A uniform chain of mass `m` and length `llt(piR)/(2)` is placed on a smooth hemisphere of radius `R` with on of its ends fixed at the top of the sphere.
(a) Find the gravitational potential energy of chain assuming base of hemisphere as reference.
(b) what will be the tangential acceleraion of the chain when it starts sliding down.
(c ) If the chain slides down the sphere, find the kinetic energy of the chain when it has slipped through an angle `beta`.

Explanation

(a) First, take an element of angle `d theta` at angle `theta` as shown
`Mass//length=(m)/(l)`
`dm=(m)/(l)Rd theta`
`du=dm gy=(mgR^(2))/(l)costheta d theta`
Angular subtendent by chain at the center
`alpha=(l)/(R )` `U=(mgR^(2))/(l)int_(0)^(alpha)cos theta d theta`
`=(mgR^(2))/(l)|sin theta|_(0)^(alpha)`
`=(mgR^(2))/(l)|sin theta|_(0)^(alpha)`
`=(mgR^(2))/(l)sin alpha=(mgR^(2))/(l)sin((l)/(R ))`
(b) Tangential force on the element:
`dF_(t)=dmgsintheta`
`F_(t)=(mgR)/(l)int_(0)^(alpha)sintheta d theta=(mgR)/(l)|-costheta|_(0)^(alpha)`
`=(mgR)/(l)(1-cosalpha)`
`a_(t)=(F_(t))/(m)=(gR)/(l)[1-cos((l)/(R ))]`
(c ) when the chain has slipped angle `beta:`
`U'=(mgR^(2))/(l)int_(beta)^(alpha+beta) costhetad theta=(mgR)/(l)|sintheta|_(beta)^(alpha)`
`=(mgR^(2))/(l){sin(alpha+beta)-sinbeta}`
By the conservation of energy
`0+U=K'+U'impliesK=U-U'`
`K'=(mgR^(2))/(l)[sin((i)/(R ))+sinbeta-sin((l)/(R )+beta)]`