A thin circular coin of mass 5 text{ g} and r | vedclass

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(A) Given: Mass $m = 5 	imes 10^{-3} 	ext{ kg}$,Radius $R = \frac{4}{3} 	imes 10^{-2} 	ext{ m}$,Impulse $J = \sqrt{\frac{\pi}{2}} 	imes 10^{-2} \

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(A) Given: Mass $m = 5 	imes 10^{-3} 	ext{ kg}$,Radius $R = \frac{4}{3} 	imes 10^{-2} 	ext{ m}$,Impulse $J = \sqrt{\frac{\pi}{2}} 	imes 10^{-2} 	ext{ N-s}$,Distance $r = \frac{2}{3} 	imes 10^{-2} 	ext{ m}$.
Linear impulse $J = mv \implies v = \frac{J}{m} = \frac{\sqrt{\pi/2} 	imes 10^{-2}}{5 	imes 10^{-3}} = 2\sqrt{\frac{\pi}{2}} = \sqrt{2\pi} 	ext{ m/s}$.
Time of flight $T = \frac{2v}{g} = \frac{2\sqrt{2\pi}}{10} = \frac{\sqrt{2\pi}}{5} 	ext{ s}$.
Angular impulse $J_	heta = J \cdot r = I_d \omega$,where $I_d = \frac{1}{4}mR^2$ is the moment of inertia about the diameter.
$J \cdot r = \frac{1}{4}mR^2 \omega \implies \omega = \frac{4Jr}{mR^2} = \frac{4 	imes (\sqrt{\pi/2} 	imes 10^{-2}) 	imes (2/3 	imes 10^{-2})}{5 	imes 10^{-3} 	imes (4/3 	imes 10^{-2})^2} = \frac{8/3 	imes \sqrt{\pi/2} 	imes 10^{-4}}{5 	imes 10^{-3} 	imes 16/9 	imes 10^{-4}} = \frac{8/3 	imes \sqrt{\pi/2}}{80/9 	imes 10^{-3}} = \frac{8}{3} 	imes \frac{9}{80} 	imes 10^3 	imes \sqrt{\frac{\pi}{2}} = 30 	imes \sqrt{\frac{\pi}{2}} 	ext{ rad/s}$.
Total angle rotated $	heta = \omega T = (30 \sqrt{\pi/2}) 	imes (\frac{\sqrt{2\pi}}{5}) = 6 	imes \sqrt{\pi/2} 	imes \sqrt{2\pi} = 6 	imes \sqrt{\pi^2} = 6\pi 	ext{ radians}$.
Number of rotations $n = \frac{	heta}{2\pi} = \frac{6\pi}{2\pi} = 3$.

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