Question

Question 3 &4 on
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Answer 1

Answer on Question #56545-Physics-Mechanics-Relativity

Question 3 has not seen.

(iv) A man of mass $m_1$ stands on the edge of a horizontal uniform disc of mass $m_2$ and radius $R$ which is capable of rotating freely about a stationary vertical axis passing through its centre. The man walks along the edge of the disc through angle $\theta$ relative to the disc and then stops. Find the angle through which the disc turned the time the man stopped.

$$\left( \frac{2m_1\theta}{2m_1 + m_2} \right)$$

Solution

According to the conservation of momentum law:

$$m_1 \omega_1 R = \frac{I_{disk} \omega_2}{R} = \frac{m_2 \omega_2 R^2}{2R} = \frac{m_2 \omega_2 R}{2} \rightarrow \omega_1 = \frac{m_2 \omega_2}{2m_1}.$$

The angular velocity of man relative to the disk is

$$\omega = \omega_1 + \omega_2 = \frac{m_2 \omega_2}{2m_1} + \omega_2 = \frac{2m_1 + m_2}{2m_1} \omega_2.$$

Thus,

$$\omega_2 = \frac{2m_1}{2m_1 + m_2} \omega.$$

The angle for disk is

$$\phi = \omega_2 t = \frac{2m_1}{2m_1 + m_2} \omega t = \frac{2m_1}{2m_1 + m_2} \theta.$$

Answer: $\frac{2m_1}{2m_1 + m_2} \theta$.

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