Question

A chain of length $l$ and mass $m$ lies on the surface of asmooth sphere of radius$R>l$ with one end tied to thetop of the sphere. (a) Find the gravitational potentialenergy of the chain with reference level at the centre of the sphere. (b) Suppose the chain is released and slidesdown the sphere. Find the kinetic energy of the chain,when it has slid through an angle $\theta$. (c) Find the tangential acceleration $\frac{dv}{dt}$ of the chain when the chain starts sliding down.

Answer 1

Since the sphere is smooth, friction doesn't come into play. As such, we can safely assume the entire mass of the chain to be concentrated at its center of mass.

The chain subtends an angle of, say, $\varphi$ at the centre. Then due to symmetry, its centre of mass will lie at an angle of $\frac{\varphi }{2}$ from the upward vertical.

Thus by definition, $l=R\varphi \Rightarrow \varphi =\frac{l}{R}$

By applying simple trigonometry, we find that this point is at a height $R\cos \left(\frac{\varphi }{2}\right)$ from the centre. Thus, the gravitational PE of the chain from the reference (datum) level of the centre of the sphere will be:

$PE=mgR\cos \left(\frac{l}{2R}\right)\to \text{(A)}$

We retain our earlier result about assuming the entire mass at the centre of mass, Assuming height descended by the centre of mass as $h$, and applying the work-energy theorem we get:

$mgh$ = Final KE - Initial KE

Through fairly-trivial trigonometry, it can be shown that:

$h=R\left(\cos \right(\frac{\varphi }{2})-\cos (\theta +\frac{\varphi }{2}\left)\right)$, which can be further simplified to:

$h=2R\sin \left(\frac{\theta }{2}\right)\sin \left(\frac{\varphi +\theta }{2}\right)$

Thus, final KE = $2mgR\sin \left(\frac{\theta }{2}\right)\sin \left(\frac{\varphi +\theta }{2}\right)\to \text{(B)}$

Tangential acceleration $a_t$ of the chain will be given by applying Newton's 2nd law to the centre of mass at the point of release. For the given FBD, tangential acceleration would be along the 'x' direction. Thus:

$m{a}_t=mg\sin \left(\frac{\varphi }{2}\right)$

$\therefore a_t=gsin\left(\frac{\varphi }{2}\right)\to \text{(C)}$