Question

A chain of length $l$ and mass $m$ lies on the surface of a smooth sphere of radius $R>l$ with one end tied to the top of the sphere. Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through an angle $\theta .$

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

Answer 1

The correct option is A $\frac{m{R}^{2}g}{l}[\sin \left(\frac{l}{R}\right)+\sin \theta -\sin (\theta +\frac{l}{R})]$

The angle the line connecting the top end to the center of sphere forms with vertical is $0$.

The angle the line connecting the other end to the center of sphere forms with the vertical is $\frac{l}{R}$.

The gravitational potential energy of an infinitesimally small element of the chain with respect to the center of sphere is

$\left(\frac{m}{l}Rd\varphi \right)g\left(Rcos\varphi \right)$

Thus the total gravitational potential energy of chain initially=${\int }_0^{l/R}\frac{m}{l}{R}^{2}gcos\varphi d\varphi$

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

The gravitational potential energy of chain after falling through angle $\theta =$ ${\int }_{\theta }^{\theta +l/R}\frac{m}{l}{R}^{2}gcos\varphi d\varphi$

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

Thus the gain in kinetic energy of the chain = Loss in potential energy
$\Delta KE=U_i-U_f$
$\Delta KE=\frac{m{R}^{2}g}{l}[sin\frac{l}{R}+sin\theta -sin(\theta +\frac{l}{R}\left)\right]$