A disc of mass 25 kg and radius 0.2 m is ro | vedclass

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(C) Given:Mass $M = 25 \ kg$,Radius $R = 0.2 \ m$,Initial angular velocity $\omega_0 = 240 \ r.p.m. = 240 	imes \frac{2\pi}{60} \ rad/s = 8\pi \ rad/

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$\pi \ N$

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(C) Given:Mass $M = 25 \ kg$,Radius $R = 0.2 \ m$,Initial angular velocity $\omega_0 = 240 \ r.p.m. = 240 	imes \frac{2\pi}{60} \ rad/s = 8\pi \ rad/s$.Final angular velocity $\omega = 0 \ rad/s$,Time $t = 20 \ s$.The moment of inertia of the disc is $I = \frac{1}{2}MR^2 = \frac{1}{2} 	imes 25 	imes (0.2)^2 = 0.5 \ kg \cdot m^2$.The angular acceleration $\alpha$ is given by $\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - 8\pi}{20} = -0.4\pi \ rad/s^2$.The retarding torque is $	au = I|\alpha| = 0.5 	imes 0.4\pi = 0.2\pi \ N \cdot m$.Since the torque is applied tangentially,$	au = F 	imes R$,so $F = \frac{	au}{R} = \frac{0.2\pi}{0.2} = \pi \ N$.

$A$ disc of mass $25 \ kg$ and radius $0.2 \ m$ is rotating at $240 \ r.p.m.$ $A$ retarding torque brings it to rest in $20 \ s$. If the torque is due to a force applied tangentially on the rim of the disc,then the magnitude of the force is:

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