Man walking on a rotating platform

Given: platform $M = 40$ kg, $R = 1.5$ m (disc), man $m = 60$ kg, initial speed $N_1 = 20$ rpm. No external torque acts about the vertical axis (the platform rotates freely), so angular momentum is conserved.

Moments of inertia

• Platform: $I_p = \tfrac12 MR^2 = \tfrac12(40)(1.5)^2 = 45.0$ kg·m²
• Man at the centre: contributes $\approx 0$. Initial $I_1 = 45.0$ kg·m².
• Man at the rim: $mR^2 = 60(1.5)^2 = 135$ kg·m². Final $I_2 = 45 + 135 = 180.0$ kg·m².

Final angular speed

$$I_1\omega_1 = I_2\omega_2 \;\Rightarrow\; N_2 = N_1\frac{I_1}{I_2} = 20\times\frac{45}{180} = 5.00\ \text{rpm}$$

In rad/s: $\omega_1 = 2.094$ and $\omega_2 = 0.524$ rad/s.

Change in kinetic energy

$$T_1 = \tfrac12 I_1\omega_1^2 = \tfrac12(45)(2.094)^2 = 98.7\ \text{J}$$ $$T_2 = \tfrac12 I_2\omega_2^2 = \tfrac12(180)(0.524)^2 = 24.7\ \text{J}$$ $$\Delta T = T_2 - T_1 = -74.0\ \text{J}$$

Reason

Angular momentum is conserved, but kinetic energy is not, because the man does internal work (muscular effort and friction between his feet and the platform) while he moves radially outward. The system's energy decreases by 74.0 J; that energy is dissipated, mainly as heat in the man's body and at the feet. Since $T = \tfrac12 I\omega^2 = H^2/2I$, an increase in $I$ at constant $H$ always reduces $T$.

Answer: New speed = 5.0 rpm; kinetic energy decreases by 74.0 J.