# Is this angular momentum question correct?

1. Aug 5, 2010

### Lachlan1

1. The problem statement, all variables and given/known data
A ball travelling in a straight line, colides with the end of a pole on a centre pivot, ie, the pole initially has inertia given by equation (ML^2)/12.After the colision, the ball sticks to the pole and the two rotate together. What is needed to be found is the angular speed

use the variables m for mass of ball, d for length of pole, omega symbol for angular speed, v for speed of ball before colision.
2. Relevant equations
cross product of vectors for ball initially are sued to generate the balls angular momentum.
so, angluar momentum=(displacement vector) *(momentum vector) = L = r*p
i take this as L=-d/2 *m*v
this is also angular momentum intital, as the conservation of momentum is used to calulate the resulting angular speed. the final angular momentum = inertia *omega
final inertia is equal to the sum of the two moment of inertias about the axis. this is
mr^2 + (ML^2)/12

3. The attempt at a solution

using
-d/2 *m*v=(omega)(mr^2 + (ML^2)/12)
rearange to isolate omega results in
omega= [-d/2 *m*v]/[(mr^2) + (ML^2)/12]

is this right though? im not sure ive done the cross product correctly.

2. Aug 5, 2010

### Filip Larsen

Assuming that the velocity of the ball is perpendicular to the pole when it hits, it looks ok to me.

Though, you probably want to express r and L (in ML^2) in terms of d to avoid confusing yourself later (especially since you also used L to mean angular momentum).

3. Aug 5, 2010

### Lachlan1

yeah, ok. thanks for your help