# Homework Help: Angular Momentum - Find angular velocity of a hanging bar after a collision

1. Mar 2, 2010

### cheinbokel

I think I've gotten it but figured it would be best to get a second opinion because I feel like I made a few leaps of faith. I sincerely appreciate your time. <3

1. The problem statement, all variables and given/known data
A 1.0 kilogram object is moving horizontally with a velocity of 10m/s, when it makes a collision with the lower end of a bar that is hanging vertically at rest. For the system of the bar and object, linear momentum is not conserved but kinetic energy is. The bar, of length l = 1.2m and mass m = 3kg, is pivoted about its upper end. Immediately after the collision, the object moves with speed v at an angle (theta) relative to its original direction. The bar swings freely, reaching a maximum angle of 90 degrees with respect to the vertical. The moment of inertia for the bar about the pivot is $I = (ml^2) / 3$. Ignore all friction.

Synopsis:
1.0kg object moving horizontally at 10m/s.
1.2m, 3kg bar $I = (ml^2) / 3$ suspended about its upper end.
Object hits the bottom of the bar in a glancing collision.
Bar then pivots up to 90 degrees with respect to the vertical.
Object then deflects to an angle $\theta$ below the horizontal at a velocity v.

Diagram:

Questions:
2. Relevant equations
The all-important: $F = ma$
Conservation of angular momentum: $L_i = L_f$
Conservation of kinetic energy: $K_o = K_ol + K_b$
Angular momentum around a point: $L = mvr sin(\theta)$
$\tau = F(lever arm) = I\alpha$
$L = I\omega$

3. The attempt at a solution
Enclosed in quotes to make it easier to see.
Thanks so much!

Last edited: Mar 2, 2010
2. Mar 2, 2010

### ideasrule

I haven't checked all your numbers, but your method seems good to me.