Inertia, torque, angular momentum, energy - billiard ball? answer: a. h b. 2R/5
ID: 2300638 • Letter: I
Question
Inertia, torque, angular momentum, energy - billiard ball?
answer: a. h
b. 2R/5
please show work, and help on others
A billiard ball al rest is given a sharp, horizontal impulse by a cue, acting at a distance h above the midpoint of the ball, as shown at right, In terms of h, find the ratio of the ball's spin angular momentum to its linear momentum p immediately following the impulse, For what value of h will the ball roll without slipping immediately following the impulse, in terms of R, the ball's radius? Suppose the ball is on a table with friction. If h = 4R/5 and Vo is the speed of the ball immediately following the impulse, what is the speed of the ball by the time it starts rolling without slipping? How much kinetic energy does the ball lose to frictional heating, for the experiment described in (c)?Explanation / Answer
b.
Take a line on the billiard table, passing through the lowest point of the ball. Let this line be AB.
All the forces on the ball pass through the ball's lowest point and hence through AB.
Therefore torques of all forces about AB = 0
Therefore angular momentum about AB is conserved.
Initial linear speed = v0
The point where impulse is applied is at height r+h above AB.
Therefore, initial angular momentum of the ball about AB
= m v0 (r + h)-----------------(1)
Final linear speed = v0
Finally the ball rolls without skidding. In that case, angular speed = linear speed/radius
Therefore, final angular speed w = v0/r
Moment of inertia of the ball about AB is I = 2/5 mr^2 + mr^2 = 7/5 mr^2
Therefore total final angular momentum of the ball about AB
= I*w
= (7/5)mr^2 * v0/r
= (7/5)m*v0*r------------(2)
Angular momentum about AB is conserved. Therefore, from equations (1) and (2),
m*v0*(r+h) = (7/5)*m*v0*r
Dividing by m*v0,
r+h = (7/5)*r
Or h = (7/5 - 1)*r
Or h = 2/5 r
Proved