ENGR 214
Chapter 16
Plane Motion of Rigid Bodies:
Forces & Accelerations
All figures taken from Vector Mechanics for Engineers: Dynamics, Beer
and Johnston, 2004
1
Equations of Motion for a Rigid Body
Equations of motion:
 F  ma
G


 M G  HG
remember:
H G = angular momentum about G
HG  IG
IG
= mass moment of inertia about G
H G  I G   I G
M
G
 IG
2
Plane Motion of a Rigid Body
Motion of a rigid body in plane motion is completely defined by
the resultant & moment resultant of the external forces about G.
Free-body diagram
F
x
 max
F
y
 may
M
G
 I g
3
Sample Problem 16.1
At a forward speed of 30 ft/s, the truck brakes were applied, causing the
wheels to stop rotating. It was observed that the truck skidded to a stop in
20 ft.
Determine the magnitude of the normal reaction and the friction force at
each wheel as the truck skidded to a stop.
4
Sample Problem 16.1
v 2  v02  2a  x  x0 
0   30   2a  20 
2
a  22.5 ft s
Free-body diagram:
 F  ma
 F  ma
M  I 
x
Gx
FA  FB  m  22.5
y
Gy
W  mg  N A  NB
G
G
7 NB  4FB  4FA  5N A
But FA   N A , FB   NB
Unknowns: N A , N B , 
 N A   NB  m  22.5
N A  NB  mg
  0.699
N A  0.35mg , NB  0.65mg
5
Sample Problem 16.2
The thin plate of mass 8 kg is held in place as shown.
Neglecting the mass of the links, determine immediately after the wire has
been cut (a) the acceleration of the plate, and (b) the force in each link.
6
Sample Problem 16.2
 F  ma
t
t
mg cos30  mat
at  g cos30
at  8.50 m s2
F
n
 man
FAE  FDF  mg sin 30  0
FAE  FDF  39.24
M
G
 I
 FAE sin 30  250    FAE cos 30 100  
 FDF sin 30 250    FDF cos 30 100   0
Solving:
FAE  47.9 N
tensile
FDF  8.7 N
compressive
7
Sample Problem 16.3 (SI units)
25.4 cm
15.24 cm
2.268 kg
4.536 kg
A pulley weighing 5.44 kg and having a radius of gyration of 0.203 m is
connected to two blocks as shown.
Assuming no axle friction, determine the angular acceleration of the
pulley and the acceleration of each block.
8
Sample Problem 16.3
25.4 cm
I  mk 2  0.2246 kgm2
note:
For the entire system:
15.24 cm
 M  I
 mB g  0.1524   mA g  0.254 
0.2246  mB  0.1524 2  mA  0.254 2  


G
4.536 kg
2.268 kg
  2.37 rad / s2
5.44 kg
aA  rA
 0.36 m / s 2
aB  rB
 0.6 m / s 2
2.268 kg
4.536 kg
9
Sample Problem 16.3
Obtain the tension in the chord
10
Sample Problem 16.4
A cord is wrapped around a homogeneous disk of mass 15 kg. The
cord is pulled upwards with a force T = 180 N.
Determine: (a) the acceleration of the center of the disk, (b) the
angular acceleration of the disk, and (c) the acceleration of the cord.
11
Sample Problem 16.4
F
x
 max
0  max
F
y
ax  0
 may
180  (15)(9.81)  15a y
a y  2.19 m / s 2
+  M G  I
I  12 mr 2  1.875 kgm2
180(0.5)  I  1.875 
  48 rad / s 2
12
Sample Problem 16.4

acord   a A t  aGt  a A G
 2.19   0.5 48

t
 26.19 m / s 2
13
Sample Problem 16.5
SOLUTION:
• Draw the free-body-diagram equation
expressing the equivalence of the
external and effective forces on the
sphere.
A uniform sphere of mass m and radius
r is projected along a rough horizontal
surface with a linear velocity v0. The
coefficient of kinetic friction between
the sphere and the surface is k.
Determine: (a) the time t1 at which the
sphere will start rolling without sliding,
and (b) the linear and angular velocities
of the sphere at time t1.
• Solve the three corresponding scalar
equilibrium equations for the normal
reaction from the surface and the linear
and angular accelerations of the sphere.
• Apply the kinematic relations for
uniformly accelerated motion to
determine the time at which the
tangential velocity of the sphere at the
surface is zero, i.e., when the sphere
stops sliding.
14
Sample Problem 16.5
SOLUTION:
• Draw the free-body-diagram equation expressing the
equivalence of external and effective forces on the
sphere.
• Solve the three scalar equilibrium equations.
 Fy   Fy eff
N W  0
 Fx   Fx eff
N  W  mg
 F  ma
  k mg 
a  k g
 M G   M G eff
Fr  I 
5 k g
2 r
NOTE: As long as the sphere both rotates and slides,
its linear and angular motions are uniformly
accelerated.
 k mg r  23 mr 2 

15
Sample Problem 16.5
• Apply the kinematic relations for uniformly accelerated
motion to determine the time at which the tangential velocity
of the sphere at the surface is zero, i.e., when the sphere
stops sliding.
v v 0  a t v 0   k g t
5 k g 
t
2 r 
   0  t  0  
a  k g

5 k g
2 r
At the instant t1 when the sphere stops sliding,
v1  r1
 5 k g 
v0   k gt1  r 
 t1
2 r 
5 k g 
 5  k g  2 v0 
t

1 

2 r 
 2 r  7  k g 
t1 
2 v0
7 k g
1  
1 
5v 
v1  r1  r  0 
7 r 
v1  75 v0
5 v0
7 r
16
Constrained Plane Motion
Motions with definite relations between acceleration components
17
Constrained Motion: Noncentroidal Rotation
Rotation of a body about an axis that does not pass through its mass center
at  r , an  r
M
O
2
 IO
18
Frictional Rolling Problems
Wheels, cylinders or spheres rolling on rough surfaces
r
G
y
P
W
x
F
N
P  F  maG
N W
Fr  I G
Unknowns: F , aG , N , 
19
Rolling Motion
• For rolling without sliding:
x  r  a  r
• For rolling with no sliding:
a  r
F  s N
For rolling with impending sliding:
F  s N
a  r
For rolling and sliding:
F  k N
a, r are independent
• Geometric center of an
unbalanced disk:
aO  r
Acceleration of the mass center:
aG  aO  aG O

 
 aO  aG O  aG O
t

n
20
Rolling Friction
Rolling Friction
• Rolling an object on a soft surface (grass or sand) requires
more effort than rolling it on a hard surface (concrete)
• No sliding in either case
• No object is truly rigid
• Rolling causes continuous deformation of both surfaces giving
rise to internal friction
Pr
Wa
a
P  W   rW
r
Coefficient of rolling resistance
Steel on steel: 10-3 - 10-6
Tire on concrete: 0.01 – 0.015
Sample Problem 16.6
mE  4 kg
k E  85 mm
mOB  3 kg
The portion AOB of the mechanism is actuated by gear D and at the
instant shown has a clockwise angular velocity of 8 rad/s and a
counterclockwise angular acceleration of 40 rad/s2.
Determine: a) tangential force exerted by gear D, and b) components
of the reaction at shaft O.
23
Sample Problem 16.6
F  m a m a
F  m a m a
M  I   I 
x
E
Ex
OB OBx
y
E
Ey
OB OBy
G
E
OB
Rx  3  40  0.2  24 N 
Ry  (3  9.81)  (4  9.81)  F  3  (8) 2  0.2
mE  4 kg
k E  85 mm
mOB  3 kg
  40 rad s2
  8 rad/s
Ry  F  107.07
0.12 F  4  (0.085) 2  40  ( 121  3  0.42  3  0.22 )  40
F  62.97 N 
Ry  170.04 N 
24
Sample Problem 16.7
If pin B is suddenly removed, find the angular acceleration of
the plate and the pin reactions. Very similar to sample
problem 16.2, please read on your own.
25
Sample Problem 16.8

A sphere of weight W is released with no initial velocity and rolls without
slipping on the incline.
Determine: a) the minimum value of the coefficient of friction, b) the
velocity of G after the sphere has rolled 10 ft and c) the velocity of G if
the sphere were to move 10 ft down a frictionless incline.
26
Sample Problem 16.8
For rolling:
M
C
a  r
 I C
 mg sin   r  
2 2

mr  mr 2  
5

7
g sin   r
5
F
x
 max
mg sin   F  m( r )
But F   N   mg cos 
For a sphere:
IG 
I C  I G  mr 2 
mg sin    mg cos   m( r )
5
sin    cos   sin 
7
 5
 cos   sin  1  
 7
2 2
mr
5

2 2
mr  mr 2
5
2
tan   0.165
7
27
Sample Problem 16.8
a  r 
5
g sin   11.5 ft / s 2
7
(uniform)
v 2  v02  2a  x  x0 
 0  2 11.5 10   230
v  15.17ft s
For no friction, no rolling (pure sliding)
mg sin   ma
a  g sin   16.1 ft / s
v 2  v02  2a  x  x0 
 0  2 16.110   322
v  17.94ft s
28
Sample Problem 16.9
A cord is wrapped around the inner hub of a wheel and pulled
horizontally with a force of 200 N. The wheel has a mass of 50 kg
and a radius of gyration of 70 mm. Knowing s = 0.20 and k = 0.15,
determine the acceleration of G and the angular acceleration of the
wheel.
29
Sample Problem 16.9
Do we have pure rolling or rolling + sliding?
If we assume pure rolling:
M
C
 IC
 200  0.04    0.245  50  0.12  
  10.74 rad s 2
a   r  10.74  0.1  1.074 m s 2
I  mk 2  50 kg 0.70 m 2
 0.245 kg  m 2
F
x
 max
F  200  50 1.074
F  146.3N
F
y
 may
N W  0
N  mg  50 1.074  490.5 N
30
Sample Problem 16.9
• Compare the required tangential reaction to the
maximum possible friction force.
Fmax   s N  0.20490.5 N   98.1 N
F > Fmax , rolling without slipping is impossible.
Without slipping,
F  146.3 N N  490.5 N
• Calculate the friction force with slipping and solve the
equations for linear and angular accelerations.
F  Fk   k N  0.15490.5 N   73.6 N
F
x
 max
200  73.6  50 a
M
G
a  2.53m s2
 IG
 73.6  0.100    200  0.0.06
  0.245 
  18.94 rad s 2
31
Sample Problem 16.10
SOLUTION:
• Based on the kinematics of the constrained
motion, express the accelerations of A, B,
and G in terms of the angular acceleration.
The extremities of a 4-ft rod
weighing 50 lb can move freely and
with no friction along two straight
tracks. The rod is released with no
velocity from the position shown.
• Draw the free-body-equation for the rod,
expressing the equivalence of the
external and effective forces.
• Solve the three corresponding scalar
equations for the angular acceleration and
the reactions at A and B.
Determine: a) the angular
acceleration of the rod, and b) the
reactions at A and B.
32
Sample Problem 16.10
SOLUTION:
• Based on the kinematics of the constrained motion,
express the accelerations of A, B, and G in terms of
the angular acceleration.
Express the acceleration of B as



aB  a A  aB A
With aB A  4 , the corresponding vector triangle and
the law of signs yields
a A  5.46
aB  4.90
The acceleration of G is now obtained from
 


a a G  a A  aG A where aG A  2
Resolving into x and y components,
ax  5.46  2 cos 60  4.46
a y  2 sin 60  1.732
33
Sample Problem 16.10
• Draw the free-body-equation for the rod, expressing
the equivalence of the external and effective forces.
• Solve the three corresponding scalar equations for the
angular acceleration and the reactions at A and B.
 M E   M E eff
501.732  6.93 4.46  2.69 1.732  2.07
  2.30 rad s 2
1 ml 2 
I  12
1 50 lb
2


4
ft
12 32.2 ft s 2
 2.07 lb  ft  s 2
I   2.07
50
4.46   6.93
ma x 
32.2
50
1.732   2.69
ma y  
32.2
  2.30 rad s2
 Fx   Fx eff
RB sin 45  6.932.30
RB  22.5 lb
 Fy   Fy eff

RB  22.5 lb
45o
RA  22.5cos 45  50  2.692.30
RA  27.9 lb
34