Question

A circular platform is mounted on a frictionless vertical axle. Its radius R=2 m and its moment of Inertia about the axle is 200 Kg m^2 .It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 m/s relative to the ground. Time taken by the man to complete one revolution is
1) π s
2) 3π/2 s
3) 2π s
4) π/2 s

Answer 1

A circular platform is mounted on a frictionless vertical axle. Its radius $R = 2 \, \text{m}$ and its moment of inertia about the axle is $200 \, \text{Kg} \, \text{m}^2$. It is initially at rest. A $50 \, \text{kg}$ man stands on the edge of the platform and begins to walk along the edge at the speed of $1 \, \text{m/s}$ relative to the ground. Time taken by the man to complete one revolution is

1) $\pi$ s

2) $3\pi / 2$ s

3) $2\pi$ s

4) $\pi / 2$ s

The law of conservation of angular momentum states that when no external torque acts on an object or a closed system of objects, no change of angular momentum can occur. Therefore:

$$L_{\text{system}} = \text{const} = L_{\text{man}} + L_{\text{platform}} = 0, \quad \Rightarrow L_{\text{man}} = -L_{\text{platform}}$$ $$L_{\text{man}} = m v R$$

$m$ – mass of the man, $v$ – speed of the man relative to the ground, $R$ – radius of the platform.

$$L_{\text{platform}} = I \omega$$

$I$ – moment of inertia, $\omega$ – angular velocity

Therefore, $\omega = \frac{L_{\text{platform}}}{I} = -\frac{L_{\text{man}}}{I}$

So, speed of man relative to the platform equals:

$$v_{pl} = v - \omega R = v + \frac{m v R}{I} R = v \left(1 + \frac{m R^2}{I}\right)$$

Therefore, time taken by the man to complete one revolution is

$$t = \frac{2 \pi R}{v_{pl}} = \frac{2 \pi R}{v \left(1 + \frac{m R^2}{I}\right)} = 2 * \frac{2}{1 (1 + 1)} \pi = 2 \pi \text{ s}$$

Answer: 3) $2\pi$ s