PHY 53
Summer 2010
Assignment 10
1.
Two identical point masses, each of mass m, are attached
by a massless horizontal rod of length  as shown. They
are rotating at angular velocity ω about a fixed axle
ω
perpendicular to the rod, at distance d from its center.
a.
What is the moment of inertia of the system about
the axle? Ans: m
(
1 2

2
)
+ 2d 2 .
(
1 2

2
)
+ 2d 2 ⋅ ω .
b.
What is the magnitude of the angular momentum? Ans: m
c.
If the system consisted of both masses at the CM, rotating about this axle,
what would be the angular momentum? Ans: 2md 2ω .
2.
d.
What is the angular momentum of the actual system about its CM. Ans:
1
m 2ω .
2
e.
Verify that the theorem that Ltot = rCM × Mv CM + L(about CM) is obeyed.
The same two masses and rod are now made to rotate about
the axle with the rod making angle α with the axle, and the
ω
axle passing through the CM.
a.
α
At the instant shown, find the horizontal and vertical
components of the angular momentum. [Use
L = r × mv for each mass.] Ans:
Lx = −Lcos α , Ly = Lsin α , where L = 12 m 2 sin α ⋅ ω .
b.
What is the moment of inertia I of the system about the axle? Ans:
1
m 2 sin 2 α
2
.
c.
Show that L = Iω .
d.
If the axle has no friction it cannot exert a torque along its length. Discuss
conservation of angular momentum for this system.
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PHY 53
3.
Summer 2010
A child of mass m is dropped from rest (gently) onto the rim
of a rotating carousel, a horizontal circular platform of
moment of inertia I about its axle. As shown from above,
the carousel was rotating at angular speed ω 0 before the
•
child landed on it at distance r from the axle.
a.
What was the kinetic energy of the carousel before
the child arrived? Ans: 12 Iω 02 .
b.
What is conserved in the “collision” of the child and the carousel?
c.
After the child comes to rest relative to the carousel, what is the angular
I
speed of the system? Ans:
ω0 .
I + mr 2
d.
What is the kinetic energy? Ans:
e.
The child crawls toward the axle, stopping at distance r/2 from it. Now
what is the kinetic energy? Ans:
f.
1
2
1
2
I2
I + mr
2
ω 02 .
I2
2
I + mr / 4
ω 02 .
Explain the change. That is, what work was done and by what agent?
4.
Two ice skaters, of mass m and 2m, are holding the ends of a
massless rope and skating around in circles, with the CM of the
system motionless. When they are as shown (from above) the
heavier skater lets go and moves off to the right with speed v.
Comment on the validity of each statement.
m
2m
v
a.
Total angular momentum about the center of the rope is
conserved.
b.
The CM of the system moves to the right.
c.
The energy of the system is reduced.
d.
The second skater moves in a straight line to the left with the same speed
as the first one.
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PHY 53
5.
Summer 2010
Shown from above is an object of mass M lying on a
frictionless horizontal surface. It is pivoted about a vertical
frictionless axle through the point shown, and it has moment
of inertia I about that axle. A small ball of mass m moving as
shown with velocity v 0 collides with the object, at distance d
below the pivot. The collision is elastic, and the ball moves
straight back.
d
v
a. What quantities are conserved in the collision?
b.
What is the angular speed ω of the object of mass M after the collision?
c.
m
M
[To simplify the algebra, let α = I /md 2 .]
Ans: ω =
6.
•
2v0
.
d(α + 1)
What is the final speed of v the ball? Ans: v =
v0 (α − 1)
α +1
In the previous problem, we will examine the change in linear momentum during
the collision. The CM of the object is at distance R below the pivot.
a.
What is the velocity of the object’s CM immediately after the collision?
2Rv0
Ans: vCM =
.
d(α + 1)
b.
What is the change in horizontal momentum during the collision? Ans:
2v0
ΔPtot =
(MR − I /d)
d(α + 1)
c.
What external force causes this change?
d.
Find the value of d that will make this force (and therefore the change in
horizontal momentum) zero. Ans: d = I / MR .
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PHY 53
7.
Summer 2010
The bowling ball problem in Assignment 9 can be
analyzed using conservation of angular
momentum. Initially it is moving as shown with
CM speed v0 , but not rotating, so it is slipping on
•
P
the alley surface. Eventually friction will slow
down its CM speed vCM and increase its angular speed ω until it is rolling with
vCM = Rω . We wish to find these final values.
a.
We consider torques and angular momentum about a point on the alley
surface, such as P. Show that the total torque due to the forces on the ball
is zero about that point. b.
Consider the total angular momentum about that point. What is it initially
when the ball is not rotating? [Use Ltot = rCM × Mv CM + L(about CM) .] Ans:
mRv0 , clockwise.
8.
c.
What is it finally, in terms of v? [Use the rolling condition and the value of
the moment of inertia I = 52 mR 2 .] Ans: 75 mRv .
d.
Use the result of (a) to determine the final values of vCM and ω .
Some questions about vehicles and rotational stabilitys.
a.
A truck is shown from the rear as it is rounding a
curve. The center of the curve is the the left,
•
indicated by the arrow. The CM of the truck is
indicated by the dot. As the speed of the truck is
increased, friction with the road is sufficient to keep it from slipping, but it
rolls over. Discuss the condition under which this happens, in terms of the
direction of g eff in the reference frame of the truck.
b.
A bicycle rider is moving to the left as shown The
system’s CM is indicated by the dot.. He applies his
brakes very forcefully, and finds himself and his
bike suddenly rotated about the point of contact
of the front wheel with the road. Explain.
c.
•
Explain why a bike rider must lean himself
and the bike toward the center of a curve to round it. Use the direction of
g eff in the reference frame of the bike in your explanation.
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PHY 53
9.
Summer 2010
Shown is a classroom demonstration system, consisting of a leaded bicycle wheel
with an axle attached to a string. The wheel is set into fast rotation with the axle
horizontal, and then it is suspended by the string as shown.
T
T
L0
L0
Perspective
T
L0
Side
Top
Instead of falling, the wheel turns in a horizontal circle with the point where the
sting attaches to the axle as center. The counter-clockwise spin of the wheel has
angular momentum L0 along the axle as shown. The length of the axle is R. The
system does not move vertically, so the tension T is equal to the weight mg.
a.
Refer to the top view. What is the torque τ (magnitude and direction)
produced by T about the center of the wheel?
b.
In time Δt this external torque will produce a change ΔL = τ Δt in the
angular momentum of the system. On the top view, draw an arrow
indicating the total angular momentum L = L0 + ΔL .
c.
Because L0 is large, the axle turns to match the direction of L. This turning
is precession. Which way (in top view) does the axle rotate?
d.
Through what angle Δθ does the axle turn in time Δt ? [Use small angle
approximations.]
e.
What is the angular speed ω p = Δθ / Δt of the precession, in terms of L0 ,
m, g and R?
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