Question

A chain of length $l$ and mass $m$ lies on the surface of smooth hemisphere of radius $R>L$ with one end tied to the top of the hemisphere. The gravitational potential energy of the chain is

• A. $\frac{mg{R}^2}{3}sin\left(\frac{L}{R}\right)$
• B. $\frac{mg{R}^2}{2}sin($ell{\,}i$R)$
• C. $\frac{mg{R}^2}{4}sin($ell{\,}i$R)$
• D. $\frac{mg{R}^2}{1}sin($ell{\,}i$R)$

Answer 1

The correct option is A $\frac{mg{R}^2}{3}sin\left(\frac{L}{R}\right)$

Let us consider a small element which makes angle ${}^{\prime }d{\theta }^{\prime }$ at the center :

$\therefore dx=\left(\frac{m}{L}\right)Rd\theta$

a). Gravitational potential energy of ${}^{\prime }d{m}^{\prime }$ with respect to center of the sphere,

$=\left(dm\right)gR\cos \theta$

$=\left(\frac{mg}{L}\right)R\cos \theta \theta$

so, total $PE={\int }_0^{1/g}\frac{mg{R}^2}{L}\cos \theta dQ$

[ $\alpha =\left(\frac{L}{r}\right)$ L angle subtended by the chain at the center. ]

$=\frac{m{R}^{2}g}{L}\left[\sin \theta \right]\left(\frac{L}{R}\right)$

$=\frac{mRg}{L}\sin \left(\frac{L}{R}\right)$

Hence, the answer is $\frac{mRg}{L}\sin \left(\frac{L}{R}\right).$