Consider the situation as shown in the figure. If `m_(1)gtm_(2)`, I is the moment of intertia about its axis of rotation, `R` is the radius of pulley. The objects are released from rest separated by a vertical distance `2 h`. Find the translational speed of the objects as they pass each other. Assume no slipping.

Explanation

When objects cross each other, their speed is same, the angular speed of pulley is `omega`.

The block of mass `m_(1)` has gone down by `h`, block of mass `m_(2)` has gone up by `h`.
Loss in the gravitational potential energy of `m_(1)`
`=` gain in gravitational potential energy of `m_(2)`
`+` gain kinetic energy of (block `+` pulley)
`m_(1)gh=m_(2)gh+(1)/(2)m_(1)v^(2)+(1)/(2)m_(2)v^(2)+(1)/(2)Iomega^(2)`
`(m_(1)-m_(2))gh=(1)/(2)(m_(1)+m_(2)+(I)/(R^(2)))v^(2) [omega=(v)/(R)]`
`v=sqrt((2(m_(1)-m_(2))gh)/(m_(1)+m_(2)+(I)/(R^(2))))`