A uniform chain of length $l$ and mass $m$ lies on the surface of a smooth hemisphere of radius $R(R>l)$…

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 the hemisphere as shown in the figure. Gravitational potential energy of the chain with respect to the base of the hemisphere is
  1. $\frac{m g l}{2}$
  2. $\frac{m g R^2}{l} \sin \left(\frac{I}{R}\right)$
  3. $\frac{m g R^2}{l} \sin \left(\frac{R}{l}\right)$
  4. $\frac{m g l^2}{R} \sin \left(\frac{l}{R}\right)$

Solution


We have, $\quad \frac{h}{R}=\sin \theta$ or $h=R \sin \theta$ Also, $d l=R d \theta$ Mass of $d l$ length of chain $=d m=\frac{m}{l} \cdot d l$ PE of $d m$ mass $ =d U=d m g h=\frac{m g h}{l} \cdot d l=\frac{m g h}{l} R d \theta=\frac{m g R^2}{l} \sin \theta d \theta $ So, PE of complete chain is $ U=\int_{\pi / 2}^{\pi / 2-\theta} d U=\frac{m g R^2}{l} \sin \left(\frac{l}{R}\right) $

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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