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The moment of inertia of a solid sphere ...

The moment of inertia of a solid sphere about an axis passing through its centre of gravity is `(2)/(5)MR^(2)` . What is its radius of gyration about a parallel axis at a distance 2R from the first axis ?

A

`(5)/(2)R`

B

`5R`

C

`sqrt((12)/(5))R`

D

`sqrt((22)/(5))R`

Text Solution

Verified by Experts

The correct Answer is:
D
Using the theorem of parallel axes,
`I=I_(c)+Mh^(2)" where h"=2R and I_(c)=(2)/(5)MR^(2)`
`therefore I=(2)/(5)MR^(2)+M.(2R)^(2)=(22)/(5)MR^(2)`
`I=MK^(2)=(22)/(5)MR^(2)`
`therefore K=sqrt((22)/(5))R`
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