First slide

Concept of potential energy

Question

A chain of length l and mass m lies on the surface of a smooth hemisphere of radius R>l with one end tied to the top of the hemisphere. Taking the back of hemisphere as reference line, the gravitational potential energy of the chain is

Difficult

A

$m g l$
B

$\frac{m g l}{2}$
C

$\frac{m g R}{l} \sin (\frac{l}{R})$
D

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

Solution

$\begin{array}{l} U = \int d m g y \\ = \int \frac{m}{l} R d θ g R \cos θ \\ = \frac{m R^{2} g}{l} [\sin θ]_{0}^{l / R} \\ U = \frac{m g R^{2}}{l} \sin (\frac{l}{R}) \end{array}$

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