A thin uniform rod is pivoted at O. It rotate | vedclass

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(B) The torque $	au$ required to keep the angular velocity $\omega$ constant is given by the rate of change of angular momentum $L$. $L = I\omega$. S

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(B) The torque $	au$ required to keep the angular velocity $\omega$ constant is given by the rate of change of angular momentum $L$. $L = I\omega$. Since $\omega$ is constant,$	au = \frac{dL}{dt} = \omega \frac{dI}{dt}$. The moment of inertia $I$ of the system (rod + insect) at time $t$ is $I = I_{	ext{rod}} + I_{	ext{insect}}$. $I_{	ext{insect}} = m r^2$,where $r = vt$ is the distance of the insect from $O$ at time $t$. So,$I(t) = I_{	ext{rod}} + m(vt)^2 = I_{	ext{rod}} + mv^2t^2$. Differentiating with respect to $t$,we get $\frac{dI}{dt} = 2mv^2t$. Therefore,the torque is $	au = \omega (2mv^2t) = (2m\omega v^2)t$. This shows that $	au$ is directly proportional to $t$ for $0 \le t \le T$. At $t = T$,the insect stops,so $\frac{dI}{dt} = 0$,and the torque becomes zero. Thus,the graph should be a straight line passing through the origin for $0 \le t \le T$ and then drop to zero. This corresponds to the shape shown in graph $B$.

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