A thin rod AB of length L and mass m is held | vedclass

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(A, C) Let the rod have length $L$ and mass $m$. The center of mass is at $L/2$ from $A$.When the rod makes an angle $	heta$ with the horizontal, the

Vedclass · Accepted Answer

the angular acceleration is proportional to $\cos 	heta$

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(A, C) Let the rod have length $L$ and mass $m$. The center of mass is at $L/2$ from $A$.When the rod makes an angle $	heta$ with the horizontal, the center of mass has descended by $h = (L/2) \sin 	heta$.By conservation of energy, the loss in potential energy equals the gain in rotational kinetic energy:$mg(L/2) \sin 	heta = \frac{1}{2} I \omega^2$, where $I = mL^2/3$.$mg(L/2) \sin 	heta = \frac{1}{2} (mL^2/3) \omega^2 \Rightarrow \omega^2 \propto \sin 	heta \Rightarrow \omega \propto \sqrt{\sin 	heta}$.Since the speed of end $B$ is $v = \omega L$, we have $v \propto \sqrt{\sin 	heta}$. Thus, option $A$ is correct.For angular acceleration $\alpha$, the torque about $A$ is $	au = mg(L/2) \cos 	heta$.Using $	au = I \alpha$, we get $mg(L/2) \cos 	heta = (mL^2/3) \alpha$.This implies $\alpha \propto \cos 	heta$. Thus, option $C$ is correct.Note: Both $A$ and $C$ are correct statements.

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