# Document

```ENGR 214
Chapter 17
Plane Motion of Rigid Bodies:
Energy & Momentum Methods
All figures taken from Vector Mechanics for Engineers: Dynamics, Beer
and Johnston, 2004
1
Principle of Work and Energy for a Rigid Body
T1  U 1 2  T 2
T1 , T 2 
initial and final total kinetic energy of rigid body
U 1 2 
total work of external forces acting rigid body
A2
Work done:
  s2
U 1 2   F  d r    F cos  ds
A1
s1
 
 
 
For a couple: dU  F  d r1  F  d r1  F  d r2
 F ds 2  Fr d 
 M d
2
If M is constant:
U 1 2   M d 
1
 M  2   1 
2
Kinetic Energy of a Rigid Body in Plane Motion
Plane motion
combination of translation & rotation
G
G
T 
vG
1
2
mv 
2
G
I G
1
2
2
For non-centroidal rotation:
I G 
T 
1
2
mv 

1
2
 IG  mr
2
G
1
2
2
2
T 

1
2
1
2
m  r  
2
1
2
I G
2
2
I O
2
3
Principle of Work and Energy: System of Rigid Bodies
T1  U 1 2  T 2
Can be applied to each body separately or to the system as a
whole
T1 , T 2  sum of kinetic energies of all bodies in system
U 1 2 
work of all external forces acting on system
Useful in problems involving several bodies connected
together by pins, inextensible chords, etc. because internal
forces do no work & U 1 2 reduces to the work of external
forces only.
4
Conservation of Energy
T1  V1  T 2  V 2
Example: rod is released with zero velocity from horizontal
position. Determine angular velocity after rod has rotated .
0.5 l
T1  0
V1  0
T2 

I 2
1
2
m v2 
1
2
m  12 l  
2
1
2
2
2
V 2   m gl sin 
1
2
1
2

1
12
ml
2

2

1 ml
2

2
2 3
5
T1  V1  T 2  V 2
0
1 ml
2 3
2
  12 m gl sin 
2
 3g

 
sin  
 l

1 2
6
Power
Power = rate at which work is done
 
dU
Power 
 F v
dt
For a rotating body:
Power 
dU
dt

M d
 M
dt
7
Sample Problem 17.1
A 240-lb block is suspended from an
inextensible cable which is wrapped
around a drum of 1.25-ft radius
attached to a flywheel. The drum and
flywheel have a combined moment of
inertia I=10.5 lb ft s2. At the instant
shown, the velocity of the block is 6
ft/s downward.
Knowing that the bearing friction is equivalent to a couple of
magnitude 60 lb ft, determine the velocity of the block after it has
moved 4 ft downward.
8
Sample Problem 17.1
T1  U 1 2  T 2
T1 
T1 
m v1 
2
1
2
1 240
2 3 2 .2
6
2
1
2
with v 
I 1
2
 6 
  1 0 .5  

2
 1 .2 5 
1
r
2
 2 5 5 .1 2 ft  lb
with x  r
U 1 2  m gh  M   2   1 
 4 
U 1 2   240   4    60  

 1.25 
 768 ft  lb
T2 
1
2
m v2 
2
1
2
I2
2
2
 v 
2
2

v 2  10.5  2   7.09 v 2
2 32.2
2
 1.25 
1 240
1
255.12  768  7.09 v 2
2
v 2  12.01 ft s
9
Sample Problem 17.2
m A  10 kg
k A  200 m m
m B  3 kg
k B  80 m m
The system is at rest when a moment M=6 Nm is applied to gear B.
Neglecting friction, a) determine the number of revolutions of gear B
before its angular velocity reaches 600 rpm, and b) tangential force
exerted by gear B on gear A.
10
Sample Problem 17.2
T1  U 1 2  T2
T1  0
U 1  2  6
T2 
But  B  2.5 A
B 
600  2 
 62.83 rad s
60
T2 
1
2
1
2
 A  62.83
2.5
 25.13 rad s
2
1
2
I B B
2
 I  I   2 .5  2   2
B
 A
 A
I A  m A k A   1 0   0 .2   0 .4 k g  m
2
2
2
I B  m B k B   3   0 .0 8   0 .0 1 9 2 k g  m
2
1
I A A 
2
2
T2  164.23 N m
6  164.23    27.37 rad  4.35 rev
11
Sample Problem 17.2
For gear A:
T1  U 1 2  T 2
T1  0
A  B
T2 
1
2
rB
 27.37
rA
1
2.5
I A A 
2
1
2
 0.4   25.13   126.3 N m
2
U 1 2  F r A  F  0.25  10.95  2.74 F
T1  U 1 2  T 2
0  2.74 F  126.3
F  46 N
12
Sample Problem 17.3
A sphere, cylinder, and hoop, each having the same mass and radius,
are released from rest on an incline. Determine the velocity of each
body after it has rolled through a distance corresponding to a change of
elevation h.
13
Sample Problem 17.3
T1  U 1 2  T 2
T1  0
Friction force in rolling does no work
v  r
T2 
1
2
IC 
2
1
2
IC
U 1 2  m gh
v
2
r
2
T1  U 1 2  T 2
m gh 
v
2 m ghr
IC
2

1
2
IC
v
2
r
2
2 gh
I
1  O2
mr
14
Sample Problem 17.3
• Each of the bodies has a different centroidal
moment of inertia,
v
2 gh
I
1  O2
mr
Sphere :
IO 
2
5
mr
2
v  0.845 2 gh
C ylinder :
IO 
1
2
mr
2
v  0.816 2 gh
H oop :
IO  mr
2
v  0.707 2 gh
NOTE:
• For a frictionless block sliding through the
same distance,
  0, v 
2 gh
15
Can also be solved using
conservation of energy:
T1  V1  T 2  V 2
T1  0
V1  m g h
T2 
1
2
IC
2
V2  0
m gh 
1
2
IC
2
16
Sample Problem 17.4, SI units
1.524 m
0.305 m
A 13.608-kg slender rod pivots about the point O. The other
end is pressed against a spring (k = 315.212 kN/m) until the
spring is compressed 25.4 mm and the rod is in a horizontal
position. If the rod is released from this position, determine its
angular velocity and the reactions at the pivot as the rod passes
through a vertical position.
17
Sample Problem 17.4
T1  V1  T 2  V 2
T1  0
V1 
1
2
kx 
2
1
1
2
 315.212  10  25.4  10 
3
3
2
 101.68 N m
I 
1
12
ml
2
 121 m l 2  m r 2   2 


T2 
1
2
I O 

1
2
 121 13.608  1.524 2  13.608  0.457 2   2


2
 2.738
1
2
2
V 2  m gh  13.608  9.81  0.457
 61 N m
101.68  2.738  61
2
  3.85 rad / s
18
Sample Problem 17.4
To get pin reactions:
M
F
 I O
0  I O
 0
 m aGx
R x  m  r 
Rx  0
O
x
W
F
y
 m aGy
W  R y  m r
2
R y  m g  m r
2
 1 3 .6 0 8  9 .8 1  1 3 .6 0 8  3 .8 5  0 .4 5 7
2
 4 3 .3 6 N
19
Sample Problem 17.5
Each of the two slender rods has a
mass of 6 kg. The system is released
from rest with b = 60o.
Determine a) the angular velocity of
rod AB when b = 20o, and b) the
velocity of the point D at the same
instant.
20
Sample Problem 17.5
SOLUTION:
• Consider a system consisting of the two rods. With
the conservative weight force,
T1  V1  T 2  V 2
• Evaluate the initial and final potential energy.
V1  2Wy 1  2 58 . 86 N  0 . 325 m 
 38 . 26 J
V 2  2Wy 2  2 58 . 86 N  0 . 1283 m 
 15 . 10 J

W  mg   6 kg  9 . 81 m s
2

 58 . 86 N
21
Sample Problem 17.5
• Express the final kinetic energy of the system in terms
of the angular velocities of the rods.

v AB   0 . 375 m 


Since v B is perpendicular to AB and v D is horizontal,
the instantaneous center of rotation for rod BD is C.
CD  2  0 . 75 m  sin 20   0 . 513 m
BC  0 . 75 m
and applying the law of cosines to CDE, EC = 0.522 m
Consider the velocity of point B

v B   AB    BC  AB
 BD  

v BD   0 . 522 m 
For the final kinetic energy,
I AB  I BD 
2
1
12
ml
2
1
12
m v AB 

1
12
 6  0 . 375  2
 1 . 520 
1
12
2
6 kg 0 . 75 m 2
I AB  AB 
T2 
1
2


1
2
1
12
2
m v BD 
 0 . 281  2

1
12
1
2
 0 . 281 kg  m
2
2
I BD  BD
 6  0 . 522  2

1
2
 0 . 281  2
2
22
Sample Problem 17.5
• Solve the energy equation for the angular velocity,
then evaluate the velocity of the point D.
T1  V1  T 2  V 2
0  38 . 26 J  1.520 
2
 15 . 10 J
  3.90 rad s

 AB  3 . 90 rad s
v D  CD 
  0 . 513 m 3 . 90 rad s 
 2 . 00 m s

v D  2 . 00 m s
23
Principle of Impulse and Momentum
For a rigid body in general plane motion: F  m a G
F m
d
dt
 vG 
Fdt  m d  v G 
M G  I G
t2
t2
 F dt  m  v
G2
 vG 1 
m vG 1 
 Fdt  m v
G2
t1
t1
Can be split into 2 components (x and y)
M G  IG
d
dt
t2
 
M G dt  I G d   
M
G
dt  I G  2  I G  1
t1
t2
I G 1 
M
G
dt  I G  2
t1
24
Principle of Impulse and Momentum
For non-centroidal rotation
t2
I O1 
M
O
dt  I O  2
t1
25
Conservation of Angular Momentum
• When no external force acts on a rigid body or a system of rigid
bodies, the system of momenta at t1 is equipollent to the system
at t2. The total linear momentum and angular momentum about
any point are conserved,


L1  L 2
 H 0 1   H 0  2
• When the sum of the angular impulses pass through O, the
linear momentum may not be conserved, yet the angular
momentum about O is conserved,
 H 0 1   H 0  2
• Two additional equations may be written by summing x and
y components of momenta and may be used to determine
two unknown linear impulses, such as the impulses of the
reaction components at a fixed point.
26
Sample Problem 17.6
m A  10 kg
k A  200 m m
m B  3 kg
k B  80 m m
The system is at rest when a moment M=6 Nm is applied to gear B.
Neglecting friction, a) determine the time required for the angular
velocity of gear B to reach 600 rpm, and b) tangential force exerted by
gear B on gear A.
27
Sample Problem 17.6
I A  m A k   1 0   0 .2   0 .4 k g  m
2
A
t2
I G 1 
M
G
2
2
d t  I G  2 I  m k 2   3   0 .0 8  2  0 .0 1 9 2 k g  m 2
B
B B
A  B 
1
 B  600 
t1
 25.13 rad / s
2.5
2
 62.83 rad / s
60
Gear A:
0  F trA  I A   A  2
F t  0.25    0.4   25.13 
F t  40.21 N s
Gear B:
Solving yields: t  0.87 s , F  46.16 N
28
Sample Problem 17.7
v1
A uniform sphere of mass m and radius r is projected along a rough
horizontal surface with a linear velocity v1 and no angular velocity. The
coefficient of kinetic friction is  k .
Determine a) the time t2 at which the sphere will start rolling without
sliding and b) the linear and angular velocities of the sphere at time t2.
29
rolling
Sample Problem 17.7
t2
Along x:
m vG 1 
 Fdt  m v
G2
t1
m v1   k m gt  m  2 r
v2   2 r
v1   k gt   2 r
t2
Rotation about G: I G  1   M G dt  I G  2
t1
0   k m grt 
 k gt 
2
5
v1 
5
 2r
5
mr 2
2
r 2
Solving:
7
2
v1  r  2 
2 
5 v1
7 r
2
5
 2r
v2   2 r 
5
7
v1
t
2 v1
7 k g
30
Sample Problem 17.8
SOLUTION:
• Observing that none of the external
forces produce a moment about the y
axis, the angular momentum is
conserved.
Two solid spheres (radius = 3 in.,
W = 2 lb) are mounted on a spinning
horizontal rod ( I R  0.25 lb  ft  s 2 ,
 = 6 rad/sec) as shown. The balls are
held together by a string which is
suddenly cut. Determine a) angular
velocity of the rod after the balls have
moved to A’ and B’, and b) the energy
lost due to the plastic impact of the
spheres and stops.
• Equate the initial and final angular
momenta. Solve for the final angular
velocity.
• The energy lost due to the plastic impact
is equal to the change in kinetic energy
of the system.
31
Sample Problem 17.8
SOLUTION:
• Observing that none of the
external forces produce a
moment about the y axis, the
angular momentum is
conserved.
• Equate the initial and final
angular momenta. Solve for
the final angular velocity.
Sys Momenta1 + Sys Ext Imp1-2 = Sys Momenta2
2  m s r1 1 r1  I S  1   I R  1  2  m s r2  2 r2  I S  2   I R  2
2
 2  1
m s r1  I S  I R
2
m s r2  I S  I R
I R  0 . 25 lb  ft  s
 1  6 rad s
IS 
2
5
ma
2

2
5
2

 2  2
2 lb
2


ft   0 . 00155 lb  ft  s
2 

 32.2 ft s   12 
2
2
m S r1
 2  5 

    0 . 0108
 32 . 2   12 
2
2
m S r2
 2   25 


  0 . 2696
 32 . 2   12 
 2  2 . 08 rad s
32
Sample Problem 17.8
• The energy lost due to the
plastic impact is equal to the
change in kinetic energy of the
system.
 2  2 . 08 rad s
 1  6 rad s
I R  0 . 25 lb  ft  s
2
2
m S r1  0 . 0108 lb  ft  s
T  2
I S  0 . 00155 lb  ft  s
2
2
2
m S r2  0 . 2696 lb  ft  s
2
12 m S v 2  12 I S  2   12 I R  2  12 2 m S r 2  2 I S  I R  2
T1 
1
2
 0 . 275  6  2  4 . 95 ft  lb
T2 
1
2
 0 . 792  2 . 08  2  1 . 71 ft  lb
ΔT  T 2  T1  1 . 71  4 . 95
 T   3 . 24 ft  lb
33
Eccentric Impact


u A  n  u B  n
Period of deformation
Period of restitution

Impulse   P dt

Impulse   R dt
As for particles:
e  co efficien t o f restitu tio n 
Same relation applies for rigid bodies
Note: velocities are along line of impact
 Rdt
 Pdt

 v B  n
 v A n
 v B  n   v A  n
  v A  n
  vB
n
 e   v A  n   v B  n 
34
Eccentric Impact
If one or both of the colliding bodies rotates about a fixed
point O, an impulsive reaction will develop
35
Sample Problem 17.9
A 0.05-lb bullet is fired with a horizontal velocity of 1500 ft/s into the side
of a 20-lb square panel which is initially at rest. Determine a) the angular
velocity of the panel immediately after the bullet becomes embedded
and b) the impulsive reaction at A, assuming that the bullet becomes
embedded in 0.0006 s.
36
Sample Problem 17.9
t2
Impulse & momentum:
m vG 1 
 Fdt  m v
t2
t2
I G 1 
G2
t1
M
G
dt  I G  2
I A 1 
M
A
dt  I A 2
t1
t1
For entire system:
x components:
m B v B  Ax  t  m p v 2
 0 .0 5 
 20
1
5
0
0

A
0
.0
0
0
6





x



 3 2 .2 
 3 2 .2

 v2

m B vB
y components:
0  Ay  t  0
Then:
 1412   0 
I A  IG  m 
Ay  0
 0.05 
 20 

 1500   A x  0.0006   
  3.50 
 32.2 
 32.2 
A x   259 lb
IA 

1
6
9
12
I A 2

2
w ith
 m p b  m  129 
 m B vB
but
Solving:
b
2
 
14
12
v2   2
IG 
 16  m p b 2
2
2
2
   16  m p b  m  129    2


 192 
 2  4 .6 7 rad s
v2 
 192   2
 3 .5 0 ft s
37
Sample Problem 17.10
A 2-kg sphere with an initial velocity of 5 m/s strikes the lower end of an 8-kg
rod AB. The rod is hinged at A and initially at rest. The coefficient of
restitution between the rod and sphere is 0.8.
Determine the angular velocity of the rod and the velocity of the sphere
immediately after impact.
38
Sample Problem 17.10
t2
Impulse & momentum:
m vG 1 
 Fdt  m v
t2
t2
G2
I G 1 
t1
M
G
dt  I G  2
I A 1 
M
A
dt  I A 2
t1
t1
m s v s  1 .2   m s v s  1 .2   I A  
I A  I G  m R  0.6 
where
IG 
1
12
mL 
2
1
12
2
 8  1.2 
2
 0.96 kg  m
2
v R  r     0.6   
 2   5  1.2    2  v s 1.2    121  8  1.2 2  8  0.6 2   
+
v B  v s  e  v s  v B 
1.2     v s
 0.8  5 
4   1 .2     v s
12  2.4 v s  3.84  
Solving:    3 . 21 rad/s
v s   0 . 143 m s
39
Sample Problem 17.11
A square package of mass m moves down conveyor belt A with constant
velocity. At the end of the conveyor, the corner of the package strikes a
rigid support at B. The impact is perfectly plastic.
Derive an expression for the minimum velocity of conveyor belt A for
which the package will rotate about B and reach conveyor belt C.
40
Sample Problem 17.11
• Apply principle of impulse and momentum at impact (just before & just after
impact)
t2
I B1 
M
B
dt  I B  2
t1
 m v1   12 a   0 
I B 2
IG 
2

 2  
  IG  m 
a   2

 
2


 

 m v1   12 a  
2
3
ma 2
2
IB 
v1 
4
3
1
6
1
6
ma
2
ma 
2
2
4
ma 
2
2
3
ma
2
a 2
41
Sample Problem 17.11
• Apply principle of conservation of energy (just after
impact until maximum height)
T 2  V 2  T3  V 3
h 2  GB  sin  45   15  


2
2

T2 
1
2
I B 2

1
2

2
2
3
ma
2

2
2

1
3
ma 2
2
2
V 2  Wh 2
a sin 60   0 . 612 a
(solving for the minimum 2)
T3  0
V 3  Wh 3
1
3
2
2
2 
h3 
2
2
a  0 . 707 a
2
ma  2  Wh 2  0  Wh 3
v1 
4
3
3W
ma
 h3
2
a 2 
4
3
 h2  
3g
a
2
 0 . 707 a  0 . 612 a  
a 0 .2 8 5 g a
0 . 285 g a
v1  0.712
ga
42
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