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

By impulse – momentum theorem : J=MVCM
$\Rightarrow V_{CM}=\frac{J}{M}=\frac{\frac{\pi}{2}}{100\times\frac{5}{1000}}=\sqrt{2\pi}$

Total time of journey=$\Delta t=\frac{\sqrt{2\pi}}{5}$
By angular impulse-momentum theorem,
$J\times\frac{R}{2}=[\frac{MR^2}{4}]w$

$w=\frac{j\times\frac{R}{2}}{\frac{MR^2}{4}}=\frac{J}{MR}\times2$

$=\frac{\frac{\frac{\sqrt{\pi}}{2}}{100}\times2}{\frac{5}{1000}\times\frac{4}{3}\times\frac{1}{100}}=2\times75\sqrt{2\pi}$ rad/s

Number of rotations = $\frac{w.\Delta t}{2\pi}=30$
\Rightarrow$$ n=30