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.  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.

\[\therefore dm = \rho \left( \frac{m}{L} \right) Rd\theta\]

Since,

\[K . E . = \frac{1}{2}\text{m}\nu^2 \]
\[ = \frac{\text{m}R^2 g}{L} \left[ \sin \left( \frac{L}{R} \right) \right]\]

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

\[\left( \frac{1}{2} \right) \times 2\nu \times \frac{d\nu}{dt}\]

\[ = \frac{R^2 g}{L}\left[ \cos \theta - \frac{d\theta}{dt} - \cos \left( \theta + \frac{L}{R} \right)\frac{d\theta}{dt} \right]\]

\[ \therefore \left[ R - \frac{d\theta}{dt} \right]\frac{dv}{dt}\]

\[ = \frac{R^2 g}{L} \times \frac{d\theta}{dt}\left[ \cos \theta - \cos \left( \theta + \frac{L}{R} \right) \right]\]

\[\left( \text{ because } \nu = R\omega = R\frac{d\theta}{dt} \right)\]

\[ \therefore \frac{d\nu}{dt} = \frac{Rg}{L}\left[ \cos \theta - \cos \left( \theta + \frac{L}{R} \right) \right]\]

When the chain starts sliding down, \[\theta = 0^\circ\]

\[\therefore \frac{d\nu}{dt} = \frac{Rg}{L}\left[ 1 - \cos \left( \frac{L}{R} \right) \right]\]