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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

solid sphere

B

hollow sphere

C

disc

D

ring

Text Solution

Verified by Experts

The correct Answer is:
C
Let I bt the moment of inertia of the body. Then total `KE=1/2mv^(2)+1/2I omega^(2)`
or `KE=1/2mv^(2)+1/2 I (v^(2))/(R^(2))" "(omega=v/R)`
According to energy conservation loss in KE =gain in PE.
or `1/2(m+1/(R^(2)))v^(2)=mgh=mg((3v^(2))/(4g))`
Solving this we get `I=1/2mR^(2)`
i.e. the solid body is a disc.
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