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An object of radius R and mass M is roll...

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)

A

`(2)/(5)mR^(2)`

B

`(mr^(2))/(2)`

C

`mR^(2)`

D

`(3)/(2)mR^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
When the body rolls, it has both translational and rotational Kinetic energies.
Translational K.E. `=(1)/(2)mv^(2)`
and Rotational K.E. `=(1)/(2)I omega^(2)=(1)/(2)I(v^(2))/(R^(2))`
`therefore` Total energy at `A=E_(A)=(1)/(2)mv^(2)+(1)/(2)(Iv^(2))/(R^(2))`
`=(1)/(2)v^(2)(m+(1)/(R^(2)))`
AT B it has only P.E. = mgh
`=mg.(3v^(2))/(4g)=(3mv^(2))/(4)=E_(B)`
By the principle of conservation of energy
`E_(A)=E_(B) therefore (1)/(2)v^(2)m+(I^(R^(2)))=(3)/(4)mv^(2)`
`therefore m+(I)/(R^(2))=(3)/(2)m therefore I=(mR^(2))/(2)`
This is the M.I. of a disc. `therefore` The body is a disc
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