RECTILINEARTRANSLATION
Art.
starting fromnst 8peed8up to 40 ft per
witha
1013. An automobile runs 8t this speed for a time, and finally
sect,
per
ft
4
of
oceeleration
sec'. If the total distance 'traveled ig
ft
ft,
with a decelerationof 5
of the
two stations mile apart in a
1014. A trsin travels between decelerates at 8 ft per sect, minimumtime
and
starting from
accelerates
41 sec. If the train
second
station,
the
at
what
stop
i8 its
a
to
coming
the first station and
um
this top speed?
at
travel
it
does
long
How
speedin mph?
Ana. Maxv
60mph
velocity of 60 mph in the same direction.
1015. Two cars A and B have 8
are applied to car B, causing it to decelerateatthe
250ft behind B when the brakes
time will A overtake B, and howfarwill
constantrgte of 10 ft per sec'. In what
each car have traveled?
accordingto the
fundazæntal
and acceleration
dv
a — ¯
• AB an
at
let it be required to determine the
eX6mpIe,
velocity and
if
the
æc,
motion
3
acceleration
iB
definedby the
bodyafter
relation8 5 t + 4 of a
e, 8
beingin feet and t in seconds.
Differentiating, we obtain the equations of
velocity and
acceleration
and
a
Henceat t = 3 sec,
V=
automobile leaves the gasolinestation
gasolinestation. Two seconds later, another
ft
per
sect-. How soon will the second
6
of
rate
constant
the
at
and Bccelerates
Ana. t 16.7
first?
the
overtake
mobile
with
8
constant
ground
acceleration of 4 ftper
1017. A balloon rises from the
site
sec'. Five secondslater, a stone is thrown vertically up from the launchkig
What must be the minimum initial velocity of the stone for it to just touchthebal.
loon? Note that the balloon and stone have the same velocit) at contact.
104. Rectilinear Motion with Variable Acceleration
When bodies are acted upon by variable forces, they move with variable
acceleration. To determine the kinematic equations of motionin ach
d
cases,it is necesary to apply the given data to the differential equations
ways,
many
in
vary
kinematics(Art. 9—3). Since the acceleration may
no generalequations can be stated 88 was done in the case of constant
acceleration(Art. 10-2). At most we can only indicate the proceduretobe
we have three principal variables, 8, v, and a, related by b
t as in the following box.
8
a
Each of •theseprincipalvariables may be expressed in terms of theW'
or they may be expreed in terms of each other or even a combinBti0D
the other. 12t us considerhere the
to
simpler combinatiOnS•
N)'
Cau r: The displacementis iven in
i.e.,
terms of the time;
find v and a.
This is the gimp" cue and
is eaaily solved by succeöe
tigé$
difeßD
dt
24 t
(3) 2 = 113 ft
a = 24 X 3 = 72 ft per sec*per sec
velocity of 45 ft per secpasse
1016. An automobilemoving at 8 constant
mon
digplocement with re8peCtto the time
v
definitions of velocity
total time required.
followed.
R#tilinær Motion
10-4J
11:. The acceleration is expressed
to find v and 8.
Ana.
in terms of the
time; i.e., a
It),
The general procedure is to start
with Eq. (9—2)written
do= a dt and integrate to find the
velocity in terms of the in the form
Wemay now apply Eq. (9—1)
time.
written in
the form = v dt
integrateto determine the displacement
and likewise
in
terms
of the time.2
thetime between these two
Eliminating
relations will give an
equation between velocity
As an example of case 11,
let it be required to
anddisplacement of a
determine the velocity
kody after 2 sec, if the
lationa = 2 t, a being
motion
is defined by the rein feet per second2and
that8 = 4 ft and v
t in seconds,and if it
is known
2 ft per sec when t
Applyingdv - a dt
and integrating between
the given limits, we have
2tdt
whence
or
dif e nowreplace the
variable v, just found in terms
erentialequation
of the time, in the
givenlimits.
d8 v dt and again
proceed
to
between the
This gives
f:u-f:
•y thia
8Pecial
ease
(t:
1) dt or [8k=
1
8
motion with eonstant accelerationis a variation
procedureOfWas
followed in Art. 10-2.
Il.
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PROBLEMS
have
we
lb.
丶
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rounding
eig
0.366
0
乛
friction'
i'
0.8.
0.6
·
tan
Ⅱ
|
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PROBLEM
ILLUSTRATIVE
have
gpeed)
ofinguffcient
0
decreasing tan
mo
~coemc
From
Applying
1142.1141.
0
,0
冖
一
0
rate
~
十
the
0
which
一
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definition
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fric
inclined0 runmng
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mph,
11
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一
CURVILINEAR
CVBVILINSAY
TBANSLATION
(Chap
acceleration. in curvilinear
components of the
and tangent to the path
Dd tangential
respectivelynonnsl
are
(Art 11-5) accebration by
tool
are ßlBted to the
Tbe
componentof B&leration,
eQresd
12—1.Rotation.' Definition and Characteristics
Rotation is defined as that motion of a rigid body in
which the particles
by
dv
Rotation
by
The tangential com.
dirdicn of the velocity.
onlyto the changing
of the acceleration, given
chapterXll.
(11-4)
of the velocity.
depends solelyon the change in magnitude
translation (Art. 11—6),obtained
The kinetic equations of curvilinear
arefrom the motion of the center of gravity,
move in circular paths with their centers on a fixed straight line that
is
CBIIedthe axis of rotation. The planes of the circles in which the particles
move ore perpendicular to the axis of rotation.
Figure 12—1shows a body which is free to rotate about the fixed
axis O.
81
(11-0
The N and T referenceaxes are chosen through the center of gravity of the
body normal and tangent to its path. Dynamic equilibrium is cbtained by
applying inertia forces of magnitudes — — and — at acting through the
center of grcvity oppositein direction to cn and a; respectively. These
forces are known as the centrifugal inertia force and the tangential inertia
force.
Problems involving forces acting on vehicles rounding a banked curve
solvedby adding the inertia forces, thereby creating dynamic equi•
librium (Art. 11—7).If the vehicle hag
no motion along the plane of banking, force summationparallelto
this plane will determine the necessary
resisting force. In the ideal case
when no reaistance is necessary, the angle
uf banking is determinedfrom
the relation
(11-8)
When friction is considerd,
the maximum and
fa.rnd from the equation
minimum velocities
(11-0, Il-ID)
FIG. 12—1.
— All particles and lines have the same angular displacement.
If the radiug to any point A is permitted to rotate through 0 radians, point
A moves through the arc distance sr = rß. Since the body is rigid, angle
440Bcannot change; hence the radius to any other point B will also rotate
through e radians and point B will move through the arc distance = rz0.
From this we conclude that all particles of a rotating body have the same
angular displacement (i.e., the same 9) although their linear movements
(i•e.,81and 82)vary directly with their distances from the axis of rotation.
It should be observed that the angle between the line AB and its subsequent position A 'B' is also equal to e. Consequently we define angular
diePZacement
as the angular distance swept through by any line in a rigid
Y• The units of angular displacement may be radians, degrees, or
•revolutions,but radian measurement is preferred in order to correlate
di8placement with linear displacement.
811
(Chap.
thBt,
It wag shown
811the particles
of tronB18ti011
have
in
linear velocity, and linear
linear displacement,
of
values
choracterigtic; all the particles
identical
rotation has B similar
of
motion
angular velocity, and angula
tion. The
values of
have the same
in
be def.ned the next article.) We shall
acceleration. (These
depends upon the fact that the angular displacement
seethat this conclusion
particles.
is the same for 811the
that the linear values of displacement, velocity
note
It is importantto
body are not the same; they vary directly
and accelerationin 8 rotating
from the axis of rotation.
with the distance of the particle
12—2.KinemaücDifferential Equadons of Rotation
Consider8 pulleyfree to rotate around an axle O under the action of a
weight W suspendedfrom cord wound around the pulley. Assume that
the weight descensls & ft, as shown in Fig.
B
12—2.This will unwind from the pulley a
length of cord equal to 8 ft 80 that point B
on the rim will rotate to occupy the position
of point A. The angular distance 0 through
which the pulley rotates is obviously subtended by radii drawn to points A and B.
The relation between the linear displacement
of the weight and the angular displacement
(in radians) of the pulley is given by the
equation
I W I
FIG. 12—2. Relation between
liæar
cneular displace•
NotÆthat r is the constant
If we differentiateEq. (c) with respectto
Kinematic Differential
The common unit is radians per second (rad per sec) but other
units such as
degrees per second (deg per sec) and revolutions per
minute (rpm) are
also used.
Rewriting Eq. (b) as
(c)
and differentiating with respect to the time gives
dv
represents the time
do
(d)
dv
The expression — in Eq. (d) represents the time rate of'change of the mag-
nitude of the velocity. It is preferableto denote this accelerationby at
because it not only represents the linear accelerationof the weight but is
also the•tangential acceleration of a point on the rim of the pulley. The
expression represents thé time rate of change of angular velocity and,
by analogy, will be defined as the angular accelerationa, according to the
followingequation:
do
(12-2)
The common unit is radians per second per second (rad per sec2),but other
units are sometimes used, e.g., revolutions per minute per second (rpm per
sec), etc. Equation (d) may now be rewritten
a, = ra
v2
radius of rotation. It will be remembered thBt
pulley. The term
313
Since v = rw, the normal accelerationof any point on the rim of the
pulley is given by
Ft = r at
point on the rim of the
Equations of Rotation
(12-1)
the time t, we have
repreeenting the time rate of change
of displacement, was defined as v, the
linear velocity of the
weight. In thig
case, v must also be the linesr velocity
of a
of change of anodar
Art. 12—21
are the kinematic differentialequations
Although Eqs. (12—1)and (12—2)
Ofrotation, 8 third convenient relation may be found by eliminatingdt.
rote
(12-8)
tile
of rotation
It will be helpful to summarize these differential equations tabulated i
CBIled
and hence, by
anguLr velocityanddispLeement
analogy, will be
be
by the symbol o. Thus the
velæity at any instant is
&åned by the
equatiÖn
and COmpare them with Bimilar expressions for rectilinear
on the next
page.
motion
[Chap.
314
Art.
Rotation
Rectilinear Iviotion
d2s
rncnt
o.dØ= ad0
vdv= ads
Rotatin
trith Constant
Angular Acceleration
315
When they are arranged in this fom, the
'student should have no difficulty in remembering the equations of rotation with
eonstant angular
c celeration. Analogous to rectilinear motion,
is
8
the
velocity; w,' the final angular velocity after time I and initial angular
angular displace-
ds
dv
12—31
they are therefore mathein the symbols used;
These relations differ only
into each other by the
can be transformed
matically identical.
tions deduced above, viz.:
s = re
9.
ILLUSTRATIVE
PROBLEM
1201. The initial angular velocity of the compoundpulley
B in Fig. 12—3is (j
rd per séc counterclockwise and weight D is deceleratingat the constant
rate of 4
ft pet sec2. What distance will vejght A travel before comingto rest?
(124)
an =
a vector quantity; e.g.,
Note that the engular displacementof a body is
or counterclockwise.
the direction of rotation may be either clockwise
multiplying
by
9
by the scalar
of
terms
in
defined
Since. and a have been
factor —,it followsthat Wand a are also vector quantities. Their direction
depends upon the direction of 0. We shall therefore use the following rule:
The
sense of positive 8 determines
the sense of positive
and
positive
a.
This
agrees with the previouslyestablished convention for translation; i.e., the
sense of positive s determines the sense of positive v and positive c.
The "right-hand rule" is used to represent graphically the vectors of 9,
o, and a.. This rule appliedto e, for example, states that the vector is
directed along the axis of rotation as indicated by the extended thumb of
the right hand when the fingersare curled about the axis of rotation to
corespond with the direction of rotation.
12—3. Rotation
Combining these relations into one continuousequation, we obtain
= 38B =
-—,assuming constant BCceleration,gogivev = vo+ at. Since the
form of the kinematic differential
equations of rotation are mathematically
equations of rectilinearmotion,integration identical to the respective
rotation
in
of
simüar equations
will yield similar results. These
results are tabulated below.
Rectilinear Motion
Rotation
(related by)
00
2
Solution: To correlate the given data, we start by findingthe kinematic relations
between the bodies. Using Eq. (12—4)and denoting by E any point on the
connectingB and C, we obtain
= 20B = 39C, SD= 1.5Oc
= 30B,
Constant Angular Acceleration
In Art. 10—2
we integrated the equation a =
V02
FIG. 12-3.
e
lap
(n)
The velocity and acceleration relations between the bodies'are obtained by
changingthe symbols in Eq. (a) since they wouldbe obtained from v = rc and
write
Ta which have the same mathematical form as s = re. Therefore we may also
an
3 aB
(c)
300B = 3 X 6 = 18 ft per
Considernow the motion of A.. From Eq. (b),
2
—
from Eq. (c), aA 3 aD = 3(-4) — 12 ft per sec . IVe now obtain
13.5ft
o = (18)2
'be fotlll(l from Eq. (0)
If desired, the displtcernettt of the other bodies can
sec and
ROTATION
(Chap.
316
PROBLEMS
from rest at the
diameter accelerates components of constant rate of
in
ft
6
flywheel
the acceleration
1202. A
of
normal and tangential
the
Compute
sec.
rpm per
10 sec.
a
flywheel after
particle on the rim of the
machine has a speed
wheelon a brakeshoe testing
the rim has traveled
1203. The rim of a 50-in.
after
rest
dropped. It comes to
60 mph when the brake is
acceleration
angular
and the num.
are the constant
linear distance of 600 ft. What
coming to rest?
ber of revolutions the wheel
makes in
3.10 rad per seq2; 0
45.9rev
rest to asöeed of 900 rpm and then immediately
1204. A gear is accelerated from
time is 10 sec; determine the total number
decelerated to a stop. If the total elapsed
both acceleration and deceleration
of revolutionsof the gear. Assumethat
Hint: Sketch an U-tcurve.
magnitude.
same
the
constant but not necessarilyof
1205. When the angular velocity of a 4-ft diameter pulley is 3 rad per sec,the
total accelerationof a point on its rim is 30 ft per gec2. Determine the angular
Am. a
12 rad persect
accelerationof the pulley at this instant.
1206. Determine the horizontal and verticalcom.
B ponents of the acceleration of point B on the rimof
the flywheel shown in Fig. P-1206. At the given
position, o.)= 4 rad per sec and a = 12 rad per sect,
450
Rotation with Variable
1208. A pulley has a constant angular acceleration of 3 rad per sec2. When the angular velocityis
2 rad per sec, the total acceleration of a point on the
rim of the pulleyis 10 ft per sect. Computethediameter of the pulley.
Angular Acceleration
317
1209. Thestep pulleys shown in Fig.
P-1209are connected
by a crossed
the angular acceleration of C is 2 rad per sec2,
what time is required for belt. Ii
A to travel
ISOft from rest? Through what distance will D
move whileA is moving 240 ft?
1210. Repeat Prob. 1209 if the radii of pulley
1211. The rod DO in Fig. P-1211 rotates
Ans.
t = 6.7 sec; 8b =
B are changed to 30 in.
in B vertical plane about a horizontalaxis
8t O. At the given position, end B has a O •
downward vertical component of Velocity
of 6 ft per sec and also a downwardvertical component of acceleration of 9 ft per
sec2.rCompute the angular accelerationof
12
Ans. a = 24 rad per sec2D; an = 32.8
ft per sect
FIG. P-1211
12—4.Rotation with Variable Angular Acceleration
Since the differential equations of rectilinear motion and of rotation are
mathematically identical, we need observe merely that the technique fm
handling the case of rotation with variable angular acceleration is identical
to that developed in Arté. 10—4and 10-5. One example will suffice.
ILLUSTRATIVE
PROBLEM
A body rotates according to the relationa = 3 + 4, displacement being
measur d in radians and time in seconds. If its initial angular velocity is 4 rad per
sec and the initial angular displacement is zero, compute the values of and 9 for
the instant when t = 3 sec. Solve analytically and graphically.
Solution: Rewriting Eq. (12—2)as dw = a dt and integrating between {liegiven
limits, we have
And. d = 4 ft
(3 t2 + 4) dt
da
Applying Eq. (12—1)in the form dB =
(a), and integrating
gives
fod9=J:
4
(a)
dt, substituting for
its value from Eq.
(P + 4 t + 4) dt
Substituting t = e sec in Eqs. (a) and (b), we have
= (3)3+4 X 3 +4 = 43 rad per gec Ang.
and P-1210.
and 18 in.
5
= 13+ 4 t
FIG. P-1209
180 ft
rod BO and the total acceleration of point A.
both clockwise.
1207. Repeat Prob. 1206 except that a is changed
to 10 rad per sect counterclockwise.
FIG.P-1206and P-1207.
Mt. 12—41
O
4
X
+ 4
3
50.25
IChn!).
ROTATION
Let us now check this solution
means of the motion curves shownI
in
equation
Fig. 12—4. Using
the change in
to m. (10—6),
General Kinetic' of'
Rotation
Art. 12-5)
•319
Determine the number of revolutionsthrovgh
whichn pulleywill rotute
rest, if it' angular acceleration ig increageduniformlyfrom
zero to 12 rad per
•and then uniformly decreased
to 4 rad per F,ec'duringthe next
during, see
3
O = 23.2 rev
velocity is
The angular acceleration of a flywheeldecretpeguniforrnlyfrom
4
4
4 x 3+
sec
Adding this value to øo 4 gives
43 rad per. seqat t 3 sec as before.
The area under the (0-tcurve is
divided into parts shaded to correspond
to the simi}arly shaded subdivisionsof
43
27 g
3rd degree
12
4
the a-t curve. From an equationsimwe then obtain
ilsr to Eq. (10—7),
4
505
el
12 + 27
39 rad per
12
X 27 — 50.±5rad
If preferred,
mgy be computed
by applying an equation similar to Eq.
(10-8) to the area under the a-t curve.
Thig gives
FIG. 10 —4. —
Motion curve.
X 3) +Qx3 X
= Ol(t2 ¯ tl)
(Area)
• (2)
X 3) = 50.25radCheck
*hich is identical term for term with the preceding computation although based onB
different cont•ept.
Finally, a comparison of these computationg obtained from motion curves willbe
found to be identicalterm for term with those of the calculus solution.
PROBLEMS
1213. The rotation of a pulley is defined by the relation 0 = 2 e 30 e + 6'
where 0 is measured in radians and t in 'seconds. Compute the values of angular
velocityand angularaccelerationat the instant when t = 4 sec.
sect
324 rad per
Ane.
= 272 rad per sec; a
1214. The rotation of a flywheelig governed by the equation o.' 4Vt; u)is in
the
radians per second and t is in seconds. 0 = 2
rad when t
I sec. Compute
values of 0 and a at the instant when = 3
sec.
Ana.
1216. A body-rotates accordingto the
per second and t ig in second8.
ompute the vaiues
O — 13.21 rad; a
relation a = -2
4 rad per seö and 0 iBzero when t
and at the i,nctant
when t
2 see.
per
Compu
the
the initial angular velocity and the numberof revolutiongmadeduring
6-sec interval, ,
An8.
— 12 rad
28.6 re-,'
General Kineticg of Rotation
12—5.
The discussion will be limited to the case in Whichthe rotating body is
symmetrical with respect to the plane in which the center of gravity moves.
The rotating body iBrepresented by it* projection u•ponthis plane of motion,
and the point in which the axis Ofrotation intersects this plane is defined
the center Of rotation. Furthermore, we assume that all forces lie in the
plane of motion except for a few exceptional cases which we consider in
Art. 12-7.
Select referencé axes through the center of rotation. The line joining
thé center of gravity ahd the center of rotation will be called the axis,
andthe line through the center of rotation perpendicularto N will be called
the T axis. The axis of rotation will be called the Z axis. We shall then
an originat the
havethree mutually perpendicular axes, N, T, and Z,
center of rotation. (See Fig. 12—5.)The Z axis is stationary, but N and T
rotate about Z as the body rotates. The rotating axes are called the normal
and tangential axes since they sre respectivelynormal and tangent to the
path of the center of gravity.
which is constrained
Consider now the rigid body of weight JVin Fig. 12—5
to rotate about a horizontal axis. The bction of the external forces
instantaneous
P2, as well as the bearing reaction F, give the body the
shown. The
a
valuesof the angulsr velocity End angular
vector R) giv •s
resultant of these applied forces (indicsted by the ddshed
a).d
=
the gravity center G an accelerstion whosecomponents
of grsvity (Art(
at ia. From the principle of the motion of the center
to it by the
related
is
9-8), we know that R acts in the direction of d and
of this
equationR —d. Taking normal and tangentisl components
vectorequation, we obtain
1.154 rad per
per
a is in radiuns
8 rad
sec' to 2 radspersec2 in 6 æc at which time it,8angular velocity is 42 rad per B.ec.
or
(Chap.
R0TATtoN
Art.,
tJerø;ru; Kinetica of
12-51
Rotation
321
the general equationg of
In gummnry,
rotation of a symmetrical b«ly
to æt OfN, T, and Z reference axed
are
with regpect
external forceg, including
gpply to 011
iT
and
'N
positive in the directions of and
where the summations
and
O,
F
found by equating the moment
the boring reactionof the resultsnt force is
The moment
(12-5)
(12-6)
(12-7)
A convenient rule for signs iBto take 2M and ET 08 positive in the directionsof and at. The direction of will always be toward the center of
rotation; that of is determined from the sense of a. The sense cf
igpositive in the initial direction of rotation, that is, in the sense of a.
dw
Unlike
translation,
however,
the resultant force R does not pass through
the gravity center G. To-determine the positionP where it intersects the
N axis, we apply the principle that the moment of a resultant force is equal
to the moment
sum
of its components.
of R, only
the tangential component (i.e., — ia) has moment effect about the axis
of rotation: Denoting its moment arm by q, we therefore obtain
z
ra.q =
FIG. 12-5.
forces for all
suz of the impressed forces to the moment sum of the effective
The moment of the effective force on any typical particle A with
ra. Thismomentisdue
only to the tangential componentof the effective force. The normal cornpont•gt ré passesthrough the axis of rotation and hence has no
moment about it. Equaling the moments of impressed forces about the
axis of rotation (thereby eliminating the generally unknown bearing re8Ction)to the moment sum of all effective forces gives
dW
ra q = — ks2a
r2 - and has been defined as the mass moment of inertia
with respect to the Z axis (see Art. 8-13). Note that a is placed outside
sign because -is independent of the' positiÖri Of the particle •
whence
(12-8)
impressed forces)
Therefore the resultant of effective forces (or of all the
from the x.xisof
through a point P on the N axis at a distance of —grotation.
or
the
=
Since =
in which k, is the rsdius of gyration with respect to the
axisof rotation, we may write
respect to the axis of rotation is giver by r —
where I, =
Of the two components
intersects the normal
Point P at which the resultant of the applied forces
be
is called the center Ofpercussion. The center of percus;ion may
forces
effective
thoughtof as equivalent
the "center Ofgravity" of the
is the
cigravity
center
actual
the
as
metingon all particles of the body, just
body.
the
of
particles
Pointthrough which passes thö resultant weight of
(Chap.
ROTATION
322
Ofmotion, remembet that
the equations
Rotation
12-6. CentroidaJ
of
rotation is centroidal rotation, in
of gravity. Consequently
through the center
passes
the, center of
the axis of rotation the center of rotation and
gravity
of rotation reduce to
thc distance between
equations
the
zero (i.e., i = 0), and
one of the
becomes
(12-9)
The bar sign indicates that
Centroidal Rotation
Nil
both 217 and are taken with respect to a cen_
troidal axis of rotation.
for centroidal rotat,ion, the resultant
Since both 2N and ET equal zero
couple of magnitude 2M. , The Converse
of the impressed forces always is a
system applied to a body reduces to
force
of this observation is that if the
rotation. The. resultant
a couple: the body will undergo a centroidal
is no fixed axis .through
there
if
couple will create centroidal rotation even
the center of gravity. We shall use this observation later in Art. 13-6.
PROBLEMS
ILLUSTRATIVE
1218. The pulley assembly shown in Fig. 12—68weighs 161 1b and has a ceniroidul radius of gyration of 2 ft. The blocks are attached to the assembly by cords
wrappedaround the pulleys. Determinethe acceleration of each body and the
of
and 2M are taken
motion.
The
323
plus in the direction
moment of inertia ig given by
161
ää¯ä
20 ft-lb-æca
For tbe 200-1bblock, which moves down, take the axis as positive downward.
Art. 12—2,the linear acceleration of
the 200-1bblock is expressed
In gccordance with
angular
acceleration of the pulley by the relationa,
the
of
in terms
= ra or, in this
a. Applying the equation of translation, we have
case,al = '2
200
200 —T
200
322
(a)
Since the 100-1bblock moves upward, take the X axis as positive upward. Froth
ra, we have a2 = 3 a. Applying thc equation Oftranslation give,
the relation at
- 100 100
-
EX = — a
32.2
C2 32.2
The equation of rotation applied to the pulley assemblyis taken positive In the
senseof rotation; hence
2 Tt-3Tz=20a
(c)
Multiplying Eq. (a) by 2 and Eq. (b) by 3 and then adding Eqs. (a), (b), and (c)
to eliminatethe tensions, we obtain
tension in each cord.
200 X 2 —ioo X 3 = 24.84a + 27.95a +20a.
a = 1.38rad persec2 Ans.
w=1611b
Substituting the value of a in Eqs. (a) and (b) gives
161
2
'at = 2a = 2.76ftpersec
a2 = 3a = 4.14ftper sec'
= 182.9 1b
Ane.
= 112.8 1b
The reaction on the support may be foundby applying
Ro— 161 —
1001b
2001>
az=3cz
(a)
1001b
(b)
2001b
of
FIG. 12-6.
Solution: The momentsof the
weightS
give an urr
about the center Of rotation
balanced moment in a clockwise
down
sense. The 200-1bblock will therefore move the
while the 100-1bblock rises. The
F
direction of motion of each body BD of each part of the system showing
writing
can now be drawn as in
Fig. '2-6b. •In
— T2=O
RO= 456.71b'
of inertia
has a centroidal mass moment
The rotating drum in Fig. 12—78
the
instant
the
At
0.25.
is
ft-lb-see. The coefficient of friction 8t the brake
sec. What is the
per
is applied, block B has a downwardvelo:ity of 20 it
of 10 ft?
distance
a
in
eons
brake force P required to stop blockB
12—7b.We first
Fig.
in
shown
systemis
of
part
The
each
FBD
Of
and
linear acceleration of block P from the
o%
given data.
exc. x 10
—20ft per sece
Centrcidal Rotation
art.
(Chap.
What torque applied to the cylinder bf prob. 1220will raise the we*hv
of 12 ft per BOC'?What will be the total bearing reacucn?
occelerotion
with on
Am. M = 578ft-lb; R = 454.61%
During the operation Ofa punch pre88,it8 flywheel decelerates uniformly
rpm in 1 sec. The rim of the flywheelweighB12881b,its ingide
400 rpm to
are 56 in. and 60 in., and it is attached to its hub by 6 spokes.
outside diameters
force is developed between the rim and each spoke during
shearing
WhBt avercge
Av. F = 3501b
interval?
the 1 Dec
mass momcntof
12B. The compound pulley in Fig. P-1223 a centroidal
161-1bweight.
the
supporting
the
cord
tension
in
the
Find
ft-lb-sec'.
inertiB of 20
= 20 E-lb- gec
(W -3221b
IF=2ft
R/eec
(b)
1001b
(a)
FIG. 12-7.
From the kinemgticrelation between
acceleration of the pulley.
For block B, the X axis being taken as positive in
the downward directionof
motion
100
100- T = —z
X (-20)
32.z
is positive in the direction of rotation; this gives
10 X (—10)
Solving Eqs. (a) and (b), we have
A _ Wiz
3ä21b
i = 12 ft-lb-sec
= 01
565.6X 1 + 141.4X
4P = O
N
565.6
1b
4
Fig.
for the compound pulley shoqn in
1224. Determine the time required
37.3
Ans.
t
=
starting from rest.
P-1224 to reach a speed of 600 rpm
to give
weight of A in Fig. P-1224be changed
1225. Tu what value should the
sec2?
it a downward. acceleration of 9 ft per
determine the
P-1226 is descendingfreely,
IE6. If the weight shown in Fig.
1b is applied.
100
=
after 8 brake force P
1b
T=
tension in the cord both before and
after,
1b;
65.2
Before, T =
Ans.
brake.
of
Neglectthicknes
F = 141.41b
141.4 = 0.25 N
tF=fNl
Finally, from the FBD of CD, we obtain
3001b
FIG. P-1224 and P-1225.
FIG. P-1223.
T = 162.11b
From the friction relationship,
3
-
611b
a = —10 rad per sect
—20 = 2 a
For the pulley,
f = 020
the block and the pulley, we find the angular
t
r = 40 ft-lb-sec
2
3'
i'
Ans.
P = 159.1 1b
PROBLEMS
1220. A weight of 96.61bis fastened to a cord which is wrapped around
1=020
f 0.20
3
B_solid
cylinder of 3 ft radius weighing322 lb. The cylinder rotates about its
'
centroidalaxis. Computethe angular accelerationand the tension in the cor
also the total bearing reaction.
ä001b
FIG. P-1226 and P-1227.
4
FIC. P-1228.
(Chap.
Art. 12—71 Non-Cenlroidal Rotation. Dynamic
clockwise 120 rpm, solve for
is rotating
the brake
p-122G
in Fig.
rest in 5 see.
0.20.
1227. If the drum bring the system to
is
brake
the
at
kinetic friction
force P required to
3b2ra6kc
coefficientof
Fig. P-1228.
to block A in
to be Gin. thick.
cord
the
in
B.
1228. find the tensionpulley supporting weight
floating
the
of
eight
= 3001b
the t,
Equilibrium
W-4831b
= 1.414'
P-2001b
W-386.41b
, 80.5
f = 0.20
i'
Ws3221b
4
FIG. P-1231
fw=3221b
12—7. Non-Centroidal Roution.
= 1.732ft
f = 020
and P-1232.
Dynamic Equilibrium
In applying the equations 2N = iØ2,ET ——ia and 2M, =
to
rotation,
we
non-centroidal
must be very careful to choose the positive
3
4
FIG. P-1229.
'I'lje
cord attached to block A in Fig. P-1229.
1220 Compute the tension in the
60.51b
=
Ans. TA
is 0.20.
tit of kinetic friction under both blocks
that will permit the 400-1b block
123.. Determine the maximum qcight of A
Ans. WA = 1991b
tcyide without tipping over.
senses of the N, T, and M summations to agree with the positive senses of
an, at, and a respectively. These equationS are used with a free-body
diagram which shows only the applied forces. Gee Illus. Prob. 1233below.)
An alternate solution using dynamic equilibrium permits axes to be
oriented at will and moment summations to be taken about any center
without any restrictions as to sign. When the equations of dynamic
equilibrium are used, they •refer to B frec-body_diagram which includes
b014the applied forces and the inertia forces.
h=3'
A condition of dynamic equilibriummay be created, as in the case of
translation, by imagining that an equilibrant is applied which is equal,
Opposite, and collinear to R, the resultant of the applied forces. Unlike
translation, this equilibrant does not act through the gravity center;
instead,it can balance R only by acting through the center of percus-
6"
fk = 0.30
sion P.
644 1b
re = 12 in.
This is shown in port (a) of Fig. 12—8where the equilibrant is represented
by its components — iw2and ra which are directed respectively opposit(
FIG. P-1230.
1231. In the system in Fig. P-1231, block A has a downward velocity of 48 ft
per sec at the instant the brake is applied. What is the tension in the cord between
and B after the brake is applied? How far will block A have moved 2 sec after
the brake is applied? Neglect thickness of brake.
ft
8
T = 488.811)
62.6
1232. Assumethe maximumstrength'of the cord supporting block A in Fig•
P-1231is
1band of that joiningdrums B und C is
is
1200 lb. If the brake
plied too suddenly, one of these cords will fail. Which
at
and
one will it be
304 1b
brake force P?
Ang.
to dn and at. This representation of dynamic equilibrium is not especially
COnvenient
since we must remember that P is located a distance q
fromthe axis of rotation.
k}
The best state Ofdynamic equilibrium is shown
In Part (c) which consists of the components Of the inertia force acting
throughthe gravity center C plus an inertia couple. It is derived from
(a) by adding a pair of equal opposite
forces of magnitude
Fa at G
ROTATION
N VI-CentroidalRotation.
Art.
(Chap.
incrtia force does not act through the
328
Dynamic Equilibrium
329
center of gravity
although it8
Buchungymmetrical
rotation ig the
of Blender rod rotating at an angle
with
the 8xi8of rotation in
12-9. In part (B) the centrifugal inertia
force actg radially outward Fig.
from
tudøig Always
rd.
One example of
z
created by inertis forces acting at center of percusFIG.12-8.— Dynsmic
Bt gravity center plug inertia couple in (c).
fore
inertiB
by
don in (a) or
of these balanced forces permits us to
as shown in part (b). The addition
force at G plus B couple. The
replace — Fa ECtingBt P by gn equivalent
(a)
FIG. 12—9.
— Centrifugal inertia forces acting on a rod
couple has the moment
(b)
inclinedto the axis of rotation.
the axis of rotation through the centroid of the triangularly
since by the transfer formula, the expression
__ Jc2
Md2 =
is
i whichis the centroidalmoment of inertia. The inertia couplela
representedby the directed arc and may be applied anywhere on the body
sincethé moment of 8 couple iBindependent of the moment axis.
Summarizing,we create dynamic equilibrium in rotation by the following
forces— rd and •equals—
(also called the
trifugal inertiaforce) actitig radially outward through the center of gravity
opposite to the direction of an.
called the
2. Apply the reversed tangential effective
force Fa
the
tangential inertiaforce) acting through the center of gravity opposite to
direction of at.
to
3. Apply the inertia couple
opposite
anywhere on the body acting
the sense of a.
centrifug
The only exception to these rules concerns
the location of the
in
bodies
inertia force
which are unsymmetrical
with respect to
cf motion described by the center Ofgravity.
In these cases, the
distributed
02. In part (b) the rod is dis-
placed 8 distance a from the axis of rotation so that the
centrifugal inertia
force consists of two parts acting as shown and
determined by
ILLUSTRATIVE
PROBLEMS
procedure:
1. Applythe reversed normal effective force —
=
1233. A uniform Blenderrod, 8 ft long and weighing•96.61b,rotates in a vertiel
Planeabout a horizontal axis 1 ft from its end. When it is in the horizontal posit%a
shown in Fig. 12—10,ita angular velocity is 4 rad •persec clockwise. Mllat is then'
its angular acceleration
the bearing reaction 8t A?
Solution I: In thig first solution, we shall not use dynamic equilibrium. Accord-
the FBD in mg. 12—10
showsonly the externallyapplid force In applying
the equations of rotation to thig non-centroidalrotation, be careful to take 2N,
2T; and 2M
positive in the respectivedirectionsof an, d', and a.
We start by computing the moment of inertia IA about the axis of rotation at A.
Ringthe transfer
formula, we obtain
+
--:Æ
322
- 16+27
ROTATION
3.30
Art. 12-71
4 rnd/sec
Non-Centroida1R0ta1in.
Applying the equation of
dynamic
Dynamic
equihbrium,
3'
w = 96.61b
FIG. 12-10.
The value of a is foundby applying Eq. (12-7). This moment summation about
the axis of rotation eliminates the unknown bearing reaction. Thus we find
a = 6.74rad per sec2
43a
96.6X 3
Am.
The reaction at A has been resolved into the normal and tangential cornponents
R, and Rt. Taking positivesummationsin the directions of üi and at, wehave
96.6
Since
Rn = 1441b
is directed downward,ET is also positive downward. Hence we have
966
96.6
R
Solution11: In this solution we create dynamic equilibrium by applying tbe
centrifugaland tangential'inertic forces at the gravity center G and add the inertia
—ra = 9a 1b
ra
I
16æft-lb
— rw 2 = 1441b
3'
W- 96.61b
FIG. 12—11.
—Dynamic equilibrium.
couple la
shown by the dashed vectors in Fig, 12—11.Eucli of these inertiB
component8act respectively opoosite to an,
at, and
-
w
O
Here we
can show
one advantage
of
moment summation about the gravity dynamic equilibrium by finding Re from o
center G. This
(EMC -O)
166.74)
96.6
a. Their values are:
144 1b
96.6
(3) a
32.2
3
We
An8.
can also use force summations
to obtain-
R, - 35.91b
R,
Thence,'as before, the total bearing
resetionis R = 148.5lb.
1441b
1234. A turntable rotating
in a horizontal plane about a vertical a-xisO
carries
bent bar weighing 16.1 1b
per ft attached to it St A and forced to rotate with
it by
a smooth peg at C.
At the instant shownin Fig. 12—12,
w = 4 Nd per sec and
a 6 rad per sec* both
clockwise. Determine the forcesacting at A and C.
Solution: The equations of rotation are unwieldy
in this problem so we resort to
dynamic
equilibrium which permits 8 free choice of a.xesand moment eenters.
Instead Of locating the
center of gravity of the bent bar, it is more convenient to apply
the inertia components acting as shown
in Fig. 12—13
Bt•the gravity center of each
segment.
The values of
the inertia foreesare
For AB:
For BC:
16.1 x 6
16.1 x 4
= 961b
16.1 x 4
= 361b
32.2
322
32.2
i$a ft—lb
- 35.9
1b
R, + 9(6.74)- 96.6 = O
[EH - 01
16.1
xo
9 a 1b
moment gurn-
moment gummation iB really
øolutionbecause the gum of thea identicalto
La u,ed in the
preceding
+ 16a 43a
gum of the
couple and the
before. Actually
moment of the
tangentialinertia force
(1 + Mit)a which,
is
by the
formula, La. Hence
the moment equation of rotation
either
or a
may be used to find a, depending moment gummationof dynamic
equilibrium
on one'Bpreference.
dynamic equilibrium
The real advantageof
when want
other
the N and T directiona,
take force
or a moment
than the
of rotation.
summation about some
other
willbe more apparent
in the next
Rt = 35.91b
Combiningits components, the total bearing reaction ie
find a from
ring reaction.
96.6 X 3 - 3(9a)
la is the gum of the inertia couplesacting on each
inertia couple
nt. Iu
value is
ROTATION
(Chap.
332
—
(4)2 (0)
12 32.2
Art. 12—7]
1238.
la -70 ft-lb
equilibrium. The value of P iB
equations of dynamic
the
apply
now
We
summstion about A.
termined from 8 moment
4 rad/eec
I F ra = 361b
A' = 6 rad/øec
3'
3'
2'
Erw2 961b
turntable 2 ft away from
on o horizontal of rotation, ag
oxi8
8hownin
the vertical The coeffcient of
friction
Fig. P-1238.
the
turntable
and
iB
0.40.
him
between
Btart8
from
re8t
and
acturntable
If the
celerotegDt the rate of rad per sect, how
many seconds will elapge before he Btart8
the angle 9 of the dito glide? Determine
161-1bman
o
9
rection in which he will glide.
Am.
— 5.08 sec; e — 4.450
FIG. P-1238.
to frame rotating about
vertical aü 88 in Fig. P-1239. Showthat the angle
37
between the rod and the axis ig definedby cog9 =
A,
Wrw2=961b
1240. A uniform slender rod weighing96.6 1bis fastened to the rotating frame
in Fig. P-1240 by 8 smooth hinge at A and a horizontalcord at B. The frame
FIG. 12-13.
FIG. 12-12.
Equilibrium
A mon weighing 161 1b i8 geatH1
1239. A uniform Blender rod ia hinged
I —Fa —361b
2'
Nqn-Centroidal Rotation. Dynamic
48.51b AN.
+ 36(6)- 96(2) - 96(3) + 70 = o
= 01
Using force summations directed along the perpendicular components of the
rotates about its vertical axis at constant speed of 4 rad per sec. Find the tension
in the cord and the horizontal and vertical componentsof the hinge reaction.
reaction 8t A, we obtain
= 01
+ 48.5- 96 —36 O
—96 + 36 = O
Az = 87.51b
= 601b
from which the total reaction at A is found to be
106.21b Am.
PROBLEMS
1236. A 3220-1bflywheelis fastened to the midpoint of a shaft 6 ft long. The
Venterof gravity of the flywheeliB0.01 in. from the axis of rotation. The flywheel
rotates at a constant speed of.1800rpm. Determine the maximum and minimum
values of the bearing reactions at each end of the shaft.
1301b
Am. Max. R 30901b;Min. R
1236. A uniform slender rod 6 ft long that weighs 64>41b ig guepended verticblly
Bt one end. A horizontal force of 321b is applied at the. midpoint of the rod. Deter
be
vaine the horizontal reaction of the axis on the rod. Where should the force
regctioD
applied to make the horizontal centerOf
zero? (Thig point iB called the
percusgion.)
8
rod in Fig•
1237. The uniform slender
FIG. P-1237.
supported
P-1237 weighs 96.61b and is
the support
knife edges atl and B. Determine the
reaction at the instant after
af B ia
suddenly removed.
FIG, P-1239.
2'
FIG. P-1240and P-1241.
st which the cord in Prob. 1240
1241. Determine the speed of rotation in rpm
will have tensile force of 200 lb.
weighing11b per ft is fastened al its
12'L A uniform slender rod b ft long and P-1242. The rod is attached to the
midpoint to horiiontbl shaft as shown in Fig'. distance L ft apart. Compute the
shaft midway between two bearings A and B
rotating 8t rad per sec.
dynamic reactions Bt and B when the shaft is
•yb3w2
sin
Ans. RA = RB•
24 gL
ROTATION
(Chap.
Nan-Centroidal Rotation.
Dunamic Equilibrium
Art. 12—7)
bearing
the
reaction at A an instunt
componcntsof
after it is Teleasedfrom rest at
position.
givcn
Am.
49.01b;AD- 133.41b
1248. Thc systcm shown in Fig. P-1248consistsof a circular
disk
to
bur, The assembly rotate; in
a vertical plane about a horizontal
end of L uniform
given position, the angular
axis Lt A, At the
velocity is 4 rad pcr sec. Compute
of the bearing reaction.
magnitude
the
48.3
32.21b
14--..-.........—L'
1b
f-02
FIG. P-1243 and P-1244.
go-5 rad/sec
Fro. P-1242.
shown in Fig. P-1243rest
the weights and positions constant
speed. The
a
194S. Two blocks hBVing
about its vertical axis at
upon frame whichrotates >locks and the frame is 020. The weight and friction
the
the blocks start to slide?
eficient of friction between
at wh8t kpeed in rpm will
neglected,
being
of the pulley
this instant?
is the tension in the cord at
Ans. n = 31.4 rpm; T = 212.5
1b
B
W=193.21b
1244. Repeat Prob. 1243if the weights of the blocks
are interchanged.
1245. Three bars, each 2 ft long and weighing 9.661b,
are pinnedtogether to form the equilateral frame shownin
Fig. P-1245. They rotate in a horizontal plane abouta
vertical axis at A. ltVhat torque is required to cause an anis the reaction
C gu18racceleration of 12 rad per sec2?
rpm?
38.2
of
frame reaches speed
P-1245. at A when the
FIG.
A = 20.81b
ft-lb;
Ans. M = 21.6
rests on a
1246. The bent bar shown in Fig. P-1246 weighs 16.1 1b per ft. It
Compute
A.
horizontalsurface and rotates about a vertical axis through
What
sec2.
per
the torque required to cause a counterclockwise acceleration of 6 rad
sec?
per
rud
3
are the X and Y componentsof the reaction at A when the speed is
19811)
Ay
= 15 1b;
-4r,8. M = 568ft-1b•,
3
w-3221b
FIG. 2-1249.
1249. At the instant shown in Fig. P-1249, the body B has a clockwiseangular
velocity of 5 rad per sec. The horizontal cord joining A and B passes over a weight-
lessand frictionless pulley. Determine the horizontal and vertical componentsof
the axle reaction
Ans. Rh = 30
at 74.3
= 381.11b
1260. The rotating assembly shown in Fig. P-1250consistsof an unbalanced
pulleyto which is bolted a uniiorm rod carrying a sphere at its end. The pulley
rotatesabout a horizontal axis at Z and
has a I ft radius of gyration about its W: 64.41b)
Smooth
gravity center G. Show that the mass
momentof inertia about Z •is Is = 45.5
ft-lb-sec*and then compute the angular
accelerationof the pulley and the tension
fw- 64.41b
in the cord.
IL =6ft
— 4.64 rad per sect; T
137.5 1b
c
FIG. P-1246 and P-1247.
1251. At
in Fig. P1250,the the instant shown angular
system has a clockwise
Velocityof 4
rad per sec. Using the re-
Wc 32.21b
Diam.= 2
FIG. P-1248.
is frß'
1247. The bent bar shown in Fig. P-124fiweighs 16.1 Il) per ft
X
rotate in a vertical plane about u horizontal u.xi$at
tinc
Compute
19321b
= u.21b
Pro)), 1250, •compute the horizontal
to
vertical components of the bearing reactionat
Z.
Rå
1b: R,
378.5 1b
FIG. P-1250 and P-1%1
(Chap.
ROTATION
wt - 200 1b, are
%
- 1001b and
P-1252. Compute the valuea of
Fig.
in
AB shown
and rotating in vertical
horizontalshaft 1 ft from the shaft
rotating
the
concentrated
balanceweights
336
Summary
337
at —ra
(12-4)
are related by
These equations
1
The initial direction of rotation is taken as the positive
sense of 0; this
also determines the positive sense of and a.
From the above relations it is easy to determine the equations of rotation
with constant angular acceleration (Art. 12—3).These equations are tabuloted below.
Rkctilinear Motion
Rotation
(related by)
Wt=1001b
24"
18" w *2 2001b
w
=
8 = vot + å at2
v2= v02+ 2 as
FIG. P-1252.
the dynamic effects of WI aud W2. WI:at are
through A and B that will balance
weights measured from the plane containing
the angular positionsof the balance
0
= 26.60; WB = 231 1b, 0B = 77.5
Ans. WA = 1681b,
AB?
WIandaxis
SUMMARY
is the motion of 8 rigid body in which the particles
Rotation (Art. 12—1)
line calledthe
move in circular paths with their centers on a fixed straight
axis of •rotation.
distance
Angulardisplacementis measuredin radians by the angular
swept throughby any radius of or line in the rotating body.
a = ra
dv
vdv=ads
d2s
a
do
= dt2
(12-2)
at2
= 002+ 2
throughthe center of rotation perpendicularto the N axis; the Z axis
coincideswith the axis of rotation.
The general kinetic equations of rotation (Art. 12—5)are given by
The kinematic
d 20
+
Problems involving variable acceleration (Art. 12—4)are solved by inte„
grating the kinematic differential equations of rotation. The technique is
similar to that developed in Arts. 10—4and 10—5.
Problems involving the rigid body motion of rotation are referred to a
set of reference axes rotating with the body. The N axis is chosen to pass
through the center of grivity and the center of rotation; the T axis passes
(12—5)
characteristicsof rotation are as follows: All particles have identical
values of angular displacement, angular velocity, and angular acceleration'
The linear valuesof displacement, velocity, and acceleration vary directly
with the distanceof the particle from the axis of rotation.
with
The kinematic differential equations of rotation are tabulated below
similar equations for rectilinear motion.
Rectilinear Motion
Rotation
+ at
ET
(12-6)
(12-7)
In the case of centroidal rotation, F = 0 and the above equations reduce to
(12—9)
Dynamic equilibrium in rotation (Art. 12—7)
is created by applying the
Inertia forces
— Fw2and — ra acting the 'ugh the center of gravity directed
re SPectively
opposite to dn and at, plus an inertia couple la opposite in
senseto a applied anywhere on the bådy. Note that I is the centroidal
moment of
inertia.