A solid cylinder of mass 2 kg and radius 4 cm rotating about its axis at the rate of 3 rpm. The torque require

Question

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

Accepted Answer

Using the work-energy theorem: $W = \frac{1}{2}I(\omega_f^2 - \omega_i^2)$. Here $	heta = 2\pi$ revolutions $= 2\pi 	imes 2\pi = 4\pi^2 	ext{ rad}$. Initial angular velocity $\omega_i = 3 	imes \frac{2\pi}{60} 	ext{ rad/s}$. The work done by torque is $W = -	au	heta$. Substituting values: $-	au(4\pi^2) = \frac{1}{2} 	imes (\frac{1}{2}mr^2) 	imes (0^2 - \omega_i^2)$. Solving this yields $	au = 2 	imes 10^{-6} 	ext{ N m}$.

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