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 `llt1/2piR`.

Explanation

Let us consider an element of length `ds` at an angle `varphi` 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-lambdads cos varphi=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=mw_t`
`(T+dT)+lambdadsgsin varphi-T=lambdadsw_t`
or, `dT+lambdaRdvarphigsin varphi=lambdadsw_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
`lambdagRunderset(0)overset(l//R)intsinvarphidvarphi=lambdaw_tintds=lambdalw_t`
Hence `w_t=(gR)/(l)(1-"cos"(l)/(R))`
As `w_n=a` at initial moment
So, `w=|w_t|=(gR)/(l)(1-"cos"l/R)`