An object of radius R and mass M is rolling horizontally without slipping with speed v . It then rolls up the hill to a maximum height `h = (3v^(2))/(4g)` . The moment of inertia of the object is ( g = acceleration due to gravity)

Options

A. `(2)/(5) MR^(2)`
B. `(MR^(2))/(2)`
C. `MR^(2)`
D. `(3)/(2) MR^(2)`

Answer

B

Explanation

The kinetic energy given to the object at the base of inclined plane is transformed to PE at height h .
Thus , applying conservation of energy
Initial KE = Final PE
`(1)/(2) Mv^(2) + (1)/(2) I omega^2 = Mgh`
`(1)/(2) Mv^(2) + (1)/(2) I((v^(2))/(R^(2))) = Mg ((3v^(2))/(4g))`
`implies (1)/(2) M + (1)/(2) (I)/(R^(2)) = (3M)/(4)`
`implies (1)/(2R^(2)) = ((3)/(4) - (1)/(2)) M`
`implies I = 2 ((1)/(4)) MR^(2) = (1)/(2) MR^(2)`