A block is released on the convex surface of a hemispherical wedge as shown in Fig. Determine the displacement of the wedge when the block reaches the angular position `theta`

Explanation

Applying work energy theorem

`W_("gravity")=/_\K'+/_\K_C`…………..i
`/_\K'=` change in `KE` w.r.t centre of mass from and `/_\K_C=` change in kinetic energy of centre of mass w.r.t ground.
Let `v` is the velocities of ball wilth respect to wedge and `V` is the velocity of wedge.
Linear mometum will be conserved in the horizontal direction . Hence `MV=(vcostheta-V)`
`implies V=(mvcostheta)/((m+M))` ...............ii
The velocity of centre of mas will be zero in horizontal direction but the velocity of centre of mass will change in vertical direction .
`(v_(cm))_y=(mvsintheta)/((m+M))` .............iii
Now applying WE theorem
`-mgR(1-costheta)=1/2muv^(2)+1/2(m+M)v_(cm,y)^(2)`
Here `mu=(mM)/((m+M))` and `v_(cm,y)=(mvsintheta)/((m+M))`
which gives `v=sqrt((2(M+m)gR(1-costheta))/((M+Msin^(2)theta)))`
and from eqn ii `V=sqrt((2m^(2)gR(1-costheta)cos^(2)theta)/((M+m)(M+Msin^(2)theta)))`