A wheel is rotating with an angular speed of | vedclass

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(C) Initial angular speed $\omega_1 = 20\ rad/s$.Final angular speed $\omega_f = 0\ rad/s$ at $t = 4\ s$.Angular retardation $\alpha = \frac{\omega_1

Vedclass · Accepted Answer

$30$

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(C) Initial angular speed $\omega_1 = 20\ rad/s$.Final angular speed $\omega_f = 0\ rad/s$ at $t = 4\ s$.Angular retardation $\alpha = \frac{\omega_1 - \omega_f}{t} = \frac{20 - 0}{4} = 5\ rad/s^2$.Angular speed after $t' = 2\ s$ is $\omega_2 = \omega_1 - \alpha t' = 20 - (5 	imes 2) = 10\ rad/s$.Work done by the torque is equal to the change in rotational kinetic energy:$W = \Delta K = \frac{1}{2} I (\omega_1^2 - \omega_2^2)$.$W = \frac{1}{2} 	imes 0.20 	imes (20^2 - 10^2)$.$W = 0.1 	imes (400 - 100) = 0.1 	imes 300 = 30\ J$.

$A$ wheel is rotating with an angular speed of $20\ rad/s$. It is stopped to rest by applying a constant torque in $4\ s$. If the moment of inertia of the wheel about its axis is $0.20\ kg\cdot m^2$,then the work done by the torque in two seconds will be .......... $J$.

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