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.

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