Question
a chain of length L and mass m lies on the surface of a smooth sphere of rad. R>L with one end tied to the top of sphere . A.find the grav.potential energy of the chain w.r.t. Centre of sphere . B.suppose the chain is released and slides down the sphere .find k.e. Of the chain when it slid through an angle @ . C.find the tangential acc. dv/dt of the chain when chain starts sliding down.Joshi sir comment
consider a small segment of chain at angle β from top, its length = Rdβ and mass = mRdβ/L
its potential energy = [mRdβ/L]gRcosβ now integrate it within the limits 0 to L/R
for getting the answer of the second part calculate potential energy twice once between the limits 0 to L/R and then between the limits θ to θ+(L/R)
then difference of potential energy = KE of the chain. solve it ?
for getting third part differentiate v obtained in the second part with respect to t, dθ/dt will be ω and it will be v/R
finally put θ and t = 0 for getting the answer.