Phy 121 - Assignment 8
A. 1. a. Yes. Since the wheel is a rigid, solid object, you can’t have parts of it
making more revolutions per minute than other parts.
b. No. The farther from the center you go, the greater v is. You can see this
from v = r
2.
B. 1. 30 km/hr. The question says both cars have the same
ω. So, by v = rω, half the r means half the v. (Or,
without the formula, to get around a track only half as
long in the same time means half the speed.)
2.
C. 1. (This can all just be copied from the formula sheet.) s = r, vt = rω, at =
r. Restriction: You must use radians.
2. As the wheel goes down, potential energy turns into kinetic energy:
Energy when released = Energy when 15 cm lower.
mgh = ½ I ω2
Since it’s a disk, I = ½ mR2 = ½ (.6 kg)(.2 m)2 = .012 kg·m2
(.6 kg)(9.8)(.15 m) = ½ (.012) ω2
.882 = .006 ω2
147 = ω2
ω = 12.1 rad/s
D. 1. The underlying idea here is that tangential acceleration comes from
changing the velocity’s magnitude, while centripetal acceleration comes from
changing the velocity’s direction. So, the answers are:
a. Give the car a constant speed. (The direction changes as it follows the circle.
But at = r and  = 0 because v’s and ω’s magnitudes are constant.)
b. Not possible. (No centripetal acceleration means no change in direction,
which contradicts the question’s statement that the car is on a circular track.)
2.
E. 1. . Less. ω = v/r. To stay in contact without slipping, the surfaces of both
gears must have the same speed, v. With v the same, the larger r goes with the
smaller ω.
2. More in B. The average distance to the mass from the axis is greater.
3.