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

`PE = KE_("roll")`
`mgh = (1)/(2)mv^(2)[1+K^(2)//R^(2)]`
`g xx(3v^(2))/(4g)=(1)/(2)v^(2)[1+K^(2)//R^(2)]`
`(3)/(2)=1+(K^(2))/(R^(2))`
`(1)/(2)=(K^(2))/(R^(2))`
`K^(2)=(R^(2))/(2)` `I = MK^(2)`
`= (MR^(2))/(2)`