Question

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

Answer 1

Let us consider a small element, which makes angle 'dθ' at the centre.

$\therefore dm=\rho \left(\frac{m}{\mathrm{L}}\right)\mathrm{R}d\mathrm{\theta }$

(a) Gravitational potential energy of 'dm' with respect to centre of the sphere

$$\begin{gathered} =\left(dm\right)g\mathrm{R}\cos \mathrm{\theta } \\ =\left(\frac{mg}{\mathrm{L}}\right){\mathrm{R}}^2\cos \mathrm{\theta }d\mathrm{\theta } \end{gathered}$$


$$\begin{gathered} \therefore \mathrm{Total}\mathrm{gravitational}\mathrm{potential}\mathrm{energy},E_{\mathrm{P}}=\underset{0}{\overset{\mathrm{L}/\mathrm{R}}{\int }}mg\frac{{\mathrm{R}}^2}{\mathrm{L}}\cos \mathrm{\theta }d\mathrm{\theta } \\ {E}_{\mathrm{P}}=\frac{m{\mathrm{R}}^{2}g}{\mathrm{L}}\left[\sin \mathrm{\theta }\right]\left[\mathrm{As},\mathrm{\theta }=\frac{\mathrm{L}}{\mathrm{R}}\right] \\ {E}_{\mathrm{P}}=\frac{m{\mathrm{R}}^{2}g}{\mathrm{L}}\sin \left(\frac{\mathrm{L}}{\mathrm{R}}\right) \end{gathered}$$

(b) When the chain is released from rest and slides down through an angle θ,

Change in K.E. of the chain = Change in potential energy of the chain

$$\begin{gathered} =\frac{m{\mathrm{R}}^{2}g}{\mathrm{L}}\sin \left(\frac{\mathrm{L}}{\mathrm{R}}\right)-\int \frac{g{\mathrm{R}}^2}{\mathrm{L}}\cos \mathrm{\theta }d\mathrm{\theta } \\ =\frac{m{\mathrm{R}}^{2}g}{\mathrm{L}}\left[\sin \left(\frac{\mathrm{L}}{\mathrm{R}}\right)+\sin \mathrm{\theta }-\sin \left\{\mathrm{\theta }+\left(\frac{\mathrm{L}}{\mathrm{R}}\right)\right\}\right] \end{gathered}$$

(c) Since,
$$\begin{gathered} \mathrm{K}.\mathrm{E}.=\frac{1}{2}m{\nu }^2 \\ =\frac{m{\mathrm{R}}^{2}g}{\mathrm{L}}\left[\sin \left(\frac{\mathrm{L}}{\mathrm{R}}\right)\right] \end{gathered}$$

Taking derivative of both sides with respect to 't', we get:

$$\begin{gathered} \left(\frac{1}{2}\right)\times 2\nu \times \frac{d\nu }{dt} \\ =\frac{{\mathrm{R}}^{2}g}{\mathrm{L}}\left[\cos \mathrm{\theta }-\frac{d\mathrm{\theta }}{dt}-\cos \left(\mathrm{\theta }+\frac{\mathrm{L}}{\mathrm{R}}\right)\frac{d\mathrm{\theta }}{dt}\right] \\ \therefore \left[\mathrm{R}-\frac{d\mathrm{\theta }}{dt}\right]\frac{dv}{dt} \\ =\frac{{\mathrm{R}}^{2}g}{\mathrm{L}}\times \frac{d\mathrm{\theta }}{dt}\left[\cos \mathrm{\theta }-\cos \left(\mathrm{\theta }+\frac{\mathrm{L}}{\mathrm{R}}\right)\right] \\ \left(\mathrm{because}\nu =\mathrm{R}\omega =\mathrm{R}\frac{d\mathrm{\theta }}{dt}\right) \\ \therefore \frac{d\nu }{dt}=\frac{\mathrm{R}g}{\mathrm{L}}\left[\cos \mathrm{\theta }-\cos \left(\mathrm{\theta }+\frac{\mathrm{L}}{\mathrm{R}}\right)\right] \end{gathered}$$

When the chain starts sliding down,
$\mathrm{\theta }=0$$°$
$\therefore \frac{d\nu }{dt}=\frac{\mathrm{R}g}{\mathrm{L}}\left[1-\cos \left(\frac{\mathrm{L}}{\mathrm{R}}\right)\right]$