JEE Advanced 2023 Paper 2 · Q08 · Angular Momentum
A thin circular coin of mass $5$ gm and radius $4/3$ cm is initially in a horizontal $xy$-plane. The coin is tossed vertically up ($+z$ direction) by applying an impulse of $\sqrt{\dfrac{\pi}{2}}\times 10^{-2}$ N·s at a distance $2/3$ cm from its centre. 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: $g = 10$ m·s$^{-2}$]
Reveal answer + step-by-step solution
Correct answer:30
Solution
Let $J=\sqrt{\pi/2}\times 10^{-2}$ N·s, $M=5\times 10^{-3}$ kg, $R=\tfrac{4}{3}\times 10^{-2}$ m, $d=\tfrac{2}{3}\times 10^{-2}$ m, $g=10$ m/s².
Linear: $v = J/M = \dfrac{\sqrt{\pi/2}\times 10^{-2}}{5\times 10^{-3}} = 2\sqrt{\pi/2}=\sqrt{2\pi}$ m/s. Time of flight $T = 2v/g = \dfrac{2\sqrt{2\pi}}{10} = \dfrac{\sqrt{2\pi}}{5}$ s.
Angular: moment of inertia about a diameter (thin disc) is $I = MR^{2}/4 = \dfrac{(5\times 10^{-3})(4/3\times 10^{-2})^{2}}{4} = \dfrac{20}{9}\times 10^{-7}$ kg·m². Angular impulse $J d = I\omega$: $$\omega = \dfrac{Jd}{I} = \dfrac{(\sqrt{\pi/2}\times 10^{-2})(\tfrac{2}{3}\times 10^{-2})}{\tfrac{20}{9}\times 10^{-7}} = 300\sqrt{\pi/2}\ \text{rad/s}.$$
Rotations: $n = \dfrac{\omega T}{2\pi} = \dfrac{300\sqrt{\pi/2}\cdot \sqrt{2\pi}/5}{2\pi} = \dfrac{60\sqrt{\pi^{2}}}{2\pi} = \dfrac{60\pi}{2\pi} = 30$.
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