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(B) Let the angle subtended by the chain at the center of the hemisphere be $\alpha = l/R$. Consider an element of the chain of length $dl$ at an angl

Vedclass · Accepted Answer

$\frac{m g R^2}{l} \sin \left(\frac{l}{R}\right)$

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(B) Let the angle subtended by the chain at the center of the hemisphere be $\alpha = l/R$. Consider an element of the chain of length $dl$ at an angle $	heta$ from the base. The height of this element from the base is $h = R \sin 	heta$. The length of the element is $dl = R d	heta$. The mass of this element is $dm = \frac{m}{l} dl = \frac{m}{l} R d	heta$. The gravitational potential energy of this element is $dU = (dm)gh = \left(\frac{m}{l} R d	hetaight) g (R \sin 	heta) = \frac{mgR^2}{l} \sin 	heta d	heta$. The chain extends from the top $(	heta = \pi/2)$ to an angle $	heta = \pi/2 - \alpha = \pi/2 - l/R$. Integrating $dU$ from $\pi/2 - l/R$ to $\pi/2$: $U = \int_{\pi/2 - l/R}^{\pi/2} \frac{mgR^2}{l} \sin 	heta d	heta = \frac{mgR^2}{l} [-\cos 	heta]_{\pi/2 - l/R}^{\pi/2}$ $U = \frac{mgR^2}{l} [-\cos(\pi/2) - (-\cos(\pi/2 - l/R))]$ $U = \frac{mgR^2}{l} [0 + \sin(l/R)] = \frac{mgR^2}{l} \sin \left(\frac{l}{R}ight)$.

$A$ uniform chain of length $l$ and mass $m$ lies on the surface of a smooth hemisphere of radius $R$ $(R > l)$ with one end tied to the top of the hemisphere as shown in the figure. Gravitational potential energy of the chain with respect to the base of the hemisphere is

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