Question

A chain of length $\ell$ is a 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 a of each element of the chain when its upper end is released? It is assumed that the length of the chain $l<\frac{1}{2}\pi R$.

Answer 1

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, $dT+\lambda Rd\phi g\sin \varphi =\lambda ds{w}_t$

If we sum the above equation for all elements, the term $\int dT=0$ because there is no tension at the free ends, so

$\lambda gR{\int }_0^{1/R}\sin \phi d\phi =\lambda w_t\int ds=\lambda l{w}_t$

Hence $w_t=\frac{gR}{l}(1-\cos \frac{1}{R})$

As $w_n=a$ at initial moment

So, $w=|w_t|=\frac{gR}{l}(1-\cos \frac{1}{R})$