# Projectile hits rod hanging from pivot.

1. Dec 7, 2013

### BOYLANATOR

1. The problem statement, all variables and given/known data
A thin, uniform bar, 2, long and weighing 90N is hanging vertically from the ceiling by a frictionless pivot. It is struck by a small 3kg ball, 1.5m below the ceiling, initially travelling horizontally at 10 m/s. The ball rebounds in the opposite direction with a speed of 6 m/s.

2. Relevant equations

Lbefore = Lafter

L = Iω

Irod = $\frac{1}{2}$MR2

Ipoint = MR2

3. The attempt at a solution

At the point of impact the ball can be thought of as a particle in circular motion about the pivot with radius 1.5m and tangential velocity 10m/s.
The change in velocity is 16m/s. So the effective change in the angular momentum of the ball is

ΔLball = Iball Δωball = $\frac{IballΔv}{Rball}$

This is equal to the change in the angular momentum of the rod (opposite direction):

ΔLrod = Irod Δωrod = $\frac{IballΔv}{Rball}$

Insert values for I and rearrange :

Δωrod = (2*mball*rball*Δvball) / (mrod*rball2)

Last edited: Dec 7, 2013
2. Dec 7, 2013

### BOYLANATOR

I obviously used the fraction syntax incorrectly, please let me know what I did wrong. Thanks

3. Dec 8, 2013

### ehild

The formula for the moment of inertia of the rod is not correct.

ehild

4. Dec 8, 2013

### haruspex

itex gives up if you put non-itex codes like [s u b] inside. Use ^ for sup and _ for sub.

5. Dec 8, 2013

### ehild

As haruspex said, _ for sub, but enclose subscript between curly thingies {}

$ΔL_{ball} = I_{ball} Δω_{ball} = \frac{I_{ball}Δv}{R_{ball}}$

written as

ΔL_{ball} = I_{ball} Δω_{ball} = \frac{I_{ball}Δv}{R_{ball}}

ehild

6. Dec 8, 2013

### BOYLANATOR

Ah yes. I should stick to deriving the moments of inertia, my memory doesn't serve me well. Thanks.