AS chain of length l and mass m lies o the surface of a smooth sphere of radius `Rgtl` 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 chin is released and slides down the sphere. Find the kinetic eneergy of the chain, when it has slid through an angle `theta` c. find the tangential acceleration `(dv)/(dt)` of the chain when the chain starts sliding down.

Answer

A, B

Explanation

Let us consider a small element which makes angle `dtheta` at the centre.
`:. dm=rho (m/L)Rdtheta`
a. Gravitational potential energy of dm with respect to centre of the sphere
`(dm)gRcostheta`
`=((mg)/L)R^2costheta dtheta

`:. total G.P.E. =int_0^(L/R) mg R^2/L costheta dtheta`
`=(L/R)`
`=(mR^2g)/L[sintheta][theta=L/R]`
`=(mR^2g)/Lsin(L/R)`
b. when the chain is released from rest and slides down thrugh an angle theta, the K.E. of the chain is given by K.E. =change is potential energy.
`(mR^2g)/L sin (L/R)-int(gR^2)/L costheta dtheta`
`=(mR^2g)/L[sin(L/R)+sintehta-sin{theta+(L/R)}]`
c. Since `K.E. =1/2 mv^2`
`=(mR^2g)/L[sin(L/R)]`
taking derivative of both sides with respect to t,
`(1/2)xx2xxx(dv)/(dt)`
`=(R^2g)/L[costheta-(dtheta)/dt)-cos(theta+L/R)(dtheta)/(dt)]`
`:. [R-(dtheta)/(dt) (dv)/dt`
`=(R^2g)/Lxx(dtheta)/(dt)[costhet-cos(theta+L/R)]`
`(because`v=Romega=R (dtheta)/(dt))`
`:. (dv)/(dt)=(Rg)/L[costheta-cos(theta+L/R)]`
When the chain strrts sliding down,
`theta=0`
`:. (dv)/(dt)=(Rg)/L[1-cos(L/R)]`