Concepts Of Physics MCQ Edition [Volume 1]PhysicsWork and Energy
A chain having mass m and length l rests on the surface of a smooth sphere of radius R > l . One end of the chain is fastened to the top of the sphere. Assume the chain is released and begins sliding down the sphere. Determine the kinetic energy of the chain when it has slid through an angle .
Options
- AmRg [ ( l R ) + - ( + l R ) ]
- BmR^2 g l [ ( l R ) - + ( + l R ) ]
- CmR^2 g l [ ( l R ) + - ( + l R ) ]
- DmR^2 g l [ ( l R ) + - ( + l R ) ]
Correct answer
C. mR^2 g l [ ( l R ) + - ( + l R ) ]
Step-by-step solution
Let the linear mass density of the chain be = m l . Consider an element of the chain of length dl = R d at an angle from the vertical. The mass of this element is dm = R d . Taking the center of the sphere as the reference level, the potential energy of this element is dU = (dm)gh = ( R d )g(R ) = R^2 g d . Initially, the chain lies between = 0 and = l R . The initial potential energy of the chain is: U_i = ₀^ l/R R^2 g d = R^2 g [ ]₀^ l/R = R^2 g ( l R ) When the chain has slid through an angle , it lies between =