# Simultaneous conservation of linear and angular momentum

Suppose there is a ring in the space where there is no gravity. The width of the ring is $r$ which is negligible compared to its inner radius $R$. The ring is in horizontal position. Now imagine a human starts to move on this ring. Before the start of the motion angular and linear momentum are zero and system is isolated. When the human walks on the ring linear and angular momentum must remain zero.

$$m_{human}*R^2*\omega_{human} = -M_{ring}*R^2*\omega_{ring} \rightarrow \omega_{human}=-\frac{M_{ring}}{m_{human}}*\omega_{ring}$$

There is no problem with this part. Now we consider the linear momentum. The linear momentum of the ring after the beginning of motion is zero because for every element of mass there is another element of mass moving in the opposite direction. However, the linear momentum of the human which can be considered a point mass is not zero. The magnitude of the linear momentum of the human is $m_{human} \cdot \omega_{human} \times R$. So the total linear momentum is not zero despite the fact that the initial linear momentum was zero and the system is isolated.

If there is not anything wrong I have to conclude that $\omega_{human} = 0$. What am I missing here?

## 1 Answer

You forget the force at the pivot. For what you described to happen, there must be an external force at the center, which provides no torque about the center. So angular momentum is conserved, but linear momentum is not. If there is no such a force, then it is the center of mass, which is somewhere between the man and the center, that will be stationary, but not the center. The motion will be such that the total linear and angular momentum remains zero, of course.