Question

A chain of length $l$ and mass $m$ lies on the surface of a smooth sphere of radius $R>l$ with on end tied to the top of the sphere. Find the tangential acceleration $\frac{dv}{dt}$ of the chain when the chain starts sliding down.

• A. $\frac{Rg}{l}[1-\cos \left(\frac{l}{R}\right)]$
• B. $\frac{Rg}{l}[1+\cos \left(\frac{l}{R}\right)]$
• C. $\frac{Rg}{l}[2-\cos \left(\frac{l}{R}\right)]$
• D. $\frac{Rg}{l}[2+\cos \left(\frac{l}{R}\right)]$

Answer 1

The correct option is A $\frac{Rg}{l}[1-\cos \left(\frac{l}{R}\right)]$

The length of chain is $l$ and radius of circle is R

So the angle made by chain at the center is $\varphi =\frac{l}{R}$

$d{E}_i=dm\times gh=\frac{mg{R}^2}{l}cos\varphi d\varphi$

Initial potential energy $E_i=\int d{E}_i={\int }_o^{l/R}\frac{mg{R}^2}{l}cos\varphi d\varphi$

$E_i=\frac{mg{R}^2}{l}sin\frac{l}{R}$

Let the chain slides by $\theta$

So

Final potential energy $E_f=\int d{E}_f={\int }_{\theta }^{l/R+\theta }\frac{mg{R}^2}{l}cos\varphi d\varphi$

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

So change in PE is equal to KE

$KE=\frac{m{v}^2}{2}=E_i-E_f$

On putting the value we get

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

on differentiating wrt time

$2v\frac{dv}{dt}=\frac{2g{R}^2}{l}[cos\theta -cos(\theta +\frac{l}{R}\left)\right]\frac{d\theta }{dt}$ ...(1)

Also , $\frac{d\theta }{dt}=\omega$

and $\omega R=v$ (2)

on putting 2 in 1

$\frac{dv}{dt}=\frac{gR}{l}[cos\theta -cos(\theta +\frac{l}{R}\left)\right]$

since the moment chain was released $\theta =0$

$\frac{dv}{dt}=\frac{gR}{l}[1-cos(\frac{l}{R}\left)\right]$