A section of steel pipe of large diameter and relatively thin wall is mounted as shown on a flat-bed truck. The driver of the truck, not realizing that the pipe has not been lashed in place, starts up the truck with a constant acceleration of 0.5 g. As a result, the pipe rolls backward (relative to the truck bed) without slipping, and falls to the ground. The length of the truck bed is 5 m.
(a) With what horizontal velocity does the pipe strike the ground?
(b) What is its angular velocity at this instant?
(c) How far does it skid before beginning to roll without slipping, if the coefficient of friction between the pipe and ground is 0.3?
(d) What is its linear velocity when its motion changes to rolling without slipping?
Answers
Answer:
AP Physics C. Rollin’ & Slippin’ RUs
Here are some worked-out, illustrative examples.
1. Halliday and Resnick, Physics, 1st Edition, 1967, p 317, Chapter 13, #17:
A billiard ball, initially at rest, is given a sharp impulse by a cue. The cue is held horizontally a distance h above the
centerline as in Fig 13-14. The ball leaves the cure with a speed vo and, because of “forward English,” eventually
acquires a final speed of 9
7
vo. Show that h =
4
5R, where R is the radius of the ball.
F
h
R
Solution. Break up the motion into two pieces: during the hit, and after the hit.
During the hit: Initially there is both rolling and slipping. The impact of the cue imparts an impulse F ∆t = m∆v,
or
F ∆t = mvo
(It is possible that the friction also imparts an impulse to the ball, but we are going to ignore this!) At the same
time, there is a torque τ exerted on the ball by F, which imparts an “angular impulse” τ∆t. We have
τ∆t = Iα∆t = I∆ω = Iωo
and the torque is itself given by τ = rF sin φ = F h That is,
F h∆t = Iωo
Dividing this equation by the second equation above gives
h =
Iωo
mvo
The moment of inertia of the sphere is 2
5mR2
, so we get
h =
2 ωoR2
5vo
which can be solved for ωo;
ωo =
5voh
2R2
After the hit:
Now friction alone is acting to change both the center of mass velocity and the angular velocity. The ball is
spinning with an angular velocity faster than v/R, so the contact point of the ball is moving to the left, and
friction, believe it or not, acts to the right, in order to slow the rotation. At the same time, then, it increases the
center of mass velocity! (A similar trick is used by “fast bowlers” in the game of cricket: the “bowler” (who pitches
the ball) is allowed to bounce the ball off the ground en route to the man at bat, so the top spin is converted into
increased speed. Probably skillful tennis players can do the same.) Suppose this friction acts for a time ∆t
0 until
the slipping is converted into rolling without slipping. Then we have
f∆t
0 = m∆v = m
9
7
vo − vo
=
2
7mvo
for the center of mass velocity, and
fR∆t
0 = −I∆ω = −I
9
7
vo
R
− ωo
using the fact that the final ω is just the final v/R, since weve attained rolling without slipping. (The extra
negative comes from the slowing down of ω.) Substituting for ωo the value
Explanation: