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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.01
Problem Solving Session 9
Torque and Rotational and Translational Motion Solutions
Section
______________________
Table
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Group Members
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Hand in one set of solutions per group.
IC-W11D3-1 Group Problem 1 Collision with Hanging Pivoted Rod Solution
An object of mass m and speed v0 strikes a rigid uniform rod of length l and mass mr
that is hanging by a frictionless pivot from the ceiling. Immediately after striking the rod,
the object continues forward but its speed decreases to v0 /2. The moment of inertia of the
rod about its center of mass is I cm  (1/12) mr l 2 . Gravity acts with acceleration g
downward. For what value of v0 will the rod just touch the ceiling on its first swing?
Solution:
We begin by identifying our system which consists of the object and the uniform rod. We
identify as well three states:
State 1: Immediately before the collision
State 2: Immediately after the collision.
State 3: The instant the rod touches the ceiling when the final angular speed is zero.
and two state changes:
State 1  State 2
State 2  State 3
We would like to know if any of our fundamental quantities, momentum, energy, angular
momentum about some point are constant during these state changes.
We start with State 1  State 2. The pivot force holding the rod to the ceiling is an
external force acting at the pivot point S . There is also the gravitational force acting on
at the center of mass of the rod and on the object. There are also internal forces due to the
collision of the rod and the object at point A .
The external force means that momentum is not constant. However the external force is
fixed and so does no work. However, we do not know whether the collision is elastic or
not so we cannot assume that mechanical energy is constant. If we choose the pivot point
S as the point in which to calculate torque, then the torque about the pivot is
r
r
r
r
r
r
r
r
r
 sys
 rS ,S  Fpivot  rS , A  Fo,r  rS , A  Fr ,o  rS ,cm  Fr ,g
S
r
r
The external pivot force does not contribute any torque because rS ,S  0 . The internal
forces between the rod and the object are equal in magnitude and opposite in direction,
r
r
Fo,r  Fr ,o (Newton’s Third Law), and so their contributions to the torque add to zero,
r
r
r
r
r
rS , A  Fo,r  rS , A  Fr ,o  0 . If the collision is instantaneous then the gravitational force is
r
r
r
r
parallel to rS ,cm and so rS ,cm  Fr ,g  0 . Therefore the torque on the system about the pivot
point is zero,
r
r
 sys
 0.
S
Thus the angular momentum about the pivot point is constant,
r
r sys
(Lsys
)

(
L
) .
S 1
S 2
In order to calculate the angular momentum we draw a diagram showing the momentum
of the object and the angular speed of the rod in the figure below.
The angular momentum about S immediately before the collision is
r
r
r
(Lsys
)  rS ,0  m1v 0  l( ĵ)  m1v0 î  lm1v0k̂ .
S 1
The angular momentum about S immediately after the collision is
r
lm v
r
r
r
(Lsys
)  rS ,0  m1v 0 / 2  I s 2  l( ĵ)  m1 (v0 / 2) î  1 0 k̂  I s 2k̂ .
S 2
2
Therefore the condition that the angular momentum about S is constant during the
collision becomes
lm v
lm1v0k̂  1 0 k̂  I s 2k̂ .
2
We can solve for the angular speed immediately after the collision
2 
lm1v0
2I s
.
By the parallel axis theorem the moment of inertial of a uniform rod about the pivot point
is
I S  m(l / 2)2  Icm  (1/ 4)mr l 2  (1/ 12)mr l 2  (1/ 3)mr l 2 .
Therefore the angular speed immediately after the collision is
2 
3m1v0
2mr l
.
For the transition State 2  State 3, we know that the gravitational force is conservative
and the pivot force does no work so mechanical energy is constant.
Emech,2  Emech,3
We draw an energy diagram with a choice of zero for the potential energy at the center of
mass. We only show the rod because the object undergoes no energy transformation
during the transition State 2  State 3.
The mechanical energy immediately after the collision is
Emech,2 
1
1
I S 22  m1 (v0 / 2)2 .
2
2
Using our results for the moment of inertia I S and  2 , we have that
2
Emech,2


3m12 v0 2 1
1
1
2 3m1v0
2
2
 (1 / 3)mr l 
  2 m1 (v0 / 2)  8m  2 m1 (v0 / 2) .
2
2m
l

r 
r
The mechanical energy when the rod just reaches the ceiling when the final angular speed
is zero is then
1
Emech,3  mr g(l / 2)  m1 (v0 / 2)2 .
2
Then the condition that the mechanical energy is constant becomes
3m12 v0 2
8mr
1
1
 m1 (v0 / 2)2  mr g(l / 2)  m1 (v0 / 2)2 .
2
2
We can now solve this equation for the initial speed of the object
v0 
mr
m1
4gl
.
3
IC-W11D3-2 Group Problem 2
A drum A of mass m and radius R is suspended from a drum B also of mass m and
radius R , which is free to rotate about its axis. The suspension is in the form of a
massless metal tape wound around the outside of each drum, and free to unwind. Gravity
is directed downwards. Both drums are initially at rest. Find the initial acceleration of
drum A , assuming that it moves straight down.
Solution:
The key to solving this problem is to determine the relation between the three kinematic
quantities  A ,  B and a A , the angular accelerations of the two drums and the linear
acceleration of drum A . One way to do this is to introduce the auxiliary variable z for
d 2z
the length of the tape that is unwound from the upper drum. Then,  B R  2 . The
dt
dz
linear velocity v A may then be expressed as the sum of two terms, the rate
at which
dt
the tape is unwinding from the upper drum and the rate  A R at which the falling drum is
moving relative to the lower end of the tape. Taking derivatives, we obtain
d 2z
aA  2   A R   B R   A R .
dt
Denote the tension in the tape as (what else) T . The net torque on the upper drum about
its center is then  B  TR , directed clockwise in the figure, and the net torque on the
falling drum about its center is also  A  TR , also directed clockwise. Thus,
 B  TR / I  2T / MR ,  A  TR / I  2T / MR . Where we have assumed that the moment
of inertia of the drum and unwinding tape is I  (1/ 2) MR 2 . Newton’s Second Law,
applied to the falling drum, with the positive direction downward, is Mg  T  MaA . We
now have five equations,
B R 
d 2z
d 2z
2T
2T
,
a

  A R,  B 
, A 
, Mg  T  Ma A ,
A
2
2
dt
dt
MR
MR
in the five unknowns  A ,  B , a A ,
d 2z
and T .
dt 2
It’s easy to see that
 A  B .
Therefore
aA   B R   A R  2 A R .
The tension in the tape is then
T
 A MR
2

a A MR Ma A

4R
4
Newton’s Second Law then becomes
Mg 
Ma A
 Ma A .
4
Therefore solving for the acceleration yields
aA 
4
g
5
This result is certainly plausible. We expect a A  g , and we also expect that with both
drums free to rotate, the acceleration will be almost but not quite g .
IC-W11D3-3 Group Problem 3 Sections L05-07
A bowling ball of mass m and radius R is initially thrown down an alley with an initial
speed v0 and backspin with angular speed  0 , such that v0  R  0 . The moment of
inertia of the ball about its center of mass is Icm  (2 / 5)mR2 . Your goal is to determine
the speed vf of the bowling ball when it just starts to roll without slipping. What is the
speed vf of the bowling ball when it just starts to roll without slipping?
Solution: We begin by coordinates for our angular and linear motion. Choose an angular
coordinate  increasing in the clockwise direction. Choose positive k̂ unit vector
pointing into the page. Then the angular velocity vector is defined to be a
r
d
   z k̂ 
k̂
dt
and the angular acceleration vector is defined to be
r
d 2
   z k̂  2 k̂ .
dt
Choose the positive î unit vector pointing to the right in the figure. Then the velocity of
the center of mass is given by
dx
r
v cm  vcm,x î  cm î
dt
and the acceleration of the center of mass is given by
d 2 xcm
r
a cm  acm,x î 
î .
dt 2
The free body force diagram is shown in the figure below.
r
r
At t  0 , when the ball is released, v cm,0  v0 î and  0   0k̂ . The velocity of a point
on the rim of the wheel that is in contact with the ground is then
r
r
r
v contact ,0  v cm,0  v tan,0  (v0  R 0 ) î
(1)
so the wheel is sliding forward with respect to the ground and hence the friction force on
the wheel opposing the motion is kinetic friction and acts in the negative î -direction.
Our “rule to live by” for rotational motion is that
r
r ext dL S
S 
dt
(2)
In order for angular momentum about some point to remain constant throughout the
motion, the torque about that point must also be zero throughout the motion.
Recall that the torque about a point S is define as
r
r
r
 ext
  rS ,i  Fiext
S
i
r
r
where rS ,i is the vector form the point S to where the ith force Fiext acts on the object. As
the ball moves down the alley, the contact point will move, but the friction force will
always be parallel to the line of contact between the bowling bowl and the surface. So, if
we pick any fixed point S along the line of contact between the bowling bowl and the
surface then
r
r
r
r
 S , f  rS , f  fk  0
k
k
r
r
because rS , f and f k are anti-parallel. The gravitation force acts at the center of mass
k
hence the torque due to gravity about the point S is
r
r
r
 S ,mg  rS ,mg  mg  dmgk̂
where d is the distance from the point S to the contact point between the wheel and the
ground.
The torque due to the normal force about the point S is
r
r
r
 S , N  rS , N  mg  dNk̂
with the same moment arm d . Because the wheel is not accelerating in the ĵ -direction,
from Newton’s Second Law, we note that
mg  N  0
Therefore
r
r
r
 S , N   S ,mg  d(mg  N )k̂  0
Hence, there is no torque about the point S and the angular momentum about the point
S is constant,
r
r
(3)
L S ,0  L S , f
The initial angular momentum about the point S
is due to the translation of the center of mass and rotation about the center of mass
r
r
r
r
L S ,0  rS ,cm,0  mv cm,0  I cm   m R vcm,0k̂  I cm 0k̂ .
(4)
The final angular momentum about the point S has both a translational and rotational
contribution
r
r
r
r
L S , f  rS ,cm, f  mv cm, f  I cm f  m R vcm, f k̂  I cm z, f k̂ ,
(5)
When the wheel is rolling without slipping,
vcm, f  R z, f
(6)
and also Icm  (2 / 5)m R2 . Therefore the final angular momentum about the point S is
r
L S , f  (m R  (2 / 5)m R) vcm, f k̂  (7 / 5)m R vcm, f k̂ .
(7)
Equating the z-components in Equations (4) and (7) yields
m R vcm,0  (2 / 5)m R 2 0  (7 / 5)m R vcm, f ,
(8)
vcm, f  (5 / 7)vcm,0  (2 / 7) R 0 .
(9)
which we can solve for