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The moment of inertia of the disc used i...

The moment of inertia of the disc used in a torsional pendulum about the suspension wire is `0.2kg-m^2`. It oscillates with a period of 2s. Asnother disc is placed over the first one and time period of the system becomes 2.5s. Find the moment of inertia of the second disc about the wire.

Text Solution

Verified by Experts

The correct Answer is:
`0.12`
As `T = 2 pi sqrt((I)/(C))`
Where I = M.I. of the system
C = Torsional constant
` :. 2 = 2 pi sqrt(( 0.2)/(C))`
`rArr (2 pi)/( sqrt(C)) = (2)/(sqrt((0.2))` . . . (i)
Now when new disc is loaded with another disc, then
`T = 2.5 = 2 pi sqrt((I.)/(C)) = (2 pi)/(sqrt(C)) sqrt(I.)`
`rArr 2.5 = (2)/(sqrt(0.2))` from (i)
`rArr 6 . 25 = (4)/(0.2) xx I.` `:. I. = 0.315`
Now required moment of inertia
`= 0.315 -0.2 = 0.115 ~~ 0.12 kg - m^(2)`
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