Question

A chain length, $l$, is placed on smooth spherical surface of radius, $r$, with one of its ends fixed at the top of the surface. Length of chain is assumed to be $l<\frac{\pi r}{2}$. Acceleration of each element of chain when upper end is released is:

• A. $\frac{lg}{r}(1-\cos \frac{r}{l})$
• B. $\frac{rg}{l}(1-\cos \frac{l}{r})$
• C. $\frac{lg}{r}(1-\sin \frac{l}{r})$
• D. $\frac{rg}{l}(1-\sin \frac{l}{r})$

Answer 1

Answer: (B)

$\frac{rg}{l}(1-\cos \frac{l}{r})$

Consider a small element dl making an angle $d\theta$

$dm=\frac{m}{l}dl=\frac{m}{l}rd\theta$

Force acting on the element,

$dF=\left(dm\right)g\sin \theta =\frac{mrg}{l}\sin \theta d\theta$

So, $F=\frac{mrg}{l}{\int }_0^{\alpha }\sin \theta d\theta$, where, total angle $\alpha$ = $\frac{l}{r}$

$\therefore F=$ $\frac{m}{l}rg(1-cos\alpha )\frac{m}{l}rg(1-cos\frac{l}{r})$

or, $a=\frac{rg}{l}(1-cos\frac{l}{r})$.