Question

A thin circular coin of mass 5 gm and radius \(\frac{4}{3}\) is initially in a horizontal XY - plane. The coin is tossed vertically up (+Z direction) by applying an impulse of \(\sqrt{\frac{\pi}{2}\times10^{-2}}\) N-s at a distance of \(\frac{2}{3}\) cm from its center. The coin spins about its diameter and moves along the +𝑧 direction. By the time the coin reaches back to its initial position, it completes 𝑛 rotations. The value of 𝑛 is ____. [Given: The acceleration due to gravity 𝑔 = 10 ms⁻²]

Answer 1

Correct Answer: 30

Impulse-Momentum and Angular Impulse-Momentum Calculations 

By impulse – momentum theorem: \( J = M V_{CM} \)

Therefore, the center of mass velocity \( V_{CM} \) is: \[ V_{CM} = \frac{J}{M} = \frac{\frac{\pi}{2}}{100 \times \frac{5}{1000}} = \sqrt{2\pi} \]

The total time of the journey is: \[ \Delta t = \frac{\sqrt{2\pi}}{5} \]

By the angular impulse-momentum theorem: \[ J \times \frac{R}{2} = \left[\frac{M R^2}{4}\right] w \]

Solving for the angular velocity \( w \): \[ w = \frac{J \times \frac{R}{2}}{\frac{M R^2}{4}} = \frac{J}{M R} \times 2 \]

Substituting the known values: \[ w = \frac{\frac{\frac{\sqrt{\pi}}{2}}{100} \times 2}{\frac{5}{1000} \times \frac{4}{3} \times \frac{1}{100}} = 2 \times 75 \sqrt{2\pi} \, \text{rad/s} \]

The number of rotations is: \[ n = \frac{w \times \Delta t}{2\pi} = 30 \]

Therefore, the final result is: \[ n = 30 \]