Physics Help?

Question

A thin, uniform, metal bar, 2.5 long and weighing 70 , is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.6 below the ceiling by a small 3-kg ball, initially traveling horizontally at 13 . The ball rebounds in the opposite direction with a speed of 6 m/s.

Find the angular speed of the bar just after the collision

Answer 1

We could be a lot more help if you'd include all the units in your question. Like, is the bar 2.5 M long, 2.5 feet long, 2.5 cm long, etc.

Answer 2

This is a conservation of (angular) momentum problem. That is, the fundamental equation is:
initial momentum = final momentum
where momentum = mass x velocity.

But because of the mix of linear and angular motion, it is probably best to use the impulse model:
http://en.wikipedia.org/wiki/Impulse

Initially, the momentum was due to the ball. It came in with velocity v0 and left with v1. So it saw a delta in momentum (i.e. an impulse) of Mb(v1 - v0) of 19Mb. Therefore the bar has to have experienced a change in momentum of -19Mb to keep the final momentum the same as the initial.

So now we apply angular momentum. We can approximate the force exerted by ball on the rod by any value we want, as long as force x time product has the right value.

We know where the force is applied and hence the torque it exerts on the bar. We also know the moment of inertia of the bar. That gives us the angular acceleration of the bar. Add in the time, which we know, and we can determine the angular velocity.

(Note that velocity under constant acceleration is:

w = at where
w is the angular velocity
a is the angular acceleration
t is the time

and since a = T/I where T is torque and I is moment of inertia, and since T = Fr where F is the force and r is moment arm, we have

w = rFt/I

this shows that it is the product of F and t that count.)