4. (25 pts) A 20 g ball at the end of a string moves in a circle on a horizontal frictionless table, as shown
below. Initially, there is 10 cm of string between the ball and the hole in the table and the ball moves
at 4 m/s.
a) (5 pts) What is the initial angular velocity of the ball?
ω=v
r=4
0.1= 40 rad/s
b) (5 pts) How about the angular momentum of the ball?
Since we’ll need the moment of inertia to compute angular momentum, we find it first,
I=mr2= (0.02)(0.12) = 0.0002 kg ·m2
Then the angular momentum is
L=Iω = (0.0002)(40) = 0.008 kg ·m2/s
c) (5 pts) Some of the string is pulled through the table such that there is only 6 cm between the ball
and the hole. What is the net torque on the ball as this is happening?
Since the tension in the string acting on the ball is towards the center of the motion, the angle between
the force and the vector ~r is 0◦. Thus there is no torque on the ball during this motion.
d) (5 pts) Compute the new angular speed of the ball.
The moment of inertia has changed since the ball is rotating closer to the center of rotation.
I= (0.02)(0.062) = 0.000072 kg ·m2
Then, since angular momentum is conserved (no net torque), we have
0.008 = (0.000072)ω
ω= 111.1 rad/s
e) (5 pts) Was the energy of the ball conserved during this problem? Briefly explain your reasoning.
Since both the translational and rotational velocities change, the energy of the ball is not conserved!
While the string does not exert any torque on the ball, it does do some work on the ball as it is pulled
towards the hole. This work accounts for the increase in kinetic energy.
