Physics Quiz 9

[James Kuszmaul]

Originally Posted by Oblarg

If anyone's interested, here's a somewhat similar, slightly tricky question I recall from a classical mechanics course:

A bowler releases a bowling ball. Initially, it is sliding at velocity v. Due to friction with the ground, it begins to roll. What is the velocity of the ball once it is rolling?

Hint: Not all origin locations are created equal.

Well, once the bowling ball is rolling, you will have an angular momentum:
L = I * omega
and a momentum:
p = m * v_f
where v_f = omega * r
This angular momentum must result from some constant force of kinetic friction, f_k acting over a time, t, at the radius r such that:
f_k * r * t = L
This same force of friction will have created an impulse, which slowed down the ball as a whole:
f_k * t = m * v_f - m * v
I * omega / r = m * (v_f - v)
I * v_f = m * (v_f - v)
and so:
v_f = -mv / (I - m)
if we take the bowling ball to actually be a perfect, uniform sphere, then:
v_f = -mv / (.4 * m * r² - m)
so
v_f = -v / (.4 * r² - 1)
Which looks really weird to me, so I have no idea if it is correct.
Edit: Figured out two issues. I forgot a negative sign on the change in translational momentum and I switched where the r should go when converting between angular and linear velocity.
Starting from:
I * omega / r = -m * (v_f - v)
omega = v_f/r
I * v_f / r² = -m * (v_f - v)
v_f = mv / (I / r² + m)
Assuming uniform sphere:
v_f = mv / (0.4 * m * r² / r² + m)
v_f = v / (0.4 + 1)
v_f = 5 * v / 7