Q63 of 101 Page 132

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 of the chain when the chain starts sliding down.


The gravitational potential energy of the chain is


The chain is released and slides down the sphere


The tangential acceleration of the chain is


Given


The length of the chain is l, mass of the chain is m, and the radius of the sphere is given as R.


Formula Used


The formula for the total energy in terms of kinetic and potential energy is given as



where


The is the total energy in terms of kinetic and potential energy, m is the mass of the object, g is the acceleration in terms of gravity and l is the length of the object, is the angle of exit.


Explanation


(a) Let the angle formed by the chain is =


The length of the chain =


Therefore, the angle is written as


The length of the chain in terms of radius is


The force derivative of the chain is given as


The potential energy is calculated as


The P.E. after integration is



(b) The kinetic energy and the potential energy of the chain is equivalent to



The initial potential energy is calculated as



The change in the potential energy is


.


(c) Now to find the tangential velocity we use the equation of



After reduction the value of the is



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