A boy of mass m stands at the edge of a uniform disc of mass M and radius R rotating at angular frequency $\omega$ about an axis passing through its centre and perpendicular to its plane. The boy walks towards the centre of the disc and stops when he is at a distance r from the centre of the disc. If friction is neglected and r = R/2 and M = 10 m, the angular frequency will be

Refer to fig

Since no external torque acts as the boy walks from
P to Q, the angular momentum of the disc + boy
system is conserved, i.e

              $I \omega =I\text{'}\omega \text{'}$               (1)

where I = moment of the system when the boy is at
P, I' = moment of inertia of the system when the boy
is at Q and $\omega \text{'}$ = new angular frequency.

             $$\begin{gathered} I=I_0+m{R}^2 \\ =\frac{1}{2}M{R}^2+m{R}^2 \end{gathered}$$

$$\begin{gathered} \Rightarrow I=\left(\frac{M}{2}+m\right)R^2 \left(2\right) \\ I\text{'}=I_0+m{r}^2 \\ \Rightarrow I\text{'}=\frac{1}{2}M{R}^2+m{r}^{2 } \left(3\right) \end{gathered}$$

Using (2) and (3) we have

$$\begin{gathered} \left(\frac{M}{2+m}\right)R^2\omega =\left(\frac{M{R}^2}{2}+m{r}^2\right)\omega \text{'} \\ \Rightarrow \omega \text{'}=\frac{\left({\displaystyle \frac{M}{2}}+m\right)R^2\omega }{\left(\frac{M{R}^2}{2}+m{r}^2\right)} \end{gathered}$$                  (4)

Putting r = R/2 and M =10m in (3) we get $\omega \text{'}=\frac{8\omega }{7}$