# When is the angular momentum = rp?

the formula for angular momentum is: $\vec{L}=\vec{r}\times\vec{p}=J\vec{\omega}$

That makes sense to me. But there is also this one dimensional formula: $L=rp$

Somehow, I don't really get when that works.

Example:

Lets assume we have a billiard ball like this: How big is the angular velocity directly after the hit, if we hit it with the impuls $\Delta p$ at h.

We use angular momentum conservation: $L_{before}=L_{after}$

Getting: $(h-a)\Delta p=amv_0 \ \Rightarrow \ v_0=\frac{(h-a)\Delta p}{am}$

With $\omega_0=\frac{v_0}{a}$ we get $\omega_0=\frac{(h-a)\Delta p}{a^m}$

Which is wrong. But I can't see why.

Edit: I guess it is because $v_0$ is the velocity of the "middle" and not on the surface? [sorry for the english]

• For cross products which are perpendicular to one another (orthogonal), it reduces to scalar multiplication. Reasoning: $\vec{L} = \vec{r} \times \vec{p} = \vec{r}\vec{p}\sin{\theta}$ then at perpendicular angles, $\sin{\theta = 90°} = 1$ and $L=rp$ – bleuofblue Jan 28 '17 at 19:10

The one dimensional formula you mentioned is just because $r$ and $p$ are happen to be orthogonal. Note that the directions of $L$ and $\omega$ are always, in your plane case, pointing orthogonal against the paper or screen. (Sorry I'm not a native speaker, hope you can understand.) And yes your $v_0$ should be the velocity in the place $(h-a)$, since your left side of the equation is $(h-a)$.