Two smooth balls of mass `m_1` and `m_2` connected by a light inextensible string are at the opposite points of horizontal diameter of a smooth semi cylindrical surface of radius R. If `m_1` is released, find its speed at any angular distance `theta` moved by `m_2`.

Explanation

Let the ball `m_2` moves through an angle `theta`. The mass `m` will fall through a distance `h_1=Rtheta`.
The ball `m_2` rises through a height `h_2` as,
`h_2=Rsintheta`.

The change in gravitational potential energy of `m_1` is
`DeltaU_1=-m_1gh_1=-m_1gRtheta`
(since `m_1` loses its potential energy as it falls down).
The change in gravitational potential energy of `m_2` is
`DeltaU_2=m_2gh_2=m_2gRsintheta`
(since `m_2` gains potential energy as it rises up)
`implies` The total change in gravitational potential energy=
`DeltaU=DeltaU_1+DeltaU_2` (i)
`implies DeltaU=-m_1gRtheta+m_2gRsintheta=gR(m_2sintheta-m_1theta)`.
The change in KE of the system `(m_1+m_2)`,
`DeltaK=1/2m_1v^2+1/2m_2v^2=((m_1+m_2)v^2)/(2)` (ii)
where v=speed of `m_1` and `m_2` at the positions as shown in the figure.
From the principle of conservation of energy we obtain,
`DeltaK+DeltaU=0` (iii)
Using (i), (ii) and (iii), we obtain,
`1/2(m_1+m_2)v^2-gR(m_1theta-m_2sintheta)=0`
`impliesv=sqrt((2gR(m_1theta-m_2sintheta))/((m_1+m_2)))`