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68 ••
Picture the Problem. Because there are no external
forces or torques acting on the system defined in the
statement of Problem 67, both linear and angular
momentum are conserved in the collision and the
velocity of the center of mass after the collision is the
same as before the collision. Kinetic energy is also
conserved as the collision of the hard sphere with the
bar is elastic. Let the direction the sphere is moving
initially be the positive x direction and toward the
top of the page in the figure be the positive y
direction and v and V be the final velocities of the
objects whose masses are m and M, respectively.
Apply
conservation of
linear momentum
to obtain:
Apply
conservation of
angular
momentum to
obtain:
pi  pf
or
mv  mv'MV '
(1)
Li  Lf
or
mvd  mv' d  121 ML2
(2)
Set v = 0 in
equation (1) and
solve for V :
Use conservation
of mechanical
energy to relate
the kinetic
energies of
translation and
rotation before
and after the
elastic collision:
Substitute (2) and
(3) in (4) and
simplify to
obtain:
Solve for d:
mv
V' 
M
(3)
Ki  Kf
or
2
2
1
1
mv

MV
'
 12
2
2
(4)

m 12m  d 2 
 2 
1

M
M L 
d L
M m
12m
1
12

ML2  2
70 ••
Picture the Problem Let
the zero of gravitational
potential energy be a
distance x below the
pivot as shown in the
diagram. Because the
net external torque
acting on the system is
zero, angular
momentum is conserved
in this perfectly inelastic
collision. We can also
use conservation of
mechanical energy to
relate the initial kinetic
energy of the system
after the collision to its
potential energy at the
top of its swing.
Using conservation of
mechanical energy,
relate the rotational
kinetic energy of the
system just after the
collision to its
gravitational potential
energy when it has
swung through an
angle  :
Kf  Ki  Uf  Ui  0
or, because Kf = Ui = 0,
 Ki  U f  0
and
d


2
1
I


Mg
 mgx 1  cos  

2
2


(1)
Note: They are using ½ I 2
instead of ½ mv2 because this
time, it’s not just a pt mass, it’s
a rod,  it has an “I” (I is
different than for a point mass)
due to the fact that the distance
is different  PE is different.
Note: They are using d/2
because that is the com for the
uniform rod. What is cos 600?
(0.5 )
Note: 1 – cos  comes from the
triangle drawn as shown: x is
the radius, thus, both the
original vertical length and the
hypotenuse are = x. The length
of the dashed line will then be x
– x cos . When you later factor
out x, you get
1 – cos .

Apply conservation of
momentum to the
collision:
Li  Lf
or
0.8dmv  I


 13 Md 2  0.8d  m 
2
Solve for  to obtain:
0.8dmv
 1
2
2
Md

0
.
64
md
2
(2)
Express the moment of
inertia of the system
about the pivot:
I  m0.8d   13 Md 2
2
 0.64m  13 M  d 2
 0.640.3 kg   13 0.8 kg 1.2 m 
2
 0.660 kg  m 2
Substitute equation (2)
in equation (1) and
simplify to obtain:
Solve for v:
d


 Mg  0.8dmg 1  cos  
2


0.32dmv 

I
2
v
g 0.5 M  0.8m1  cos I
0.32dm2
Substitute numerical values and evaluate v for  = 60 to
obtain:
v
9.81m/s  0.5 0.8 kg   0.8 0.3 kg 0.50.660 kg  m  
2
2
0.32 1.2 m0.3 kg 
2
72 ••
Picture the Problem. Because the net external
7.74 m/s
torque acting on the system is zero, angular
momentum is conserved in this perfectly inelastic
collision.
(a) Use its definition
to express the total
angular momentum
of the disk and
projectile just before
impact:
L0  mp v0 b
(b) Use conservation
of angular
momentum to relate
the angular
momenta just before
and just after the
collision:
L0  L  I
Express the moment
of inertia of the disk
+ projectile:
Substitute for I in
the expression for 
to obtain:

and
L0
I
I  12 MR 2  mp b 2

2m p v 0 b
MR 2  2mp b 2
(c) Express the
kinetic energy of the
system after impact
in terms of its
angular momentum:
m v b
L2
Kf 
 1 p 20
2 I 2 2 MR  mpb 2
(d) Express the
difference between
the initial and final
kinetic energies,
substitute, and
simplify to obtain:
E  K i  K f
2


m v b

2
p 0
MR 2  2mpb 2
 12 mp v02 
m v b
2
p 0
MR 2  2mpb 2
2


m
b
p

 12 m v 1 
 MR 2  2m b 2 
p


2
p 0
80 •
Picture the Problem The angular momentum of the particle is
  
given by L  r  p where r is its position vector and p is its
linear momentum. The torque acting on the particle is given


by τ  dL dt .
Express the angular
momentum of the
particle:
Evaluate

dr
dt
:
   

 
L  r  p  r  mv  mr  v

 dr
 mr 
dt

dr
 6tˆj
dt
Substitute and simplify to

find L :


Find the torque due to the
force:

 

L  3 kg  4 m  iˆ  3t 2 m/s 2 ˆj
 6t m/s  ˆj

72.0t J  s  kˆ

 dL d
72.0t J  s  kˆ
τ

dt dt
 72.0 N  m  kˆ


90 ••
#90 Explanation / justification. You cannot just treat
the disks inside of the cylinder as point masses. Each
point in / on the disk is a different distance away
from the axis of rotation.
You can, however, use calc to solve a problem like this.
Think of it as  r m, for each of the r’s in / on the disk.
OR
You can also think of the resultant I as Icm + other I’s
(specifically, the disks), which is the justification for
use of the parallel axis theorem.
Picture the Problem. Let x be the radial distance each disk
moves outward. Because the net torque acting on the
system is zero, we can use conservation of angular
momentum to relate the initial and final angular velocities
to the initial and final moments of inertia. We’ll assume
that the disks are thin enough so that we can ignore their
lengths in expressing their moments of inertia.
Use conservation
of angular
momentum to
relate the initial
and final angular
velocities of the
disks:
Solve for f:
Li  Lf
or
I i i  I f  f
f 
Ii
i
If
(1)
Express the
initial moment of
inertia of the
system:
I i  I cyl  2I disk
Express the
moment of inertia
of the cylinder:
I cyl  121 ML2  12 MR 2

 121 M L2  6 R 2


 121 0.8 kg 1.8 m   60.2 m 
 0.232 kg  m 2
2
2

Letting
 represent the
distance of the
clamped disks
from the center of
rotation and
ignoring the
thickness of each
disk (we’re told
they are thin), use
the parallel-axis
theorem to
express the
moment of inertia
of each disk:
With the disks
clamped:
I disk  14 mr 2  m 2

 14 m r 2  4 2

1
4

0.2 kg 0.2 m 2  40.4 m 2 
 0.0340 kg  m 2
I i  I cyl  2 I disk

 0.232 kg  m 2  2 0.0340 kg  m 2

 0.300 kg  m 2
With the disks
unclamped,  =
0.6 m and:
Express and
evaluate the final
moment of inertia
of the system:

I disk  14 m r 2  4 2

1
4

0.2 kg 0.2 m2  40.6 m2 
 0.0740 kg  m 2
I f  I cyl  2 I disk

 0.232 kg  m 2  2 0.0740 kg  m 2
 0.380 kg  m 2

Substitute in
equation (1) to
determine f:
0.300 kg  m 2
8 rad/s 
f 
2
0.380 kg  m
Express the
energy dissipated
in friction:
E  Ei  Ef
Apply Newton’s
2nd law to each
disk when they
are in their final
positions:
2
F

kx

mr

 radial
 6.32 rad/s
 12 I i i2 

1
2
I f  f2  12 kx2

Note: This should really say
F radially is caused by the force on the
spring. That part is from Hooke’s
Law. You are using N2 on the RHS
only. They then use the eq to solve for
k.
Solve for k:
mr 2
k
x
Substitute
numerical values
and evaluate k:
2

0.2 kg 0.6 m 6.32 rad/s 
k
0.2 m
 24.0 N/m
Express the energy
Wfr  Ei  Ef
dissipated in friction:
2
 12 I ii 

1
2
I f f2  12 kx2

Substitute numerical values and evaluate Wfr:
Wfr 
1
2
0.300 kg  m 8 rad/s 
2
2


 12 0.380 kg  m2 6.32 rad/s   12 24 N/m0.2 m
2
2
 1.53 J
84 ••
Picture the Problem Because the net torque acting on the system is zero; we can use
conservation of angular momentum to relate the initial and final angular velocities of the
system.
Using conservation of angular
momentum, relate the initial and
final angular velocities to the initial
and final moments of inertia:
Li  Lf
or
I i i  I f  f
Solve for i:
f 
Express Ii:
I i  101 ML2  2 14 m 2
Express If:
I f  101 ML2  2 14 mL2
Substitute to express  f in terms of  :
f 

Express the initial kinetic energy of
the system:
Ii
I
i  i 
If
If
1
10
1
10





 2

mL 
ML2  2 14 m 2
2
ML
2
L2 
M  5m
M  5m
Ki  12 I i 2 

1
4
2
1
20
ML
2

1 1
2 10


ML2  2 14 m 2  2

 5m 2  2
Express the final kinetic energy of the system and simplify to obtain:
K f  12 I f f2 

1 1
2 10


ML2  2 14 mL2 f2 
2


1
20
1
20

2 
 M  5m 2 
L  
ML2  5mL2 
 M  5m






1
20


ML
2

 5mL2 f2
2

2  
  ML  5m  
L  2
1 
20 

M  5m





 ML2  5m 2 2  2


2
2 
 ML  5mL 
85 ••
Determine the Concept Yes. The net external torque is zero and angular momentum is
conserved as the system evolves from its initial to its final state. Because the disks come to the
same final position, the initial and final configurations are the same as in Problem 84. Therefore,
the answers are the same as for Problem 84.