Is this angular momentum question correct?

[Lachlan1]

Homework Statement

A ball traveling 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.

Homework 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

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? I am not sure I've done the cross product correctly.

 

[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).

 

[Lachlan1]

yeah, ok. thanks for your help