Question

A rod of length 'L' and uniformly distributed mass 'M' per unit length is undergoing rotational motion about a perpendicular axis passing through its mid-point. If the angular speed of the rod changes from '0' to 'ω' in 't' seconds, the applied torque is -?

Answer 1

Initial angular velocity=0
Final angular velocity=ω

Using equation of circular motion-

$ω = 0 \times t + \frac{1}{2} \alpha {t}^{2} \\ 2ω = \alpha {t}^{2} \\ \alpha = \frac{2ω}{ {t}^{2} } \\ \\ torque = I \alpha \: \\ I - moment \: of \: inertia \\ here \: I = \frac{m {l}^{2} }{12} \\ \\ so \\ torque = \frac{m {l}^{2} }{12} \alpha \\ = \frac{m {l}^{2} }{12} \frac{2ω}{ {t}^{2} } \:$
Hope it will help you.