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

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(D) The initial angular velocity is $\omega_0 = 240 \ r.p.m. = \frac{240 	imes 2\pi}{60} \ rad/s = 8\pi \ rad/s$. Using the rotational kinematic equa

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$\pi$

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(D) The initial angular velocity is $\omega_0 = 240 \ r.p.m. = \frac{240 	imes 2\pi}{60} \ rad/s = 8\pi \ rad/s$. Using the rotational kinematic equation $\omega = \omega_0 - \alpha t$,where the final angular velocity $\omega = 0$ at $t = 20 \ s$: $0 = 8\pi - \alpha(20) \Rightarrow \alpha = \frac{8\pi}{20} = 0.4\pi \ rad/s^2$. The torque $	au$ is given by $	au = I\alpha$,where $I = \frac{1}{2}MR^2$ for a disc. Given $M = 25 \ kg$ and $R = 0.2 \ m$,$I = \frac{1}{2} 	imes 25 	imes (0.2)^2 = 0.5 \ kg \cdot m^2$. The torque is also $	au = F \cdot R$,where $F$ is the tangential force. Equating the two expressions for torque: $F \cdot R = I \alpha \Rightarrow F = \frac{I \alpha}{R} = \frac{0.5 	imes 0.4\pi}{0.2} = \frac{0.2\pi}{0.2} = \pi \ N$.

$A$ 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 in newton is

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