PHY 53
Summer 2010
Assignment 9
1.
A heavy cylindrical container is being rolled up an
incline as shown, by applying a force parallel to the
incline. The static friction coefficient is µs . The
cylinder has radius R, mass m and moment of inertia
about its symmetry axis I.
F
d
a.
Draw the cylinder and the incline and make a free-body diagram showing
all the forces on the cylinder and where they are applied.
b.
Assume the motion up the incline is at constant speed, and that d > R .
What direction is the static friction force? How do you know?
c.
Assuming µs is large enough, or θ is small enough, to allow the cylinder
to roll without slipping, how large must F be? Ans: F =
2.
θ
R
mg sin θ .
d
In the previous problem:
a.
What is the best value for d, i.e., the value that lets F be as small as
possible?
b.
Show that the minimum value of F is 1/2 the force one would need to
push the cylinder up a frictionless incline of the same angle θ .
c.
Suppose the person applies this minimum force to roll the cylinder up the
incline. What is the largest angle of incline that can be used, in terms of
µs ? Ans: tan θ = 2 µs .
1
PHY 53
3.
Summer 2010
This situation is like one in Assignment 3,
except that now the pulley’s inertia must be
taken into account. The incline is frictionless,
the string is ideal, but the pulley has radius R
and moment of inertia I about its frictionless
axle. Call T1 the tension in the string attached
I
M
m
θ
to M, and T2 the tension in the string attached to m.
a.
If m is large enough to fall with acceleration a, what is a? Ans:
m − Msin θ
a=
g
m + M + I /R 2
b.
Find the difference between the tensions, T2 − T1 . Ans:
T2 − T1 = (I /R 2 ) ⋅
4.
m − Msin θ
m + M + I /R 2
In the situation of the previous problem, answer the same two questions:
a.
If the incline has friction, with kinetic coefficient µ k . Ans:
a=
b.
m − M(sin θ + µ k cosθ )
m + M + I /R
2
g , T2 − T1 = (I /R 2 ) ⋅
m − M(sin θ + µ k cosθ )
m + M + I /R 2
g.
If the incline is frictionless but the axle of the pulley has friction, which
causes a torque τ f opposite to the angular velocity of the pulley. Ans:
a=
(m − Msin θ )g − τ f /R
m + M + I /R 2
,
T2 − T1 = (τ f /R) + (I /R 2 ) ⋅
g.
(m − Msin θ )g − τ f /R
m + M + I /R 2
2
.
PHY 53
5.
Summer 2010
You are to analyze how a car gets its energy from the torque the engine applies to
the drive wheels. A torque supplies power according to the formula P = τω
(derived from P = Fv for the power supplied by a force).
Shown is the car, which is accelerating to the right.
Its kinetic energy consists of three parts: the CM
2
energy 12 MvCM
, the rotation energy of the front
wheels
1
I ω2,
2 1
and the rotation energy of the rear
wheels
1
I ω2.
2 2
We assume the wheels have the same radius R and are not
slipping, so we have vCM = Rω . Let the front wheels be the drive wheels. A
clockwise torque τ 0 is applied to them by the engine. This makes the point of
contact of those wheels tend to slip backwards, so there is a static friction force f1
to the right on them.
a.
What is the total torque on the front wheels? [It must be clockwise if the car
is accelerating to the right.]
b.
The rear wheels must also have a clockwise angular acceleration so they
don’t slip. What direction is the static friction force f 2 from the road on
those wheels?
c.
The friction forces do two things: they exert torques on their wheels and
together they give the external force that accelerates the CM. Write the
three equations expressing these things. [Be sure to include τ 0 in the
torque on the front wheels.]
d.
Now calculate dK /dt , where K is the total kinetic energy, and show that
dK /dt = τ 0ω . This proves that the car’s increase in energy comes from the
engine’s torque, even though that is internal to the car.
3
PHY 53
6.
Summer 2010
The refrigerator shown is being pushed by a
horizontal force F applied along a line through its
geometric center, which is also its CM. The static
friction coefficient between the refrigerator and the
floor is µs . The dimensions of the refrigerator are
w
F
•
h
as shown, and its mass is m. a.
If the refrigerator does not tip over, how
large must F be to make it slide? Ans:
F > µs mg .
b.
Consider torques about the bottom right corner. For what value of F will
the refrigerator be just on the verge of tipping over, if it doesn’t slide?
[The normal force from the floor must act effectively at some point on the
bottom surface of the refrigerator.] Ans: F = (w / h)mg .
c.
Draw a line, starting at the CM, along the direction of the sum of F and
mg . Show that if that line passes to the right of the bottom right corner,
the refrigerator will tip over.
d.
What is the minimum value of µs such that the refrigerator will slide
rather than tip over. Ans: µs < w / h .
7.
Suppose the refrigerator in the previous problem is
resting on the bed of a truck that is accelerating to
the left as shown. The coefficient of static friction
with the bed of the truck is µs as before.
a
•
h
a.
Discuss the ways in which this situation is
like the one in the previous problem.
b.
Find the minimum value of a that will make
the refrigerator slide if it doesn’t tip over.
Ans: a > µs g .
c.
Find the value of a for which it is on the verge of tipping over, if it doesn’t
slide. Ans: a > (w / h)g .
[The easiest approach is to use g eff .]
w
4
PHY 53 — Summer 2010
8.
Duke Marine Laboratory
The bowling ball shown starts down the alley with
CM speed v0 , sliding but not rotating. The ball has
mass m, radius R, and moment of inertia about its
center 52 mR 2 . The coefficient of kinetic friction
with the alley is µ k .
a.
What is the acceleration a (magnitude and direction) of the CM while the
ball is sliding? Ans: a = µ k g , to left.
b.
What is the angular acceleration α (magnitude and direction) about the
CM while the ball is sliding? [What direction is the torque?] Ans:
α = 52 µ k (g /R) , clockwise.
c.
Write the equations for the CM velocity v(t) while the ball is sliding. Ans:
v(t) = v0 − µ k gt .
d.
Write the equation for the angular velocity ω (t) while the ball is sliding.
[Take clockwise to be positive.] Ans: ω (t) = 52 µ k (g /R) ⋅ t
9.
The ball in the previous problem will stop sliding and roll instead when its linear
and angular velocities obey the rolling condition.
2v0
.
7 µk g
a.
At what time will that happen? Ans: t =
b.
What is the CM speed when the ball rolls? Ans: v = 75 vo .
c.
How far did the ball slide before rolling? Ans: x =
1
12v02
.
49 µ k g