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 D ν D T of the Chain When the - Physics

Numerical

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.

Solution

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, $\theta = 0^\circ$

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

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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 8 Work and Energy
Q 63.3 | Page 137