A small ball is placed at the top of a smooth hemispherical wedge of radius `R`. If the wedge is accelerated with an acceleration `a`, find the velocity of the ball relative to wedge as a function of `theta`.

Options

A. `sqrt(2R[g(1-sintheta)+aRcostheta])`
B. `sqrt(2R[g(1+costheta)-aRsintheta])`
C. `sqrt(2R[g(1-costheta)+aRsintheta])`
D. `sqrt(2R[g(1+sintheta)-aRcostheta])`

Answer

C

Explanation

Here we analyze the motion of ball with respect to wedge. As we are analyzing the motion of the ball from accelerating wedge, we impose a pseudo force `mararron` the ball.
The work done by the pseudo force is

`W_(ps)=max`
The work done by gravity is `W_(gr)=mgy`
The work done by normal reaction force is zero as there is no relative sliding between the ball and the wedge in the direction of normal reaction.
Then, the total work done is
`W=max+mgy` (i)
KE: The change in kinetic energy of the ball relative to wedge is
`K=1/2mv^2` (ii)
where v is the velocity of the ball relative to wedge.
W-E theorem relative to wedge:
`W=DeltaK` (iii)
From Eqs (i), (ii), and (iii), we have
`max+mgy=1/2mv^2`
Substituting `x=R sin theta` and `y=R(1-cos theta)`, we have
`v=sqrt(2{gR(1-costheta)+aRsintheta})`