Question

A chain of length $l$ is placed on a smooth spherical surface of radius $R$ with one of its ends fixed at the top of the sphere. What will be the acceleration $w$ of each element of the chain when its upper end is released? It is assumed that the length of the chain $l<\frac{1}{2}\pi R$.

• A. $w=[1-\cos \left(\frac{2l}{R}\right)]\frac{Rg}{l}$
• B. $w=[1+\cos \left(\frac{3l}{R}\right)]\frac{Rg}{l}$
• C. $w=[1-\cos \left(\frac{l}{R}\right)]\frac{Rg}{l}$
• D. $w=[1-\cos \left(\frac{4l}{R}\right)]\frac{Rg}{l}$

Answer 1

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

Let us consider an element of length $ds$ at an angle $\phi$ from the vertical diameter. As the speed of this element is zero at initial instant of time, it's centripetal acceleration is zero, and hence, $dN-\lambda ds\cos \phi =0$, where $\lambda$ is the linear mass density of the chain. Let $T$ and $T+dT$ be the tension at the upper and the lower ends of $ds$. We have from, $F_t=m{w}_t$
$(T+dT)+\lambda dsg\sin \phi -T=\lambda ds{w}_t$
or, $dT+\lambda Rd\phi g\sin \phi =\lambda ds{w}_t$
If we sum the above equation for all elements, the term $\int dT=0$ because there is no tension at the free ends, so
$\lambda gR{\int }_0^{\frac{l}{R}}\sin \phi d\phi =\lambda w_t\int ds=\lambda l{w}_t$
Hence $w_t=\frac{gR}{l}(1-\cos \frac{l}{R})$
As $w_n=a$ at initial moment
So, $w=|w_t|=\frac{gR}{l}(1-\cos \frac{l}{R})$