A body of mass `M` and radius `r`, rolling on a smooth horizontal floor with velocity `v`, rolls up an irregular inclined plane up to a vertical height `(3v^(2)//4g)`. Compute the moment of inertia of the body.

Explanation

The total kinetic energy of the body
`E=E+E=(1)/(2)Mv^(2)+(1)/(2)I omega^(2)`
`rArr E=(1)/(2)Mv^(2)[1+(I//Mr^(2))] " "["as" v = r omega]`
When it rolls up on an irregular inclined plane of height `h = (3v^(2)//4g)`, its KE is fully converted into PE, so by conservation of mechanical energy
`(1)/(2) Mv^(2)[1+(1)/(Mr^(2))]=Mg [(3v^(2))/(4g)]` which on simplification gives `I=(1//2)Mr^(2)` . This regult clearly indicates that the body is either a disc or a cylinder.