A solid cylinder of mass 2; kg and radius 4; | vedclass

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Question

(A) Given: Mass $m = 2\; kg$,Radius $R = 4\; cm = 0.04\; m$,Initial angular velocity $\omega_0 = 3\; rpm = \frac{3 	imes 2\pi}{60} = \frac{\pi}{10}\;

Vedclass · Accepted Answer

$2 	imes 10^{-6}\; Nm$

Vedclass · Answer

(A) Given: Mass $m = 2\; kg$,Radius $R = 4\; cm = 0.04\; m$,Initial angular velocity $\omega_0 = 3\; rpm = \frac{3 	imes 2\pi}{60} = \frac{\pi}{10}\; rad/s$.Total angular displacement $	heta = 2\pi 	ext{ revolutions} = 2\pi 	imes 2\pi = 4\pi^2\; rad$.Final angular velocity $\omega = 0$.Using the kinematic equation $\omega^2 = \omega_0^2 - 2\alpha	heta$:$0 = (\frac{\pi}{10})^2 - 2\alpha(4\pi^2)$$8\alpha\pi^2 = \frac{\pi^2}{100} \Rightarrow \alpha = \frac{1}{800}\; rad/s^2$.The moment of inertia of a solid cylinder is $I = \frac{1}{2}mR^2 = \frac{1}{2} 	imes 2 	imes (0.04)^2 = 0.0016\; kg\cdot m^2$.The torque required is $	au = I\alpha = 0.0016 	imes \frac{1}{800} = 16 	imes 10^{-4} 	imes \frac{1}{8} 	imes 10^{-2} = 2 	imes 10^{-6}\; Nm$.

$A$ solid cylinder of mass $2\; kg$ and radius $4\; cm$ is rotating about its axis at the rate of $3\; rpm$. The torque required to stop after $2\pi$ revolutions is

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