A wheel rotates about its geometric axis at a | vedclass

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(C) Initial angular velocity $\omega_i = 60 \ rpm = \frac{60 	imes 2\pi}{60} = 2\pi \ rad \ s^{-1}$.Final angular velocity $\omega_f = 0 \ rad \ s^{-

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$\frac{\pi}{15} \ Nm$

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(C) Initial angular velocity $\omega_i = 60 \ rpm = \frac{60 	imes 2\pi}{60} = 2\pi \ rad \ s^{-1}$.Final angular velocity $\omega_f = 0 \ rad \ s^{-1}$.Time taken $t = 1 \ minute = 60 \ s$.Moment of inertia $I = 2 \ kg \ m^2$.The angular acceleration is given by $\alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 2\pi}{60} = -\frac{\pi}{30} \ rad \ s^{-2}$.The required torque is $	au = I\alpha = 2 	imes \left( -\frac{\pi}{30} ight) = -\frac{\pi}{15} \ Nm$.The magnitude of the torque required is $\frac{\pi}{15} \ Nm$.

$A$ wheel rotates about its geometric axis at a speed of $60 \ rpm$. If the moment of inertia of the wheel about this axis is $2 \ kg \ m^2$,what torque is required to stop its rotation in one minute?

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