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Question
Find out the increase in moment of inertia I of a uniform rod (coefficient of linear expansion α) about its perpendicular bisector when its temperature is slightly increased by ∆T.
Solution
Let the mass and length of a uniform rod be M and l respectively.
Moment of inertia of the rod about its perpendicular bisector. `(i) = (Ml^2)/12`
The increase in length of the rod when temperature is increased by ∆T is given by `∆l = l.α∆T` .....(i)
∴ New moment of inertia of the rod `(I) = M/12 (l + ∆l)^2`
= `M/12 (l^2 + ∆l^2 + 2I∆l)`
As the change in length ∆l is very small, therefore, neglecting `(∆l)^2`, we get
`I^' = M/12 (l^2 + 2l∆l)`
= `(Ml^2)/12 + (MI∆l)/6`
= `l + (MI∆l)/6`
∴ Increase in the moment of inertia `∆I = l - I`
= `(MI∆l)/6`
= `2 xx ((Ml^2)/12) (∆l)/l`
`∆I = 2*I α∆T` ......[Using equation (i)]
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