A man of mass $m$ stands at the edge of a horizontal platform (mass $M$, radius $R$, moment $I = \tfrac{1}{2}MR^2$) rotating at $\omega_0$ (no friction at axle). He walks to the **centre**. The new angular speed of the platform is:

Conservation of angular momentum (no external torque). Initial: $L = (\tfrac{1}{2}MR^2 + mR^2)\omega_0$. Final (man at centre, $r = 0$): $L = (\tfrac{1}{2}MR^2)\omega_f$. So $\omega_f = \omega_0 \cdot (\tfrac{1}{2}M + m)/(\tfrac{1}{2}M) = \omega_0(M + 2m)/M$.