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.

Answer

`(Mr^(2))/2`

Explanation

The total kinetic energy of the body
`K=K_(T)=K_(R)=1/2Mv^(2)+1/2Iomega^(2)`
`=1/2Mv^(2)[1+(M/(Mr^(2)))]` [as `v=romega]`
when it rolls up an iregular inclined plane of height `(h=3v^(2)/4g)`, its `KE` is converted into `PE`, so by conservation of mechanical energy
`1/2MV^(2)[=1+1/(Mr^(2))]=Mg[(3v^(2))/(4g)]`
Which on simplification gives `I=(Mr^(2))/2`. This result clearly indicates that the body is either a disc or a cylinder.