A solid ball rolls down a parabolic path ABC from a height h as shown in figure. Portion AB of the path is rough while BC is smooth. How high will the ball climb in BC?

Explanation

As path `AB` is rough, the ball will have both, translational and rotational `KE` at `B`.
Applying conservation of mechanical energy between `A` and `B`, we have
Loss in `PE` in going from `A` to `B =` Gain in `KE` of translation + Gain in `KE` of rotation
`mg h = (1)/(2) m upsilon^(2) + (1)/(2) I omega^(2)`
`= (1)/(2) m upsilon^(2) + (1)/(2)((2)/(5)m r^(2))omega^(2)`
`= (1)/(2)m upsilon^(2) + (1)/(5)m upsilon^(2) = (7)/(10) m upsilon^(2)`
`upsilon = sqrt((10 gh)/(7))` ..(i)
As the path `BC` is smooth, at `C`, there will be no change in rotational `KE`, but translational `KE` will become zero.
Applying conservation of mechanical energy between `B` and `C`,
`mg h' = (1)/(2)m upsilon^(2) = (1)/(2)m((10gh)/(7))`
`h' = (5)/(7)h`