A thin circular coin of mass 5 gm and radius is initially in a horizontal -plane. The coin is tossed vertically up ( direction) by applying an impulse of N - s at a distance from its center. The coin spins about its diameter and moves along the direc

Question

A thin circular coin of mass 5 gm and radius $\frac{4}{3} cm$ is initially in a horizontal $x y$ -plane. The coin is tossed vertically up ( $+ z$ direction) by applying an impulse of $\sqrt{\frac{π}{2}} \times 10^{- 2}$ N - s at a distance $\frac{2}{3} cm$ 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 $g = 10 {ms}^{- 2}$ ] [2023]

Smart Achievers · Accepted Answer

(30)

From impulse--momentum theorem,

$J = M V_{C M} \Rightarrow V = \frac{J}{M} = \frac{\sqrt{π / 2}}{100 \times \frac{5}{1000}} = \sqrt{2 π} m/s$

Total time taken, $t = \frac{2 v}{g}$

$= \frac{2 \times \sqrt{2 π}}{g} = \frac{2 \times \sqrt{2 π}}{10} = \frac{\sqrt{2 π}}{5} s$

By angular impulse--momentum theorem,

$J \times \frac{R}{2} = I_{C} ω = [\frac{1}{4} M R^{2}] ω$

$∴$ $ω = \frac{J \times \frac{R}{2}}{\frac{M R^{2}}{4}} = \frac{J \times 2}{M R}$

$= \frac{\frac{\sqrt{π / 2}}{100} \times 2}{\frac{5}{1000} \times \frac{4}{3} \times \frac{1}{100}} = 2 \times 75 \sqrt{2 π} rad/s$

$∵$ $θ = 2 π n = ω t \Rightarrow n = \frac{ω t}{2 π}$

$∴$ $n = \frac{2 \times 75 \sqrt{2 π} \times \frac{\sqrt{2 π}}{5}}{2 π} = 30$

Solution

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