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← JEE Advanced 2023 · Paper 2
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JEE Advanced 2023 · Paper 2, Physics question 8

A thin circular coin of mass 5 gm and radius 4/34/3 cm is initially in a horizontal xyxy-plane. The coin is tossed vertically up (+z+z direction) by applying an impulse of π2×10−2\sqrt{\dfrac\pi2}\times10^{-2} N-s at a distance 2/32/3 cm from its center. The coin spins about its diameter and moves along the +z+z direction. By the time the coin reaches back to its initial position, it completes nn rotations. The value of nn is _____.

[Given: The acceleration due to gravity g=10 m s−2g = 10\ \mathrm{m\,s^{-2}}]

A thin horizontal coin of radius 4/3 cm (radius marked from the centre). A vertical upward impulse is applied at a point 2/3 cm from the centre (dashed arrow from below). Axes: z up, x and y horizontal; g acts downward.
Answer and worked solution

Official answer: 30

Data: m=5×10−3m=5\times10^{-3} kg, R=43×10−2R=\frac43\times10^{-2} m, J=π/2×10−2J=\sqrt{\pi/2}\times10^{-2} N s, lever arm r=23×10−2r=\frac23\times10^{-2} m.

Translation.
v=Jm=2π/2=2π m/s.v=\frac Jm=2\sqrt{\pi/2}=\sqrt{2\pi}\ \text{m/s}.
The time to return is T=2vg=2π5T=\dfrac{2v}{g}=\dfrac{\sqrt{2\pi}}{5} s.

Rotation about a diameter.
I=mR24=5×10−3×169×10−44=29×10−6 kg m2,I=\frac{mR^2}4=\frac{5\times10^{-3}\times\frac{16}9\times10^{-4}}{4}=\frac29\times10^{-6}\ \text{kg m}^2,
ω=JrI=23×10−4π/229×10−6=300π/2 rad/s.\omega=\frac{Jr}{I}=\frac{\frac23\times10^{-4}\sqrt{\pi/2}}{\frac29\times10^{-6}}=300\sqrt{\pi/2}\ \text{rad/s}.

Rotations.
n=ωT2π=300π/2⋅2π/52π=60π2π=30.n=\frac{\omega T}{2\pi}=\frac{300\sqrt{\pi/2}\cdot\sqrt{2\pi}/5}{2\pi}=\frac{60\pi}{2\pi}=30.

Topics:Angular momentum and its conservationMoment of inertiaMotion in a straight line
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