# Angular momentum conservation, mass striking a rotating rod

## Main Question or Discussion Point

I recently ran across an experiment which I needed to explain because it appeared to violate energy conservation. After looking over it further, I became interested for a slightly different reason, but first I'll present the idea.

The experiment involves a mass $0.75 m$ striking a rod which is lying flat and pivoted in the center, the rod having a mass $m$ and total length $2r$. As the rod is pivoted, linear momentum is not taken to be conserved, but angular momentum and energy are conserved. As such, I would expect that the point at which the mass must strike the lever so as to exactly stop and transfer all of its energy to the lever could be calculated using (m for striking mass, r for rod):

$$m_{m}r^{2}_{m}\omega_{m} = \frac{1}{3}m_{r}r_{r}^{2}\omega_{r}$$ and $$\frac{1}{2}m_{m}r^{2}_{m}\omega_{m}^{2} = \frac{1}{6}m_{r}r_{r}^{2}\omega_{r}^{2}$$

On solving, this yields $r_{m} = \frac{2}{3}r_{r}$ for the $0.75 m$ mass, or equivalently, the mass must strike the rod 2/3 r from the pivot in order to stop completely when colliding.

First, this result seems interesting to me because it implies that no matter how fast this mass is traveling, it will completely stop as long as it hits the rod at this distance from the pivot. Is it generally true that for a given mass striking a rod in such a manner, it is only the distance from the center which matters and not the velocity?

Secondly (perhaps related), is this reasoning false? The experiment claims that the mass stops perfectly when it strikes the lever at a distance 0.40r from the pivot, rather than at 2/3 r, which was why it claimed that energy was not conserved.

Last edited by a moderator:

Related Classical Physics News on Phys.org
kuruman
It is generally true that if a mass moving with speed $v_0$ collides elastically with an equal mass at rest, the moving mass will stop and the target mass will move with speed $v_0$ regardless of the value of $v_0$. Here you have the rotational equivalent of that, in a sense.
You have made a mistake in that the moment of inertia of a rod pivoted about its center is $I_{cm}=\frac{1}{12}mL^2$, where $L$ is the length of the rod. If you use that, you should get the correct answer which is $r=\frac{1}{3}L$ not $0.40~L$. You could have seen that your answer was incorrect because the rod extends only up to distance $\frac{1}{2}L$ on either side of the pivot so $r=\frac{2}{3}L$ means that the mass misses the rod.