Question

A piece of clay, mass m, is thrown with a velocity v and strikes a rod and sticks to it, causing the clay-rod system to rotate counterclockwise.

In another scenario, a rock, also mass m, is thrown at the same velocity v at the same rod, but bounces off. Again, the rod rotates counterclockwise.

Assuming that the rod has a length L, and is bolted at the end by a frictionless pivot, in which case will the rod move off with a greater angular speed? Justify your reasoning.

Answer 1

In the case of rock, the rod will move with greater angular speed because that collision is more elastic and exchanges more kinetic energy with the rod.

Initial momentum of the clay = mv
Momentum of the rod + clay system just after the collision = (m + M) V
Where M is the mass of the rod and V is the initial velocity of the clay + rod system
Since the momentum is conserved,
mv = (m + M) V
V = mv / (M + M)

In the case of rock,
Initial momentum = mv
Final momentum just after the collision is - mv' + MV'
Where v' is the velocity of the rock after the collision and V' is the velocity of the rod just after the collision.
Since momentum is conserved,
mv = - mv' + MV'
V' = (mv + mv') / M

It can be easily see that V' > V, ie the rod has greater angular momentum in the case of rock