Take the sphere as reference frame and conserve enegry.
mgR + 0∫θ(ma)(Rdθ)cosθ = mgRcosθ + 12mv2
substitue a = g
Akash Anand Why you are not considering the gravitational forces on small particle?
Sourish Ghosh I did. I conserved potential energy. Am I going wrong somewhere?
Akash Anand No..you are absolutely correct :)
Aniq Ur Rahman Howddugt dz trm 0∫θ(ma)(Rdθ)cosθ ???
Soumyadeep Basu Thanks.
Sourish Ghosh @aniq that is the work done by the force 'ma'. Note: 'ma' is the pseudo force we must consider while taking the sphere as the frame of reference.