Question

A uniform round board of mass $M$ and radius $R$ is placed on a fixed smooth horizontal plane and is free to rotate about a fixed axis which passes through its centre. A man of mass $m$ is standing on the point marked $A$ on the circumference of the board. At first the board and the man are at rest. The man starts moving along the rim of the board at constant speed $v_0$ relative to the board. Find the angle of board's rotation when the man passes his starting point on the disc first time.

Answer 1

Let man move with speed $v_0$ then momentum conservation

$u=v_0\left(\frac{m}{M}\right)$

angular rotation of board is $\omega$

$\omega =\frac{u}{R}$

time taken by man to complete revolution

$t=\frac{2\pi R}{v_0}$

Now angle turn by board is $=\omega t$

$=\frac{u}{R}\times \frac{2\pi R}{v_0}=2\pi \frac{u}{v_0}=2\pi \left(\frac{m}{M}\right)rad$