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A thin uniform rod, pivoted at $O$, is rotating in the horizontal plane with constant angular speed $\omega$, as shown in the figure. At time $t=0$, a small insect starts from $O$ and moves with constant speed $v$, with respect to the rod towards the other end. It reaches the end of the rod at $t=T$ and stops. The angular speed of the system remains $\omega$ throughout. The magnitude of the torque $(|\vec{\tau}|)$ about $O$, as a function of time is best represented by which plot?
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Verified Answer
The correct answer is:
Angular momentum, $|\mathbf{L}|$ or $L=I \omega$ (about axis of rod) Moment of inertia of the rod-insect system. $I=I_{\text {rod }}+m x^{2}=I_{\text {rod }}+m v^{2} t^{2}$ Here, $m=$ mass of insect $\therefore \quad L=\left(I_{\mathrm{rod}}+m v^{2} t^{2}\right) \omega$
Now $|\tau|=\frac{d L}{d t}=\left(2 m v^{2} t \omega\right)$ or $|\tau| \propto t$
i.e. the graph is straight line passing through origin.
After time $T, L=$ constant
$\therefore \quad|\tau| \text { or } \frac{d L}{d t}=0$
i.e., when the insect stops moving, $L$ does not change and therefore $T$ becomes constant.
Now $|\tau|=\frac{d L}{d t}=\left(2 m v^{2} t \omega\right)$ or $|\tau| \propto t$
i.e. the graph is straight line passing through origin.
After time $T, L=$ constant
$\therefore \quad|\tau| \text { or } \frac{d L}{d t}=0$
i.e., when the insect stops moving, $L$ does not change and therefore $T$ becomes constant.
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