Question

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

• A. $\frac{m g l}{2}$
• B. $\frac{m g R^{2}}{l} \sin \left(\frac{l}{R}\right)$
• C. $\frac{m g R^{2}}{l} \sin \left(\frac{R}{l}\right)$
• D. $\frac{m g l^{2}}{R} \sin \left(\frac{l}{R}\right)$

Answer 1

Answer: (B)

Given,
Mass of chain is $m$
Length is 1
Radius of sphere is $R$
$
\begin{array}{l}
\Rightarrow h = R \sin \theta \\
\Rightarrow dl = R d \theta
\end{array}
$
The mass of dl length of the chain $dm = mdl$
Potential energy.
$
\begin{array}{l}
dU =( dm ) gh \\
\Rightarrow d U =\frac{ mdl }{1} gh =\frac{ mghdl }{1}=\frac{ mgh ^{2} \sin \theta d \theta}{1} \\
\Rightarrow dU =\frac{ mgR ^{2} \sin \theta d \theta}{1}
\end{array}
$
On Integrating
$
\begin{array}{l}
U =\int_{\theta}^{\frac{ x }{2} \frac{ mgR ^{2} \sin \theta d \theta}{1}} \\
U =-\frac{ mgR ^{2}}{1}[\cos \theta]_{\theta}^{\frac{\pi}{2}}=-\frac{ mgR ^{2}}{1}\left[\cos \left(\frac{\pi}{2}\right)-\cos \theta\right] \\
U =-\frac{ mgR ^{2}}{1} \cos \theta=\frac{ mgR ^{2}}{1} \sin \left(\frac{1}{ R }\right) \\
\text { Hence, potential energy is } \frac{ mgR ^{2}}{1} \sin \left(\frac{1}{ R }\right)
\end{array}
$