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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 \(\sqrt{\frac \pi 2 \times 10^{-2}}\) N - s at a distance 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]

1 Answer

+1 vote
by (49.1k points)
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Best answer

By impulse – momentum theorem : 

J = MVCM

Also, by angular impulse – momentum theorem :

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