MECHANICS (II)
DR./ A. HEMAID
MECH. ENG. DEPT.
Sample Problem (1)
The double gear shown
rolls on the stationary
lower rack; the velocity
of its center A is 1.2 m/s
directed to the right.
Determine
(a)
the
angular velocity of the
gear, (b) the velocities
of the upper rack R and
of point D of the gear
A) ANGULAR VELOCITY OF THE GEAR.
vA = vc + vA/c
vA/C = r1ω
vA = 1.2 m/s and
r1 = 150 mm = 0.150 m
1.2 m/s = (0.150 m)ω
ω = 8 rad/sec.
A
VA = 1.2 m/s
A
VC = 0
VA/C = r1 ꞷ
ꞷ
C
C
B) VELOCITIES.
vR = vB = vA + vB/A = vA + ωk x rB/A
= (1.2 m/s)i + (8 rad/s)k x (0.100 m)j
= (1.2 m/s)i + (0.8 m/s)i = (2 m/s)i
vR = 2 m/s .
Velocity of Point D
vD = vA + vD/A
= vA + ωk x rD/A
= (1.2 m/s)i + (8 rad/s)k x (-0.150 m)i
= (1.2 m/s)i + (1.2 m/s)j
vD = 1.697 m/s at 45•
.
Sample Problem (2)
In the engine system shown, the crank AB has a constant
clockwise angular velocity of 2000 rpm. For the crank position
indicated, determine (a) the angular velocity of the connecting
rod BD, (b) the velocity of the piston P.
SOLUTION
Motion of Crank AB.
The crank AB rotates about point A. Expressing vAB in
rad/s and writing vB = rωAB, we obtain
vB = (AB)vAB = (3 in.)(209.4 rad/s) = 628.3 in./s
vB = 628.3 in./s at 50°
Motion of Connecting Rod BD. We consider this motion as a general plane
motion. Using the law of sines, we compute the angle b between the
connecting rod and the horizontal:
vD = vB + vD/B
vD/B = 495.9 in./s
vD/B = 495.9 in./s at 76.05°
vD = 523.4 in./s = 43.6 ft/s
vD = 43.6 ft/s
vP = vD = 43.6 ft/s ◀
Since vD/B = lωBD, we have
495.9 in./s = (8 in.) ωBD
ωBD = 62.0 rad/s
PROBLEM 2 SHEET 3-02
Small wheels have been attached
to the ends of rod AB and roll
freely along the surfaces shown.
Knowing that wheel A moves to
the right with a constant velocity
of 1.5 m/s, determine (a) the
angular velocity of the rod, (b) the
velocity of end B of the rod
Given: vA = 1.5 m/s,
Required (a) the angular velocity of the rod,
(b) the velocity of end B of the rod
A
vA = 1.5 m/sec
60o
20o
B
Taking A as a base point
vB = vA + vB/A
vB
vB/A
20o
60o
70o
60o
vA
20o
vB/A
vA
vB
Taking A as a base point
vB = vA + vB/A
y
vA
20o
60o
vB/A
20o
vB
x
∑y component = 0
vA sin 60 = vB/A sin 50
vB/A = vA sin 60/ sin 50
= 1.5 * 0.5 / 0.766 = 1.696 m/sec
But vB/A = ω * 0.720
ω = 1.696 / 0.750 = 2.26 rad /sec
∑x component = vB
vB = vB/A cos 50 + vA cos 60
= 1.09 + 0.75 = 1.84 m/sec
INSTANTANEOUS CENTER OF
ROTATION IN PLANE MOTION
It is the point which all the points of the body in general plane
motion rotate around it instantaneously
To determine the I.C. , the direction of the velocities of two
different should be known.
Draw perpendicular to each velocity to meet at I. C.
C will be at ∞
vB always directed vertically downward.
vA always directed horizontally to the right
Draw two perpendicular to the direction of the
given velocities to meet at C
Given vA you can get ω
vA
vA


AC l cos
And vB
SAMPLE PROBLEM (4)
The double gear shown rolls on the stationary lower rack; the velocity
of its center A is 1.2 m/s directed to the right. Determine (a) the angular
velocity of the gear, (b) the velocities of the upper rack R and of point
D of the gear
vA = rAω
1.2 m/s = (0.150 m)ω
ω = 8 rad/s i .
Velocity of Upper Rack. Recalling that vR = vB,
vR = vB = rBω
vR = (0.250 m)(8 rad/s) = 2 m/s
vR = 2 m/s .
Velocity of Point D. Since rD = (0.150 m)√2 = 0.2121 m,
v D = rDω
vD = (0.2121 m)(8 rad/s) = 1.697 m/s
vD = 1.697 m/s at 45
PROBLEM 2 SHEET 3-02
Small wheels have been attached to
the ends of rod AB and roll freely
along the surfaces shown. Knowing
that wheel A moves to the right with a
constant velocity of 1.5 m/s,
determine (a) the angular velocity of
the rod, (b) the velocity of end B of
the rod
vA = 1.5 m/sec
A
70o
60o
20o
B
30o
60o
I
AB
IA
IB


Sin60 Sin50 Sin70
AB  Sin50 0.75(0.766)
IA 

 0.663m
Sin60
0.866
AB  Sin70 0.75(0.94)
IA 

 0.814m
Sin60
0.866
VA    IA
1.5    0.663
1 .5

 2.26rad / s
0.663
VB    IB  2.26(0.814)  1.84m / s
EXAMPLES
Determine the moment of
inertia and the radius of
gyration of the shaded
area with respect to the x
axis and y-axis.
The beam AB supports two
concentrated loads and rests
on soil that exerts a linearly
distributed upward load as
shown. Determine the values
of wA and wB corresponding to
equilibrium.
•The
composite body shown is
formed by removing a semiellipsoid of revolution of semimajor axis h and semi-minor axi
a/2 from a hemisphere of radius
a. Determine
The y coordinate of the centroid
when h = a/2,
The ratio (h/a) for which Ῡ = 0.4a
EXAMPLE No. (6) SHEET (3)
A thin semicircular plate has a radius
a and a mass m. Determine the mass
moment of inertia of the plate with
respect to (a) the centroidal axis BB′,
(b) the centroidal axis CC′
perpendicular to the plate.
that is
Mass (m) = ρtA
Imass = ρtIarea = (m/A) Iarea
I AA' area  I DD 'area  I dia
A=(1/2) πa2
1  4  1 4
  a   a
2 4  8
m
m 1 4 1 2
I AA' mass  I DD 'mass  I AA'area 
 a   ma
1 2 8  4
A
a
2
2
1 2  4a 
2
I BB '  I DD '  m( AC )  ma  m 
4
 3 
I BB '  (0.25  0.1801)ma 2
I CC '
1 2
 I AA'  I BB '  ma  0.0699ma 2
4
I BB '  0.0699ma 2
I CC '  0.320ma 2
EXAMPLE NO. (8) SHEET (3)
The area shown is revolved about
the x axis to form a homogeneous
solid of revolution of mass m using
direct integration, express the mass
moment of inertia of the solid with
respect to the x axis in terms of m
and h.
x rh

a 2h  h
xh =a(r-h)
xh + ah =ar
h
x
a
r
r-h
2h-h
2h
EXAMPLE NO. (10) SHEET (3)
Determine the mass moment of
inertia of the 0.9-lb machine
component shown with respect
to the axis AA′.
h
0 .2 1


h  2 .4 0 .6 3
3h = h + 2.4
h = 1.2 in
h
0.2
2.4
0.6
EXAMPLE N0. 2 SHEET (4)
Rod AB moves over a small
wheel at C while end A moves
to the right with a constant
velocity of 25 in./s. At the
instant shown, determine (a)
the angular velocity of the rod,
(b) the velocity of end B of the
rod.
AC  (10)  (7)  12.2in
2
2
I.C.
10
  tan
 55o
7
AC
 cos 
IA
1
ω
AC
12.2
IA 

 21.3in
cos  cos 55
IC  AC tan 55o  12.2 tan 55  17.4in
VA    IA

VA
25

 1.175rad / s
IA 21.3
BC  AB  AC  20  12.2  7.8in
IB  ( IC ) 2  ( IB) 2  (17.4) 2  (7.8) 2  19.07in
VB    IB  1.175 19.07  22.4in / s
θ
θ
VA = 25 in/s
7
  tan
 35o
10
1
β
VA
VC / A    AC  VA cos(90   )
 12.2066  25 cos(55)
  1.175rad / s
VC  VA cos   25 cos 35  20.479in / s
β
90-β
VA
VB/A
  tan 1
VB / C   BC 1.175  7.7934


 24.09 o
VC
VC
20.479
VB / C    BC  1.175  7.7934  9.157in / s
VB / C
9.157
VB 

 22.43in / s
sin  sin 24.09
Inclination of VB equals β + Φ = 35
+ 24.09 = 59.09o
β
VA
VB/C
VB
VB/A
Φ
VC
90-β
β
VA
EXAMPLE N0. 4 SHEET (4)
Knowing that at the instant shown
the angular velocity of rod BE is 4
rad/s counterclockwise, determine
(a) the angular velocity of rod AD,
(b) the velocity of collar D, (c) the
velocity of point A.
ω1 = 4 rad/s
VB = ω1 x BE = 4 x 0.192 = 0.768m/s
IB = 360 sin 30o = 180mm = 0.18m
ID = 360 cos 30o = 311.77mm = 0.31177m
VB = ω2 x IB
ω2 = 4.27 rad/s
VB 0.768
2 

 4.27 rad / s
IB
VA
ω1
0.18
VD = ω2 x ID = 4.27 x 0.31177 = 1.33m/s
IA  ( AD ) 2  ( ID) 2  2( AD )( ID) cos 30
VB
30
ω2
VD = 1.33 m/s
IA  (0.6) 2  (0.31177) 2  2(0.6)(0.31177) cos 30  0.365
I.C.
A
VD
VA = ω2 x IA = 4.27 x 0.365 = 1.48m/s
VA = 1.559 m/s
I
30
0.31177
D
ω1 = 4 rad/s
VB = ω1 x BE = 4 x 0.192 = 0.768m/s
VB
VD = VB + VD/B
VD/B = ω2 x BD = 0.36 ω2
60
VD
VA
30
VD / B
VB
0.768


 1.536m / s
sin 30
0 .5
ω1
VD/B
30
VB
ω2
VD/B = 0.36 ω2 = 1.536
ω2 = 4.267 rad/s
VD = VD/B cos 30 = 1.536 cos30 = 1.33 m/s
VD = 1.33 m/s
I.C.
VD
VD/B
VA = VB + VA/B
V(A)X = VB + VA/B cos 60
VA
V(A)X = 0.768 + ω2 (0.24) cos 60
V(A)X = 0.768 + 4.267(0.24) cos 60 = 1.28 m/s
ω1
30
V(A)Y = VA/B sin 60 = 4.267(0.24) sin60= 0.8869 m/s
2
2
V A  V AX
V AY
 (1.28) 2  (0.8869) 2  1.557 m / s
1 0.8869
  tan
 34.7 o
1.28
VA/B
VB
ω2
I.C.
VD
VA/B sin60
VD/B
60
VB
VA/B cos60
Absolute and relative acceleration
in plane motion
aB = aA + aB/A
(aB/A)t = α x rB/A
(aB/A)t = rα
(aB/A)n = -ω2rB/A
(aB/A)n = rω2
aB = aA + αxrB/A - ω2 rB/A
A
A
A
aA
B
aB
aA
=
B
+
aA
aB
aA
aB/A
(aB/A)n
(aB/A)t
(aB/A)n
B
(aB/A)t
Sample Problem (3-6)
The center of the double gear has a velocity of 1.2 m/s to the right
and an acceleration of 3 m/s2 to the right. Recalling that the lower
rack is stationary, determine
(a) the angular acceleration of the gear,
(b) the acceleration of points B, C, and D of the gear
SOLUTION
a. Angular Acceleration of the Gear.
vA = -r1ω
1.2 m/s = -(0.150 m)ω
ω = -8 rad/s
aA = -r1α
3 m/s2 = -(0.150 m)α
α = -20 rad/s2
α = αk = -(20 rad/s2)k ◀
b. Accelerations. The rolling motion of the gear is
resolved into a translation with A and a rotation about A.
Acceleration of Point B.
aB = aA + aB/A = aA + (aB/A)t + (aB/A)n
= aA + α x rB/A - ω2rB/A
= (3 m/s2)i - (20 rad/s2)k x (0.100 m)j - (8rad/s)2(0.100 m)j
= (3 m/s2)i + (2 m/s2)i - (6.40 m/s2)j
aB = (5m/s2)i – (6.40m/s2)j
a B  (5)  (6.4)  8.12m / s
 6 .4
tan  
 1.28
5
o
  52
2
2
2
Acceleration of Point C
aC = aA + aC/A = aA + αk x rC/A - ω2rC/A
= (3 m/s2)i - (20 rad/s2)k x (20.150 m)j - (8
rad/s)2(20.150 m)j
= (3 m/s2)i - (3 m/s2)i + (9.60 m/s2)j
ac = 9.6 m/s2
↑
Acceleration of Point D
aD = aA + aD/A = aA + αk x rD/A - ω2rD/A
= (3 m/s2)i - (20 rad/s2)k x (20.150 m)i - (8
rad/s)2(20.150 m)i
= (3 m/s2)i + (3 m/s2)j + (9.60 m/s2)i
= 12.6m/s2 i + 3m/s2 j
a B  (12.6) 2  (3) 2  12.95m / s 2
12.6
tan  
 4 .2
3
  76.6
o
SAMPLE PROBLEM (7)
Crank AB of the engine system has a constant clockwise
angular velocity of 2000 rpm. For the crank position shown,
determine the angular acceleration of the connecting rod BD and
the acceleration of point D.
SOLUTION
Motion of Crank AB.
ωAB = 2000 rpm = 209.4 rad/s,
(constant)
αAB = 0
aB = rω2AB = (5/12ft) (209.4 rad/s)2 = 10,962 ft/s2
aB = 10,962 ft/s2 at 40°
MOTION OF THE CONNECTING ROD BD.
VBD = 62.0 rad/s
β = 13.95°
(aD/B)n = (BD)ω2BD = ( 8/12 ft) (62.0 rad/s)2 = 2563 ft/s2
(aD/B)n = 2563 ft/s2
at13.95°
(aD/B)t = (BD)αBD = ( 8/12 )aBD = 0.6667αBD
(aD/B)t = 0.6667αBD at 76.05°
aD = aB + aD/B = aB + (aD/B)n + (aD/B)t
[aDG] = [10,962 at 40°] + [2563 at 13.95°] + [0.6667aBD at
76.05°]
Equating x and y components, we obtain the following
scalar equations:
x components:
-aD = 210,962 cos 40° - 2563 cos 13.95° + 0.6667aBD sin
13.95°
y components:
0 = 210,962 sin 40° + 2563 sin 13.95° + 0.6667aBD cos
13.95°
Solving the equations simultaneously, we obtain
aBD = 19940 rad/s2 and aD = 19290 ft/s2.
aBD = 9940 rad/s2 ◀
aD = 9290 ft/s2 ◀
SAMPLE PROBLEM (3-8)
The linkage ABDE moves in the vertical plane. Knowing that in the
position shown crank AB has a constant angular velocity ω1 of 20
rad/s counterclockwise, determine the angular velocities and angular
accelerations of the connecting rod BD and of the crank DE