# Calculating the angular momentum

1. Apr 7, 2015

### marcnn

(Based on a problem from the 59th Polish Olympiad in Physics, 2009)

Let's consider a ball of mass $m$, radius $r$ and moment of inertia $I = 2/5mr^2$ approaching a wall with linear velocity $v_0$, rolling without sliding with angular velocity $\omega_0$. It collides with a wall.

Its collision with the wall is very short, so we may consider only the forces acting between the wall and the ball (neglecting the gravity, the floor's reaction and the friction between the floor and the ball).

Thus the angular momentum wrt to the axis of ball's tangency to the wall is conserved, so
$$I'\omega' = \mathrm{const}~~~~ (1)$$
where $I'$ is the moment of inertia wrt to that axis and $\omega'$ - angular velocity wrt to that axis.

Why is the formula (1) equivalent to
$$I \omega + m v_y r = \mathrm{const} ~~~~ (2)$$ where $\omega$ is the angular velocity of the ball wrt to the mass center, $v_y$ the vertical component of the mass center velocity.

Last edited: Apr 7, 2015
2. Apr 7, 2015

### Halo CX

I guess $$I'$$ is your moment of inertia, not the angular momentum as you are saying. There is a also a little ambiguity about the axis wrt to which $$I'$$ is calculated.

3. Apr 7, 2015

### marcnn

Yep, I made a typo :)

$I'$ is calculated to wrt to the axis which goes through the point of tangency to the wall and is parallel to the floor, I guess. The official solution of the corresponding problem isn't clear either :)