re4ctions at B and D and the couple applied to
;i"",
;;;;;;;;;;;r""t
t;
'orSnal
force and shear force at section :lu"r tt'" slender rod'
;[i!,:
i ik*,,,
-a
f,
,L
leJ-
sEa ! ;.
!
!
r'ldWt-*
l,l
e,f 7FE
ort,t,,
lg- - *il*
ro
'z
atI
--l+
.
{t
La.J**,rrit-u*{;
I k-Lle -*i
J
(()
$:,
(at
of bodies l 2 t'r'r'!- ground:
'
Solutioa G) f*t t!r,4tand sh. Q2 be the angular velocities and acceleraLions
4r : q;r L
gfr =.i'r i
93 : t4r f cr3i =rr1 i+-ri
,izz = ri;.r+.i'z i + tlr x t tz L=tirr i * rile j * rer1o2 k'
(u
(2)
Tte FBD of body 2 is sborvn in Fig-83.7b, where C is its ce-nme of mass and lf is the momeo' Tllil
'-"
of motion of body 2:
the s[a[t- The 6 unknowns 81, E:, Fr, Fz, F3, M 31re determined fronr the 6 equations
6s
_..
JJ
JJ
J
J
''l-J
.-l
J
J
]J
:J
:J
U
r*J
U
r'l
,tl
-J
J
-lJ
J
-J
-'J
J
J
k:
E= trru"oa a eo;pS\tiatlo-qp foa C-'-foi-* O on the
[-
f
lJ
[J'
E'
H
H'
H
L
L
L
1-.
L
l-
tJ
lJ
t.
.
l
]JJ
L)
r
u
lj,-,
l-
gC)
+ dlli + &ar(6 + d) - i'{;'"I L
- - ir; r, i + l,-t1tt2L - c,,i(D
x (rri+rzi+rsE)
Mc= ii+eir'i-Di) x (&i+B3H+ ?iLl+ei)
(6& - it& - 6rr) k
- (-683 + 6r.3) i+ (M + iLr.g+ !z&)!+
witlr I!, = o. tc = Ig' = mL2 lt2'
2 at c are parallel tb z,g,z axes
l;:'
ll,'.
(3)
(4)
The principal axes of body
Forbody2eqs(1)-(2):+t')s:Qt'tdv-Lt2''tu'=O'&t=i1'j"=it2'"t)'=t't1<'t2
body 2 for C yietd:
The Euler's equations of motion f-br
Mc, = 19,a" - Uf" - I!.\,,,r,,t.
Mcy - Suau -99"- €,\,.-'
Iy[c,=t,6,-(C,- $"\-'uv.
+
(Fa-8")A=O
(5)
=
+
u + \t(Rs+ Fa) = nL2&2f L2
6(Er - Fi - iLF? = mL2sYu2t6
(6)
(7)
F'--nec+(fta+rr-,r,e1ii&!+(Ra+ra)E='n[-]r'!f,!+{"tp*-410+d)}i+{t'rr(6+d)-}62'}LI
Er * .Fr = m(g'Vtit'Itt '
B',+Fr-m(g'Pt|tlzl'
+
:
'
{')
..
Ft-ttt-tllttzL-u1(b+d)l
R" + Fs : m[dr1(6 + dl *;lLlzl
The solution of eqs(5)-(10) vields M = lmLl?fuf,l3 -
t'O'
.
tilr(6 + dll
- ''"Llzl
l 2',,p2l.2lcb -ctiL(b + d;lzbl' 8s = f3 = ]rn[ti1(0
dlfztJ; f' = molp:f, -:'11(6+ d)|
Ft= i*lc -.trLlz-:,"'tru2Lz/ra+tillo *
R1= \m[g -.!r1z
when ri2
+ d)
0'
=
p;g-Ii3'7c'
Notc that a torque M is needed even
rvith centre of orlss C' is shor*rn i1
mass
of
rod
of.the
-r)/tr
(b) The FBD of part AE
'o(L
(d 6)i + !(l + "1i
i-ointr'O, f are on bodv 2 with aC' -- +
gc. = go+ 4 t OC' +r'z x
="
(11)
+ il luz(L+')]k
=l\.trtr * .li+ lrrrr(t+') -'i1o +d)lj + ['".r(6 (12)
W.= CR-rG -.')!xEn
l;c" =0'Ifr'
='nfi'=m(L'-tl3lt?,L'
Thegrincipalaxesofbody AEatc.".:p1.ll"ltot,r,:-oowith
as forsodf"2- yield:
r'rt'dl""
C.';ilh
fo'
A'E
body
for
motion
of
"t-t
The Euler's equations
Mc., - t.a, - ({, - Ili)Pt""t, =o
rfa"llzL
Mc.y - €, a" - g9, - lllY',', = m(L M"., = ry; a, - t*; - $")n,', = n{L - rlsuYo2l6L
(13)
+ W. = rrr(L -.)"(d'zi+ ?*tPzL),ll2L
(g& x OC'l
r-[m(I-"]lLlsc'
:+
-[*(, - t)l LlgL+4* = Pn(L - 'll
l--,1'
lj.-1
fixed'
. lb = so * cz-x 9C + * " !*x
-'
L-
dgid cxtension o-!!oar 2 remains
G
Ll9x.'
t
I
t
-;=
(tuapt(2L-,)-&'?(6+d))LI/6r (r5)
Mc.+i@-dix& = rn(f,-r)2ttr?(2.t;;)-&,1(ii;)$
is the normal
e the ! comPonent of !P
is in the -idirection' Henc
at
cross-section
the
io
Ient beins .he i
'{'
The normal
tht ihear force- The twisting mot
together-consiitt'to
j
components
and
!
of ea together
force and the
sirice tlrc i ana'b ilponents
n,ment
b;;di'u
tht
is
z,eto.
Qailself
componeni of Qp is
.""",o"r" the bending moment'
vertical shafr, at O (Fig-E:t-Sa).
mass at C is hitr; a a
axis at a
pxa-rple 3.8 'A body of mass rn *ith centre of
roaft is- rlated about the fixed'verticat
ti",-i?r,.i*
,r"
o
ilr,".-,h
*o*"rrr"ii
tts principal
".
"rrig to tlte vertical' Find
oc has a consta"i il'.ri""aun
J..uvv
such that
consranr rate c.r os-"
"-
,K,rry-=iffi;:"
"""'ffii;;;;
+=I'
N"L(-l .,,.n "-"LNi:
';,{o'L
(b)
. ,' to)
L^
i.,
lt
-:
ti
rls equ tio*s for
.
Solutionsincethe.bodyhasplanemotion'thesolutio"-t:t::O*tnedusingl'Euler
moiion fcr non-principal a-'ies'
principai axes or Z' -o*""t equation
for plane
any tuomenf, in the direc{'ion
rrirrs"joir* does nor,r-tert
tlre fixed point O'
r- Trre FBD of the body is sfterrn i. Fig.ffi;;--,*,.
from 3 Euler's Jquations ior
att"'*iita
L"
Cr,Cs'g
3-unkno.'os
The
axisof the hinge
*"utut"i-"i-J'ut'o"at tt'
The angurar velocity r,r and angular
ut:it":rit:O(t)
trJ3=-L)cosA"av=:-'sin0' d:=0'
u:g+
(2)
tt=t9=-{^rcos0i+r'rsirroj'
Mo =Gti+ Crl-nrghsindk
TlreEuler,sequat,ionsofmotionforthebodyforo.usingeq{l).(2}.r,ield:
Mo, = I?,6, - (19, - $rl,,-o
sin0 = 0,
i-e..,
,0=0,
or
or
0=t,
{3}
ft.=S:
r+
+
*
Mo,=19i,-(8-*.Yu-.
l{s, = I?uau - 1t!, - f,Yo'-'
.
Cz=O
- rrrg& si t0 = {€, - iff }-:
sin 0 cos 0
#)
,a
\j
t5)
-L'
coso = *st41;?r- {?,1'"
or.
0 =cos-![rrrgt71
€r' 8)t"!
- 19"|l'
!
'if a'? > WgnlT?v
Tlre third solution exists provided -L' < imshlt'fi.- !?1"-l: does not oierr
m"-*,"1r in .be directiirn
r,ils.ioinr
z- The FBD of the body is showu io Fig.Hi"-aretu"
equations for the 6-xed po'int
determined from 3 moment
Mt,Ms,e
unkno*ns
3
The
a:ris.
hinge
the
of
oc'rs
O- The axis of rotation is the noa-principal
1]:
.
9"=g=-cc9i+singi
= g= -sin0!- c1s0!
8r=O
9,: L +
sc
*r
'nt=-cosd' a'=sin0'^ nt=0
0
s' = -sin0' s' = -cosO; s' =
since k is principal a'tis at O'
+
*
- ry^ = t!"-* + I?r""n" + f;'s'n' =le..-
,i'. = *rq, - mglrsin o4* Megs
67
-l
rfl)sinOcose
'
I
t
E-
I.:,
Mt=:
-+
Mor=f;r&-$rz1y{oz = tSa + 8*"
i. *
'+
Mo"= t&'
MoL - 'r3*
)
--)
)
Equarion (?) ror 0 is
sir-e"s'e
,-,,
(?)
(7)
$yf
-r',, -row2r*g.*e
- raghsino ==(e,-
(s)
Ms=A
rheT-".::(:l-.y';::::;::":;:f::;,3;H';r:l*-issupporred
* Td teagth tr' (a) Find
uniro,m srenire*oi oi-*
force"shear
the norrnal
1u1 ri*a
on
ff;3,',::r.
i"*t
"tfi$;.lffiilffi;"
ex.rtea;;;t
load
au"
|oom'
-theof thJb.om-dt-d'
rhe rension i' tt " c.biJ "oa
t+rytion
f
*"-""'';[i
beading
ana
momenr
twisring
,2t4t
. lef
l_(J20,'T
(
force,
, i offi,*,
j\eft;i.il"
_ffio,
;W
t
o$r-.
'.4:8N7"
.Q,E I '.-=Q+t
'-rt-rr
to ;
C\'
'"
with the a-tis
so rhat a-.ris y is coplanar
aL o are choseu
z'y,z
inertil
of
axes
(
its centre of
is
principal
T
rvhere
'rs
Solution (a) Tlre
boom shorrn in Fig'83'9b'
au,
The
axis"iri" ,,, trre directio* of trre hinge axis- Tbe 6 unknorvns
of the rod and the vertical
*"*.",
any
exert
uot
and 3 Euler's
mass. The hinge joint does
i-notio. of the-boom: L=',.gc
n"r",n"'l]j";;of
of the boom are
R1,R2,R3.,C1,Cz,Tareietcrmined
augular acceleraLion r':
*;"0
*ili,
TLe
o.
para
noa
the
equations for
""*,*
L
l-.
L.
L.
L.
r--
L
L
L
L
L
.rtbl
I tal
(1)
u' =osin0' t'tt =0'
<" --!^'e:c"cc0i*trsin0j + "t'=t'tcos0-'
-'' =arsiu0' t'r' =0'
t't:i9=ricosg!+cisin0j * i':ocosl'
I
)
ly 2 sit'h
gI.=|li'
(2)
usine eqs(L) aqd (2)'
tb: k) + ,o-x &+ ex {gx QC)
'
+ r"2 cos0i- - !)
= llsin o(-"'! sin0 i
rng(lrsin0)lL
b= G i*cai +ffLcosl -
''
(r)
(4)
C
:
qithl;o'=0,lf;,=I!;:nlz|3and.ec(4},tlreEulecsequatiorrrofrrrotlotlfiortlreb:omforoyield:
I
11i[or=8",b"-(g-$1-'t''
8t)",
Mo. = I+'., -{*-
+
+
'g;'=''!ml1;"itta
TLasa- lmgr'sin0 -
ro
[:'mgs +
lmf,lt^'zsro e*el
i
(6)'
(?)
'
(8)
j;nrlsip0(+rzsin0!+r12c0!-t'r!)
;,tiiri* R8k+r(-si.0i+cm0i-)*rng(-cos0i-"']'aq:
0+31Di-lmf,sino&!
sirio} !+! m sin 01c,! f, cm
+ & = Rr i+a,i+Rs k : !n9(1+cm2 o)lz.oe 0-}m t*12
where B is the totat reactiolr force from
the tringe joint on the boour'
6g
E
I
(b) The FBD of pait'-r{8 of the-rod of mass m(L - {l L *lith centre of inass D is shown io Fig-E}$c.
i"i"o o.D ateoothu'EooH;ith O2.= +(t+r)i,
-i.,,. .,. b,:9o,1 ex'OD* otx(gxgP)
''i':
(s)
*U,*a)sind(-ru2sin0!+r.,2cos0j-&!i''
M^o : An + \fg,- a)coork- *tZ -z)1 x {a
(10)
The principat axes of body AB aLD are parallet to c,y,: aries n'ith I!. = O, 1f,, = I!,: m(L - zl3f t2L.
F=rn(l -dL)eD
:
= ln(L-z)?sind(-,.,?sin li+uzcali-Au
(11)
The Euler's equations of motion for body AB tor D, with ..;,s,..;.z,. . . same as for the b6om, yield:
Mo, - t?*" - U?" - I!),*t"u, =a
'
+
l{o, = t"Drio" -(e. - I!'b,-. = n(L- r)3asinp/tzt
0' ca
M o, =T!.b, - (I!. - tfll-,., - tu(L - ,c)3,.r3
| t2L
"in
M-o=.i'n(f, -zfsiao(<irj +o2 cos? \)/\2L= ea * lqt -r)cosd!- j(tr -,a)i x fo
0
(12)
F'n is obtained frori eq(f l) and thea C* is obtained from eq(12). Ttre rrormal to the cross-secti6o at .e is
in the - ! direction. Hence the ! component of {p is the normal force aod thp i *d k components together
constitute the shear force. The tsisting nromeut being the ! conrponent of AR is zero. Eence Ga itsetf is
the bending monentAs in Example 3-8, this problem cal also be solved using the nroment gquations for plane nrotion'Blairrple 3.10 L gytoscogic gtir,dcr (Fig.E:I.10a) consists of a'uruller'or grinder in the form of, a disc
of radius r and mass zn, which is rnade to rotate abouc a fixed vertical a-xis ai a constaot rate f)- The disc
rolls wit^hout slip. Neglect frictiooal force on it iu the radial directiou. The disc can rotate freely on the rod
about its a:ds. Find ihe force exerted by the grinder on the pan- Find also the reaction of the pirr aL O on
the l;shi rcd-OA- d in
tF y't,i
;'
s qfu.zEg
.
*K1rYK''-*K#
d;ffi-r.r\'.;
i.toVr.
c, Y,
\]_
/
.
\\J
f
-'-.
-tgt //,> ! .
-t*W6,!
ds .f,- 7T t-i"sg
<h"'
Inl f3 (dr
i\A
-
J
j
J
JJ
J
J
-I
J
J
J
JJ
JJ
J
Jrl
Solution The axis e of the disc is a principal axis at point O on its rigid eiGnsion] O remains fixed iri
the ground reference. At the instant shorvn (Fig.E:|.10b), the principal a:iis y is cliosen in the vertical plarie
through e-axis. The FBD's of the disc, the rod and the rod+d'rsc, are shown in Figs.ES-10c-e- The radiat
friction force F1 - O. Cn has no component along the axis of'the 4;rsc.- Ma has no component aloag the
axis of tlre pin. The 12 unknorvns Mr,ll[z,Ct,Cz,rV,r'3,f'e(3),8(3) ate determined from 12 equatioas of
motion of the rod (6) and tod+disc (6). The angular velocity of ttre disc 2 relative to body I is aloog i say
r.r1 !- Let ur, % and rir1, ri2 be the angular velocities ald accelerations of bodiei 1, -2 rv.r.t. gro,'nS- \trIith
O B = 5i - ri, and g,a - 0 (uo slip),
ir
9r = Og: Q(-cosdi+sin0i),
cnil
t-J
'aU
'at
':I
(1)
..-l
JJ
J
JJ
,J
JJ
J
E
L.
....,'j-'*
t"---..
1 -1 "..-, '
lj,
lj,
l-'.
L
IJ
TJ
L
t_
IJ
IJ
L
L
L
1_
Il.L
IJ
u
tIJ
L
L
L
l-
L
I :.
'= (.rr
,..o, -
':+
Eqs(2)'(3)
:+
,t:''ro+ gtz x@'= -(arl-oi#"blr*6Osiu{l!= Q
+ ri1 = 0
r.rr
=A(-ug -bsiallr)
L
1-
5sing/r) (6)
.
=+ M.a=qE+0ix(,tr!*Ezi+ae!i)
W=Ma-en+&ix(-$-9
:+ Mi* Mzi.=(Cz - 6Es)i+ (Cs * eBz)k
:+ Itfi = O; Mz -- Cz- 6Ea, 6s + 6Rz - 0
(7')
For the rod*disc, Euler's equations foi trted'poini O are applicatle dnce rod is inertialeS:
-n96sin0'k'+ (OsinA -:rce0)Irtk +'M:i+'([i- "j]
' W-
- -rfsi+ (ll:-[fali+
l{6" : f,.u, -(r?r- I?,furur2 + -r.F3 = 0
x fiEitr'
[N(6sing -'rcos0) - mgDsind]!
:+
&=0 i
(8)
(e)
!t6,.=I!,b,-(*-(;l-;,-.t,
+
+ ir'(6sino-rcoe p)-mg0sin0 = mO?sind[]6rsin0-(62 + r2/4)cf]
JV = fm96sin0+mfl2si!0{}&rsin8-(63+
f plwO'117(0sin0-rcosO}
(10)
Irct the normal reactioa r1I for thestatic case (A: 0) be IVs1. Eg(10):+ I{"g: mgDsinA/(6sin0-rcos0).
:+
JYlnrsr
(ll)
: I + [sind -$l2b+2bltlex0$gr!2s .'.
For 0 = g0r,cq(11) =+ IVflrI"g- 1+ Qz{2g,Ttrusa'largevalueof Oimpiesaverlrtarge valEcof AI aud
the crushing action of tbe SdDder b better.
,t mo.,"s in a circle of tadius esiaA 1 rcte fl. Ilence q1 = -fF6sia0l.'Fot t'he rcrd*disc,
+
F = mqa
&+ (lV -mglJ =-O2&sin,!
+. .& = -O26sitr0f +(m9-Ar) J
,
n'here N is given by e,q(U).: Equatirins (?) qnd (9) imply It{ p = l{1i+ ff3i = 9'
qrd soctet joint
b:rll
p,1a-ple 3.11 (a): A uaiform bloctr of rnss m - 120 kg is supportcd at D b: a
(Fig-1p.lf a). It is subjected to aa instantaneous inipulse with P = 2'10 Ns. Find the, impulsive reaction at
iUe-support Eind the velocity of C and the aagular velocity of t\,block ju1 after-t-he applicatlduof the
instantaneous impulse. (b) Rcwork part a if the bQc\.is lupportea ii D by aierticat incxtcnsible cable for
-(i) ncw-oik
Ns.
nail b if the block is supported at D by a spring of
t*,o cases l. p 240 N.s, 2- F
=
= -2{0
jost
before
stiffness 120 kN/m. (d) The bloc! is not supprjitid'biiU b trEuslating vertiiilty{orru at 2 m/s
just
vetocitl'
-Find
point
its
.8.
lagular
table-at
(rcqr
horizontal
ncar it) of a emooth
impact with a corner
poiot
t
tras
is
rough.a^od
p*J3,if
the
after impact. The coefEcbat of rcstitution c = O.6. (e) Rcrrork
,1a"!t9
joint
is
hit
at
B
by
a
zero velocity just after impact. (O The blodi support€d At D b, ; 6all and'socket
L1
L',
L."
(5)
: *, = m(?14+62) and eqs(4), (5) vield
- -69sing/r, r.r, - Osin6, d. = 0, 6, = 0. 6" =0, r;r, - -O2sin6(cosp -
For light rcdOA:
.
(4)
;:e:;*;l;;:;,-:fi::;;l;T""ff:-eanolr)!
For disc Z, I!, - m*/2, 4
htc
(3)
It
€*5
rf'
1.>
in
-6q
,
ur
o'
'.:g-
+rtr'
'*._,
i.r99''
-s
'.5
31,/.*P* I
&l*,n)
o'rj
/+_2t
GI
t9
!-he
*:
TrW.,ffi-,ffi;'G"
t1'u t
{' t*} (t)
(z)
.lo*o
J
Sot.rtioa (a)Let&betteimpulsivereactionatD(Fig.E3.llb)and u=u'ilu.vj+u,kbetheangular
t
,n,
velocityjustafterapplicationof theinstantaneousimpulse. LHo=Ia,.E, yields ar, llc=oD*oxDC
yields uc aud mAg6 = f yiel& !. r. y, t a.e the principal aries of inettia at fixed point D with
Il, = 120(0.52+0-3?)/12+O-252]= 10.9 kg.*r,
rf, = 120(12 + o-52tll2+ 0.2s2] - 20 kg.m?.
1f;, =120(0.12+t?11t2= 10.9 kg-m?,
(1)
larg,o= DAx (-60i- 240j + 120\) = (0-5i+0.15$ x {-60i:240i + 120E)
= 36!-69!- 12OL Nm-s,
(2)
The angular impulse-moment of momeutum relation for fixed point D.yields
/.g.o: f.ogo +. I!,.,L+ Iflutr!+ I!,tr' L= 10.9.2'!* l0.g,y j + 2&', E = 36i- 69j - 12t)&
+ .,. - 36/103 = 3'303 rad/s. ov = -69/10'9 = -6330 rad/s' ot' = -120i20: -6 rad/s
=+ r.r - 3.303!-6.330j - 6k rad/s
+ oc: b +u x DC= 1a-SOai- 6.330j :6!) x (-O.i6j) = -1.5i- 0-8258k m/s
+
lfrya Je -OOi-240j+ 120k :+
-J
:-l
--l
:.l
.-J
--J
J
.J
The principal axes at C are e,y, s axes with C,{ = 0-5i+ 0.25j + 0.15! m,
Ij,,= 120(0.52 +032)/12=3.4kg.m,, If,, = 120(0-e3 +f2;7tZ= 103 kg-nf,
I!, = 1fr(12 + o -52\I 12 = l2-5 kg-*'.
ite "ogUl"t
-
-.1
A= -r20i+ 24Oi- 219-1k N-s
-(b) kt the instaotaoeous impulse applied by the cable be ' i (!I. 2 O).The FBD is shown in Fig-E3-llc'
Larrs.-'GAx (-60i-l'i*r20$
-'l
-l
The impulse-momentum relation yields
n&pa, - !
!
- (30+0.15P)i-69i+(15-05P)L,
rJ
-/
Lr
(3)
(4)
impulse-moment of mornentum relation A fl+ = Irrrr. for centre.of mass Cryields
- €.-,L+ Ifru"!+ r9,.,k - 3-4r, i+ 10.9r.r, j * 12.&r, k - (30 + 0-15P) i - 69i + (15 - 0-5P) k -'
:+ o. - (30 + 0.18P)/g.4 rad/s, o, = -69/10.9 = -6.330 rad/s, u,, = (15 - O.5F\l:r2-5 rad/s
(5)
=+ r^, = (30 * 0.15P)/3.41i - 6-330 j + (15 - 0.5P)/12.5lk radls
(6)
lb = I!m= [-60i+ (f - Pli+ 120\]/120 m/s
(7)
uDy = acy = (f - F'11LZO
9-e =yr+ glx CD = tb + sx 0.25j =)
: I0 and eq(7) yields f = F - 240 N-s > 0u5r, =
then il^
taut, than
thai +1,the ^-Lrcable ---.i-o
remains iarrl
l. F =240 Ns. r{ssumc rl.-i
m/"'
Eeuce the assumption is v'alid and eqs(S), (6) yield gr = 19.41 [- 6.330i- 8-4 [. rad/s, g" = -0'5 i-t- L
2. F =-240N.s.r{orr-.thatthecableremainstaut,thenup, - 0 andeq(?)yields f = F'= -24$ Ns /
0. Eerice the assumption is not valid. Ttus the cable becomes slacL. with T = 0. Equations (5)' (6) yield
g=_ -1-?65i- 6.330i+ 10.8k rad/s, sc = -0.5!* 2i+ t m/s.
7t
-l
l
il
r
-J
-J
-J
J
J
J
J
-_-J
r
-J
sJ
I
since its stiffncss is.8'mte- Henee
l.j
t.
IJ
L:,
l:
t_.
IJ
IJ
L
(.),,.Tb9.:PIl$.G,:.*-qbl:.r*rO;;#'Provide anv impulsive reaction
#ti""'.ui-!qs(s),191 *ltu f = 0'
ii'.1;J;ir;;
"H:T":
**1ol lrorizinfr .Atil ::',t"": t1 .,''T:l'"J"
er 'rt'Po
(d, l're [n€ :ffi
ff
j::1t':', being gt,cb, gq1. The FBDf is sLorvn in
begre
(,- .. f,ranaa
l,o r,t, be .the values just0'2sj
1:-yr*1
T"
Y:
+ 0'15 k m, oc = 05 ! * 0'25 j - 0' 15 L m' Hence
;;';;fu ;;
'l
*-;::
?;:;r: .i;,- @= -o.si-
e i m/s,
m /s,
ar = 0;
0. * = o!"i' d =' ,LL+ ursj + ,1 g
t^'
gc = !E= -2j
{p = * + d x @=(o-r*rl +0.25e,)ia (oi -O't*n,-0'5ur1}i+(O'5-i -0'2&.,1)k
The 4 unknorr ns u',,u!r,4' u| are obtained using
conservation of nroment of momentum
m5) abour n*"1ryr"i d""iraaorr wirh o, and the coefficienl.of resritution
+
+
(8)
(9)
llio (- Ha*OC x
*::-t_
_
OC x rr.elt =0 *OC x rQc
Eb = Itc + €.t'.i+ tfirt'ri+ IgyiL+
(0.5!+0.25j - 0.15h} x 120(-2j)
3.4u,,i+l0.9oii+ 12.5a,: k + (0.5i+ oj'i] 0.15$ x 120uij =
(10)
ruf = -10'59 - 529au!"
:+
!: i-*'.* o"; : -36
':+
10-9.ri=O
i:
u|=o
(11)
a'o--o'Lfu''-0'5tI=-0'6(-2X13)
obr=-0'6ue, +
(12) + o'' .stibsrituting q,,,d,'fromeqe(10), (12) in eq(13) vields ui = -1-2X7 "tlt-a1^::"(q'
3'662k tad/s'radls, tl,= -i-UU, ."alt' U"o* lc-= -L'237i */t' i' = -4'0'tli-4-041
point
0 is consen'ed s'ith 0
about fised
(") The FBD is shown ia FigB3-lre- The momenL of ulomentunr
(0'5'0'25'
jt'"t dter impact' since g! = g = sb- Usiog EC =
being a poiqt of the rigid Utai
.l oa*'
]t'' .
13.6 kg-,.u, Ifl = A- 120(0'5X025) = -15
+0-1521
e. uL-0='e(ue.-i)
;
=
Ijt, = 12d(0:S2+0J?)/12+O2sa
l2(0'?5)l
It=0ks'*''
rfl = 120[(0 .32 +Lzllr2+Gls"+0-52] ={3'6
:ill':4'5'kssr2
9 &gro"
If; = lz}f(f+ 0.52)/12+0-52 +0,5?l = 50 kg'nrz' t?' =0 - 120(-0'15X05) =:
120k,
[r€}[!4J
:
ilo =9.+ QC y mgc: (0.5i+ 0.25j - o.15$ x 120{_2i1 -36i-t+l
- [i: *j * ][;il =[-il] + =[riff]
lfi]
=+
gr= -3'6't3i+0'3?63j - 2-I3h md/s
case of e = 0 for whi& only the-cornpone:t "'-""^: 9'
Note thae this case ofg! - I does not comespond to the
DB = 0'5i-0'3i+0'1L'
(f) TheFBDof thesysteaof masspointand btocl(m*m1) isslacrsg in Fig'E3'llf'
* = +uif+'i-tl _,'.
=Q, e8 =-zi+i-: L m./s,
s't
- !8 - 6 * ! * ry:(-l i+'i'i+ oi !) x'(osi - 0'3i+ $'
uif
c8
I
,
'':10:.;r;'+o-&al)i+(0'5r.t'}-'0:la';li-(0'34+0'&ailE '
The mogrcnt of momentum gD of the systcm of mass
(14)
-(15):
point and the block is conserved:
g'o= ED + t*4i+1*r'r!+I!,u',L+Dg x nir* = 0+LExmttb
=+ro.erli.+r0.cr;L+2erl&+10S1-03j+0'1Llx2(uli+ui!+ui$
!:
1
f.0.9ar!
- O2tti - 0-6o! : 0-'{
: ,::ro.*.r! +o2o'r- "1 = LL
!:
tu" + 0.6u! + d, = 4-2
., ;...r:;1 ixi.: i tj .Lil;.'.] l+I,i -'
x 2(:2!+ i- k)
= (0-5i- o'3j +0'1k)
u{ = 0.03669? + 0.018349ui + O'OSSOapu!
=+
0'0183't9u! * 0'091?a3o!
=+ . -'" =0-9-t5046 + u!, = -0-01 - 0.03"1- 0'05u'
.\.r,''
1'-'
(16)
(1?)
(18)
,
=...
1. Thelindof impactisidong!e.,---[. Eence ot,=t,'g','='-1.i-u..gt=uli+i-k ttecoefficieotof
restitution relation Yields
:+ a'k't'r+0'&'1 -l/'--0'6(0+2) (19)
ob,-'6. = -e(aar-'*)
(20)
eqs(16)-(rS) .+ uL=O, @, =-0.036697-0-01gea9uf, o! =-0-06-0-03ui
.using eq(20) in (19) : i, - 1.165? m/s, eq(20) =+ ari = -0.05809 rad/s, ar! = -0-09497 czd/s
+ y' = -O:05809i-0.09497! radls
2- Eor sticliing together * = {s- Eguation (15} +
v', = O-Lt t'o* o,*":,
io =o-*'t',- 0-tari' 1"' = -0'3''t -A'lt',
Substituting from eq(21) in first part ofeqs(16)-(18) yields
l-rt.t 0.3 -0.r1 [r.,'-l I-o-sl
+
L_tl
*r rdtil
LrlJ= Lj;il
y'
-
l.',1
=
(21)
[0'03455 I
l:iJ L;nu:.1
--0-03455i+0.05168j - 0'00965{f, rad/s
(g) , The FBD's of the block, rna.ss-point, and mass-poirrt*block ..g 5[esn in Fig-E3-1I5' ,Bet P be the
fi.ted point, in space coincident rrrith B-
'
CB=0.5i-0.05i+0.1!m,rc.=-0.5i*0.05j-0.l!nr
gc=98 =3!-2j m/s, ga=O lb= -2i+ !- ! m/s' 'i ='',L+t'ri+t'ri k
Forblock Ep- I{p +
8".!i+tfr-'ri+/9"-'.1+ PC-xmls-g+PC xrnk
{22)
e3)
l- The liue of imPact g' = i :+
$=o.!-2jm/s,
Q4l
*=u',i+j-km/s-
The b unknowrrs or,o'ri*t'.rtt'r,ul ate deterrnined f,rom eq(23), cou*rrration 'of notuentum of tr + r'zr ifl
a-dfuection and the coefrcient of rcstitution relation. Usingeqs{22} and {24) in eq(23! yields
3.Ut t+lo.qrik+12.er1h+(-o-si+0-05j-0.1t')x120(u"i-2i)=(-0.5i+0.05i-0-tux120(3i-2i)
.
!:
j:
'
k;
p,for m*mr:
la:*
:+
e:
:+
+
3.aw!-24=-24
=
10.gr; -12a,--36
:+
:+
:+
!")'_ =0
F
(25)
*:'r: -3.3028 + l-1009u, (26)
t;', = -L.44 * 0-48u;
{27)
12.*ri+120-6u"=102
(28)
.L20a,+2o',- 120(3) +2(-2)
uf = 178 - 60u,
+ d. x cB =.o,i- 2i+ (-li+.i,i+-lH x (0-5i- 0-05j + 0-1k)
(2e)'
u" * 0.1tri, + 0-05(,1 = t-13409.u, - 030228
'oBz =
:+ (1.13409o, -0.'10228) - (I78-60u") - -0'6(3+2)
ob.-oL, - -c(as, -re,)
a- 2.8691 m/s and eqs(26),(2?) yield u| = -0.L442 cadls, u!,= -0'06283 rad/s
'tr' =
: -O-1442j - 0.06283k rad/s
2- t€t gL - uri+lryi+urL. The 6 unknowns ufrru'r,tt'r,us,uyru: are determioed from eq(23) and
conservration of momentum g for m * nr1. Equation (23) yields
?.*,f.L+LA-9uiL+tZ.&,t' k+(-O.Si+0-05j-0.1$x 120(u" !+u, j-+u. k) = (-0.5i+0.05 j-0-l $x120(3i-2j)
-I3
J
.JJ
,l
J
J
J
J
J
J
rJ
J
JJ
J
J
JJ
I
k,
-
LJ
'
,--_l
* = ls =
=
"
0'tr'{
(tr, * 0.1c.{ + 0'0&.,:) i+ {o' -
+ o'st"')i+
- o's'"i) L
t* '
. li:
=(0.07772*r-r340eo'*a-24a,-o-s5o@;r$:r*l!;i,,Xi}iir.f;T:11"(33)
L_'
u
L
t-
Usi:rg eq(33),
m4 +
g
the conservation of momentunr 1
1
+
L
tJ
L
l--l--.
L
L.
mr yretds
+ 3il5Z9o'+ 0'1764?u') = -238
j:
120u' * 2(4'?859 + 0-240"
k:
120o.* 2(2-Ow3-0'55045o'+0'l?645u',+2'8405u')=.i
,i,#:^
;*,
[Il
[[-i.roos o'352s ,li,]'l
125'68 J lu'
-
rnl* = rny4 *m15 for rn *
120(3i- 2j) + 2(-2i+ i- $
120(u'i* p'i+ u' B + 24 =
* o'24rtu - 0'55046u') = 356
*'to'*,li'J*1'13409o'
l20u'
i'
+
L
k
=
,,*::^ l
tr_ilfkl [[l= [;li:;:,J
:=:ffi."l:::,t?l:[i' rad/s'r'r' =o'r881 rad/s'' = ;o'"t- 1*::t:t
J;
*o n"
ace'
bo:t, out. in space.
Exampte 3.12 A cuboidal
na,.
one face lralf
has oue
ope,,.
oPsl'-l1i.ig.Ea.rz.)
*"^'i-"t""t intothe c'on$guration i[
.,o-rhereisnoexternarmoment"::lI'*i*ft=i*:il::1,"",$i::*;::"
;tffi':i::;::ii:T::T":'T;i:il;l"J*veroci'lva'clhisi:mii
,-
rrc':
7
z
&
Solution fr"t'? Ut the centre
IJ
l'
l"
L
IJ
L
rJ
:a-O
(cl
zo' (e\
Zo-
l_-
e
conhgurar*'ra of mass in the: .oloig,r."aior
+
tEcJ = trn![el = trc[o "olr
Es
0
r*t o
Fug
(r)
F . - tG - : r rG,.rl
='i:""::]::*1.*f,Lt
*:::*nea u1 aao11;ms
. c"' = 9
" deletion)
mr = lz(ir+q(u')Jo = lOo?c :*lu.t
surface
bethe mass per unit
box
:
atea of ttre uox'-1 =
:
i.
j
:
2-
;
eatdAEKF
'?t3
- :1Ut .T',:"i-"E
=o26
*
planeyzisaplaneofmassWlmltrratc
l*-
)
)
Ir--i
)
L-' o
'J
Ir.\
l-
,
. !1,,=-',*'111'1009u'-5'5046u' (32)
"'d''=8-16+o'48u'+43ur
L2.5u!.-60or-6u'=102 *'" ('Lii
(0'5i- 0'05i+ 0'1E)
-;i+'t'^u)x
r+
o'i+
x
vL + d. CE=o'!+
B:
U
+.,
10sc.r!+60ur-tfi.: -39
j:
L
L-
(30)
:
'3'5294u' -l!1647o';
lr. ,n ';*turr* 12'"---24 + l, o::-i::::
(31)
--t'
;ffiHJ{r,*.;;"i*
iirapriacipalasisofincrtiaitG
decornposition ant nara{H
fl lo*:1o' -
u
o
+
I!'=0'
:
cc'=u'l*'719(iJ Sol2o'
Gcz=bz-'rc =9(i-!)a/20'
GCt=..(i- $o/zo '
*s(Gc")' tG c"l'
lf;, =-mr(ec0 (Gct)' - *'(@"@'o(tg o nl {
= - 10o2o(-
"
| zo\(a | 2ol- (
";(6t-*l'ol
.r+
- o"
I
- 1s4/20} =
z,"l c I to
:'
rf -i4
ilf
;Y
-\rlA:
t.:-d
jv
,;Full box:
{''
+ A, :3o4o * l0o2o[02 + (al2}l"] = l2laaat4l
f;. - -azalr.zllz+(9al2Ql2l= -343aaof l2A0
S. : a2 ol(az + 01 ! 12 + (tsalror?l= t283o{a/1200
E; : (l2lI 40 - 343! l2O0 + r2$/ 1200)oro = 457 o / L20
Hc = (8?j * 457\'loaousll20
paoel ABDE:
panel ADKF:
_!{
t-
aa
oPen box:
ec(1) +
;; t- L- il,- o-a.tr. vr
just after closur' eof lid. For aclosed
Let t t: ar" i* u, !* urk be the angular velocity
principal a-tes of inertia at its centre of mass G rvith
Ql
box, z,y,: are the
$, -- If,o =2l2a2o{a2+(2a)2}ll2l+2[2o2o{(2o1?ltz+(alz)?}}+z[axa{a21tz+o271=37oaa/6'
gr=Il"ut,1i--f;lrrj+I.c.r.,.k=I7aac(u,i+-rj)/6*laaou'k
tny --'8t"tol740,
*'. = 45?/360vo, :? t=
.
-\
-.-.t
$' - :3n'o
(3)
i
a{
J-\
_.r_
iEToto/i40)i + ({57@s1360) !
,' -c ,r}j;xc,
\
i
.
f;q
I
i.;f- 1>/)
- ;volume, each of mass m- Find the drag force on the satellite. \r-=
l{!(Jg3\(fi1'=u/g
'\------l
th+g
restitutron e.
o[ restitution
1:
T- - - -- -fY€ Assume frictionless collisions rvith coefficient of
sphere aucl
tlte
[o
frame
i[ertial
an
solution r4re neglect chauge in speed o[ sate.llite due to irnpact. Atiach
by the satellite in
Blarnple 3.13 A spherical sateltitc of radius 8.translates rvitir
velociry uoi in a rarefied atmosphere having rr molecules per uuit'
. <
I
and {3) Sield
Since tlre'eriternal momenf, is zero, II.c = fl;' llelce eqs(2)
-,. : 0,
qq
\
volutne zrR2rts srrept
consider velocities rv-r.i. this frame- The motecules in the cylindrical
of the collisions Pet unit trme'
unit tirne make impact with it (Fi8-E:!-13). The drag fotce equals the impulse
The impulse l, on it is expressed in
L{olecule 1 moving with to maties impact rsith line of irnpact along'g,.
terrns of its velqcity componeots just before ap-d just after irnpact:
u'r. = -eugcc9. t'1. = rt1, = euosin0
ula.= uoccr!
f1 rn(ul- -urn)e, = -(I*c)rnuscos0e,.
=
of these impulses ori the
The opposite impulse !2on the sphere is Iz = (1+c)mr5cos0g,"- The componeuts
in the ditection i add
sphere in directions oorrrrol to ! due to all impacts cancels out. leheri'as€oulPo[ents
up. Tlre component I, of !2 in direction i'is
t'_'a-
_r-<l
'.::.-,'__._
1.
,|
.",--r
l-.ti '
I" = (I*c)rnue.*AJ(-cos0)= -(l +rlnruecos?d
Let JV be the number of molecules hitting_ the sphere per unit titne ar angle 0, betrveen angles
,lV = a x (volume swept b1r the arc between 0 atd 0
* d0 in unit time)
= n x (volume of cylindtical ih"ll of .udius Esin d, thicliuess
n[2r(E sin 0) (coe 0 A dlr]ool =
!1
0 and' 0 + d0'
cos 0(ftd0) and length us)
.!-
a R2 o,sin I es 0 d0
c€ 0d0'
The impulse per pnit time due to the i*p""t oi'rV *"f"1ks- is .:\'I" = -2r(l + clmnffvisin?
by
git'eu
rl2,
is
from 0 Lo
Hence the drag force D, equal to the total impulie per uuit tinre for 0 ranging
*12
1
' :' .',. '
'
D.:-Zr(t+c)1a1n113l"'"tr.*. iA;,-2"q1+c)mnp?r,i1-|cci{g}fl/?=-}:r(1+e)ma.Ezu!
-
:
' +*r
rD
:!
-
I
Ll.
tj
l;,
,i,Y ht
aogular velocitv
tZry -trt'*1
"t''1'; 6i.fo 9*B {(;:;)j:-o
\-1ff1;
slarnpre 3.14iDisc-rof iass
and its centre has " ""t;"'L!il1;nt-*:"Yn
z attached to a sPrins
i 1*;Tf:#: V
\|{ )"1^
GFIC-iti
B
Hfji"i'3i"n"oir'iiilillii;il*i.*
ill*"::
''JY"nu"-"lrrl'
*ith'Jffi;
transr,tioe
.":[
r"-t
and b
lHilj;II:'lJ:"J':is
Gv'€rI\y'il - /'"t
'
*t:'ffi*'.xi'i'::5'lu.";fii:i:-':ii:iff";il::
T'"n'l;,.
surface. rhe impact
of
i,t,,
;t Y"H:t 1[":J:::'";;;;; 0'?'
=
?U I ;1'hema-'rimum compression
'. *-'t.dY.,:
smooth with coefficient "it*"*tt""
-fr lult, i1^e"- /%
*o":n
subsequen'
thespringin
:t'"""rsrduringirnpacr,instantaneous
:::I},'#jfl:tjilff*:
and the
;5T*-#r"*1r::H:X.*"*t
qrnL
o",tiilri;
,
ou disc ";;i;;;ngjnrubsequeor
impurse on
-:t:..':y;i;1ff.'P'ins
motionvelocities
-;"::"",,
:+ sino" =
-::6- The
of-tbe-spnng----]--*r-r* o" r4b) and cos60;a/5
= 0'8
=
--rnaximum extension
Ij
lj
l-,
rJ
IJ
I*L
t':
0't
and' cos
,
.
along
d;;';*)
Solutioa The line oi i*n"o is just * titu i""t t*
t-'-:t are gi'en by
U'*"
veiocities
of centrq *a 'ogt'lo
or' j l0sing=6 m/s' 'dl="t1 =2taAfs
L0co0=*nll"'
u'r.
=
=urr
ttL=1'2kg,
=-3 m/s' u\ =t'4 =0
rtt
m/s;
=,-5sin0
:
= -rl
tttz=05 kg, .u!,r=t'?r
L
-5€060
and conservation of'fo-
olr^,rttzo are determrt
L
rj
L
L.
L
L
I--
,L^-r'ro= -e{u2* - uto)
.', -
,
??"
*
tal rnornentum:
-::
mroio*m*Ln=rlr1o1o.**'-:''+l2u"o+0'6r'!*=l'2(6)+0-6(--3):5'a(2)
eq"trr,(ir-+,:; =;;
lnstantaneous impulse
r/-
;r"'1;1_:;,j.-1;,1',
on disc 2 is Jv'q'
The mor.ion of disc 2 aftcr lnpact
lltt" r ,. .., . ..
talies place under a
-
iffi; - ;;:':1;;"
t"
f6lss 3,s rvell
--,:" force which is a couservativeenergv of disc
"ott"n";:;;;
* ianical
sr.:*:ff.;;;**"n'ls'Ius'ia*erimpac'l''Lrre
0s
r:*::l*"**:;,[T.*TIffi
1
*. ..u'rJilffi ;-;;; = aa+
l-o.z *
tq
".a
spring extcnsion,6,
+
-
-l'l:h: *E'16go nr/s'
- -1'129'
=4(-cos0g -si*og)
*t*'*"
st2:'12g,,.49.'=?'2(-sinde'
'
,i= -t'izm/s' u! = 8-16 m/s'
p*el
In the maxirnum compressed
51 = -0.1 rn' *t""i"'j'"t'*ro""ti"t
a r1
r 0'{ - 0't = 0-3
of tlrc spring' e" = B' ' =
t'ol'
t'aoalv' i" this pcition be
' .'-'
*+
IIo' =ttr2rluct = ta2tlo!4
*
0'2 =
ur.
=
' 'Q'=
m a,.nd spring extensio*
I'6'32
= lo'rz
O'6(8'16)103
0'6(8'16)/0J-=
m/s
rn1
:fr+V1=f*V,:.1*,1o,i4,1+ite1=!::,,:+.,i1+}t8,
..;,,.
.'.,..
}(0.6X1632f1}t(0.r)"=.}(06)tq.ro?.+r..rzzJ+}[(0.?)?
=+ . &=39?0 N/m:3'9?0kN/m
*
.1.=
t* ,r,"l-onr exteosioa u" b -rtt " =.q'j}[oit-t
'-
6(8'16)/(0'{ + 6) = 4s96/(0'4
"':::1
-"1t0-ulto:+esg6'?/io'4+5H+
Ho,=@*o=,.*'{dn
;)
f-
i
u-l
u,,
t'--\
!
IJ
r
'T+v
=f +v *
0 and circumfe*ntial
t'elocitv uc:
+ 6)
i(3{0}61='
=lionltt,iu'+t't221+}lrszoyo:)2(4)
66J +nsa-ze: - 1s9.9, -.--::-;
+ 39?05{+317
* (6+0-1)[39?A5t +r,tg52+ 15?'86 --tz,s..fl =-o
1?5i,9--=,8'"
=+ 39?06!'+ 2ng62+ 15?'86-- "= "ifil t'i17.538 = o
(5)
,
- ''i' r'Jg
:
"i
;
:
'":83')::
']16"+i
-'=:t .''
'
al
I -r
ll
fi'<
compressed position 5 = -0'1 is a toot
where (5 + 0.1) has bei& fgtofed out from eq(4) since the ext_:me
Cardau's method,' is 6 = 0'2005' which is
of eq(a)- The only real rqot of the cubic *"oiio" (5), obtained by
,-<
g' (b) ti'ewotk part a for the
[he plate at angle
coefficient of restitution e = 0'4 and 2' if the rod imbeds in
the motion of the rod just afLer impact
plate rvith
''---- the same motion just before irnpact- (c) Find
uvv r'e'v
vr free
L@E
case of
fs
;nrooih rvith c = 0'4'
()Jt
i*O1rct-F.t
in part a if the plate is massivc, i-e., m2 ) mr and
\-+v^
ll1:
-
{
<
2-tls
Ai"
r-il;.-
, ,i L lJ*a i"l
oo-:-ri'es,
f-Y1-o{
:'
(cl
Solution (") t
*
t*
0'24 m'
u'rr: it. - -5sind = *3 m/s. t'la = -icos0 : -{ m/s' ri = 0'3cc's0 =
:O-38 m, ''' = 0-4 rn'
oL = -2 radfs, o.,: = 10 radr/s, dr'= 0.16 m, @
0-2? + 0-05?J - 0'016 ks'nr?
r9; = 0.1(0.6)3/12 = o.oo3 kg-rn" 19- = 02[0.6? + 0-f ]/12 +
[{o' for rod L' ttb' - Ho' for rod+plate and
The 3 unknos'ns ut'1,t,\,u". are deternrined using f/i- =
grouud coirrcident. q{th '{'
coefficient of restitution relation, r,ihere D is a fixed point of
For rod ,
:+
For rod+platc .
IIb, == HD,'
0.003o',
:+
tr":ri-
rnlu!,r11 = f9't, -trl1tr1'11'
:+ a.,i = 8ui* +30
I!,r;'2 + r9.'"ri + rrrlr:'1.d1 - *ur+ I?:-, + r7r1u1'd1
- 0.lui.(0-2a) = 0.003(-2) - 0'l(-4X0'24)
IIb, - Ho,,
:+
:+
0.016a,1+o.OOi1tu',* * 30) + 0.1ui-0.r6) = oI)16(10) +
+
.u',
{2}
ei
U'Bn-u'en= -e(uso-tA;)
+
ui. = -0.9?143 m/s,
and
:+
-!'ro -
eq(Z) +
:
-
{3)
14\
t)'8,, ='-tt t'z = 0-a(-25u!* ) =,-,oti
P. = .n2rr':0i4 x'10"-'4 m,/S.
:-
0'003{-2) + 0'r(-4xo'16)
= -2-btt'1n
tu/.l = ola + (-lrr - Jt + (-2X0-24) = -4'48 rn/s
a'tn =u'r* +r.ri11 = .r'ro * (Sui" +30)(O.il;: 2'92u'ri+7'2
'n
(t)
:
<
l2-92oi^ *72) = -0'4(4:+ 4'48)
.a!r=2-429 rzd'ls.
at O for rod+plate
2. -i (fig.Sf.rsc) is obtained from Hb,: Hs. for rod+plate' Let m'o'i' about z-axiskg'*''
be
ro- P = Ifl+[I9,. +m1(di +a{f --0.016+0.003+0.t(0-163 +0'38?)=0'036
:+
lo,.t'o = I?r-z+ I?.'., + m1o1od1 - m1u1rd2
Hb, = Ho.
+ ca! = 5.667 rad/s
0.03&ri = 0-016(10) + 0'003(-2) +o'l(-{XO'16) - o'l(-3X038)
For rod+plate:
=+
(b) 1- The FBD's are shown in Fig'Eil'l5d'
+
e.:, = ao +.r2 k x OCz = l0 L x (0'2 i + 0'OSi) = -O'5 i + li ^1"
,i = orr='-0.5 m/s, u2, = 2 m/s. I9,' = 0'2(0'61 + o'l211tz = 0'00?5 k8l-"
71"
t
*:
:
:
f^,,
f.:
r_
L
L
L
L
L
LL
IJ
po
whlie D'.o afixed
fc tod*plate *a *"fiA"i:&o=t6otloo'retat1cn,
:,.t1:
-..1.i.:!
::
::,':l
':
'-l
30
:+
as in eq(f), for rod 1 Eb. : ED,
=8ui' +
:+
E'p.: Ee.
t::yi-rn2u!.(0'2) = *-' - m2o2'(02)
For plate 2,
+
:+
,
.:---
0-00?*.,1-o-2ui.(0.2) =0.00?5(10)-0.2(2X0'2)
tri : -0-6666? + 5.3333uin
(5)
(6)
ollo : ?2o +{.r?(0-2) =2 + 10(0.2) = 4 rnfs,
:+ rnlolrr + m2t/2n = ntlvto * rn2t'3o
p!. = -0.5u,1'
0.lo!o *o2l/,, = o.l,(-4} + 0.2(2) +
tL - 'S^^ - -0-4(u3o - urn)
p.'r: p,
e
+
:+
Using eqs(3)da) and
+
(2-06674 - 0-13334) - (2.9,2ol' +7'21 ---0'4({+4'48}
2-066?of. - 2.92oi,: 3-9413
,i" = -o.oozo
[2-066?(-0.5) - 2-92]u'1" - 3.9413
,4:0-{985 mfs,' . qrl'= 1'992'rad/s
*
usingeq(7)
,Eqs(?),(6) Yield
(7)
(8)
*7"
2- The 3 unkuorvns u!,d2.,Q Gig,E3-l5e) are determined using P', = P" l'" = pr| H'8. - E2' tot
rod+plate, where .B is a point fixed in Sround and coincident, rvith c:-
l{
IJ
+
u
uc. = -b.i - 4j m/s, !c, = -0-5i+ 2!ni/s, {c"= 'i'i+ 4vi
(ui" - O-O{:ri)j
*, = *,+-lk x CzCr:4-i+ u1, i+ -lL x (-0-04i + 0.33j) = (u!, - 0'33r{)i4
u1r: -3 m/s, ul, -4 m/s, p2. = -0'5 m/s. tr1, = 2 m/s'
(e)
o'r"=4,-0.33o!, *: n -o-oa<.r!
"
pL=?, +
:+
plv = ps +
l--_
IJ
]J
l-
Hb.= Ee, *
L-.'i''
Ll)
f ,
13
L_--l
r,',". = - 1.3331i + O-ir-i1f O;
uj,.. = 0.01333&-.ri
(tl)
rn2sc. l
::
0'22',i)l
0.0071,i + 0-00iu'l + 0.r1'g 0a,-t'02666?t^r!) - 0'33(-l-3333'0-33(-3)l
= 0.0075(10) + 0.003(-2) + 0'lt-0'04(-4) -
+
u!r-zS36 rad/s.
(c) Since the platc is massive its 1nglrlar velocity temains unaltered- Ileuce ot' = o9" 7 O '4(10) -'A;m/s'
The 2 unknowns ,1,{to are determined ftom H'9, = HD, and the coellicient of restitution
D is a point fixed ia grouid ard coiocident uith ,{. .
As in eq(l), forrod 1
Usine eqs(3),(4)
Lj-i :
Lj"l ;
0-1(4= -O-33&{} +O2aL,=0-l(-3}*02t-0'5) +
itlTtlr* tIr=4, = tnrrrr; + n!3u2e
b.r("r, 0-04-i) + 0.2ui, = 0.1(-4) + 0.2(2) +
*, : (-1-3333 - O-22oili - 0.02666?q'i j
I?: riL + tf,',ri k + czc r x *&,1 = A:.; k + tr,? " t f
]$
Eqs(e)-([l) +
t'[J:-
mrui-+roa{, --riirur: *lztzttzt
-
+
IJ
;
l. rr;' = HD. fot platr-2.t'o=
ate4ualruow*ri:*li}fi CiaetAmined uiing f6.-= Eo. Yrodpoint
of groirnd co'incident with :{'
For rod+plate
L
I-
L-
;'
IIb. - If o,
+
r.r', = 8ui'
* 30
*1 "
-0.4(us, - urn)
_l*,:]1..=
-{
.
i
' +-.'(2.92ut;'+?.2) = -0.4(4 +4-48)
Eq(12)
=+ r./.= 30-526 rad/s
,.. -[g
relatiirn' where
(12)
u'1, = 0.06575 m.fs
<
to a thin disc of mass
lringed
B1-mple 3.16 A'iqd. Q4 pf y.nass 0-6 kg is hinged aL O t:
ptates at 40 rad/s' the
sijiona*,.lhldisc
hetd
is
rod
position,
the
given
the
In
.I.t'" hinge at is frictionless
0-f kg at / (Fig.E3.16a).
asupport and
.'j
'A
spring is unstretcbJ'
spring has a compression of 40 mnr and the rotational
Find (a)
released'
is
rod
The
the rod from hinge at oon
N-m
of
0.1
torque
frictional
constant
is
there
and
velocity of the
from o just after ,.1"".., (b) the.angular
the angular acceleration o[the rod and the reactio.
t"'
from o:'n:i_':::::i,:.t:::::o;,'$":to
rod, the angular accelerarion of rhe rod and the reaction
\'
R, ioi,,_
Y
,ph"..ffi
,,,{T>€i, 33i
;!'rtn
lot o{r=4d
e+l
h9v/-s
-l:Tl
Ad#,rF:t++T_-{f tbl
(b)
e,
;(,{=o*6,
:-q(.-N
V6= o.BCr
V^
o'Bwr
=
Solution The FBD's of the roi+disc and the disc o.l"Lo*uo
/ t'tt
,,;
ttll - T:t-"3, t', :'
Tt ,1,0;1,It,"ii;,i,
in Fig'83-l6b'
: ;, *r, ,*. = 1:1.(u1),/2 = 0.0005
;ffiH+, :T i;i;;i";;i;l ;.;l;,,;
L')r'.l,2'i'o'tttug"1"*f poriitionofthecenlreofnrass'4ofdisc2'
kg rn? Let the a,,gurar
t'elocities'oftherodandthediscbe
t'
Ma=!
+
I,o.u'='1{a-=0 *
t-'2=0 +
u3=co8scant-40rad/s
(li
the ntoutettt'
nr and g^ - 0-&'rt go' Fo: the rod+disc'
The e-xtension of the spring is 0 : 0.2 siu 0 - 0.O4
L" [or tlre
ai o, kiDetic *"'g'-1'and potential elrerg}'
of mornentu m Ho,abotrt the fi-xed axis of rotation
general position 0, are given b1'
(2)
o-sto'l)(o'o.11": o':11' + 0'o2
Eo- : r!,,,t1+ gf;u2* 0.8nr2ua) - 0-128*rr + 0-0005(40) +
(3)
+ oioom(o1:1 = O'OSe'rf + 0'4
T : l€,4 + l!m2vi+ !41-;1 = +(0.128)ri + lto.r(0.s,.'r)"
y - lkraz+|r(o-zsina-0.041?-n.r9(0.4sin0)-*zg(0.8sin0) =92+50i0-2sin0-0'0{)?-0'32ssin0(a}
(a) The FBD of rod+disc is shorvrr in Fig.E3.l6c. using "q1Z;'
are obtairred using f - I*iec- and kinematics'
ro' = No, gields ri1 ' ur = 0 arrd rRr ', R:
I{6'=!t,fg,+0.192&t=_0.1+{(0.2)+0.6g(0.a}+o.1g(0.8}:}4.,1=20$0rad/s?(5)
t6! m/sz'
* = -ul1a.a)i+6r(0-a)i= Sitnfs?, ee = -"'i1o't)i+i1(0'8)i=
+ Eri+ Eri+ ('l+0'6g+0'1e)!= o'6(8i)+0'l{l6j}
F= Imicc'
+ ltr = 0, '.R2 = -4'467 N'
yields:'r1'
configrrra*ion 2 f9r 0 =3}9'='irf6
(b) Work-enerpr relation from the configuration f ,for.0 ] 0-P
and kinernatics' The putl on
,"i"g l:imtg6,
Using.q(2), H6. =':Ms,yields dr1. &,Roare obtai*ed
dre spring for 0 = 30'is 100(0'2sin30o - 0'04) = 6 N'
(Tz+vz) -("r + Vr) - Wn"r-2
*
t(o-09&/? + 0.4) + {(r/o)'? *
r.rr :3.451 rad/s
+
50(0'2sin 30" - 0'04)2 - 032esin30"ll -
(0.{} + 50(-0'04)?l = -0'lr/6
(6)
frs. -- Ms, +
(7)
)+0'1g(0'8cas30o) '2(z.16)
O-192ir = -0.1- 100(0.2sin30" :0.04X02cos30o) *0'69(0'4cos3f
(8)
+ &1:2.772 rad/sz
m/s?'
a = -uzr(O.a)e. +6r(0-41e4= -4'?638g. * 1"10889
?1
:-
C
yo_
',,,.,...,..
='', ' " ':
. @A=-r.r?CI.g)e''+€i(o-s)+-+-9'52?5p;+2,21?6qrtils2'
'" :--; r'-; :
t'
--' - j05(:4.?638)+o.l(-g.ozis)
E=Err,.g"r. :+
- ^:
4:'L45N
-'(9)
fr'+
"
&+(0.6g+0-1g-6)sin30o
:
(10)
*
s.
+
OO1'''0"1 O'1(2J1?6)
R+ + (0-6g * O'ls'- 6) cos30" =
er- :
-* :.0't*
general configurati'on' ri1 can also
t"
oipression
coovealent
with
{f ia f. ..1,"
For this one d.oJ- system
o
1:
1
-)
0.096(2trrrirr)+[2'+50x2(0'2sin00'0a)(0'2cos0]-0329cos0\'t1=-0'1tr1
(ll)
cos0] = -0'r
+ so x 2(0'2sin0 - o'oaX0'21T0) - 0'32s
+
0-096(2;t)
[zo
+
eq(7) obtained earlier from
Equation (11) for f, = 30o -is: identicat to
o1factor
cornmoD
write the quations of
on caocelling
for rod+disc' $'e catl also
*"*"oilequation
th.
writing
of
lto..,Instead
t!.
ilo, =
to obtain
e[minat,e the r:11,ioFs 8s"fi4
t.he disc separatery
and
rod
tlre
of
f' itf, are obtained
",,a
motions
is shown in Fig-El}'t6d- N'
(c) The FBD of part og of the rod having miss rns = 03 kg
.rjng equutions of motion of aB:
. f." =m3aDi + 8'+JV-+(0'3g-6)sih30o
Fe=m3aso +
i
3,ri(o-21 :+
N - s-osg'N
E +S+(03g-6)cos$!r=::t(o'l ^=.,',i:=2'814N
/t{p,=I!,61:+ilf-0.1-2('/6}+(03g-6)(0.2cos30.}+S10.+1,=.tcg(o.al./31-t
+
,1I = 0'5954 N-m
N,F'Mcatral.obeobtainedusiagequatiotrsofrrrotiorrof(partBr{)+uisc.
the glr'en instant
Exampre 3-1? A 4-bar rinkase -*"Y ': ::::::1 ::i" ::T:i:T::'i.:':;:::':J
Find rhcsupporr reacLions at
i
'::.t_i:j", i;;:iT:""'.":J;
1)
,in
; i;;;.;;.ir.).
B &r -{Y
I r"Hji'x'I'l
?,1t Bia*-
9'
ternrs of arl'&l usiug
Solutiorr \\t first exPres t'tz,t">3,A2'@3 iu
kinenralics o[ the connectcd tinkaggs'
Pointso.r{onbodyl;.r{,B,Coabody2andD,Eonbody3havetlresanreplaneofnrotioa.Heuce
rlB=0'4jnr' 9B= 1-2!m
!r3=r,3k' !4=!'p=9''
m' OA: tn
95 =t"'3k, go=uD=-ql. -t4e =0'2j
gz:-rL,
ir =.rrL,
0-4j -'''t'(t'2)i
g8 = ee +;'i x AB = os(DB)! +
a^ =u1(OA)j-- 2 * lj = 2j'
"*.1:':.:
t1&-+ a"! x'48 = -n{(pB)i+a{DB)i
,^ = -u!(o.A)i+6r(oA)i: -4!+ alj-, ea = etj
+ -4i+ar j +ti2 k x 0.4j = -1'666??(l'2)i+'3(l'2)
+!:-4_o.4uz=_3-3333'j1w1=|.2t;4+til2=-1.666?rad/sz.til3_o.&83&.lr(2)
(3)
sc = sA-uiec+-zkx Ac= -4i+ti'tj+ti'zLx0-2j=-3'6667i+d'rj
9 (3 x 3)
g,r=2krad/s, gta:-2k,
The FBD,s of 3 links are shown io Fig-E3.17b.
il
;l"TI"" il::r,"",' "f'
,J
I
I
I
I
I
U
U
_)
24
The 9 unkuowns 8r' "''R5'61 a1e
- o.$s+ F4 = [1(1)2/3]ar
t
i
t
I
:_
;
I
determined'using
+ & = -te',ss + 0-3333&;" (4)
link3: Ittp,=IDt,t3 + -6_r.2g(0.6)-1.2R6 =l|.2(L.2)2|3]ri3.:+
& - Et -O'1, = 0'4cir
Fy - m2aai +
link 2 :
8o
1
R6=-10.886-o.*ilt
(5)
t
... :
:
: (-19.095 + 0.3333ur) -0.4c - o.&r + , 6t = 3-ZS1 Bd/s2
tis = 3.151 rzdfez, fufi5 = -12.40 N
E€(2),(4),(5) :+,
-17-83 N,
:+
:+
Bs +.8" : Q-A4444
Iink 2 1 Mc, = 19.&,
-0.2(r?" t ng) = ll.4(0.4r2 /121(-1-6667)
+
-10.886 +o..E.,
.F, = n2aca +
R5
Es(8) and (9) +
liukl:
- & - 0.4(-3.6667) = -l-4667
Es = -0-7111 N,
Fr=mracrz =+
8s :0-7555 N
l?r * Es = t[-&,i(o-S)] = -2?(0.5)
Rz
(7)
(8)
:-
(e)
* R<- s = l[.ir(0.5)] = 3-781(0.5)
ft: - Rs = 1.2[-&13(0.5]] : r-2[-1-666??(0-6)J
Iiak 3 -- r. = tn3acai +
Fy = rn3aca- :+ J?5 =-35 - L-2g = 1-2[d,3(0.6)] = f -2{3-l5l(0-6)J Fy = m1agr" 1
(6)
{'i
(10)
:+ -.Er = -2-756 N
+ E2 = 29-53 N
:+ l?r - -2.711 N
+ Es: t-641 N
I
-r
'\_
g1:mFle 3.18 Each of the trvo rvheels in ttre mechanism of Fig.E3.16a has mass m1 and a-dal radius of
gyration h- Each link OB has mass rn2 and is modelled as a thin rod of lenglh f,. The collar of mass ,n'3
:
._
slides on the fixed vertical shaft rvith i constant friction force .F- The spring has stiffness t and is contacted
by the collars rvhen the links rcach the horizon[al position- The s.vstem is released from rest at position
0 : 01- Assume that the friction is suflicient, to prevent the rvheels from slipping. Find (a) the acceleration
of the tolt". the.instant of release, (b) the velocity arrd acceteration of the collar *'tren d - d2 > 0, (c)
"t
the r-elocity and acceleration of the collar rvhen 0 = 0 and (d) The maximum compression of the spring. (e)
\l:ork ouL parts a to d if the rollers are smoolll.
T.
:'._r
l
._q
,s
b
J\
:-\
\a/
i-_j-
v
_i9
L
:.fl+r---"
>
:
l*r
dotqm
_.,=r
lF
,i
;'-r
-i._E
(di
it
tbi
Solution The work<nergy pdnciple is conveuient to use for the conrpletc s]'stem of I bars. 2 rvheels and a
collar, since this s1'stem has only one degree of freedom and the expressions of.T,Y and tft6 can be s,ritteu
i
't
'-
conveniently for the general position of the system.
B moves verticatly down and O moves horizontaily. Eence the inst,antaaeous centre of rotatircn of link OB
is at the point of interseciion 12 of [he norruat to uo - -to t e O and the normal to og = -u j at B- llence
the angular velocity r.r2 of link 2 aod os,uo can be related to u using IzB -- LcqO,IzC = Ll2, I-tO = f,sin 0:
:
__rf
:i
-j
,l
:
a
f-ul = ol(I2B) - os /(I2Ol : vc/{IzCl, :+
lal=o*.€{L.. ry = }usec0, uo = uean0 (l}
(21
$:-tsec0lL., .'
Ya = Lsift?i {;l,:r' ii=,Lce00..-.:-a . +
Body 1 rolls withsut slip =+ trr=uo/f =utatilr. ?and'[/ are expressed iU ternrs:'of,trand 0:
7.- lrm3tf +2flmpf;+ ]msefrolJ +2{lm2y1+ +(m"Lzl0)tl,3l
= i[*. * 2rn1(l + k]o/':')tarr2 0 +lm2*c2 elvz
V = tn3gLsir.fl +2nzs( |f sinOl + +&42 =(rr.1--l- nr3)glsin e * l}d,2
t+'t = Wnc :+
Tt
.!_
L
."_
,a
1
.
F-
.,|
' r\s-{l
(3)
i<
i !!-
(41'
the t+rm |ta2 Ueing included only if thespring compresses by A- Wac= -Fu. Work-energr relation is
applied in iategrated fiorm to get u and in rate fiorm to
i-
ri in the configuration of iuterest; ,Using eqs(l)-(a)
;
r--
:!
;:
_<
r
.<
',' E
[-"+2m1(1 +k|lrzltzrnze+!rn2seczrJlon+llhrnr(l+t3/r2)randsec2d+$m2secz 0tart0lv20
*(m2 +m3)grcos 06 = -Fo (5)
,,81
,--2 -'iii6.ri=;1qt
Substiiuting 0 from eq
"*a
ll'!'
."*.ayl,i iyyu,".F** ti1:," r:""
[m3+2mr(1+r3t")t"n'zai3"Iirf;.itt
iiigl*lo"
: ::, :
^',,,,,.,
,s<;c,'*$mziec'dt"n'lo2.*otL"'.
(a)Attheins0aniofrelease,0=0larrdu=C.I{eoceeq(6}yields^e.r,.
0t1
gr
rl / [''e + 2*'(.L+ *3/t') ttf + ]m3sei
1r = (ra2 +
'o3)9
from c<>nfiguration !
Applying rvork-energy relaiion
(b)
eqs(3).(a):
+.
o =12{(*z*
Wsa'-' *
=
1zi + va) - (]I'r +
orlo'+(mz*mslgL('sin92-singr)
k,k.rt,"..'2e.+$*r*";
V1l
it*r+2rnr(t+
zt 0 =01 to configurat'iou 2 aL
)tt"t0e
rns)9 - F}L(sin,, -"r"ili I {ms+2tnr(1+e3lf
from eq(6)
The acceleration'u is then obtained
i
oir substitucing 03 for d
0 :82 and using
- -Ff'(sin-gi -sin82)
+ }"'2sec? 0'}l'/'
(7}
and u fromeq(7)'
*i:"tttt*
o: ,Pt(*r'imr)g
(.) For 02 =
n- [(mz+'ta)s - F]/(ttt3* 5tn2;
Fop'6,- 0, eq(6) vields
rrheels have zero vltocitl'
Iz-is at () and hence tlre
configuratioa'
this
ia
since
*-hen the
m1 does not appear in eq(8)
3 of the syst'em (Fi€1'E3'f8e)
configuration
in
;;tt*
of
at0"ction
co.figurat'ion I and Lhe
{d} t et 5 be the t-*i*t'*
."Liioo u.t-*gr;rhe initial
verocitn
4ppryi*g';;.;;;.gy
zero
wrrore slstem.has
yields
.."fi*.ran 3 and using eqs(3) and (4)
(T3+Ve)-(fi+%)-Wnct-s
0l= -F{Lsindt +6)
rneo6
][62J-[0+(m:+ma]glsin
(e)
*ti;t::'ff
[0-2ra,2sf6l2)+
+
rlLsin,1 = 0
(*' +
(}, eqfT)
1;; - '16 -
*
(E)
j'Fl'f,ntn"'/t*
:
Yields
"r3)e
-
is the nraximum de0ection
of the spring'
;1"'f; :";;;:1",":':#itH*"il'H.:#':'3;*1""..x.*:::,T;*:ll:*""=""t'1"
l-€', tJ1 = ur' rrErrvv
*'flo;l'
of mass is zero,
enrain unaltered
arrd (9) r'.fi:"::L:::
external forces about its centre
to o1' Inciderrtally eqs(S)
at'":.:'.:"::,ti::,Xl:l:;ir11::i;;
"*ta.r."l
since tlr-rs"t".'n
"ti*t
terms involving t6 sec to zero.
rvithout slip'
pcitions t*"" fot it'" case of rolling
because u.r1 = 0 in these ttuo
^ ffi n *-------I
r"'
;;:-,;;;";'';".'::'
]iilfffi
-{[-:',-- lFil
tle:::::1.:l::: lffie,'*a'&[
"i"T:;':i;;.';.1*' ;n"-*o*'o'k :Y'::ll'":lili;'Jllii
inertia or
-P
lfQ{-i'"
;"";;;;*i'ir'"i' cen*es'r
t-)
i*:' Tn',*:::-:t *Y"'l:::
The svstem 1L3
the --ir orrorarion is
'fP
16rnrfr.
'cffi:,ffiJffi:;:;""t
\o)
\o)
!.r^.1-<
:.
,. ,. - -c,r.^
brocris*-v--Hq^i
rhe
sudinsor
-.tr' rt= v-*Q-,
i,i;;r. srops s1.s2 preventing
f::[[fJ:ii.
tt*ou"a' 6)-ttt io =-1'7'n1t"
velocity . :'tl
angulat acceleratioir of Utt CD'
The stops $1,s2 it€ simultaneousl'
find angular velociry ind
bar CD' the ttnsion
and acceleration of .A relative to
otthe block"a when
il:;ffi;;;;-cD
in the clle-an:d,
e
TJq; ' l'u+.'
.l::'" '- --9 ;T'?-t'")l"a
-
^r fi
it: "t'",::: -l-t'd
-fHL*l^
be r'rs so- thal th1,1a"timu1
|
the axis of rotation' (b) What should
uro so ".*'fr ;t
l*01*"*"*'oi uu"r A is rs? (c) What should'be
;;;
.,{
rq? Ilst9f
i|]fil Jieiust b""o-o slack when r{ has moved.out by
R,C
ibl
8a
^
1
sol.tiorr The momeirlofdheextern' ':**,': -tl'-' "1:::.3.":::::fJ"T: ;::i':"ti;::'": Hl:
T;'n:::i*:k;*;::;T:',ri::Tri!;
:,::?":JI"T:1Ti;i*il.ff
":i'ill.ff
Hi:.,T*f
:::::.JH"fi
ff;i,ff
ffi :*1;tt;f"';;i::"'",:X::"*
leb)
f SmT#;'TlllT:T;T;:"J#'tr:{;"%;;::,iil:-"P'Qrravemoveddown
f"i:$;i ilT:lffi:::;;;;;;""i1' 'l';|2=: i|2(rig
n{52ti +t}y" =
E3
-- .
a -.-.-2r-.-
Ho. =16rnrfro +2tntlc't+ 2(2m)(3rs\2t't
54mr3';o
m(52r1+2rz),'t = rn(52rfr a 2rfr)r"s =
^ 1r.
coostant
(r)
(2)
. ,--.^:..-rl
=+ Ho,=m(52rfi+2r21-+4rnrir^r=0
(ii)')i + 2[]r(r -'o)'i -4mg$.-ro)12
2
+ ;'11-*'41(zm)fl&ardt +
(3)
+21,1'ml$'t')'
T +v= !(romrfr)c,
2m9(r- ''o)
2-54ry(r'
+
1'5mi?
'ol2l'a
- rn(26rfr + ,'),''f
radial velocit'y of A be ur'
for r'= 216 be srr'g,r and
acceleration
Ho' = 0 [eq(2l'
(a) r-et the angular velocitr--and -"Y
y =*n.o.]'i 1*i"t "*''.
;ili;,
f,lo,
from
=consramr
arl,11,61 are obtained
_.0
(4]
0'9r';E - 0'9(g/re)ti'
54mrfrc^rs
:- , :.:=
Eq(l) ? m[s2rfr +2(2ro)i]-r =
i =:n(26rfr+rl)t.,fr
Usiogeq(3),f+,Il=constant+,,4zo,fr+1z,.o1?1..,f+r:6rrruf+2.5,agnfr|tg-2ttt.gti
l*r, *
The FBD,s
*
ffi'lifi'ir-4nr(2r.s)u1&,1 =0
:a' = -8,1ro1l60rs = -0;r'*53(c/ro)
in Fig.H}-lgb- The
p
position
of btocks ,{ a,,d. for-trris
"ru "r.or.,r.
btock P ;
block -.t :
F, =2*up'
f" = mG -
ri'l
:+
+
equation of
'rotio. 'of
{G}
block P
2T *2rrrg =zrn(-ilzl
t?)
T -2futttg = 'r{i-r'i(2rs}j
{8}
F6-m,,r$+z;61+At1=rn(2r6i'1-l2r;;-1)(9)
(tO)
1"-rng"j0
+
F-:mi
rz?=0Sirng' Atl=f i39mg' N2:1tr,!'
i=0'089'
))vields
eqs(?)-(lt
of
Using eqs(4)-(6), thesolution
i 8-O8g ia tlrc
arerrl = L211(grs)U3 a*d =
'
CD
tod
to
relati*e
.,{
blocli
and acceleraiion of
The veloci*
nor which i =
,*:y:::.:::"::1,.f,:
*!.1"'::
''n
ilY.Tl'ilH"ffi,".
eq(3)]'
outward radial direction'
0
J2,t s af€ obtainCd
::;[j, :."."**i*tl)l a'd r +v =coustart [usins
(11)
:!4tnilato
Eq(l) + *L52rt+2(2toi21.,n ..+,,,
nr( 2orfr +.ipl (12)
(2r:d;?Ic3 +i.5rIg1fr/ro - 2m9re =
+
,n[2ffi
con-staru.i;:
using.eq(3) , T* v=
a
-
yietds t^r6 = 0'4303(cl"l'1"
r- FBD's of blocks
Substituting ca2 Irom eq(11) in eq(12)
the
ror
rhe
r*Lu3be
19'J:'::LH:.1"ff?r"."=;""."'t ["q(r)],
l:'Y""
",,go1".,"locG
with ? =!'.4r3'are'r al
A and P are the ""*" * in Fig'E3"19b
ra\for block A and F' ='Zmap" for block P: '
(.)
:='*
e = -.it -
(13)
o3 = 0'9rs
Eq(l) =+ m[52r[ + 2(2ro)2]t'a = S4mrfrars +
(14)
:+ i =2g
+
=2n{-il2l
-2mg
block P ; F, =2tnop,
* uo=l'666?(9/r3]t/z (15)
+
-2-5mg=m[i-r'f(2rs)]
btock/ : F.=*(i-tC2)
Exalople3.20Acylinderofmassraisstation..y.,ll:.a.brllkofnrassmllringedtoafixedsupportai
is 0'2 and for contact of'the
of brock rvirrr trre cylinder
i.,
o (Fig.E3.20a). The coefficient of friction "".r*a
83
-.
}..
tr
U
u
klj
rJ
r ,-
lj
IJ
L-
L
tJ
L
lJ
L
1_IJ
l.j
l--tJ
u
IJ
lJ
t-"
u
L
tl-.
u[.,
[-=,,
[j]
1--
L.
) ..\:'a :tr:
'. .! f'-' r '
/- llrl
.":-.::'-
.t:
-il
_j.ri
crylinder.rrith tlre gtglit+5p-+ -A constaat forcc P i1.1qnti9{i"?:.th9 c1!inf1 at a cggstant height of L2R
-'ra1"9
-6f
tf,i'cyUnaer fot trvo cases: 1- ra1 : 3m,
of P for impendift'uiiiii6,ii
aboye the gound. (a), Find tUi
2- tal: 9.6m. (b) For mr = 9.6m, find the instantineoG'iititei"tio"rof C and the angulat acceleration
of tbe cylinder if the applied forcc P is greater than the .v{u9, gbtained in part a- (c) If P - 2-3rng for
at bottr
rrl1 :9-6mg, 6nd thc velocit| and position of C w[en ,tip
"PPE S--Nr
^ .-^
"i*i"
.se
ff'8
+r
-*e$
-.1
Y8'3
;+-ti
ur
'r ilffii-t.,,t'o#1
f"1-,?, ',;ffh:
3.(ffi1,,T-{ h)x tt'.(.ril}*
Sottrtion (a) The cylinder o,iit-tlve ir"piiaiiig;niriiJn "f roltiuf s'ithout:slip at either -A or B- The FBD's
of the block and the cylinder are shorvn in Fig.E3-20b. The $ un[llqrvns P,/{r,Fr,N2,f2 are deterrnined'
usiag M6, = 0 for block, 3 equations of equilibrium for the cylinder and the condition of slip at B or A.
:+
Mo. = rVr($.66)'- m1g(O.it) = O
/r! = 5m1gl,6
(1)
ltr6asslracimpendingtollingrvithoutslipatSandwitlrslipat,riwithfr=O:2/Vroppositetothe
directiron of impending slip (Fig.E3.20c).
Mg.=0
.Fr=$
.+
:+
P-A.ZNL-Fz-o
&=o
- :+
A'2-rng-/Vr=0
=i
0-2N1(2.e)-L.2RP=0
+
P=NJ3
(2)
0.1333Nt
F2 =
(3)
1- F.or m1 - 3m, eqs(1)-(E) yield rVr - 2-5ng, P = 0.8333nrg, f? = 0-3333mg, At = 3'Irng' \\'e clecl'
whether the assumption is correct:
#zNz:0-1(3.5nr9) - 0.35m9 +
f
F2l = O.3333m9 S PzNz.
Eeoce the assumption is correct- =+ lmpending rolling without slip at B occurs for P = 0-8333mg- [t should
be noted. that slip does npt necessatily occur at [he surface where the coefficient of friction is smaller2- Fot m1 = 9.6m, eqs(l)-(a) yietd N1 - 8rn9, P = 2.667mg, Fz i 1.06?m9, JtI2 = 9rn9' \1'e clcct
whether the assumption is correct
pzNi:0.1(9rn9) = 0.9m9 :+
lFrl= 1.067rn9 f.PzNz
Eence the assumption is not correct. :+ Assume that the impending moiion is rolling witt-igut slip at .A and
wit.hslipat8withFz-0.1/Vzoppositetotheditectionofimpendingslip(Fig.E:l.2M).
fr=O
M* =S
.. Fr=O
:+
_, Nr-*g:-.9*C:.Q
:+
0.84P-0.lJVr(28) =0,.,.' :
=+
P:0.1rV2 -'Fr = 0
,
PrNr=0.2(8m9)-l.6rn9
:r::' 'ii'.ri 'r:+
. :. i:: :::.:
Ifih cfcd:whether:the assumption is corect:
:+
*
JVz =
P
9m9
=225mg
' f1 !'t'35mg
: .. . -
=)
|Frl=.1.35mg S PrJVr
Eeoe the assumption is correct. + Impending rolling without slip qt .A occurs for P = 2-25m9-
8!+, ,
:
(5)
(6)
(?)
ii
'i
'I
';'.
")
.t.
.
:,
:
.,1'.
:
i
I
,]-.
'i t'
'
'
., ,i
,,'.,t,'
^.-- .,'
rolls'---lvrthout slip at A with ri =
ft)
\-, For P > 225mg,'tf,E cflintter
i, , R o are oLtained using 3 gqua'tio9L:. of motion of the cylinderl
,
*
,
F,=0 *
,A'r':tb:8mg=0
F,=ma *
t{6.: I!,,) +
Eqs(9) and (10) +
P-O'1JV?-F1 =ntz '
alL(Fig-E3.20e)' The 3 unliaovns
t,'l
(8)
.t- :
"=Y^^
Ft=P-0'9rr.g-ma
(9)
*
o=t)R=(1-6^P- 3'6ng)l3n (1O)
F.rE-O'1r1Iz R-O2PR= lrrr&z(alRl +
F1 = 0-3rng +O'4667P
:. o or*2-7861119'
(
if O.3rng + 0'4667P S O'2(S'ng) i'e''7f 2'25ng < P S
This soluiion is valid if lprl ,.rgr, ,i.e-.
,Fof P >2-Ts6mg,ttre cylinderrolls with slip at
obtained using 3 equations of motion of tlrc
+
Fr=O
F, = qa +
!{ct = I!,'b +
A'2'6'c are
A and B (Fig'E3Jof)' Ttre 3 unknos'ns
cl4inder:
(l t)
=+ Atz-9mg
(r2)
=+ o=(P -2'5mg)/n
* ' it= (1'4mgE -}'4P)lnR" {13}
N3-nr9-8nry=9
P - 0'1N? - 1-6n:9
t'6m9.R-O-lJt!:R- OSPR=
'"'"t''--
(c)Since2.25mg<P=2.3m9.<2.?86ryc'tl,uirri+ialrnotionisrolliugrvithoutsliparS{EigrE3.2og}.
I{o,-:Y1r-9-6rnd0'50] = 0
For block :
The equations of motioa <[ the c-5rlinder for
&:o
+
?
geuerat position r yield
+
At2-rng-I[s:O
( 1-{}
:\ 1 = 4.8m90/r
':
1\': = (1*
4.80/rlm9
(r5)
*ma
{16)
=+ Fr =(2-2-}-4ilblfimg
(1-09333 - 0-64612)s (1?)
0-2R(2'3rn g't = iilR"(al Rl a, a =uR=
2.3tng - 0-f iVt - P, = ina
F"=ma :+
t{s, = f,a * & B: o-lNzE .Fi = (1'106? * 0'16b/c)rne
Eqs(16) and (i?) :+
i'e'' if z ( a'7n9b'
(l:106? * 0,160/zlrng < 0'2(4'8rn96/r)
Tlris solurion is valid if lfllf ar1^r1, i-e-, if
eq(l?)"
\tlocity u1 o[c iu rh'rs position is obtained using
Tlrus slip s.arts at both contac* fior z = o-722gb.
,--,!:= (t-0e333 - ossl)g *
+
+u? = [1-09333(0'72296
- 0:41')'g d'
fo""'d'= l:,'?"e333
I+
u1 = 0'1738(sb)'tt'
-0-66) -0'6461n(0-7229/A'(|;ls
irrctuding the rvheels is rn and tlte centrc
] massof aself-propelled vehicle (Fig.E3.2la)
of g1'ratio. of t1 ' l-3 '
have uIESS rr1 , rn2 and a-tal radius
of mass is at C. The rear and front wheel assemblies
driven car'
ttrs,rvheels:.arerp, ild pt' (1) A rear'-rvheel
The static and dynamic coefficients of friction 1si
u is f,he
torque M1 ' The rvind resistauce is cu2:'where
starting from rest, m-oves without slip uader a drivi.g
the
t a*d after travelling distance z' (b) Discuss
velocity of the car- Find its acceleration and speed .i ri*"
marcimum
the
(c)
Find
slip of a rear-rvlteel driveo car'
basic method of analysrs' for motion with possible
achie=e
and the driving corque required to
possible acceleration for a rear-wheel dri,ven car without slipping
driven car x'ithour
the maximum possible acceleration for a fiour'rvheel
it- Neglect rvind resistance. (d) Find
'Mr.Mzrequired on [he rear and front wbeels to-ltrie;'e it' Neglect wind
slipping and the driving torques
driving
*" car and the angular accelerat'ious of its u''eels if the
resistance. (e) Find ;"";":i;J;;
"i
wind resistance'
torques are Mt,l{z and the wheels slip' Neglect
8s
:
J
JJ
J
JJ
J
J
J
J
J
J
J
J
J
J
J
J
J
J
J
iJ
J
J
J
J
J
J
J
J
.+v
re$
-vrQ
t-1 T*. ql
lNz
i,tq*,
F,
rPe- .-.-o
of rouitg uirhort sfip. Irence *ork-eaergy
'o' ,.r*t".?0"."". esuation'
ir" mot or the car' rront *'heers
gou"oios
rorce
.TlT;?. ':? ,]*;:;":"il:tlt:ffJt'-'t'"* oo
rriction force
rt.^ r*ant ..hels and the friction
iaput torque to- rrre rront rvheers and-the
f::::til*t::ffi;'J;ffi;;"fr:"i"**
acceleration' In ttre FBD of t5e car'
u.o''i" che desir-ed angular
on these is in the backward directio'
provide forrvard accereration- The rsork
trrc forward direction to
in
is
wheers
the
rear
on
force
frictiouar
the
the verocities of poirrcs o[
zero (such as frictiou fiorces since
citrrer
is
'
forces
internar
and
external
doae by the
and the driving
tU" wtlf.aone by the rvind fotc* ctt2
Lri
..
(such
zero
pair-rvise
contact are zero) or
a,.d angular
".""0i
of the car' and the aogular velocitl'
torque M1 onthe rear whee\. I*t, o,t t,itu" 'r'" """ulrati'on
car and rate of
& = alR- The kinetic energy of the'
*
,i*
.
implies
slip
No
=
wheels.
its
of
acceleration
given by
by all internal and external forcqs are
;;;a;.
+ tnr2&!(u/R)?]
+
1 = |(m - m1 - r",r)r'+ []r,1u1 + !nr'*!(ola]"t []'zau?
(u
m2killttllu2
(mr&?
+
+
''
(2)
. = if*
g{ - Mp+cuz(-u) =(M1lR-c'z)o
-- -r-
--..-'d
ahe acceletation is obtained using worh-cner'r.retatioi
:+
qnd' cancetrut.
cancelling
t{ QII.'
T= tiz'
in ratc foun ?=
'
rcomr1roa factor n:
(Mtf R- cu?)u
[* * (*.ei + 'r'r3]/E2l oit =
'a=(M1lR-"'?)1@*(*tt-? +n+ki)l?2l
(3)
dvldlaad separating the wariables'
tb obtain u(t) [o. coEstant Mr' iategratc cq(3) w.r.t-ztbv wtiting oI -=u auyanaad sepaqatini t&e tariables.
by writiag a
;;;;4ri
lr"
ro.'"o*tant Mr.integralc cq(3) rv'r'L
A#=ry
.- I lm +(-"t? + "r?ti)/E'zl'
t
=t / [m + (m1!] + m,Li) I R21'
l r' ffidv
.:
obtained usiqS'the'work'e'aergy relition
For eoastaot Ml, o(t)caq also be directly
being Mt(xlB}.
work done by Mrin-distance e (rotatioa z/E)
.,
in'inteigated form' the
tz-Tt=Wt-z .3 |t',"+{1rri +r'f.2k1.llR2lu2 =MplB'
For.Mr(t), o(t) is obtatuied by intcgrating eq(4)
rv"r't' timc
, = [/ u,t t] dtl I RVn+ (m1ef + $"k11I Rzl'
(b) Tte FBD's are showu in'Fig'Eii.2lc' Let O be a futed point of the
Ho, -(m1tlo1 * ir1uh1) f (mzti rz + *rYlrr] +#,1X13 =
86
grcund coincident with A.
mplr + mrti w1 * mzkloz-
(5)
lr
,ll ,<-,1
.t
.
1'q'
be deterrematic viriables a'61'&2' can
O:'
foidei;\il1;IViifr,Pz''&
8
unkaowns,
11
The
"iU.tbodn front and rear wheel assemblies of car)
'Rz'fu'\
i"di* (main
the 4
rEised usi.g I (3 x 3) eqiraiions of motioa "i'g
we shali writc ? equations iavolving-ouly
and 2 conditions of slip or uo slip tt e "rra i.-g;;et*,
For the whole car:
nrgb2- euzd=lrtoh+m*lua+mzkZ6'
Mu- - Ho,
'vl6me-ilr-Nz-t
Fc=o
For rear rrheels:
. Fr=ma
Msr. = $,'itv
+
_._
1--
_
1:l
-
trr
Mt - &R = m1klbl
p, replaces p1 iu eqs(1I) a,nd (12)' An assumption
.,..<
::i
(8)
&- F2-ctt2=ttto
.I]
ContactcoaditionatB:ifnoslipthenul:olI-'ifslipthenFl=peNv(11)
F2 = pvN2or if stip thin
if no slip then u2 = a! R'
Contact coudition at z{:
For tLe case of impending slip
<
i,,)..
-{
\
(12}
is made regarding condition
-
<r
3
ofsliporuoslip.at"orit,i.t"landB--/\rr,Nz,fr,Fsaree.teyledinterrns'ofo,6l,b2usingsuccessively
equations of (11) andldTz) La 5'ield
ir
eqs(6),(?),(g),(10). These are substirured
"-tii""o
is finatly made 'about the
forces are computed' A thecl
and'the
solved
are
These
o,riliria.
for
of the
equations
slip and rvhether the value of the mignitudes
' p.op". directions of fiiction for-ces at the contact(s) rvith
appro'lxiate
frictioaforcesatcontact(s)withnosliparelesst'hanorequalto[lrerespective.O.t"dl"Oofsiaticcoefficien[
rsit'h other
satisfied then the problem is rervorke<i
these ;;;"
If
reaction.
normal
and
friction
of
"ot
por
is gi'en b.v 'q(a)'
i* Fig.E3.21o *,y:l_lrl",:-'i* force. no:l*,."
the front rvheels rrill lose
amount since at some st'age either
florvener, a cannot be increased to unlimited
can be determined b1r Grst
Thu *"r.i*om.pcsible aceeleration
stipping.l
start
wheels
rear
the
or
contact
1olR u1 = i2 - al R'
i,ll =
findiug tbe forces as in part b. For no slip,
"'2 +'rl2k3)/R|v (13)
Ho, =(m1l;fr.l1*m1uJr1} *$n2k!u2}n2llh2) }m3uh3 = [,nn+(m1[i
-_
<
r!
il"H:"rril,:15;::*":
Forthe6uaknown'qJVr'IVa,fr,fz'd.'M1'rveusethefollowing6equattons:
NiO - mg0r = {mlr +{rar*? +'rn2*?2ttlRlo
M6. : iIs.
For the .rhole car:
ra9-lVr-A'u=0
Fr=0
:+
F, = tna
For rear wheels:
Msr, = I!,"t1
For front wheels: M6r, = I!,'b2
Ft 1I4,Nr'
condition of no slip aL B:
and condition of uo lift.off of frogt
aA i;;;g
-
(18)
nr, > g- (tg)
w\ee[s:
eq(a)J:
Substituting these in eqs(19) yields
"i-.reil
wheels if
and no E$-off of hont wheels if
:+
I-
o 1 o1= p,9bz/6[1 + m2kl/rr.}} - P,{h+
a S az = s\flh+ (m1ef + n2*lltmBl'
The ma:cimum acceleration': a = min(c1'42)'
^-{
,-
i-
i!
lt
-
._
and
t
.
+rr.zkl\lR]rollb' (20)
N1:lrnsb2*{mlr+(mlef +n2ktr')/Rla)lb, N2=[msD1-{*h+(m1&!
(21)
,/r.2hlllrzlaa
f'r': (m + tuzk:'/Rz)o, M1 = fm+ (mlei +
Ft = n2+],alY\ ,
no'slip
1
(17)
in terms of o using successively eqs(14)'(15)'(18)'(16)
The forces lvr,Nz,fz,Fr and M1 areexpressed
,u" *o1r.-"nergv retatien in rate form' i'e''
(15)
(16)
-F2-no
Mr - FrR= *1*11alL
F1
FzR = r.2klol R
tu)
i!
(m1ef + m2L1lll'rl.Bll'
=
<
g'r
:
q
'r!''
of
wheel
gom eq(21)' -The liftpff c- --';*"*
'tto(t'
is obfned
tot'
ac&lerarion
this
g3ea
being stry
i) .t Atrf :"* T::..;:;;aed ii due
a beias
of c
* the vaue or
:":
ao' to
Tbe value of Mr requir"d
*qd;;;;";i*
*ohuo ti"
teo-sheel
a
9$-'-+i;
'"
"*t".
thi direction of frictioo force on
analvsisfor;i;"!*5r-**"i:"l;:fff
ris-E3-Jld.. No'ice
u
= w viaa". ":":,'::' "'
. -'
""'i',"
lll,:f #:; :::il:;::**:"T"'T'T:;;
(mrt? +
{i[1 + Mz)/t*+
tR
L-
[rn + (m1tf +
L.r
*
mzkzz\l?2]uu = (M.1 + Mz)ol^
a=
r/.zk:2)l P.1P. (22)
-
,R:. ,...
h,E
-
;3
*#-ry
d=#
"##*,t,iT-,\
-E--
'
i-
frovidedthereil'"m*e "r'ttie
' 1t6,, = trur
For the shole car:
lt
t-t
LJ
LJ
For rear rvheels:
L-.
For ftont wheels ,
LJ
u
*
-i"-i--ii)"r",
(24)
Ft* Ft =trt?
Mt - Ftf ='rni-{c/R
+
]
(23).
'
mg-Nr-J\r'=0
[:l
*[z- FtR= m2klolB
=(rr+fz) l,m<P'{NttiI:}=
jol"- '
?hema:crmumacceteration "
arrd (2?} ujing eqs(20)
IJ
y'{rr"dim=pts
",- ^^.tit**t
' F'=o'N,-
.and (2s} for IrIr , J{z , Fr , F2,.o|
(29)
.
l{t . Mzarre obiained from eqs(26}
(*tei + mzti)}/l + '"i ri/E}
Mt = $.el&R{b * v,lrrruit+
rcl The FBD's are shor'vn in
:'^:
by eq(t The 5 -unkno,sns"I{r'JV3'a'ur1'<'r2
')'
Fig-E3'21t'- Ho: is given
.
F, =ma -.
\)
l-- rl
r.- f
:)
Fq front wheels: o"r;
t''ioz
j
mg-ltlr-N:=0
Fr=o-,
!
_)
Jqb - atg
Mo' = i{o'
For the whgle car:
l.: J
IJ
1-,
': IV'D-nrgbz='
'''
f. = t'td
Mct.= I3-"i"
h{cr, : €:a,'
'-.
l-
u
L
brirrnno*r,, Ai..lv,;Fr,F3,o,Id1'It[2:
EguationsCI3)and{2a)yi,eldlVr.Hzgivenbyeq(20).Itfollort-sfromeqs(25),(28}and(24)that
t_-
u
tu
1"
u
L
f---
_*;.*.,,,,..#
q'h$';ii""f;tt*::.i,::,,it;
or-Tt,
.. (d).
a rtr'PtNz=mo
ttllrlr+
IrErrI'
+
:
a1s
(31)
(32)
t2..-
Mr-ltNrJl= mrti&r
m*?6t
\'-- ^**'R=
t33)
\oo'
i
(35)
Snbetituting&y&2,oftomeq9(33).to-.(35)ineq(30)arrdusingNr*JV:.=rngfromed3t)yietds
'
tv'L---Mzllb- (36)
E))
tttl-- Mt
Pt(tr
JV2
'r
[rng{61-Ft(h
=
8]] + Mt* Mzllb'
fV1=tmgt6z+pt([-
aa
[t
-t
^'--l
,j-J
from eq(36)
eqs(33) and (34) usin5 lv1'il3
,,,t.rizare theu
all the wheels
"o*p.aiii;m
the rear *i*r" tral;i'r.and,with
Tbe analyses of the retardation oi. ,"r,i"u with only
reveals that a
(35)
and
equations (29)
Compirisoa of
braked, are sinilar to those presented for accilerttion'
slip on
in such *f1jt that there is nq
"
gr€t€r retardaiion of lr,gis obtained if all the wheels are brakedthere is actual
jlip on the ground yielding
are locked, then
the ground. On the other hand if all the rvheels
":J
f:,J
para*er
km rvith a verocirv orsoboo km/h
is raunched ar an arrirude
aod its minimum and
'to the earth,s surface. Find 1- the minirnum and maximum altitudes of the satetlite
tlles of travel from the
otUit and its time period' 3' the
maximum velocities, 2. the eccentricity oi itt"
velocity components
and 4'.-theradialanrl cirlumferential
perigee-and apogee to the end of the minor a-tis
and (ii) after traversing
altitude of 1600 km for the first time'
and the speed when (i) the satellite reaches an
to a malfunction
o""* ?*i,--"' (b) Work out part a il 1ue the satellite in
a polar angle of 60o ar earth's centre rro* ti*
(c) tlom
s" * iL" circumfere*tial directio''
the vdocity at launch is at an outrvard "rrgi. or
the optimunr location in
Find
of-eccenLricity 1-2'
trajectory
a
llur"
to
launch"d
be
is
to
probe
part b, a space
the objective' (d)
in its *elocity at launch to'achieve
the orbit where i[ should be launched and the "rr"r,5.
be launched from it so t'hat
U'-aA oUlivation module is to
At the maximum altitude of the sarcllite in Part
'-
;Xj:T:;t'i.T;::;erire
"ry.
-J
J
',: I
T
'J
J
J
in its subsequentorbit its minimum ai"'"""eno*theearth'siu:-.::i::i:,
taunchiirg'
viug cases: 1- circumfettntial i:"""*il?,
rockets) &r'the-'follot
required at launch:(imparted by appropriate
.f'z
t.
of the earth as 6400
2- radial launching frorl tlre satgltite. Take tlre radius,E
|^,I,
,<u nn;.:
ilit'*itlffi
il;'ffi (ffi"9*K# JJ-I
;q
k-r;=--+{YF:"r
sofntion 3'*"1*,
determine
\-
1 1fi<fLl!"| ,-7 *,'
1b; sY.
\
t,r Sl*t1 Br..
the given dita for ihis porallel }
the i basic constauts Glr{,ho,D from
laraching.
GM = sR2 =9.s1+{6400 km)? =
,.rrffi(64o0
ro - 6400 +650 =?050 km'
cMlh|= 1-16416 x'10-4 km-r
rmin = Kl|zth"+ D)-r = ?050 km,
[*ir, = rmin - E,:,650 hr.n,. . .
omin = ftolrr,mrx = 19245 km/h,
km), =520?56 x l0r2 km3/lf
oo:ooo:30&}0 km/h' L6 =16o6o = 211-5
km-!
D = lllts - GM ft:|il - o'25428x l0-{
imax = $Mfte- D)-r = 10990 km'
-, Lma:r = :*1 - ft=aigg
x 106 km3r/h
t*'
omax = Irglrmin = 30000 km/h'-
e:=':y':*:^0.::*'
=':ly,lz):
[ = c(l -_'?l't" = 8802 *:'
1 T2
1 alris (Fig'E3 : 1t'a.d
'22a)
mindr
the
of
end
zt'the
to
r*in
Q
at
perigee
P
Let ?rbe the time to travel from
.c
: (rmin +rim.,g)/Z = 9020 km
be tle time to travei from.apogee '4 at rm'a;g to 8'
!i = (area OPQ\ I $ol2) = 0'5078 h'
area'O PQ = tobf 4 - 6(oe)/2'
?i = (area OAB\ | (hol2) = 0'6?17'h'
area OAB J rabl4+ b(oe)12'
T-he launching pcitiou turus out to be at rn in'
At r = 6400 + 1600 = 8000 km :
v6 = lrsDsind
Llr'= GMlhf,+ DcosC :+ e =70'27o '
- 5062 km/h. 04 = hglr=261137 tim./h' +
At {:600 :
km'/h' +
v, = hgDsind = 4658 km/h, vo = l:,ab = 2?311
81---.
J
e = (ui + o?)rl2 =ZbStZ
tmTl'
u = (uf + v1)u2 = tl7o5 km/lt'
o
"--f
J
,lJ
I
U
r
..-J
J
J
J
"rJ
,J
-J
JJ
J
!': \
--/
L l-S
f--.
'-'t ''
The speeds
-w" u could'il;* ui; "urained using energr conserva[ion: \ma2.-GMmf
first determine 116, D foFttris ao,.pamllclloonchiog (Fig.H]'22b)
i"
"ro
= 6400 + 060 = zo50 kg1,
-'
ui,= uscc3o - 299589 km/h' .
t- :
I -
cM/h =1.16236 x t0-{ km-l
t
==:::';::;,
;::
";; = no/'*o*-
lJ
I '
;j*'
19124
-*
'- '-*-- ;'
*ft'
uo = 30000 k,.r/h' uro = uo sin 3o = 15?0'08
,-,
=.roud'o
=2LL'2LX 106 km2/h
D=l(tl\-GM/hil2+(r'"/ho)?lt/2 - o'zot8ss x lo-{ km-r
'ro
**'
;: "'J^:;'='*'-;,'j:*
km/h' e=^io1cu =o,243r
=
km/h.
umax = l'o/"min = 30186
^::Z:o,=::';*i!,,",',1,i,', ;=[X Zi^o,Y,',i',!,;]r:]::::
16'49o
bndo : (""//'o) I $/'o -GMth?)'=0'2960?' {s =
lJ
E
tr
il':'":,:'"',::::[::;';,:]#::$.",:,11,';]ili']';1i:l''ii"iH:[HffiS:::T::i
r',=..gi#jqx"1;=-. ,1$:;,,'_:?km,h
At 0 =
"; =';'"1"; :,ffi:}f] ;, = o"',-, |-ffir,?'-:'';='liij,i ),r -2a{ss km/,r
l-_
The fact that the values of u at r = 6000 km are the same
IJ
il,:r:::::L'ffi:y;e
f
in parts a and b follorvs directly from the equation
objective with the minimum Au, trre probe should
=) Gr{lh?=
+ l/69e? - (I + tlt'2\Dl
er:h2rDr..Gi't =L2.
l/rmin = GMlhl+ DL
,
+ D1 = 0.77g555 x 10{ km-r'
+ ur = ucr = h1/r*11-401164 km/h,
h1= (L.2Ct{
t
I
:':,';::xlT;:.,
...,
km/h, +
:+
lJ
tr,
L
:+ ur = ucr = h6lLL044= rgszo
l-,
L'
fj'
r-
themoduleare l1044kmand67fr) km'
1/11044
-=:::'":;,'o.*''^
,'
[r, = ul - 19124 = -154 km&'
1i:::il111[*'.]i11ilm:':II[
Lffi';*,:lt,'r;*mlm'*T1;.Tffi:
is giveo bv.
12 = 6700 km of the module
Conservation of energy
t-
l Dr\'t'= 263'129 x 106 km2/h
Au = ur -r.,1.*=40464-30186- I'02?8 km/h'
(d) l.-r"rnt"...",ttu--u*imumandminimum,radiigr!.treorbitof
Li r,o, D refer to the trajectory of the probe (Fig'83'22d)"
L
be launched ia the circumfer-
:if"*:l':X#i:l[*in*;i:;i:l':]",',1'u,;i'.1.H"15:,H;S,I;'*"*
D1/r'2
L]
I -'
L"
L
t = io.4 /cllnltgi
T=rob/(hs/2)=2j59h
o-(tr,',ir,*rmax)12=9020'5km 6-a(r-"')'t=6?90-6km'
f
tr
L
L
I
F
Y
:
o2 -- ooz
= holrz = 11044 x 19124/6700 = 31523 km/h'
MmlLL*a4 = \'tttti-GMmf tz +
lm(ul+rsrzl2l-G
q,hich'ts very large compared to laulobtained in part 1'
ea
u' : 4(F6 km/h
**--)
r,-*l
;;-I
aJ
exc-luding the fiork' is mr with ceotre of mass
I}xarnple 3.23 The mass-pf -a fork-lift tru& (FigE333a),
at Q' The fork is supported by a
U
at Ct. The mass of the fork "od t5" o"te i" -, ii "*:"."f .-4* and vertical reactions' The inertia
Ugttt nofzontal
smooth roller at A and a connection tt .B *hfu-;upport-s
just su6cient
at A, B , E,F if the fork is given
of the wheels of the truck is neglegible- (a) Find the'reaction!
(b) Find the ma:rimum deceleratioa the
one of the ground reactions tP *to'
..'iJ
t.'iJ
t:J
upward acceleration to reduce
slip nor
nei[her slips nor tips and the rvheels neither
truck can have with four-wheel brakes so ttrat th" "."t"
the centre of its
centre of mass is at height h3 above
lift-off the ground. The mass of the crate is m3 and ils
the fcrk is f:- (t)
the ground is p1 aud for the craie and
base- The coefficient of friction for the wheels and
at tlre isstant the fork has uprvard acceleration
Find the ground reactions for a rear-r,rheer driven fork-iructi
/t/r' A;ume
o1 relative to the truck and th-e.-drivt'r t"'o'l t'
tn::' **""FffTfjpfiI*
,,J
-J
:1i#-i
u
l
fffffiFffi*#,ffi"ffi
'ffiffi"H$qf*4"Hffi#j
to,
;f 1
JOo,
,;a.i.qffi
,J
J
'z *-fN,
f:*S'{'h_-_Wffi
}#, i#,:t,
tf Nr
'i{r. Na +*/
,
- e.
-}_ :"F.?('ft8'i..
^a:
u" solved
iv'l'n.."tr'oa''iit
_!-,}oJ'
N1
s.r.,tio;'dlhis probrem,t:,i,:", ="...3,J,:i,,#ra;.1.",'
for individ"'i'ot* '"oich invoh'et:T;:"'r:ltl;l"ll1r";:::"J.,lll=
Mc==00forindrvroua
for dre
rttsc, Ma
Using F =
lixed points t=r:lil
1.
1- usins
=rttg<.t
= io, forconvenient
.
ir":1i.:_=*5::*xT:HTl"ii:-Ti:J:rternar-ro,ces
'
follos'ed io 8x321'
equations- tt i.
on trrc svsreor appear ia 'lrre
"
o.vsram cao be ren'tit as F-ngc"pp'o"tt'
attanslatingsystemcaobere$'titten
3. Theequationsof motion,F:tng6, rn-=nof
act'ing at c constitute a null
g, i.e., the external-forces *a",,il. ruirh * 'itertia lin'-**"
g-, Mc:
using F = Q' [
- O for
this augmented force systenr-and
with
rvorking
^
in
consist's
E thac
method
systern- The
adt'antage'over the first trvo methods
ttt"
poiit'
to"'""itttt
any
is
A
where
this null force system,
is zero' Method 2 rvould require Proper
;;'ii11i::i.:::*
sv"t'm."b;;;
force
augmented
the
of
momeut
the same as in me|lrod 3.
moment equation rr,ould be
resulting
the
H6,,though
of
computation
acting on the
Weillustrat"*u,hod"',and3inthesolution.\1,efirs[.aPpl1,metlrod3.
er<ternal and inertia fcirces -rn;gc:
The
c'
be
fiork
the
of
(a) t et the upwaril acceleration
sysLems' Hence
in Fi5'83'23b' These are null force
t'"
wheels
(l)
and.the
fork+crate
truck,
"ho*o
was
Fr=O
Pront$rheels: MP,=O
Me.. = 0
B.ear wheels 'Whclettuck:
MF =0
; =;
Fork+crate:. MB,=g
F'=0
Fr=O
(2)
F'=.9
'
:+
JVr=[m1963+m2(9+a)(61+62*6r)l/(bz+6:)
;; *; .-'*'(o+ a)Drl / (62 + re)
''
N =-.z(g+a)bqld
Er=[
=
ft2=m2(g*c)
Jj
J
'J
J
,!
i
ir
J
-J
,J
!
::
i
-_.1
,-:{
(3)
:-
l/,\
(4)
,(\
(5)
<
(6)
-q
(7)
-i:t'--
si
.v-.<
+r
S
-!
tt-
,.. .; ^:f. Ii-:i.i.1;
!.;;
r.-i.-:::-r'=.':-:
I -1,
'
t*
' r.' i.
'r,ar!{-i
with a' 4rz becomes ryxo tot a '=
lr is observed oo*aug6-iq{ (a1 tu.t /v1 increas€ P,9 trr lecrlases
(?}fvi;vr1n Ef;e;"- be determined usirg eqs(3) to a :ting on the
{mg7/m2b1- 1)s. lyith this-viluc-ot i,
foi:ccs -m"rc(b) Lct the backward accelera6on of the. trudkll"'t,-fr" **o1. *d iuertia
l^
-.[.
lThese are null force systems' Eence :
truck, crate and the wheels ate sihown in i'ig-ea-23c.
N2=liig6,2-mz9br-(m1alr1*m2olr2! /$z+b) -'
Ms,=A
Fortruck :
t-_
L-t
(10)
with in ease
For
L*
t-.
fr = Q
For cra[e:
fv=0
l.l-
(13)
II3 ='fi13a
:+
M9, =A
L=
Fot no slip of crate, using eqs(13).(14) :
(i4)
= m3!
0 =+ b = d1l2- ahslg
Ar3D - nr3sd7l2*n3efu-
JV3
Fs ( ,raNs
msa.f l.?@3g :+ '
(15)
a-S g,zg
!i:l
(17)
a{:gd1l2hs
+
b--'|1./2'-"'A"ry',0
F6r'notippingof,crate, usingeq(15):'
c:miu(o1' Prl' lt29' gd1/Zfu\'
Eence the maximum retardatiou :
wheels are shosrn in F(-83'23d'
(.) The external and inertia forces -m;lLc;acting ori the truck and lhe
LJ
t,
L_-
These are null force systems' Eeace
L-.
L
Frontrvheels :
Mp. =0
:
Mq, =0
Rear wheels
LL-.
L-
LJ
lt
L__
u
t-t.
L.
LJ
L
LJ
L*.
L-._
_i
(18)
Fr:0
r;- LhlR
'Il9l
t20)
:+ a=MJR(m1 +mz}wholetrucli: {'=0 + (mr +&z)a=fr.*Fz
mzahzll (bz+fu] (21)
nrt = [mrgts*mz(g+atX0r*6a*&3)- rzlolrl MF. -0
(ba *03)
Q2\
JV! = [m1s6 z- rrlz(9*ar)6r * mralrr +m2oh2ll
frr
(22) is given bv eq(20)' '
where o iu eqs(21) and"!.
(3)'(5)dB)'(15) aad (21)'
we illustrate method 2 for obtaining rroment equations
Fig'El]'23e' consider points o and
(") lrt u be the upward vetocity of the [crk- The FBO'" are showtr in
the given instant'
S n*"a to ground and coincideat with .F. and B at
:+ ec(3)
Ho,=-m2u(61+02+h) + Mo'=U:'-m2o(61*6:*6s)
Ttucti:
:+ ' eq(5):
+': Mi;'='H5t = -'n2o6{ '
:'''l :
Hst J'-r,..2u6l'
fork*crate ,
Consider
(b) Leruberheforwardvelocityof theforkwith.i='-a-TheFBD:ar:.Yn,rnBie:EXlt?3ft
'
given
instant*
ai the
:
points G and K fixed to ground a.od coincident ncith E,and,D
Ttu&:HG,=m1tlIr1*tn,thz:+Mc,=iIn,=-m1aIr1-m2ah2eqr6}.:
tc(15]":'
+
'
Crate: U*"=fisah3Y*r=i{x'=-rn3o[3
velocity of the fork- Thc EBp' i''
(") lrt u be the forward. velocity of the tructi and q be the uprsardwiih F' Iet' the location of Ci from
and coincident
shown iu Fig83,23g. Cor,"iao P*t'g fixed to grotind
pr.and
iih".g,r.* instant c = 01 *62*0s:9 = lr2. For.the t14',
0 be giveu by e,y, whe re i:,,,i,:=
i
Ho,Jmsulrl{t.r.zoy-rr.2v1, :+
- -rr.
Mo'=Eo,=m1oh1}m2cr!*nzvil--r/r2a1a-m2v1i=m1clr1+m2ay+?,/l210u|-t,12.,|Z-T,.:I..:i '
':+ eq(2t)
i: .
- nt1ah1 *mzohz - rr.2r.v(b1* 62 * De)
.
L-.)
L-
(s)
ord N1 increases by the same arrcunt'
"
:
(11)
o1o1=(m162-*rtrl!!-Vrlrr*rnah:)
Fornolift-offofrearwheels,usiugeq(g): JVz20 =
(Lz)
pr$ir+N:r)l'ftar+"tz) noslip of wheels, using eq"(rg),(9i,o - (f'r+ Fzll(mr*mz) S
"Ls
'L-r JVz
^,^ decreases
Equations (S) and (9) imply that
^":-''
l
L
(8)
o=(rt +Fz)l\qt*mz)
'"E --0"-t"
LJ
LJ
'
- FY=0
t_
L_-
:.
.;
.- -- ::r=g:ri
:
l:.
ii ..::,
€-|t+''::
ga
-' -
.
"
-; ' --::l:i":"t':':-
- -::;=".r,1::
l:IJ
Qr
I
Ct
.:
lj.
lj
.
.....
a. tdrrarroNAr, MEcrrANrcs aND EQUTLTBRIUM
lj
tj
tj
q
t7
cg
(-;
1
r-l
from ar.r admissible configuration
t is called uirlual displacentent 6q(t\:
(4.1)
6q(t) = q-(i) -o(i),
,virtual, displacement since ar actual infinitesinral displacement dq(t) occurs in infinitesinral
This is called a
in zero time and is not along an-r actual
time along a paih of rnotion, ivhereas virlual displacement 6q(t) is
t--
pathofmoiion.Notation'6O'differsfromthenotation'dO''LetaforceF;(f)actatthematerialpoint
the virtual
1*O]"t-":lt"t.":
i. The positiou .,r".to. 1;(tiof point i cau be expressed in [erms 9f S(t)' Let
in thG
rvork done by F
The
'-(t)
poinr i due to virtual displacement 6q(t) of the sistem be 5r;'
;;";r*;
F'(t)'tirt'
t'
is
time
value
at
at its
virtual displacement iu zero time, keeping the value of the force conslan[
external and i*ternal forces acti'g
the
all
b-v
done
rvork
uirlzal urortdone by F;- The virtual
i;;;ilahe
denoted b1'- 6l'{r(t)'
oo th" system in virtual displacemenc 6q(r) of the system is
the path q(t) is catled [he tarialiarr
The diflerence in tl're value of an entity for tlre path q'(t) and lor
r-alues of the entity is called the f rsl
of the entity and is denoted bv 6( )- The first order diflerence in the
Hanriltoo's Pn-nciple:
uaiationof the entity and is denoted by 6(r)1 ). \'Ve norv state tlte
.A geometrically admissible motion c(r) of a syslenr betrveen prescribed confiSuratiorls {1 at tl and ri2
indicator (V-I') vanishes:
at t2 is the actual moiion of the system if the follorving variat'ional
1-.
u
[J
t:
tl-.
l-
v.r. = ['i ;.rr, * 6w)dt =.o
J
t
t
:
(4.21
r,'
srnall neighbourhood of g(t)'- l'he
for /,rbitt1rry varied nrotion of the q=i.r,. s'(r) = c(t) + 5q(t) in tlre
inertial frame'
refereace frame in rvhich Eamilton's Principle is valid is called
t-.
l-
42 LAGrtANGElq EQUATIONS FOIL DISCRETE SYSTEM
t-
Consider a discrete .system o[ a degrees of freedom rvith geueralised
u
l_
lJ
coordinaies 91 , " ' !In, rvhich are all
'
independentofeachother. Leiq=lgr gt ...- {n-l tld f bethethecolumnvectorofthegeaemlised
actjns zlr-i =I;({r'"''9'}' can be
coordina[es. The virtual work do[e lry internal and external [orce-'Fi
i
ercpressed in terms of 59:
a
i
tj
lj
a-.
6w= I8i6c; : Q;6q;, where q, =T L,'H'
I
(4-3)
,.
the ith generalised coordirrate
and sumrnation *rr"ntior. nl" u""r. used- e; is called the generalis cd forccfior
L
l-
g;. The first variation of ? is obtained as follorvs- Let 6g(t) = e4(t)' ie"
c'(t)=q(t)+ea(t),
rvith {rr) = r1(t3) = 0, and e ( 1'
T.he kinetic enerst function ?' for the vatied path is expressed as
series about e = 0, i.e-, about the original path q(t):
1*
L
t-
::
t
4.1 EAMTLToN's PRTNCTPLE '".' .:
FiA'('l
move
system
a
qLet
be
Let the ge'eralised coordinatrs to, d""".ibu the configuration of a system
t2- Consider a geometrically admissible
frc:n the configuration g-1 at time t1 to the configuration f2 at time
tir, C({z) = 'i2 (Fig'4'1)' Consider
motion q(f) of the system between these configurations such that S(tr) =
these confiSuratious such that g- (' I ) =
another geometrically admissibte motion g'(t) ;f the system bets'een
q(t)' Tlre iafnitesimal
qr, q-{lz\- f2, which is an aroi{raryratied rnotion in the smallneighbourhoodof
I
q(t) to a neighbouring configuration q'(f) at the sanre tinre
,displacement,
t-:
t_
tjt_j
tj
Q.
:
a function of e, and expanded by Taylor's
?- - ?!i'({), c'(t),fi} = r[d + €q, q+er;, {] = ?(e} ='[.=o * #1,=o{ i#l.=rt? + ':''
\.'
6rtrtd'(r), q'(t),4-"[.i(t], c(r),4 -"(.) -rl.=o = #1.=o' * i#1,=;2+-"'
E3
(")
,".i'
'*'-''
tj
t_:
LJ
tj
u
tj
t_
LJ
(
tj
t-.
tj
]j
chiio,rokj of airjr"T ii"ihn
':
a\e 6rst o.a"..L,,sos:.te-:l-.tlt- i f1'1]sfrief ';"li;'$9i€i.--..
o,ui"--
l,l't#*
zf,e srsr
term in the integrand or
"q(i)
is integrated by parts tn get
H
H
L-
(6)
a factor of r1 as in the othir two
tffia,(t)l[ - /:' tfitffi-#,-Qllrdt=8' ' .':"]'.,
(a)' trtence' the inCegrand
zero since 6(t1) = 6(t2) = 0 by
-ide O"<rjniegral term
o,o be valid for arbitrarv ni(tl
-ts
4f,r
ternrs:
t::
r ,, -
in (c) should be zero as
:
:
{
i =r,.. ., n.
0T _ n.
aldi,) - 6fi= v;
(4.s)
aotdinates itilepettent of each
and nonconservativq par ts Qr' Qi"
other'
O':*U
discrctesystcnr
Y'!U-rU
TlreSe g\r€'f,ogrcag c's egtalidrs for
a
3 #i:r*t'"'"u"*o*ve
T:ffi###";ill
:H"ff
.j6w=Qi5qt:(Qi+Q?.}6qt=j?+Qt.6ci=?x+Qi.)6c;
aud Q; = Q?'
AV
Qi=-ast
_'av
(d)
i= r....,n.
(4.6)
strbat'iunigtr Qi rromeq:ti;: -:i':':i,
_ u; ''
+
;(ad,) - aq, 6*= ^..
-
lJ
t'I
,'
t-.
lj --,
I
:
=o'
l-].
'
SuLir'*1-r.g(43);(4'4)in(4'2)andusing6s;=er;;vields'-'r'--"'.'1
*ffo'*Q;t;l,tte
tr o[ rhe
"y"i"*
rrie &hoe the Logra*gian fittctiott
u
tj
it, .
#1ffir'#Tr' ; Yrpil} = i# *'+ ff'ti'
lJ
]j
,;
*=#_X woRK
*If'u
+.2
a*d(46)
I.TLINCIPLE oF \/IRTUAL
t-trnsider a system rvhich remains
bv ,(,i, g' t) = ?(4' c' t) -
*
y(c)' llence
*rkt_#-_o,., i=,,..,n
in equilibriutn, ie-, c(t) = {-r.= iu:
6GI
* d
of the
o
*
(4i)
T = 0- For rhe
^ order aud hence the
first
i'(t) =
Vqriedparh C.(r) = C(t) + 6q(r). the generalised
" (4'2) reduces a [1"6W at =
Hamilto"titin"ipt"
0.
OtUf
*
order,
=
is of ilr"
?*,
e
knetr?energy
"".o,..1
6w =0'
j
."'f q(,r ro(,,) = 6q(t3) - 0 +.
for all virtual displacements 5g' thetl
velociiy
in equilibrium and 6l{'=0
to the
t\>nr.ersely itthesystem is initiatly
generalised coordinates q(t) corresponding
are zero. Choosing t'he origin of the
forces
of the
generalised
solution
rlr
0' Hence the
initiat .oJiriont are g(0) = 0' d(0) =
ttre
configuration,
equilibrium
:o:ttal
and' the sy:tem remains in equilibrium'
of motion for this case is c(t} = 0
uork
lben the
hsms{eneorS.L"g...rg"{"quations
'.;.16o{
of virtual work: If a sistc* is ia cguili6ritm'
principle
the
proved
have
s.e
gcoilticolly
a
fhos
uirlual dkploccmcnt b zctt' Conrterselg'
thc intcmolatd. exlental lotqcs in orbittary
clone. fC
uJrta<sieleconfiguvtiott,ofosyslenlls.cllequiliDriunrcottfgalrrtiott.!!"*,:,::"::o,""bythcexlental
is iai{iolly i* eqrilibrium:..
loirtlrat displacemed' ptooided fi]e systea
.rr.A inlc*o I lorces is zero for arbitmry
(4'8)
.
6W=0
v 6q'
Foradiscretead.o.f.(degreesoffreedonr)systemrvithiidependentgi,eq({.8)+1
:+
Qi-O'
V 6q;
6l4r=Q;6qi=0
lhe generuEsed [orlr,s ate zc,to'
ip.rforc discrele system in eguilibium' all
94
i=l'"''a
(4.e)
t
-
,HE .
ll'Ki'
ilD.J
\,
)'--
--
H-i
A
porENTrAL ENERGy
srATroNany
4.4 pRrNcrpLE ot'
-. J.{.
-.-- -:.--' - ---l-ret 6W. and 6[7," be the virtud ivorF doae by the conservative and nonconservative forces, then eq(&8) =+
6W = 6W. * 6Wn. = -iV + tVn. - A,
V 5q-
6Wn": 5V
=)
I
j']-
e
(4f0)
11 -
If alt internal and external ficrces.are conservative, then eq(4.8) .+
({-11)
v 6q.
til = o
6w = -fi :0,
+
i.e-, cbe first variation of potentiat energl is z,ero. Thus the potential energl/ has a stationary value (an
extremum) at an eguilibrium configuration. Thus priuciple of stationary potential cnerry follows I/ a
systcm srbjected.onlylocotsert;alioeforccs rsincguiliDnunt,thet6V=OY6q. Conuersely,if 6V -OV5q
and if lhe system is iniliatty in cqailibiam, lhen it remains in equilibrium-
lq
r'
;-t
For an n d.o.f- conservative system, eq(4-tl) .+
r<
:+
V
v 5qi,
5q,, :+
ilr=Y&,=o
=fr;6qi=0
i= I,---'n-
Y=o
U*=O
6V
G-rz)
-
ir
Equations (4.i2) are Lhe equalions of eqailibrium of on n d.o.f. conserttaliue system.
:
.wortK
4.5 PELINCTPLE Or RATE OF \rfRfUAt
:-
Consider a sei of adnrissible virtual velocities ri .at an adnrissible equilibrium configuration- The rate
rt. rateo[sork done s,ith these virtual velocities- Since ri.an as re{l be
of virtuat'work 6liz i, a"nn"a
".
=-
\
(4 13)
v i
,r,ir-o
-
i.e., for a syslem in equilibriurt, lhe tole of llrrtual work dol.e is zero for orbittory viriual oelocitiesNOTE: If the system is constrained and the forces of constraint (internal or exCernal) together do zero *ork
in virtual displacement consistent rsith the constrain6, then in eqs(a.8)-(4.13) the virtual rvork is only due to .
the active internal and external forces. If all the internal forces together are *'orkless, then in eqs(4-8Ha-f3)
the virtual work is only due to the e-xternal forces on the s-vs!el'rlDircct Proof ol Prittciple of Vrtua! firo* for a rigid body. l{ for a rigid body can be ecpressed as
_
lA = F -ya + W.u, + d.W = F -gcdt+ Mc.g_dt - F.drc + Mc.d4where d0 :,"tdt- Hence,..*
(4-14) !
6w- = F -6u + y"
A rigid body is in equilibrium iff f, =!,M.c- Q, provided it is initially iu equilibrium. Ilence eq(4J$=+a ,{h
rigid body is in equilibriuro iff5t4z = 0, V virtual displacements 5t, 54. provided it is initially inequilibtiu.m - 44
-
4.6 srABrLrry oF EeurLrBruuM
.=.-!
-E
:
lj,' :-
!s.
coNFrcuILATroN
W
1Ji r;
-
An equilibrium configuration of a system is said to be stable if ony smcll initial disraficncc (displace*ar/li \ of,f,o
and / r,elocity) results in a motion rvhich b timired b a small teighbotrhood of the equilibrium configuration :&
W
The equilibrium position o[ rod I in Fig-42 is stable but those of rods 2 and 3 are utrstable. Rod 3 is unstable U,rl
since a small iuitial angular velbcity causes large displacement from the equilibrium position-.
F,3.q.z
Cotditioos of stobitity o/equilibrium configuration of a corserualiue system of connectcil'igid bodi* arc:.
(r)- If the potential,energy of a conservative system of counected rigid bodies has o lbcol''iniaimua at a
position of equilibriurn, then that equilibrium configuration is stable.
\
Proof,. Choose the datlrm of V aL the equilibrium configuration 96, i.e-, 7(96) = 0. Since V has .
minimum at qo, for posiiion g in the neighbourhood of gs,
=,n
(1)
(1)
Y(c) > 7(q6) = 0.
\^
.
,
:
.{
\< -
i
Let the iaitial displacement and velociiy imparted to the sysi.em be q' and ri'. Then
'
+V(c)=?(,i') +V(c')-E(c')>0
"(,i)
E(q') has a small positive vatue
and tz(q')
since
>0
r(,i)= E(c)-V(cl
Q)
> 0 by eq(l). Equatiou (l) + there exbt aset
=)
"(,i')
ofg,int,heneighbourhoodofg9sottrat7(q,)=a(g.)andeqs(2);(3)+
T(i') =.8(s') - V(s') = 0,
9s
+
i' =A.
-'l
-t
>
__
-t
_J
.,-]
-I
l
-J
S/
I
t-
Ll-
l.
L
L
a-
ffiIt*Xy":I;1":#rI"1["#;;H.ffi;E *" equiribriirm .""rir,i.."rti* {s
kt-_
l-
t
gs' IIence V has a local minimum d go'
given by
g, the equilibrium configuration 96 is
For one d-o-f. systern with coordinate
where g' is in the ngighbou+;d of
dvl
LJ
dq lq=qo
L--
tJ
L:
orr ir aY
ffro,
dq' lc-co
=0-
(4-15)
F-tli
.t2v #v
*'o #:#='^"=';#10,."'q=qo ('t-ro)
^.r.._.--
.
bY
dt - - ci^ rs grr€I
Civenr b-t
---€-.-alinn
-. configuration
-^-a.
{r3'92s
the equilibrium
For trvo d.o.f. system rvith coordinates 91, 92.
(4'17)
0t' 0. AV
* qr =916, {? = 920.
-n
0q,
ac, - "
i'e'' if the derivat ,ives at 96 satisfy:
1z
tras a local minimum at {o'
if
stable
i.
i'""
confiEurt
The equilibrium
LL.
4.T,,NUMBERoFINDEPENDENTEQUATIoNSoE.EQUILIBRIUMoEAftIGIDUoD:
g. Altlrouglr Mg = g,it does rroc -.-ield
bod.v are: f-= o L{.:=
a
rigld
o[
equilibriunr
of
Tlre equatiotrs
.-\
tJU
L
l_
IJ
L
1-
anJzmoreequationsindependentoftlrep,"'io,'ones.T-trecumberofequationsofequitibriunr.rr'ritterrin
.olporr".rt form, for different force s1'stenrs
are as follows:
:
1.General,4=9,1'11=0(6)-Parallet'll:-a-xistli=0'Mt'--O'M^v-0(3)'
0 (3)'
4'
Coplanarineyplane "F'=0'F"=O"{f;' =
(3)'
3- Concurrenl;F=0
F' =0 (2)'
6- dpl"n".concurrentinr-yplane:I."=o'
5- collinear ll r-axis:& =0(1)O (2)'
?- Coplanar parallel, ll z-a'<is in zy plane : I1 = 0' I6a' =
0 for '4 on lg- Alt forces intersecring the same line r (d , F = 9, M t:Q (5) since M-^'9=
INDETERMINATE SYSTEM
,"O,ICALLV DETEI,JVIINATE AND STATICALLY
XN."
".^'aud the
='"
I
reaction forcbs and,couples atthesupports
'
U
u
IJ
,----
.
if the
A system is said to be statically dete -rinate
9'' Llt(f ) =
using only the equations of equilibrium [E(P) =
internal force resultants can be determined b-v
fiorcc''dcforrnation
For the laiter case'
p; otherwise it is ca.Ued .ror,.oit, itilclcnninate'
Ql for all its parts
relations are needed for compete soluLion'
IJ
l-.
I ,,
t-.
l-
be dererr.ined lry
the teacrtions if supports cannoc
if
itdeteminatc
is'cxlentally.slatically
A system
intcrual force
- using only the equaiions.of eqtlilibrium- A system is iateraally stoticollg ;il,c1e|{.inale ifi|the
systens
tqt'"tions o.!'g'uilibriunr' Stiticalll|.tinf.:,t:1"*e
resuliints cannot be detetminea UV ,r.iog oniv'ittt
timperature'
during assembly and due to rise in
are subjected to forces due to initial mismatch
in
in Pig'a'3 is preseated the fotlorving'l'al>lc: The information for statical determiT'Jc1i;";";:*t
Remarks
:
force resultants No' of Eqs'
No- o[ gq' f
Rractions
beam 1
beam 2
Rr, Rz,R: (3)
Rt, Rz,Rr' R< (a)
arch3 Rr-&(,%,86)(6)
beam
I
i
r\grr'\Yr--
L-_
u
l.J
and th't6anm"m
4 Rt, Rz,c, (3)
beam 5 Rt, Rz, r?e, Cr (4)
"t'
N' n4' s (3)
3
staticallv deterrrrinate
- N'M'S(3)
s ili
3
staticallvdet'emrinate
. 3
3
3+3
I3
3
a'r'
'v:
/v' M' S'T {4}
externally stat' indet'
3
::*'SLi".;=J:;tr;
r-1"".l{a
]J Jrtrft*iu
{
i
i
NHH{L
+=#
t ",fr'
ft-,\
f?u*t{F
:TiTk
^"a""'si
"#
IJ
"*
l-.. '
".,rX":P.**
r-- I
lj
LL.
r---^ A *hf''3!t]
P- P- p' 13\
n
-,,
96
3
-
int'ernally srat' indet'
(.)€ Ll
-S''.
wwi{,e
<
. )€
-)f:: <
,,.
.*:J[:":';fi":;;;;.;;;;';;;;;'
cootdinate direc[ion are atso '..i8;ned a.pcitive "ign lEig'a'a1'
ditecad in-ve<oordinate-direction are assigned . i.g",ii"
The
direct-ed
and thce acting on a -ve face and
,-<
.=
;
in riq.:a'sa- Its equat'iqns of eouilibriul
:.
are
.i{
f" = S(:r+32)-S('}+g"*Ac =r'0'
F,= JY{e* az) - N(r}+n-At = 0'.
ta}
(1"Ar}atr'e=0'
A[t; = it{$ +Ar}- Jr/(r)* rnry&*S(t:+A'r)Ae+
.
l' Dir'iding
load goAz from '{ and thetefiore rsLete 1Ae is tlre distance of tbe resultant of the transverse
"
inFolvitg
tcrm
the
r'
at'
values
average vdlues becorrr the
cqs{a} by Az and taking the timit as Ar - 0' the
q drops out and rt'e get
,
dY
nt
{4-19)
dA'
,=-.rr,
E- -;$-nrE=-C,
-
t''
:=
. ds
d2.u
dn
E=c-6-
E=-E-E
Eqaerions (4.19) are Lhe diffcrc.*tiol cqvalions
-r
<!q
(4-20)
f-ir!
load- For
o! cqnilibrivmof a straight beam under coplanat
f =-r.
'#--,,a
dtn
#= -r,
#--''
__
(:r'2t)
--
CoasidcrthefB{}daoirrtrnitesimalelement(Eig.4SbJofthebeanatorrgd'o'$'he.eadbcreteforce
of eguililrrium, [ = o' & =
f-.g+ Frj_ aad a discele couple ifok acts *j;:;,rr" b"..*-'Tlr.the.qo.tio*s
ol quilibittttt "t
jump
coaditians
0, lf;, = 0, of this elernerri io tt Emit as .,
- o, "i"ra
il __.
"
---<
i
N(rf,}=N(af,)=-F..'s("f)_5(16-)=-Fyo,if(:f,}-l{(rf,}_-l{o.(i1.21)
'
/
. ''
Tteiumps in ir-, S. M at a6.equal negative of f'o' Fvo' Ms' tespective$'- ,+
Geosctrical'Itcz,Vltctiot of ncrlt.Its.' E{uationsi (4'19)
-!{
fl(,,}.=rv(e1).i_|1,ou..-.-(areautrder-a.ediagram)
S(az) - S(41) = -
l8z
:
<
- -(area undl:t g.r diagram)
J,.'oo
M (rz) -rlf(21) =' j:: t * -
i<
*= -(sum of areas under 5-z and ra-e diagrams)
[=]Sif.nr=ol.
(slqcof&r-edurve)=-n, (slopeofs-rcurve}=-g. (slopeofi/-l.curve)=-(S+m), a.part
ort'e bearn
considering
bv
obtaincd
r". aiil-.. roaded seg;en* "iu".- are
si;';i;
with rf' s' tf
-:TA
a = 0 to z in Fig'(4'4b) is considgr-ed
to *. side of the sectioo at, r. If the b""", *;;;;f-.
* Irc.e.
_{
_{
"irr
sign'
in +re coordindtc direction ate also assiSned a negative
e-axis along irs
discretc and distributed loads' cl"9*
C.oosider a straighr;; JjJ;+t*.
of tlre
components
ihe
Let
plane of toading (Eig-{-'lb)'
ceuuoidat a,xis and y-*ds transrffse'to i[ in rhe
distributed
the
and
q(a)
N/m'
r be n(z) N/m and
distributed force in the e and y directions at location
N(r)' '
oiequiliblium for lhe axial normal force
equations
tl,e
momcat be m(r) N.m/m- ll'e.raot to establi.L
i
from
gf
bear'
a
element
ru(r)- ihe FBD of an
strear force S(:) and ihe bendiog;;;*
ta t*.Lx$,hich.does ror carry any discreb load, is slrown
_{
I
BEAM
4-s EQUIiBRIUM equertoNs oE STRATGHT
Alinternalcross-sectionofabeamwhmeoutrvard(fromthebeam)ao-rmalisdirectedril-1:::::::: is
is directed io tf. -y-.*'-o-':,::,1'rection
same
morneDt resulLants are assigoed the
and
force
called a -ve face. The cornponenG of the internal
in
directed
and
face
components actiag on a +ve
siga if their effect on the beam is similar- Thus ttle
in
direcLed
and
-ve
and those acting on a -ve face
+se coordinate direction are assigaed a positive sign
and
corrrponents acLirrg on a +ve face
d,
*M
t;
li
tol
rhe ransverse
,18 .J
!,,' ^
-r
.v
:-{
!e
ri - -(ortrer rorces in a dir-I, g = -(other.**:::,ri.-r' 1; -,::1':-::f"::H ii:t*''
i{
i
,,1)
*r4,
tx+ &>( )-'
l
-{
-{
-q
:
9't
:.
/jt
il
'i
i
i
i
a'
t
i
L-.-
*gn*"t are computcd and the N-t, S-z' M-z
The end values and the extrgmal values if any within "".h
results pt""""t"a eatlier is helpful for this'
diagrams are drarvn- The geometrical interpretation of the
and the constant's of intcgration
If q(r) is not a simple function, theo eqs(4-19) are directly intcgrate<i
de[ermined from the end conditions'
L
.,
L
L
t
L.
L.-
-L,
)
L-
-
Lt,
L.--
,
iiqfliry:ts; yr.l rT:_i?TIr. fi:::*j:I
I
,
:
\--
t- )
.,
[,
t-..q
t-
_)
L..t-
)
t---
L
L
)
)
_)
tI *l
Li
lr
_)
LJ. )
L_
4.10 EQUTLTBRIUM OE TRUSS
so that their centroidal a'ies are
A framervork of tiangnlarsf,apes formed by joining bars'(members)
lruss' The joints ate idealised
concilt:rcnt at the joiis a'rd the apptied looils ect otly-aurlcseToirls-is-calleda
joints for space truss' The members are idealised as
as lringed joints for plane truss and as balt aul soctet
joink- Thus alt nrembers of the truss are two force nrembers
wcighllesscompared * 1L. t6rads applied at the
o[ connection to the joints' t'et Fi be
ttith the forces necessarily acting along the tin" 5oining their points
of a joint' the forces due to members are shown
the tensile force on the ith member at its ends. i" " fiO
r:alue for Il''
equations of tire joints yields a negative
as pulls on the joints. lf the solution of the eguilibrium
then the actual force in the ith member is compressive' '
j joinG alld having c -roun{atLn consttaints' 'Tn"
c,onsider a frame made of, 6 rigid bars, connected at
j joiuts is 3j (2j for a plane fran're)' The tcrms appearing
nuqrber ofcoordinaies needed to specify position of
t.t'""' The number of constrain[ equations specifying
iu ihe parenthesis in this section refer to a plane f.*r,',u 7
z^- ,,B')" = L2al'and the number of geometrical
the fixed length of 6 bars is 6, e'g'' (ca - 16)? *(y^ -y'f +(
O".tn: d'o'f- o{the
c' e'g',
constraiat equations duc to c foundatiou constraints is
'-e =,0'-UC:0'^Tt:
n=3j -b-c(n=2j -6-")' [ngeneral' lhe
frame- lfallthecousfraintequatio*s are independent, then
Hence n Z 3i -b-c(n>2j -6-")numberof independent constraint equarions is S(0f c)a mecicaism in which some individual member(s)
is
it
then
If a - 0, then it is a rigidJrcme and if a > o,
still
of sonre nrenrber or some founda0ion coustraint
catr move as rigid body. A franre is over-rigid if renror"al
frame is
foundation constraint is calted rcilvndattL A
malies it a rigid frame rvith a : O- Such a member or
constrainr nrakest' " "t:tlTl:T
fisr-igidif removal of any member or any foundation
of
forces at j joints is 3j (2j) and the number
concurrent
The number of equations of equilibrium for
unknown fcrces,in 0 members amd from "
?
condition for a' rigid frame is that n : 0' i'r
,er of unknown forces is less than or equal
numb
the
taat
lly
dctctminc{c'is
condition for frame to be statico
"
equatiotls are arranged in a matrlr
to tlre nunrber of equilibrium equations, i-e., 6+c <3j. If the equilibrium
the unknotl'n force vector [flte+"1 and
form: [n][Ft = [P], where [A]1e;]xtc+c) is the coeffiiient matrix of
nccessary
rank [A, P] :=16 * c, nhich is therefore the
i"f,"r, I ii" U.ir""to.. Irs solution exists, if rank [A] =
ooi snffcicot condition forlhc frome to bc stat.iullg dctcnninatc'
and su-ccessive addition of
A simplc spaliol (ptator) lruss is formed Ly a basic tetrahedron (triangli) so formed is just-rigid'
joinG- The framework'
3 non-coplanar (2 non-collinear) members to form other
are independent in this ese' tf the
The constraint equations provided by the fixed length of members
motion o[the trussas ":i9:j:e91t:j:
fotrndati,on constraints are also independent, then [hese Prevent
3j:6+6(2j:6+3)'
aodc=6(c=3). Eence,thed-o-f--ofsuch.asimpletrussarea=3j-6-6-0 =+
of ttre whole ttuss'
equilibrium
of
The 6 (3) foundation reactions are dltcrmined from the 6 (3) equations
considered nerd,' The forces
Tte jcrint which is the last to be formed during the fotmation of the "lls:-:
(2) equations of equilibrium for
in ohly 3,(2) members meeti'g-at this joint Je'determined frbm the 3
and solne its 3 (2)
fi" next clnsider- the last but oni joint that was formedjoint'
the concurrerrt for""
"yrt"**!ltT:*d that
ite procedure
equations of eqtrilibrium fior the unknorvn forcesin tie3(z]membe*
which qeated that
is repeated for the last but two joinL ttiat was fiormed and the fiorces.in Q-*"*O:rs
'
all memberc' Eence'
joint are determined. This sequential procedure leads to the determina:io1 o[ forces.in
asimplctntss-roi{[6(3}ilrdcpcndcltfovndatio*co*sltlirttsisstoticallgdclctmilatc...'.
@
-ro
i
1
:
1
I
i
!
i
t
l
I
i
I
I
i
i
i
1
I
itre
Es
\t,
ry' . -\
.r-{r.
,-F*
'...r f
op-y
nn
Fq
u!
:5'
- - sls.q.b
--Cig.k.6
F; is negative'
if *: *l:T": of
Ia the jo,:nhwise'deresrinarion of meniber forces descr'rbed .{qve,joints in whic'h it appears
-anv subsequently'
of equilibriumoi
then this negitiru rrdu.l"l"Urri,r,.a in all equations
plane
tbe special case of joints 'A and I of a
consider
pnlts
Ft.
oaly
The FBD's of all joints should shorv
joint r{ yields Fz - 0 and
applied lload' f' = 0 for
truss, shorvn in Fig.4.6, rvhich are not subjecteJto
4 7' 8 are zcrc-totxe members
F! = 0, & = 0 for joint B yield Ft = 0, F.l = 0' Thus members ,r,cmbers
of a plottc {rzss' For this purpose'
o
Sometimes we are interested in littding forces irt iusl lcro
it into tro parts by cutting a member of interest
a section is taken through the truss so that it divides
aodgnfenblyoilyluoothermemberswhichott,nolconctrt./rlltuithil'TheFBD.ofthetrusstooneside
sectioaed' using the equations of equilibriunr
of this sge.lion is dra*'n shorving pulls on all the meiubers
often convenient to select A at the point of
Ml:0, F" = 0, .F, = 0, lve can n,.a ro..o i,. l **b.o. lt is
above FBD' then these
ficundation reactioos aPPear in the
concurrence of some unknorvn membe.r forces. [f
l1
l-
...:-
t'i
--.
,
-i.
p,lr directed oppcite to the impending r
I; ( prJv- For impending slipping rvithout tipping of a bod1" '=' .---.^^4
impending 1..
simultaneous imoendinq
For -i*rrtranonrrs
othe area or conract- D^-
;;;;;;";;.
".,"r.o.
tipping aod slipping, F = y,N aud .tr[ acts at a corner'
load is int:teased, tne assume
2- Ia ordet to decide rvhether tipping or slipping occurs first as 1he applied
is
'that it tips 6rst and find.4 N and .r,".t ,ui.tt* tat 'S P,/V- If the c'hecli holds' then the assumption
of slipfing-takingPlace first'
cortect- Else the assumption. is *.rong and rve ret'ork on the-.basis
a prescribed load P' we find the values P1
order to decide $'hether the systen'r is in equilibrium undet
-1* I*
tso directions' If P lies outside the range of
and Pz of the load for s,hich there is impending motion in the
ittremains in equilibrium'
Pz. then the systern does not remain in equitibriurn, btherrcisi
Pr
"nd
and no slip at another'
one
at
a body'rvith trvo cont-acts $'iih the po+sibility of slip
-A:"*:
4- C,onsider
opposite to impending'stii: direction] and
impending slip at contact 1 (say) [i-e., /V1 and Fr = P' r Nr directed
check wlether'the a-ssumption
equilibri"no slip ar iontact 2 [i-e., /Vi and Fe]- Solve th" "qrriions of
"n9 a-ssumption i'S correct and the
then the
of no slip holds, i.e., check whether l&l I FrrN2. lf Lhe check holds,
with the assumption of no
olution b over. Otherrvise, the assumption i *.ong and the problem is tervorked
'
pe,N2 directed opposite to impendiirg
and F1] and impending slip at 2 [i-e'' N: and F2 =
slip at I [i.e., /Vr
'
slip ditectionl, and finally check that lftl S P,1Nl'
including atl the fric[ion
5- For a.system of bbdies with multiple contacts, count, the number of uuknorvnsslip.equals the difference
is impending
fotces as'the unknorvns- The number of contacts at wtrich there
equations of all the bodies'
equilibrium
rhe number o[ independent
of the number:of ;;;J;;*;;;;;
stip, and employ at these contacts & = r'jffi
Assume the specific:1oJ",., .f these contacts rvith impending
.; t*n"nutn, slip as obtained from assumed kinematics' Solve the f,quations
dtuect€d opposite a il;;;;
at the contacts rvhere
of equilibtiu,m and these kinetic conditions ani thun check whether lql< t'jtntY:
has beun obtained, else r.ercork wiih
no slip is assumed at,the beginning- If the check holds then the solution
--<
'...,,<
:'r<
!\
.-
I
Thus1/rcteaclionPoothejottntalisla*gctttloacitc,leofrcdilsr=.Rsinc.
4'7 il
INYvrOf,\/rING FRICTION
4.12 rIrNTS FOR EQI'ILIBRTUM PROBLEMS
- t r--^^
-thenormalreactionNisatacornerandt,hefrictional'::""\.[A
'{:r- For tipping rvithout slipping of a body'
i,i#I;,iJ'#T',:[
I
-1
r=(P'N/P)'R=Bsina -i+)l;
\
-l
1
4.11 N.EACTION AT A JOUILNAL BEARING
journal beli-ng of the same radius (Fig'4'7)' The inevitable
consider a shafitof radius R at rlst in a short
(radial) foice,v and atangent'iatfrictilonal f11ce 1'N
clearance results in a li,e contact *'ith tle no.rrr",
to Nn *'hcre
Luthe bearing' The tot{ reaction P is at an angler'c
relative to
slip rera,Ye
about to sllp
is abouf,
rrhen the journal ts
'Tlre
.^m
tha
..'is Tlre
a-xisthe
r
tr
from
distance
,ut at sorne
some dBlance
1P
p
act radially but
o _ t*_ip, is theranSle of friction- Thus does not
j-'
b1./mourerrt M of the reaction about the a-xis is given
" 6 :- ^
;t {,;Itf I
=
:
,lI
than 3 members
equilibrium of the whole truss' If more
should be determined beforehand b1r considering the
in the solutio''
equations of equiribrium are als-o used
are cut then additional sections are made.nd irr"i.
;;=;.^;:;'.o
1<
:\
1-1
.:
i
+
i.
i
1-rrr
I
.-r
:
;
,i1-
l
?-)'
-..<
.,<
.jt
<
:
:-a
.a
99
\l
,
r''
.;:--''
- "'1
...
i<;\:
'
l ,.
lr :'i'"
Exaraple 4-1 .A .0iuck with e-ear wheel drivte:;'- i'";
(Fig.Bt-l) has a'winch-mounted al its back: 'Thii's';
winchi!: 't-'i."':'
driving torques on the rear wheels and on the
ti-"1 :'
are ?1 and ?2. Assume no slip for the rvheels"itte'
drag force
coefiEcient of friction for block 5 is p- The
c1u4 and
wind'
are
on che truck and block 5, due to
ct'c2 are
c2us where r{'u5 are their velocities and
system
constants. Derive equations of motion of this
l_-
rj
rJ
f:. (+r*q
E)(AMPLES.4
,',.1-
.ri,
l.-.
3,a>f)
' '::i'l
G.vll
F;3- e q-l
using c aud { as generalised coordinates
or the rvincrr cause rotationd-R of-.he-r*'heels
Solution The di"ptac"nr.nt r of the truck a,d tt u;.ot"ti*7 tlrervheel s k ilR and the velocity of block 5
velocity of
and displacement (z * r{) of block 5- The angular
Y- The ki'etic
tnt =ytt"* tt tlre datu.r for pote.tial energv
is (i + ra)- l{re take the initial configuration o?
and potettizl energy are given b-v
l{
l-..I{
t{
IJ
6d2]+ ]mo;' *.l*'tt +
lepa"li?'+}(rr +''ut16'+ msri'i'',;"'
T = ![mfi|+ r1(i/a]2] + ][,nzi2 + I2(i/l?)?J+ jt"rii ,*
ma'*rn5J {t' i
=![rnr * m?:+:m3 ]
/
-v' :
(ti1 'r ,n2 + .f,3'1'' nta)g'z iin O 'l- fisgr(z * rd) sitr
0
l.d]=
(1)
(2)
'..'
5 moves up by (6r -i- 166]- The
rvheels rotate b-v 6tlr-and block
rhe
5r;6o,
displaceme[t
virtual
In a
do virtual rt'ork on the rear
The driving torques T1 and 12
:
prn5gcos,.
pN
is
s
block
on
frictional force
do no I'irtual rt'ork sinpe it'
torques acting on chassis 4
opposite
Tlre
respectively-rvheels and the wincrr
frictio'ar forces on the
trlo.r 5 are c1i a.d c2(i + rp)- Trre
tra'srates- The drag forces on trre truck "rrd
Hence
of points of con0act is zero for no slip'
rc-heels do no *,ork since the velocitl.riXor + t tdl.
es 0 (62 * t 66) ; c1i6c' "z{i+
. -, 6iryo. = Tt(6r l. Rl * Tz66 - pmsg
1611166
Pmsgr cos 0 - t:(i +
+
ri)l6z
[?r
ct(i+
ci
cGd
- = W I R -pm5r
s
Qi" =vrl R- pra59co6a : c;i:'c2(i+td['.'ar
ff
dv
0z
=yl.r* rn: * m3 + ',,{ * ms
= (rnr{ rn2 * ,rr3 + "r.{ +
=lT2 - pm3'orcosf, -c3(i
* Ut + I)lR?li + 'ns"d' .# =(ra
fi =,r,sgrsino,
rn5)g sin d'
T:o.
ctz
T-s.
A.E,
': .:.::
The Lagrange's equations' of motion are:
*ff)-**T=e2",
*msr3)d { rn5ri'
*te)tl t3)
*(#\-T|-X=QZ''
i
d h)lr1li*msr6,
^
c1i - c2{i + -d)'
0
* (ror * arz * rns * rq * ms)g sin 0 = T l-R' Prr,sgcos -
[rn1 *rn2*me*m4*ms+(Ir
L
L
(Ia+m5r2)6+m5ri+m59rsin0_Tz.pm5!!tcos0.""(i+'d)"
U
U
L
L.
r,
L
:
-
A
m1 with a thin bar of mass m2
consists of a thin disc of mass
Example 4.2 The System shorvn ia Fig.E4.2
by a linear sring of,s;tiffnes
a horizontal track and is restrained
on
slip
rvithout
rotls
disc
The
it.
to
pinni:d
the d'sc' The springs arc uodeforned
-rs
connected bet*reen the rod a$d
E- A torsional spring of stiffness !r
motion'
rvhen 0 = 0, a = 0' Derive the equations of
foo
l,
Solution The frictionat fegge -on ihe rotor does no rvork since the I'elocity of
the point of contact is zerq. The displacement z of the <iisc cenf,1e aud rotation
d of the bar, imply rotation 6 - z/f-of thedisc. The angular'velociiyul of
the disc and aagular velocity ar2 of the bar are given by u1 = if R, u2 = 0
and zA = u* f,sind. The telative twist of the torsional spring'ts $ + x/R)'
The virtual displacement 6:c, 60, + 6xa = 6x * Lccsl 60, and
6Wn' : P 6z-a : P[62 + Lcos| 601 = Q'fu + Qi"60
ei" = P,
qt'= PLcosA-
5).
I
JJ
J
+
J
J
J
Eis.€tr.z
The velocity of C, the kinetic energ,' and the potential energy rvith datum at O are given by
ec=lLa .dk * Be=;!-akx (I/2)(sindi+cosdi) - (i+ ILitcosl)t-*Lisinei
T=llllrrrgzltilRl' +rrr1izl+ |nr2[(i +*Li)cosllz +(*Lisin0]'1+*(*rtz1tz'1P
= l(3m1/21- n2\i2 + {.n2l2l6)it2 + lntzLii} cos?
V - +k* + +tt(d * rl R)2 - f,m2sLcoso
ff = 6*rt2rrr.z\i+ lrr.:Idcosd,
{ = rnrLzlft+ l,n2Liccso,
X=lr1k{o+x/R)/R
ff = t ,te + x/ R)+ *rnssl sin 0,
The Lagrange's equations of motiorr are:
d r?Tt 0T aU
a\7*)-fr+*:Qi"
d r?Tt
AT
OT
AT
o0
J
JJ
-0,
J
:
= -f,m2liflsinfl.
J
J
AT AV'
;lril-*+fu=Q';"' +
(l}
:+
{3m112*mr)i+'}rn3l18ce 0-i2$n?l**a* k.(0 +z/R)lR= P
,n2L2612+ rlrn2tr(i cosd -. irisin g) + !r*2Li0 sin 0 * I;r(f + zl Rl * ]rnssDsin 0 = PLcs0'
(2)
AI
n t26p*ImzLiced+ k40 +zlR)*|m2gtrsin0 - PLcqA'
i
a
F.quatlons (1) and (2) are the equations of motion of the system,Exarnple 4.3 The system shown in Fig-D4.3 consists of a rotor n'ith axial nroment
of inertia / and two bars.pinned to it. Each bar has mass ,,1 aud the mornenLs of
inertia about principal a-xes'g.at the centre of rnass C ate'. I1.,lzz,Ias. The bars
4
>i
\
i\
d
9L
are restrained by two torsional and one linear spring. The springs are uudelorrned for
.J
0 = 0. The torque applied to the rotor is fi and an internat mechanistn ipplies lorque
7i to each bar to raise.them. (a) Derive the equations o[nrotion and shorv thaL these
{
are directly related to the.p.rinciples.,of, moment of momentum and sork-energy. (b)
(
Obtain the first integrals o[ these equations. (c) Find the angular velocity d of t\1
(d)
0?i
T1
a-.
rotor for which the bars make constant angle d6 to [[e vertical for
=
=
.\
q
iConsider the case of.free roiation of the rotor, i'e', Tr = 0' Initially = uf il = 0'
rvhen 0 - t/2. The internal mechanism applies torque to each bar iu such a way thai
the bars are lorvered ro t,1r" ,"stical configuration and 0 = 0 at 0 = 0. Find the angular
iD
velocity of.the rotor rvhen 0 = 0. and the work done by the internal mechanism to
change the configuration ofthe bars from horizontal to vertical- .
Solution Tle angrlar vetocities of rofir 1 and bar 2 are g1=, jj- and gz = 6i+0L- Hence gc' is tiven by
I/
gi = vA+ s2'x Ag -- (u a u, x o A) * ttz x &-= 6 tx Ri * (Oi + o g x o(sin d [ - cc0!)
'
= -(B + csin 0)d & + aa cosdi+ o0sin gi
o2s = (E + osino)z$z + o202.
I
tol
JJ
J
J
JJ
J
J
J
J
J
J
JJ
J
J
--.1
J
{
r
1
The exteasion
"q*i*p!ne*,j-!$0:
The ki19!ic.+ergv and potcntial
' ''
energr
:
G;G;ilr.,iaa"i"il;i.'d':*..9^n,'i1x...,.:;.-.i1:..,.'.'...
t = iI i'+ 2(i)ft{(r? + ;i'* ai.i'+ a2021 + '.!' .f" o+ rzed? sin2 e + tr.itz1
(1)
='Vli)+rrrt'o+I:zsiq?0-+m(R+asin8l16'+lls+'rar.2102
acoso * 2,L'sin? o * ktoz+z(+kcazl=
-2mg
f = -2tngacosO* |qzrsinal?
(2)
reversea gotqYo
by the torques ?i on the 01" O.:t tlrl
In a vittual displacemen t 6l,6l,r'irtual :*.1 i" done
virtuat rotation in direction l- Hence
1i actidg on the rotor do no virtuil work since the rotor has no
.
6Wn'
:Tt6i 1 46dilSt6O-+Q!:6e-+--
Q{' =Tl
Qt' =zTz'
l
:
q-
= 2lI 12* rrr cc? I * I32sif d + rn(E +
a0
{ae =ilfu+ mazli
osin 0)?]<i
,i
*OQ --0,
5=0.
o0
A - 2*orsin0* 2kL"sio?i +2:k.0
a0
(") The Lagrange's equations of motion:-'
=o'"' +
*(#)-'**#
*(#)-#-#=Qi"'
-
$avn*
r:,
*', * ,-y1-
siuof )il
(3)
=A
11".::,": :rr:::::^-a,,1r
t
Zlle*tt*210 -Kt"r- r11)$in,cos 0 +Znlft*asin8)acosa]t'3+?rnsasin
o azb,o:trz(41
a +2kL2sin2
the **T* about the fixed a'tis D0'
Equation (3) is the moment of nomentum equation fot
at 'r1'
:
Equation (4) is ttre moment of, rnomentum equation'of:::t'ty::':'axb
-
i{st = i{6,''
fo.* ih" fottowing equation ftom cqs(3],({i: eq(3)C+eq(l)0
6
I t2* 11 1 cosz 0 + Izzsinz 0+ m(E + csin0)2}d
*, tt r-'- r, [Le .in o + 2',.(R +a sin 0 )a cos alit - 1U* +,,nailii
12{
'-
1
*
+2k':1!0=T'6+tTzi
'
-(Ir"-Ir1{sinpcos.0'+2n(R+osin0lgccd}tid?*[2mgcsind+2&f,?sinzd
+daii"ioon+6qg6,-.1;;fcisosinaiz-€ru#tsin8)acodltii?
trtrlr* rr; i";r,r +,r;,sins?
i"'i' si"? + 2;u 6z'1'26;1 zff iil6' = t'Q + ?ui
+2[t$ *
i'l i
:+ fiUrn+frrcos2 0*I22si,:0*rn(8+asir,efl6'
' -- -1to +'ro91162 -
i
t
t
I
i
zmgo ccO +2&f,2 sins 0 + *ta1 =T'i +tFzit
1
form' ECuat-roT^(3) adt{5) fould
shole
*"-"otoirro-*t rm equati'on and wort-energl t"luitT for the
rhich is the work-energy relatiou for the whole systcm in the rate
have been writtcn
utr*ti ;;
;
syst€$r. lntegrating eqs(3] a,od (5]'oace, we get .. '':: '. ' ;-:. ,
a(r'tV) =Wo'
(6)
l
i
i
L,i
I
)
L.-)
L.h:+
L-
r)
'l I
t
a
.lo5l:-,
f
L--
i,'
t)
_.:
i = 0, i = a, and eqs(3),(i) yield d - constanl = fs (say] and
(c) Tr=Tz=O,0=dq'"]\
Jf,
;-r
t--.d
t
--
- lzmsasina +2bL2 sin20 *2k.01/ IUzz- r11;fi,racosa +2nr(E+asind)aerdl
I
-.<
(d)
LeL u2 be the angular velocity of the rotor rvhen the bars
are vertical. We apply eCs(6) for the
displacement from 0 = r/2 b e
to get
i
-0
2(I /2 + Iu + m,B2)utt
=+
- Z{I 12 * Izz * m(R + a\2)u1
*,n(.R + a)!q | (I /2+ f,, + mE2)
= II /2 *
t['"c - (T + Vb _ (,l+ rz1,
J^Ut lz* r- + m(E + o),|-i + zkL2 + k.(r
t t2
1
122
(7)
/2)?l + lU /z* r11 * ma\,.ti - -znzsal
(8)
The rvork done by the internal mecha.ism is given by eq(g), *,lrere a,3 is given by eq(7).
a
:
(t-S ,r,2
B l*
"ntironmentispositioncdb1,alazy.tongtypeder,iceslrorr,ninFig.E4.{a,lr-vappl.ving"M.).jlE1
t)fo
torque.tly' to the handte of the bar zlB. The mass of ihe instument i",r. und
ihu*t
.-'i'
mass of rods is I kg/rn. The pirch o;,h"
*;,
*;;;;;t;;.
;il
,i,i
X\,
";;.*
t'::'"",::t:,Y"*'*T:tion-,Sor.tilrheslfl
one
i. :,
*:::,^:.j::Y
lras
\:H
degtee
of freedorn- Let d be the ge*eralised coordinate. Forces *,hicrr ,ro
nork
act in the y-direction at poinG I to 5
'c ' i\
c.r./:\
displacenrent sd,
f,d, c,*/'..
{Fig-Ea.4b}- For viriual displacearent
tlrey-coordinatesofthesepointsofapplicationare.<
Exarnple 4-4 Atl instrument to monitor conditions iu hazardou,
1
y, = |a sin 0, yz = (a * 6) sirr d, y3 (@ + 2b +c) sin 0,
\Zzt
=
y. = (o* 2b+2c*d)sing, Us e+(o* 2b+2c*2d)sinr?.
=
7\
u oo;
*"1
j:" =^,"
dya =(a +2b+2c+d)cosdd0.
-: "! slr5
d3r5 =(o *2b+2c+2dlcosl69. f-;lf-\,'-.' -or
-, '--*Y.'
-tst.vt.cizarcosttcv.
-\m
The rotatiou 6<lof the screw ca' be ecpressed in terms of 69 b3, considering
*k#."
+
:
or
+
;1Zi
-t'.\
}t.'
:+ 6(A8r= -2asiad60 - p6gl2r :+ 6d = _4rcsinl60lp-
Applyiug the principle of virl,ual rvork 6W
:+
i.
::"'' $ir'a'{,,:li{-u
/S:--\
:':
"::.:::::i'!;^
-48 =Zac-rx,l
A
=f,Frr6y;* l{6ti =0 +
2\ag 6y1 + 4\bg 6y2 * 4\cg 6% + 4\dg 5y4 + rtug 6ys + M6O
[Ig{o' + 4b(o+6) + 4i(a +2h + cl+4d(a+26+ 2c+ d.ll
= O.
.
+ory(o * 2b * 2c -t- 2d)] cos 0 - Qtralp)Atsir 0t 60 = 0
V.)6e
M = p*10$s{o' ++6(o +0) + 4c(a * 2b,q sy * 4d(o*2b + 2c+ d)} +mg(o+ 2b *2c+?i)l
3_r
Mi
i
s:.q-{r
k.tez-e,)
{2-,,r,
The torque llf decreases as 0 increasesExam.ple 4.5 Derive the er2ression of the forie
P aod the moment M required to maintain the equilbrium of the pendulums shown in Fig-Bl-ba. Tha
springs are undeformed for 01
- 02:0-.The pendulurns are modellcd as uniform bars-) ,. ,.,
r
*[*
/ 4ra
'+",1*f,{le'-o')
K<,,,
trrI
Solution llre shall present three solurions usiug 6W - 0, 6W"c - 6V - 0, and 0Vl0q;: O: i
l- The extension of the spring is 0sinds. The rela[ive rotation betw'een the ends of the torsional spring
connected to the pendulums is (gz-0)- The forces acting on the trvo bars are shown in Fig.Bl-Sb- Note that
:
the'pair of moments due to torsiooat spring does some net work on the system during a virtual iisplacement
-:
J
)
l
-J
!
I
!*o_3
J
J
-
,
k.'\r
tj_
lj-1-^
t.
l'
u
tJ
L
tIJ
TJ
tJ
L
L
l__
1L
l_
IJ
lJ
llJ
L
IJ
L
L
l-.
L
L
L
L:
L-
L
.t 1
-
IE
IE
,ti
4' whereas
lrr.in e, ce, - rlz;,.02602,
='r'r-Jr,* +rr.o. rr, ,r; --
The virtual work of ext'ernal and internal
-
13
+ t[
502 should separately be zero'
ie-'
1
*lP,1co;02- \rrr2g/;2si 0z- k'(i02-'A)1682:O
Hence, tbe coefficients o[ 501' and'
P Lr cu A
r
P Lac<x,l2
(l)
(2)
i{ * &r(0r - 01} = 0
(3)
-
the
eqs(J) and (f)r are*respectivel-v'
** :o('|
ef the r*-rrorc'
rnomenr equ.ibriu'riqulq
-obtained
th<j,
^I:0l::lt"'
J
z
lr.
equati,on.for
equiribrium
is
Equatioa (3) vietds P and then lf
momegt
I
O'}FA'
.!.;\
'- "'' Y 6i0t and V60:
cos h *
(n1 12 * m'lg Lrsin 01 - &62 sin 0r
*
I'zsin01 k'(Az - 01) = 0
!m29
i
- -iL'sin d1 601'
6A11 F (0' - 0)150r+ [-e{02 Ae'
+ r'V, -.
-
tr
fotces on the bars
(-I;Dsin01)6vr + M
5W : P 6st* m2s 6x7* r..ts 6zs+
q
Lr cose L (m 12 + m2)s L1 sia0' - &0' ":i ::"
= fP
iLrcc d1' 6e3
t;
i.t
II,
,t
I
:i
;::'f$:;";::':::':XiltT"iilX"i::T:S'ffi1':f;::i:';:""':i'i'1d
6yt.= 6cosd1 6d1'
* Lzccgz60z, y1 = 0sin01'
,,.
n',- r.rdro, 1* L2sirt6z, 6gy= L1cm0r60r
il
."u""r"n"r.i:i
7:Hltx"lT:,f:[is:.fu
sprins and sravilv forces is orrtained
as.-,v 'using Y:
mz9(Ltcos01 * |f,2 cos02)
\*'gL'cos01 6tl:&6"sin01cos01 60t+kr(6t-rdtX6gz-rr,l*trnrg.LrsilrLrlar-rn2$(-f,1sin04-|L2sin036d3{4)
from eq(4t'the eq(t) given
vidds' after substiruaion
f,ence 6W = 6Wo" -6V - P 6w.+ M 60r-6Y70as d{ibetl
calculation'
"*ti'
. ..earlier- The test of the procedure is same
Yirtual
constant f6r the PurPose of
rz = !t(6sin9r)' + i&r(0, - 8r)' -
3- The Siven force p and moment M zretre.ied tgitlr
as
6att
1o1k
In thai sease P and M ate trea'ted as consert"ative
*2s(L1ccfi * f,tz'G'azl
v = -pyt- araflrto"indl)3 +!E (az-l-rlz - l*'gL'"T":
c661 - *rn:g& coo: (5)
(t'"
+
e'1'*gi;1e*
'n:)eLr
+
sindz)
f,e
*t(osla
= -P(.L1sin,r *
"iJ
lhc c4uilibn.,- eq.,alio,.,x ovldqi = 0
,r ,, *Tf **
'
arc co.oscraatioc, theait is most corwcnieat
to asc
!e
0v
pLrcos6l*f62sin01 cos01 - kr(02-01)+(|rn1 *m)gLtsino1 =0
=-M
00t
40
aV '
=
a
-PL2c<x,01t t'(gz -g.)+ *mtgi2sin02
do2=
- Fauations (6),(?) agree with eqs(2) and (3)'
titted up by tbe force trL from
Exaroptre 4-6 The platform h Fi8-E4'6 is
the rralue of P
E
(6)
(i)
c
t^t
i
jr-OH=O
gi:c'oA=AG
are light' Find
the pistoa of the hy&iulic cylinder' Tde hnks
- *14'
as
resoh'ed
is
E
treai A,Ur,Ur,U, as constants' Force
for given rralues of R. Mt, Mz, Mg' Tte springs
- Solutioa We
are undeformed for 0
R= -RcosCi+ Rsindj *ith
I cl"lt 12
0 - c)l/t(d - a cos 0)2 * (a sin (a
sin
0),
a
cos
C)
- [(d [cos C,sin
The potential eoergy of a constant force F
wich datum'at
rhe
*r.;;fi;:';acts ar, G- ff"
'l]" "t:Ot.:ft,"
to'r{r iswlt2and
*.H;5:;1ff:A;.:fi"rffffi;.I";J'i.;-"..*aie
t64
..
I
.d f)l
.:s.-
',i,
:ir-:.
-.... , :.
potential energf of a moint6trl/, aqtiog on body f in.positive ; direction
of body i. For the given one d.o-f- system. V is given by
is -M;0; where d; b the rctation
ii
,, {iiliJ
v = -(-W)va - (-P)" E - ?Q * Wtl2];v{- (-M'X" - 0) - (-M'zXo) - (-MeXd)
i l-l
+ 1e',[. ' : 2A - *1212 ++t ,P'- ol4l2 -(-Ecos{)eg - (Esin{}v6
l:
i
=-(-w\(zlsind+6)-(-P)(c+2acq0) -?a.yw1l2)(4osiud)-(-Mrl(". '*0)-(-M2Xo)
i
p-
i
ZWoas|-2Pasin 0 +4(Q-W1/2)acos0- lt1* t43
(l)
t2t,,[ 20 -x/21+t,,10-n141-l?ncospsing-Rasin{cosd=0
| 2wocosd + 4(8 -
I
I
t.1
I
T::l;:',;j'f;:";,1^r;j
,!
.""1!, n.Jj,',1"","',-'
in Fig'81'7a' The unstretched
Example 4.2 Inacessible objects are clamped by 'uhe nrechatristn st:orvn
on cjre,obiecl if the handle-srltfjo'*ll1 bv equal forces
is .Lo. Find the clampin*
tength of the
:plrrc
".." *
F- Neglect friction. (o)
| r .. r
t
qF-
&L
rf- ---k-
ta
*Y
.Sl--
A-'(
c ->t+- b --l--o
giten to link 1- T11e
instantaneous ceni're I: of
in Fig-pl.?b. From sytnmetry, the virtual velocity of poini B is along OB' T'he
angular velocrry o2
virtual
link 2 is obtained at the point of intcrsection of the normals to 3L and 5' The
of link 2 'rs obtained by *riting the r"elocity of l{ in trvo rvays:
''t't2
= u{lb
The componen0 of the velocity of poiut D in the direction of force P is
virtual rate of rvork Yields
6W ;
F'ri+'t-(a - Ln)u,6 - Pugafb = Q
uza = uPofb. The PricciPle of
P:1ri+*lL-
.t
I
-t
-..J
ir
--i
!-J
I
j
I
|<- a .af--.-1q|i
of virtuat rate ot *-**
U,ty a1c -- u2b '+
_i
,-J
fixed, a rircual angular velocitl'r'r1 is
in this problem, since it happens to be most.onu"ni"nt..Iteeping 0 as
are shown
virtual velocitl- ccmponencs of points C and r1 it rhe direction of the forces
.
I
,l
--'r-1
tIet the clamping force,be P- The application of Che principlb
Solution
a
I
o,tr c.sc I Yt
"/r;ffi'\
I-.J
I ,_l
Rsin{(osin8}
- (*M';1{e)* }&r,.[ a] - ..l2l2 + ][,,[0- ol4l2 - (-BcosdXacos0) The equilibrium equation TV/Oq;:0 yields P:
ry:
a0
l-,
L.
.L,
Ldlblac
U,
\l
-a
-t
.l
I
J
-,1
l
!/
-t
--l
\i--j
--J
tl
J
\l
J
,I
tl
J-t
*l
!
u
I
,]
curvature
of the top cylinder above the point of cootact is h. Let Ot,Oz be the cenires of
lo5
of the peripheqr at
I
_j
-.j
..J
..J
J
1'\Ll\l.
.: . -;::-.. rt. ::-.
/
l=-
)
L
l-
L_
t-
L---
LlLIt
L--
l'l-
1-
L---
LL.
ILJ
LJ
L-.
L_
LJ
LJ
L--
tj
LJ
IJ
l.J
LJ
LJ
L-.
TJ
LJ
u
rj
t-u
t-:
u
th" *ui"T, P<il{4?'c.
i]}=*: .*--4qonom!&tio ,0;0 (ric'F:'8a)- Mark(Fig'E4'8b)
when rhc point of
rhe points
"r"1""i
cont"cii" D with z3oti=e-T.u"olup
"*aitiolimji[es
on body 2. @nsider tL"4qp-:?*4;i*-ti""."f -ry..v,-?,.{l*3 small rotatior
BD=DA
:+ -''R2Q=:Ro'tr'7n *
:i;
$7fualR2
ihe tine Ab2Q is at *"81e
position, O1,D,O2are collineat at 4gle 0.with fhe verticat' 3nd
In the displaced
with datum at Or is
0 + d wit; the vertical- The,potestial eaergy Ir of body 2 of mass'rn
v(0\=mg(orol)ced+(o26Jcc(0+cI = ms[(fl1+'Rz)cc0+(lr-82]cos(L+ hf F.:)al
v,(0D=#=_*onEr*&}siad1(Il:a?X1+Rr/8:)sin(L+h|.R2}0|
(3)
+ RylR)z cc(t+ Rtlb)el
(r.r
v,(0) * =-me(Er *.82)c0 + -.R?Xr .-.0 as the egui[brium configuratioa' which is
using eq(2), the equilibrium cmditioa v' = 0, yields 0
y"(0)
otherwise obvious- This equi[brium pcition is stable if
(I)
) 0' i-e-' using eq(3),
if v"(0)=-mc(Er aRzl+(r.-R Xt+ ntlez|z]= '#(Er +Ez)2[RrEz /(Rt+Ez]-al > 0
i.e.'ifn<Rt&ltRr*Ez)]=ll$lR|+l/Ed;....Ki
.For rhe casetr
=.E1Br/(Er+aa), tue statrility is
0- For this case, eq(3) +-
decided by eraluating higher deriratives-liace
Y"(0) =
(5)
Vt'(0) = -rng(Es * /?:){cos d - cos (t + ful R:.)9l
Vo'(g)- me(Er + fi2)[siuf - (1+ E1/82)sin(1 + &/Riel
\'
v"'(0):o
yrttt(g) : *c(11r + E2)[cos0 - (l + Ri&zl2 cos(l * P'i.lR2]01
V'* (Al = ,lr.s{R1+ 8z)[l - (l + hl Rzltl < 0
It follows from eqs(7), (9) that'tb,e potential encr8y v does not have a local minimurn
(6)
(7)
(8)
-1s-)
at 0 = 0' alld therefore
the equilibrium configuration is onstable for this case'
equilibrium configuratioa seduces to
For the particular'cases given in Fig3l-&, the conditio"jn) :t:t*le
(]'is belorv the centre of curvature OaI . Br : oo: siable if h < Rz,Le-. if the ceuire of mass
2. Ez = oo, stable if L < Er3. Rr = -R: stable if h < Rzl(l - RzlR).
*
4. R2 = -k stable if h.< Erl(l - &lR).
'trr4
"
stableifll(0,i.e.,iftheceltreofmassisbelos'thepaecteoint'
5. 8r=0orr?z-0orEr=Ez:0:
lJ
11 !
?
(fiS'S+,S1],
;
at
A
suppo*
the
Exarnple 4.9 'The bar AB is piaaed to
c
A
tu
:.*,-,"';;%:;:Tfi;":is Ij:.jl,Ll.i at u:J:;"*;la;;]1('
-&a 5 is ^u^mY,
D
which
pioned l'o a roller
and to bar BC at B. The bar BC
guided in a vertical slot- The bars are coastrained by five springs' Find
i
".+6f' *.
\"
I
"!L1
X"
;;;; ; ;'il';'";;;; ;*.iia :cquiriirium coonsurarion fl_*aeg, -r^/
i4}u- @u
"
of thesystembecomesunstablellki'B\p1'1lt<rf|
\1,-.
Solution Inthedisplacedpositiono€thesystemGig'Ea':lb)'anglesa '" f,I 1
{y
\i*
r
and { are related by
oW,
(.)
-r
,#' 6 A
,,
zl'i
-$=a|tb
8E=Dsind:csing- :+ uetg2ld.ordertcrmsip+ug
*"'*'
&3 Lquals {t + 0 * C) - r = O*The telative rotation betwecn euas of.the-tprsioaal
' -.' I :
i
:
s"H:
potlntial energy I/ of this conseratirc slqtem,,accurate.upto 2nd order terms. is given by
v(0)
+2[]e:(asi!0)1+ ].rs(a + d)2 + *br|" + P[c cc0 + (0 + c) cosdJ
= ikLoz
-
:
+ Pfacxx? +(6+dcos(crela)
= tr*re' +-ezicsino)2 + |r"1e +allblz +^if7"otu12
. = i[*, + [.(r + o/o)' + e$2 fi1e2 + &242 sin2 d * P[a cos I + (6 + 4 "*Gflf)J -v,(01=[Br+ts(l +albl"+k*2lb?\rt-rtzc2sin2d-P[csill0+(0+c)(c/o)sin(adlbrl ''
v"(0)=[&r+ts(t *alb)z+k*2lbzl*tkza2cs20-'Pfoccxl+(0+c)(c/0)2cos(c0/!)l;.'":-t*:-i:...
-:
=.
19_6
..
,l
-t
..I
,tl'
'ing*ition'f,=o'
For the eguilibrium !o,
,:.-
v'(0);0 (o'k) ald
(,
'1
t,r
:
\
,:
-_ pla*(0+cXglq2l...,,.
alb)z. * *oo21b2l *2kza2
*"(,
*
,"toii foi*
y"(0) > 0' i-e'' if
O is stable rf
0
(
,I.,*'..r..,::
(r,
l:
=
The equilibrium configuration
t
ti
J
J
P<tte,+&:(1+,,lbl''v*4o",1021+2k2a2lll4+(6+4(a/D)']'(1)
load Po:
the equilibrium
p
the minimum vatue of at which
positiou becomes unstabre
is cated the critical
eq(u:+=[{&r+!s(l+o|b\z+kqaz|b2l+2k2a2|lfo+(b+cXc/6)2J.(2)
u,:,',J.-T:::::Jiff-:r'#"T-ril1t';t;l'ilT:"'*
orthe reg is stabre ir
lli;;.ff; I :L-,,,"";'i*l;,1;;;A);;'d;;;;";i*ongu'"iioo
I
,/i11
-- t,
ry
< Zkla i'e'' if k > mgo/Z'
If &1 = I(a = 0, &2 = 0, te = &'- " =:' '=
'
ns
i:
}k,
rnsisting of 6 li(s
equrlru'ruru t::"f":l:l::::"i.L1':;ler
f,ne equilib.rium
!-rnd the
mas\
Exarnple
Exarrrple 4.10 Find
""-'r.'"i"it ity. Each smaller links has masl
(Fig'E4'10) and discuss-thethe
lengthof
free
ot""Ja*ther
All links are uniform and th<:
lit;;;;;4m'
bigger
each
rn, and
Siil:;:Tl ,ri'iltlit*ition
d of trre svsrem, the renscrr
of the springs I;1
thisconserl,ati,esystemwiththedaturnforgravitationalpot.errtialenergy,atoisI
(1toe 8e - ccc0)2t -2m1g(acos0)
+
Y(d) = 2[]81(2ocos 0 -2l.ceLs]? ]tu
0)
J
JJ
J
@
at
'o' Sf
J
* d) - 2',.9(0'541e -^2m9(3'5ocsg):2(anr)gto*" ,r,
* mz * 4m\sacso - *::
- 1(2*''+4(2mr
t mi* 4rn)gasino
cos09)
- 2ar19(3ocos 0l - m2si(aacoso
- r'"f
v@ =;11;;';'
=
f, {a| =2(4t1 * *2}sio" 0 + coso[4(2m1 * m2 a-
4tn)ga- 2({&l + &r}a2(cos
sino=0 (ii)a'
are at ' 1' d = 0' 2' 0 =
Thus the positions of equitibr-ium
V, =O
(2)
9
+ (t).
- cos 06}|
2rnr * m: i-4rn)9 / (aEi +i&)c*cc
:
consider the stability of
if cosg3 1;W"
since this position exists only
'
mz
_
4(2,o1*
v,,(0)
1- 0
=
ao
a- if equitibrium posiiion e, oito, th."'ir'(;i
;;;"(o)"t
b. if equilibdum posiriou & $od n& "r"r,
0o < 1
(5)
these equilibrium positions' ''
o"ios"o(1}*" condude that
*r*)r;:';d;;;itJr;
"oa
poqition'
o l"'to unstable equilibrium
iL e ="-;;i
position'
a stabte equilibrium
o:
,-o
3d
o = 0 is
'
oI<0'
2. 0 - *: v"(r)=:[n(2!ry11*'+t#fid-);;;'(;
equilibdum Position'
'
(4)
aud
3-0,='at= cos-rfi2*r*mz*{rn}s/(at1+t2)o+9os0ol l. 1t
cosOo) < 0
4(2llnt* mz*am)g - 2(4&r + [z)a(l ,-.., ,,
3. 0 =0zz
(a
+*:)trGos0-cosoo]l:O
[4(2m1 *rr.z*4to\sa-'(n:
:
J
J
J
:
V'{0} --2sin0(4&r *}2)ai(cosO t']"'("*o -cc0e)l
sinf,[4(2m5l-ta?+ 4ta]so-2(4k1i
J
JJ
J
is '\r
:'# ? ; *,
[#"::i:;x::r;x;l;I,'iln
a;*;1',:!:.';
J
using eqs(3) and
(4)'
''
gol}
= 2t(4el+&2fsin'es>
o-2l4kt*kt)c2(ce 03-cos
@- qs1l* art+(z*r +mz*4m)s
t t^\nnn-t(4t.*&r
v,,{0sl=44rr+tztrfin1
* 0=risan
0
and heoee 0'= 0s b a statle equilibriur-n Pqition'
lol
:.\{
,
\
I
-
a
L-.
9^1
<.
./
ladqlYy:ttre'-zgrro force memb!* in the trusses showa in Fig.Bl.ll.
(c)
2
L
J
tt
B
6
I
!-
A
L
Solution (.) By successively drawiqg the FBD of joints B, K, D, E, J, F we conclude that members 3,
7, 5,- 16, 12, 14 are zero force members.
{b) BysuccessivelydrawingtheFBDofjoints B,C,G,F,D,E,F, I,if,.C,we.concludethatmembers
3, 5, 10,8,7, 13, 12, 15, 27,26 are zero force rnembers.
By successively drawing the EBD of joints B , C, E, G, H, I , s,e conctude that rnembers 14, I [, ?, lg,
L
(.)
18,3 are zero ficrce members.
:
p,lsample
4-12 Find the forces in aII members of the bridge truss shown in Pig.tsl-CIa.
? I50
R3
\-\--
8c
.4--
---4 !1
o.l
,L.
(o)
E
EA
'z <-vL
lr,
t-=Lrz
0
/
',
9
Q:
':.7ffiil-
F,3
:{K'
Solrrtiorn The FBD of the whole truss is shown in Fig.&t.l2b- Tlre 3 reactions from the support are
deibr:miaed from the 3 equilibrium equations. for the *hJ t..r"" under coplanar loads:
-r
:+
l{s, = l6rtr -4(20) - S(50) - 12(80) = 0,
l?r=90kN, 8z=60kN. Es=0
F" = J?s = 0.
Fc = Rrt Ez -20-50-80
=0
The forces in the membexi are determined by coasidering the equilibrium equations ofjoinG in lhesequence
A,B,C,D,E,F,b,G so that at eveqr stageonly two unknown rhembfr forces have tobe determined.
cc0= 4l@'+42)tl2:0.8,
Joiat r{:
coso- 41,'2+42lrt2=0.g701 + 'sin0=Q.g, sinc= o-24?s
& = Rr- F1siu0 = 0. F- = -Fr- Frcos€ --0 :+
F,= Fz-fr=0. fc:-80-&-0
+ Fa--F2--l20kN, Fs=-80kN-,"
foint C lf; = F1.cos0 - JIs cc 0 I fu
Ft = Fs* .Fr siu d * fs sin 0 - fe sin c = 0
"*o<,,
:+
0.8Fi.+0.9?01.F5 t!9,
0.6Fi - 0.9?01& - -f0 =+ fs = 25 kN, f5 = 103-l kN
=
Joint D
lI' = (&-&)*u,i = 0, Fr = Fz*(Fe *fa)sinc = 0 =+ f6=F6=103.1 kN, fe = -50 kN
Joint.E: F, = -so F7-(& +rs)sina= 0, f, = Fr Fro * (F5- Fe)coed =Q
I
&=-25 kN; F'ls=-80 kN
=+
Joiat F: f'=F.o-frs=0, 4:-2A-Frr =0 + Frs=-80kN, Frr=-20kN
Joint O: & = &s -l Frs * Ftzc<x,l = 0, =) Frz = 100 kN, Fy : Rz- Fl2sind 0,
=
Joiot B
:
Fr = ISS kN, F1= -\10 kN.,
:
I
a
j
I
:: I
).-rr
tcB
1
I
It
-'i-.,a= ;; r"i;l :t' :"';)'
I
1:
'-:"'-
.'
''
I
i
F*= f'scosc*(6t- F$cos010 '+ ' 0'01?3 = 0
Joint G :
2ud check eq!"t]:"
J
I
:
if rve had used exac[ arithmetic rather
The 2nd and 3rd check equations wootd have been exactly satisfied
.ftaificant dElts. Tlrcre r*'ill always be 3 equilibrium equalions *'hich
tlran roundi[g-off the numbers,to 4
since the forces in these members ate
pro*ide a check on the solution. Members 1,5,6,i, 12 are in tension
the forces in these members are negative'
posiiive- Members 2,3,4,7,g,t0, f f,'fa are in compression since
15'of tLe K-truss 2
4.13 Find-the forces in rrembers 5 and 9 of truss 1 and in members 12, 14,
Exaraple
o\i
^g*ffiffii'Eff; ,Tffi,
,"
"'
,,,t1ur?ffi#,'*"i
,o
lo
30 /--K--J|A}\*.r*btt
3s*X9.
;i
ffi;
="",
*:7(!\;l'7(_
-r"j
1 + A/i\l \l/:\ o--'-=7,\i*.i
/l\:[';fr
I.j*o,',
-Bi-(
-nl
-
iJ
YZo..P ..S'R
*ffiiii uKffi
hffi
:;Wr', ^:ffi" *i
*,
;ffi.'SKSFYy^,'r
-:,
i.
t;l;f"
%-,i?,+""
t"--t -{f-
'i
q -#\
rs - $5 $:.
:ifi,f1TTf^
-r rlnk-
!J
"t]t"-rtj['l.
Ttre reactions are computed
\ts
i"';i''=
rd;
(FS-E{-r3c):
firsr from the equilibrium equations of the truss
l'[s:12R.r -20 x4- t0x 10+10x3-30 x 3=0 =+ &1 =20
kN'
Rz-20+Fs=0 :+
-36'34
,
I
f0x4-0 :+ f6-22-5 kN
It--36'34tli''
x0'4472+10-20r'Fs=0 :+ f5:26'?5tN
Thus member 5 is';having a'tensile:,{orceio126,25'kN.
intersettion O of A;,Fe *a'*i.,g Ma =O:
It[s = 8F5 -20'x 8-30 x3*Ez x {= 0
+
t
I
i
t
+ 4f'6+30-20x4A{p :Fo x'1+30x'1*8r x4-'R: x4:0
f! f<cc0*Re*Fo*30-0'=+ 0-8944& -2A+22'5+30=O *
F, = Flsin0 *
i
I
F, = R1 + Rz -20 - 10 =0
the
is better to take appropriate sections thtoughSince fotces in a ferv membes are to be determined, it
the
part
of
the equilibrium equaiioqs of tie
truss cutting Lhe.rnernbers of intercst- 'f5 is determined from
Lruss to the left of section PP (Fig-Bl'13d):
-
iI\
I
+ .R3=-20 kN, lrcs| =2l(22 11z1r|z: OS944'
+ Rz= 10 kN' sin? =0-'4472'
F,=R3t30-10=0
\
i
I
,f5 can also be oblained py'findi$6"the''point of
t
I
8Fs- 160-90+10 x4=0':+
Fi=26'25 kN
I
I
and,Rn', cuts 4 members' Eence we have to firstfld
A section ihrough member 9. sucL as sections QQ
using Mc = 0' or find fto by
in one of these members- \ff,e can hnd Fz by taking section SS and
force
the
s'e can find r'g by considedng the
taking a section ?ll" and using M6: :0- Once Fz or Fto is determined,
Consider the equilibrium o[ part of
equilibrium of joirrt D, rvhere now ooly two unknown forces are acting'
ntomeut about the point of intersection
truss ou the right of section Uf (Fig-81.13e). lVe fiud F6 by tlking
t
t
C of the unknorvn forces Fu,Frz:
Ms = R1x 4* 10 x 3 - 10 x 2+ Fro(cos0 x 3 *sind x 2) = 0
20x4+30-20+&o(0-894{x3+0'4472x2)=0 + F16=-25'16kN
+
Considet equilibriurn of joint D (Fig-81.13f):
Fe = 25'16 x 0h472--ll'25 kN
Fy =-Frosrod - f.s = 0 +
:
:
:. :
l
ro3
-l
L
]J,
L
L
L
L
IJ
t'
L
1:
IJ
2. For the *-i;;i;,
l-..
l_-
IJ
IJ
L
not.conce*tent'
a're -^1
i'ti'i;Jiand anv three orthem
. - ro 22,
o1 bei.s
L-i-d
amarrrant
coocutrent
r8, oo 23.
* '"i'""II"tlitffiil:H*-#:;il
14, r',z2.z3 nith members
ffi ;H"[};";ffi ]'[ ";; ;;"*;
. ? ,t .-..-^ .^ -iohi of
af section
wtion
XX
XX
MB, = 0 of parr of the- truss to risht
;ii.*ffi;;;il;
"oo".io.
":$"?riff
t':
_
f11=40kN-'
Me"-18Fu*7)x 18-15x 12-15x24-15x36=0 +
forces
find forces in inner membqrs' we first fiod
This is'the procedute for outer members of K-truss. To
of the joini wh"t"-.':T meet- A section Y\'
in adjacent outer members and thea consider the equilibrium
0
at G' C'onsider equilibrium equation M4:' =
cuts members 8, 12, 16, 1? of which 12. 16, l? are coucurrent
yyle:g-a('rg0):
of part of the truss to the right of section
oonsider the equilibriup of joint
g1E:3-€h.r3[):
F, = F6- Fr+ -.Frssin0 = 0
F,=&zt.Prscos0*5=0
+
=+
fit- -31"5rkN
Frz'= 37.5 x 0-6 - 5 - 1?'5 klt
"+frt.."
If-rat:-''
L
t-ffi
IJ
L
L
IJ
tok.U-rSs
+lL-J rok*l^+lm
_>
R,,,
*
10-40-0-8Frs =0
'r,
g:6arnple 4.14 Draw the axial firrce,llril,' B-M' diagrarns:::*
t!rJ
F6-l0kN
+
i, - 18f's*20 x 18- 15 xL2-L5x24=0
!M<
TJ
L
1--
e.{ (rfg-Bn'rs8)
z.di1s --''413 + sia, = 0-8. cosd = 0.6.'A"".tioo'ke
'Tlre
The
-- --- --'*--f
o:::t-*
in Fig'81-1aa-
NB.,
r-bl
-'6[-.. r
f'1
(cl
.,, -4
l-fl-\-,.
i<-------. x
j., =r,, ,--6;i:,
I i-- 1;^, .6-1
-,'-.-.1o
..,., . ----....,--J
€=_ -.-...=:-:-u_.a-_,_.__;'',
Uott' sides- the FBD of the
supiorted-oi
Solution The reactions are detetmined firsi "fi.e the bEam is
:';',
beam is shown ia Fig.Bl-t4b- The equations of equilibrium'yigld
L
L
I
II
;
-i
t{s-108s-rt0=(10x4)x4+2Ox8=0 :+ ]?s=4kN'
Fr=R1+R3+20-(10x4)=0:+ {r:16kN'
F,=Rz*15=0 + ft2--lSkN,
r' For the first 3 ranges we draw FBD
We derive expressions of JV,S,i/ ficr five differcnt ranges of
trvo ranges
(Fig.B[.lab) of the part to tbe teft of the sectiori and for the last
to the right of the sectioir-
we dtaw FBD of the part
i
O<z<1: .IV=-ff =-'Rz:15kN, S=-Fi - -Rz=-16.kN' l{=-Me:Etz-162'
M(0+) = 0, a-('-l = 16 kN'm:+
S:-Fi --Rz=-[6kN'
IV=-ff =-Ri:15kNi
L<r12:
-..-
ilo
1
l
-r-
-.t.t
f
)
(,,
A.
tt
'l
I
2<r <6:
+
-J
-J
=i
.-J
r^l
kN-m'
t+1 =56kN'm' M(2-): ?2
74 =.-lta-gE1:f 40= fPc+{0'
kN:'
-i" ::
Irr= -F1..--Rz:l.5kN!,, . S=:S=-B:+t0(z-J)=tot':36
l
,S(2+) = -16 tlr' 5(6-) = 24 kN
I
i
a\
i
Mhasertremalvaluewhereschangessigninthissegmentfors=toz-36-0'atz=3'6m40'- $(z -2t212'
M - -M,+= Rt:.+ 40 - 10(c - 2\2]/2- 16c+
M(2+) = 72kN:m. M(6-):56 kN'rn'
*
6<z<8:
-l
M(Xf| = 84'8 kN'm'
--J
N=Fi=15kN' S=^s=fts*20=24kN'
M = Mt - E3(r0 - 'I) + 20(8 - r) = 200 -242-
?.J
M(6+) = 56 kN-m, I4(S-) = 8 kN'm'
N=fI =0,
8<e(10:
M(8+) =
-J
S=Fi =E3=4kN, y' i{'1=E3(10-:r)=40-42'
--J
8 kN.m, Ir'l(10-) = 0'
of tY' S' if for diFetent
i B'Ir{'
D rr diagrams
rr-zr'nlc are
ate drarvtt
d
in Fig'Bt'14c otttrg the expressions
and
S.F'
force,
a-tial
The
-^.,^- - rierrar.c force in u-direct-ron, or
l
J
:h:]:Hll;l;ill?;T;,;'i:,;;J;;*"::=:::::t:'",.:'",:,":r:H,'[i::ii"l#FI];ll
'i:r:::,::;i::"1;T:il#i*i#H;;il""F;:.-=iyr::.::':'t:*T,"ffi
rarving
:l*.T;
varr;s.euadrarilart*:with
,
rtrte(
diagranr
the segment tfitl't g =cons[atl[' ) varrcs
l#'":'ffa::Tf,:;'=x#:fl:::*;;;;;;;","*;:
..o,nu -r,e rrf
undegi:S-E.
of the area undes'i:S'E'
and A'l/ ctluals -te
-ve of the load on ihat portion
acts'
;;;;t
coupre
couple act's'
dis.ere
discrete
a
cidrr
6r
-S- As
o.vhere
s
"r,",,so'=isn, r
f;?t;i;i:#"":::J::ff";;;;.;;";"
iFia
F/
tS;
Cpse'
-- -r-^..,*
urhere
.4-h
-+1 ;-
l
l< z -'4l'.5 4
t.4
If
{okx , lot<*
-d-51<-
-.1
--
--*;>*."*1.},r
-
i)
4+-
6<==..^.=b=56{
_-J
'.,o1<
equ;als
\ r*-)' '1" Ji-i.
..J
N
l" z'--->1-t'5 AL
tt. {<-- z,1 - rr
N
i:-
,-_
ai1
(c')
.
i_
'ii
I
+to
. 1_I
L+l
ro-G7
-i90 A
r?.s
k-z__L'_
4jL
q f,*r,*J.3,9ffi+
!ft.-
|
. -\
,,<-x-c':rf
.-___-__F1
i <{l
|.-q
, .-/tit
,E
- ec ==$fl=[ztx4)
.rn" ror""lvst*..ri's;;o;;
., l,i
at point
- -40 kN-m'
t* " *o"'"nt f,' = 10x0'5-90x0'5 expressious
of
lone side' We derive
i!<
--, ii'!r
<<
BcDisreplaced (Fis'pt'lsb) bv its-equivalent
D: forces 90 kN and 10 kN in negative " ""d5;"it"o
1i<<
beam is supported on
The reactions need not be determined "irr""-tt "
Fig'84'15c'
N,S,M for five different ranges of r trsing FBD's in
}<<
lv! :-Mt=-42-2'212'
S=-fi =4*2t'
/f=-Fi=0,
0' M(2-\ = -12 kN'm'
5(0+) = 4 kN. S(2-) = 8 kN M(0+) =
:+
M = -*[t= -4x-2*/2+10(z-2)
N =-F"o - -?.5 kN; s = -Fi = 4*2c-lo'
2<z <6:
'
S(2+) = -2 kN, 5(6-) - 6 kN
+
10 = 0 at r = 3 rn- '
4*2cfor
S
segment
[his
=
in
sign
changes
e-ttremal value where S
O< z 12:
!
I
.-..<
-r'-
l
ilr
M has
ilI
-<
.-j
.u/-di:
li
ii
./ .l r{
';,'<
il
i:
-
ird+ -. ir,-.-+:{(6-) - '-2(!kN'm, tt(l)= -11 kNrn'
i--]
t-.\
[_,
4+2x6-10=6kN'
6<s<?: N ---F|=-zJix,(2 x-'s=-ry=
+ io1'- 2) = -6r *
6) x (c
16
*"
? <:c <
8':
LJ
I
L-
-3)
-i^ : -42M(7-\:
-26 kN'm
,1/(6+) - -z}kNrn,
10- 10: -4 kN'
lf = *4l = -?-5*90 = 825 kN' S= -Fi =4*2x62) + 10 x (' - ?) + 40 = 4z - L4
M'= -Mtr = -42- (2 * 6) x (z - 3) + 10(z -
7a =
= -Fi - 4+2x 6- 10- 10+ 10s(8+) = 6 kN' s(11-): -6 kN
S
LJ
LJ
LJ
LJ
L_-
L-.
t_
t'
L
tu
U
1.
*.
.&f has extremal value rvhere s
of N' s'.,lf for difierent
the oipressions
are drarvn in Fig.Bt-r4d using
Tbe a:rial force, s.F. and B-M. diagrams
rvith varying slope of'-g and
hnearly, s *"lo q,aJraticatty
segments- For the segment in rvhich c :ari;
E 26 kN'm altd occurs rvhere
The ma:simum magniLude of M
&f rraries cubically with varying slope of -5'
a d.iscretc couPle acts'
shown in
'Hod
and B'[{' zt-E for the bent beam
the axial force' s'F'' trvisting momeot
.
ilffiH"
v
kNlm
&'ig-Et-16a.
;'
'z
- -.-rt,
B
-6*rz1\
e1ft*/m
z^-i*,^4,%€
'-x fW 4-X}ryt
Ytrf1,*ty
"lV
zk|\
^l*o*
(c)
:t thJeam to the right of
the equilibrium
tbr
*tli'+)l
^fi
the section ai E (Pig.Bi-16b):
of
solution We consider
e
F=&*(2i-4\)+61-4L-'4i-0
;tC":e
"-
is admissible'
3
l.-.- '
t:
Li
l.
tj
L
s = 14-s- (r-8)2 = 0
-:l'l::5
*
t_:,
changessign in thissegment for
.t"a = 5 m, 10 rn ofwhich only =: 10 rn
+(z-s)212
r^ - f--?I+40-l0f:
4t-(2x's.l-(",1}]10(:.-?f10x(r-?)+40-r0(z-8}
M
-t'{a:
i
s.ii{s
8)/31 = -62 * uu
,: +*8(z
1l:
-slJ(z - 8)t(' M(10) = 1057 kN-m"''''
lf1a*1=lt nnot*' M(rl-'1= r-3'5 k}{'m'
L
L,-,
tJ
tj
1(z-8)- Llz(' -sI(" -8) - 14 - z (' - 8)''
+
&=4i-si+8LkN
x 6j * (2! - i) x (-+ll+'(!)
Me=q? + (2i- 2i- 3 k) - (2i - 4$ + (2i- 2i- B
en = 4!* r2j + 12k kN'm
is in -ve 3r-direction (a negat'ive face)'
coorpinate directions are positive:
?lhe normal to the section at E io Fig.Bt.l6b
poiniing in lhe
x (-4!) = Q
Hencc the comPonentQ------t
i
;rrc
- ComP. of F,e in -i dit' = 8 kN,
=? S.I1=rest'of thgcomP- of &=4i*8t'kN'
s, = jkN, & -8 kN, 's = (4'+82)r/2 = 8'944 kN
=
Mr= ComP. of eR io -i dit' = -12 kN'rn'
=+ B.M.:rest of the comp- af en= 4i+ 12L kN'm'
12'65 kN'm '
M, = -4 kN.m, M, = -12 kN, M = (42 +l22ltl2 =
IV
,
:
.'
rla
t
I
=:Po* *,u, -- t,^=r,Yffi!,f;
4. tu=0-1;Ir=0-53m'
tJ
ih-op
*^\ 2.17t**o
]-.
u
Y?
^\.S(r-r *:v fcv-43,"#q1,H', "}(.S"tLr +"/ N
(o-)
u
l$Nczu* 6x 1$)--,.\{6nlx J}\<,**
tr
eAS(:: rkE*-'#K';'*M-s, '6+{;
+o,
orequiribrium
srrown * .,r.*.iro.
"q,"rio.,"
,",3i?-\:;,".::r-jr"* ",o.r,lJi".toarederermine
"*,
the I unknorvns: 'rv'1.F1,N2'Fz'ivs'Fs'e'P'
H
;:';;:T."iH;;:;;.;;;"or,,,i.i..,t
];1
f :
in seaerar' i'l.
ifi;:::"J::t,1il:::f,:*t;*Xil';f:"H,:l'lJ::"J:"fr:::.'.:::'
'"
;i.*'xi-#**xl:tg:;x:"':tr:':i::ir';:l;I:"ff"loiorounacr
f
--l
I
ira
Ms=A2N7x0'4-F3x0-4:0 :+
It- N2-240-Fs-0
Fv:r\rs-320-
t
02N2=
we check rchether the assumptio-n
E
[
1f
F3:O'2N2
:+ N2-240-0'24r2=0 +
i;'rX:=lil;l];ff.
no sliq at B is corlect'
N2=300N' F3:60N'
+ A'3=360 N'
.mption is correct. The
;;;#i4.,:tl'*;i*i*;:i;;'"]:r*I,T:Ii1fi,;.:TL;11i.iffii,';;
lJ
**Xi:'ir:,Xh:*:l*.**;:?:,'T:tilY,i#'::;"[ff':I'i'lapeadiit
N'
Ar =
- t6o*Go = o
*
F,=P-02Nr-120-300=0
MB=L60x02*120x0'6*3tl0x0'4-Px[-Nre=0 +
Fy =Nr
E
I .
I
1oo
P-4{0N'
s-2'24-4'4h
tr
fff;j.":1,?$*;n*jj;,X;::nxHi;;,:::Ti#i:;;::":
1-I
and & is arbitrary- Equatioas of equilibrium
tr
]
H,i
L,
l-.
r
"**
(21
m''fi'n
r*:i**n*rii-*;:**,m;::n:Hr"::l:::::Tril;:';ffi5T:t
tr j
1-=
(i)
orrrnvnnJ-annY
=
u;==;:l::::ljJ'*300x04-Px053='
F=P-F'-120-300=0
u3
P=422'6N'
I
*],oo*,
=+
F1=2'6'N-
I
...
'
1 02' 1lhus motion impentls for P : {if2'6 N
The aisumption of rio's{ip- is o+. siuce l&l/X1 = 0'026p1
slie ai 8,_
*itr, tippiru of the blcck abour E and rolling of the cylindet wi!hou_!
Cascsgand{llFsllx:=0.15?9}t,s=0.landhencethqTul?'igoofnoslipatEiswrong.Eencethe
tbc eylinder
Kinema[ics implics 94 = 9.D = g,t and hence
cylinder has impendingllp "t B and no slip at .4.
dorawards
is shorvn in Fig'Bt'l?f rvith Fe = lrIVs = 0'1/vs
has impending translation: upwar&. The FBD
and Fr arbitrarY'
Mo = Fz x 0.4 - 0-1AIs x 0'4 = 0
&=Na-320- F2=0
F'=Nz -240-0.1'rY.=0
+ Fr - O.lNs
*
:+ .lt/3 - 320 - 0.1Ns = 0
*
=+ N2-240 - 0.1 x 355'6 = 0
.TJf
:lii
.,I'J
}J
iJ
r\! = 355.6 N, Fz = 3556 N,
5, - 275.6 N'
!
i
i'e'' the assumption of no slip at r4 is
\!e checL that l'F2l/N2 = 35'56/275'6 = 0'129 I pz = 0'2 is o'k''
it will have
and only 3 equilibrium equations imply that
correcL. The 4 unknorvns, M1 ,Fr,r,P for the btock
coodition'
this
by
provided
rI
t
I
I
point E, rvith the 4t'h equation
either impending slip or impending tipping about
rvith Fr = lrJVr - 0'2Nl
wc assttmc impenditrg slip uithout tipping. The FBD 15 5|16\\'n in Fig'Rt'l7g
t)
dow'orvard and
J
-J
J
'l
J
I
fv = Nr - 1.90 + 35.56'= 0
2?5'6'='0
, F, = P= 0.?/Yl - 120 ir's = 160 x G2* 120 x 0-6+2?5'6'x'0''4'-
p:'v l1:- lllr =0
*
P i 420:5:I*li
+
z,
-J
= L.722 i,$360h
i.e', $'hether
\ve check rvhether the assunrption of no tipping is correct,
H"n." the assumptiou of no tipping is correct' Thus
casc 3: lr = 0.5, eq(3) + r = 0-032 > 0.
slip at
impends for P = 420.5 N with sliding of the
block and traoslation of the cylinder without
J
0
160 x 0.2*120 x 0'6+275'6 x 0'4- P x 0'53 =
-J
motion
'{'
&=Nr-160+as.5o=0
.,
F,=P-fi-120-2?5-6=0
'1 = 0-069 <
i, =0-2.
+
*
P=404-23N'
JV1 = 124-{ N'
+
f1=8'03 N'
I
L
I
i
_-l
-<
--
Thus motion impends for P : 4M'?3 N
-<
rvithout slip at '4'
rvith tipping of the block about E and translation of the cylinder
roiatlcn irrespective of the va]ue of torque
Example 4.18 A disc 1, mounted on a shaf[, is prevented from
iu contact rvith it' Firid the condition
M applied in the direction shorrn in Fig.E4.18a by placing'a disc 2
Neslect weight-of disc 2'
so that disc 2 acts * u rri"tion lock for disc I'
;";:;,;;;r;&
I
tl
(3)
Cascl:}=0'53,eq(3)=+r=_0.0694(0.Hencetheassumptiooofnotippingiswrong.Tbustlre
is sho$'n in Fig'E4'l7h with Nr' ]?t
stipping' The FBD
blocL has impending tipping about point E rvithout
at .8 and F1 is arbitrary. Equations of equilibrium
Ms =
I
,-*-)N--,1
W\ffi":
W,I
-<
<
.ac=.1R
-<
at A and B' Eeoce the
solution Disc 2 is a 2-force member since it is subjlcied to only trvo reartions
tt. roration of disc 1 is prev-eutcd if disc 2
two forces must ac! along the line ,4B as showu in ris.E4.rai,
e(tan-rtsr andaitt"-tp2' FronrFig'E4'18c as20=(d-r)/('R+r)'
.{
Eence the reguired condition is
--<
.<
><i-a
doesnotslipate"na61]i"u.,if
0 = !co6-r(d-r)/(E+r)J
( min(tan-r !'t,tan-t ltz) :+
(d-fi1(R+r) > 2cos[min(tan-r p1'taa-lp2)J
I
l
:-
fitr
;i
-
a
:r{
!
Lj
:.
LJ
LJ i
LJ
LJ
!
!
:
t-
L.
Lr,
L
IJ
iN
N/(b)
(t')
I -t-
r =' '\t't'
SolutionThenormalreaction.lsw|LN/m.Assumeihattheimpend"ingsli':li:1?::.1'=::"':::
ihis assumlfioo ls not
i"-; *'"uD is as shown in Fis,E4.leb- Eowever,
;::X;";;;.*;;
bar ls rn rne s'u,c urtw
n. - F, r^L Tbe
-rB-- fi<fion
f-:+:^f^..force
ls not satisfied by the forces in Fig.Bl-l9b0g;not
Ma
equation
the
equilibrium
uus
sYu.^urru*'
=
since
slrlce
correct
correc.
:atrcr,Ieo :I ""t
1L --- :- impending
:-^--r;-rarrrinn
totation
on- Hence there is
partly
' - F:-D, ro-.-;1L
has to be distributed partly iu idirection aad
i"- idk*lt
.ninorvn.
- an
=-,,-i-nn.n
io Fig.Rl.lec wirh c
shoron
*
rso;
:r-;: |j.T:T ;ffi;J;*,il';";;";ih.
-Eguations of equilibrium:
(u
+P=(L-2alpwlL
s=P-(pwlL)(L-o)+(vrflLlo=o
;;r. t , *" ,**tble root less than -' is listed- Eencq eq(U *
L?
=0 "+ a: (1- Lld,'L
=-1'fZ
.'l.o*.;,,
with a i'ertical st'ep' is
9?r, with its front.wheels in contact
weight
of
trolley
wheel
A
four
4-20
Exampte
P rgquired
of tle trotler' Find tbe:i:rirnr 'iu -fioice
pulred by a force p (Fig-Bt-20a)- .C is tl" .",it " or mass
all surfaces t"t:T,::it-h the 8!9'1rn4
do rnount the step if tlre coefficiert of static frictionlat
"f
"5
lj
u
L
''-
c
IJ
IJ
L
lj
IJ
L
L
IJ
r-.
L.
L.
-,,
-,,
A--
fU.
k-a* -4-r:r,o -*4
F?i.r
y#'#
;TY#+, &r x%
-}'t<-15rc---+{
1_
t,
L
,i^=1u*1L\(iL-o)'12- (pwlL)o(L-a+o/2\=0':* 2a2 -4La+
p
IJ
J
t4
to)
u
L
I
=-,,*.r:
o thibrizontal
valle of
,,t9a). Find the value
^.e tH;lar'r"stn,9,1a holzgntal,table (Fig-Bt'l9a)'
4-19
Exanple
- be uiformlv
Assu,,," the *ormal t"-T::"" from'the
force p for which,o,* * t"-i*;".ion.
T:
(cl
is p'
i*i.ti""U. The weight of the bat is W. ail' the coefficient of static frictionl<-o-ri/t-
z
1 0V;/
*lI,*
--- L-.*
(b)
i
lr:,,.,
q%.=
luip by losing contact atESolution There is no reactioo at E since tt " trJn.yi-t3t*-"tt' tft"
it is a 2-force member with the tbrces actiag along
is sub-iected to only two forces through ,4.and D. Eence
fu: "l ? *u @ acting along
AD asshorsn in Fig.E4-20b- Similatly, \rheel 2 is a2-force member rvith the
l{ence the FBD of the trolley is a-s showl
BG u:shosrn in Fig-Bl-20c rvith sin0 :32A.,40O: 0.8,ccsB': 0'6:
of
zero--The'3 equations of equilibrium
in Fig-pt.2m. Notice the in[eresLiug fact that the.friction forces are
this coplanar system Yield I?1, .R2' P:
Fr=P-Ezcos0=0
& = & +.&zsin0 -W =O
-TT::
+ R2=$PlT
:+ R1'=W -4P13
+ 1500rv-3000(14/ -4Pl3l-400P=0 ' *
a
Me = 1500W - 3000Er - 400P :0
kN is hinged
Exaurple 4.21 A uniform rectangular'plare ABCE of weight 6
The
abour its ed6e AB and supporred by a light flexible cable cD.(Fig.E4-21a).
ai
hinge
the
but
'A
axis
hinge at I can only support a force normal to.the hinge
the
and
B
and
can atso support axial load- Find the reactions at the hinges /
P:5W/t2
tension in the cable.
Solution The FBD of the plate' is shown in Fi5'81'21b' The coordinates of
The
the midpoint G of the diagonal AC ate (0,0,6) + (-2,4,4)ll2 = (-l'2'5)'
reactions and the tension are given by
u- iJz-l
& = Rr i+ Rz j_+fis ki 4, = I?a !+ fts!+ fio k, T=rwJ G-O)]= t9 t_ 1i-o B,
where t = TI(CO) arid r?o ' AB -- 0 since &B L AB' The 7 unknorvns Position vecr?r, -. - , R5,t are obtbined from- Ma = g, {. = g.- &s ' AB = 0'
its tip,
tors are obtaiaed by subtracting the coordinates "i icml lrom those of
s;
.i
- i'
l,t5.
r-F
tt
5tI
Lb)
lI{
t
!o
4
.1 -'.
"-
.t'
.<
,-
AG*-:-i+Zi- krn,'e:-2i+di-ZE-,,'
ra:nj-rt*
:
&" - AB = (Eni+ /?"i+R.t) -(4i- 2L)
=a&._ 2Ra = a.
..is
(i)
E: &,q,+ Es.+ ?+6k-q
I
li ii _rk
L{a- A.G x 6k+lC x T.+ SB x a"=l_l
.,1*
l0 0 6
:.
(2)
i
4
3,_il-li 85 -21:O
l?l
EI
-
i: L2-24t*4&+2.R5:e,
j: 5-L2t-2&=0,
-':----l
k: -4&=0,
'=+ &-0,
l=0-5,
.' 2R6 -F .l?5 -.0
(3)
' Eq"(t)and(3) :+ .Es-.116-0 :+ &a=0
and
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