# Man walking on the rotating platform [closed]

The problem is:

The solid cylindrical platform with mass $M$ and radius $R$ rotates with angular frequency $\omega$. On the edge of the platform there is a standing man with mass $m$. Then man begins to walk round the edge with velocity $v$ in the opposite direction of platform's rotation. And the task is to find new frequency $\omega '$.

So, I thought that the equation for this problem is

$$(I_{plat} + I_{man}) \ \omega = I_{plat} \ \omega ' + I_{man}\ (\omega ' - \frac{v}{R})$$ where $I$ is a moment of inertia. Equation is written in the inertial system.

But with this equation the result is like $\omega ' < 0$, so it begins to rotate in the opposite direction, it looks weird (with given values of parameters). And I was told that this equation is incorrect.

So, where did I do a mistake?

## closed as off-topic by John Rennie, Kyle Kanos, Daniel Griscom, Martin, Norbert SchuchJan 8 '16 at 12:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

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• Hi and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. – John Rennie Jan 8 '16 at 11:12
• Is $\omega'$ supposed to be the new frequency of the man or the platform? Your expression looks correct for the platform... And if the man walks at a velocity that cancels $\omega'$ exactly, the platform must be rotating much faster to carry all the angular momentum. I don't see how $\omega'$ would go negative for positive $v$. Who told you that you were wrong? – Floris Jan 8 '16 at 13:18