Find out the increase in moment of inertia I of a uniform rod (coeffic

1.	Find out the increase in moment of inertia I of a uniform rod (coefficient of linear expansion α) about its perpendicular bisectors when its temperature is slightly increased ∆T.
Answer» M.I. about its axis along perpendicular bisector = \(\frac{1}{12}\) ml² When temperature increased by ∆T, length of rod increases. ∆l = lα∆T ∴ New M.I., I₁ = \(\frac{M}{12}\) (l + ∆l)² = \(\frac{M}{12}\) (l² + ∆l² + 2l∆l) Neglecting (∆l)² (very very small quantity) – I₁ = \(\frac{M}{12}\) (l² + 2l∆l) = \(\frac{Ml^2}{12}+\frac{Ml∆l}{6}\) = I + \(\frac{Ml∆l}{6}\) . Therefore, moment of inertia, increase. ∆I = I₁ − I = \(\frac{M_l∆l}{6}\) = 2( \(\frac{Ml^2}{12}\) )\(\frac{∆l}{l}\) ∆l = 2l ∝ ∆T.

Find out the increase in moment of inertia I of a uniform rod (coefficient of linear expansion α) about its perpendicular bisectors when its temperature is slightly increased ∆T.

Answer»

M.I. about its axis along perpendicular bisector

= \(\frac{1}{12}\) ml²

When temperature increased by ∆T, length of rod increases.

∆l = lα∆T

∴ New M.I., I₁ = \(\frac{M}{12}\) (l + ∆l)² = \(\frac{M}{12}\) (l² + ∆l² + 2l∆l)

Neglecting (∆l)² (very very small quantity) –

I₁ = \(\frac{M}{12}\) (l² + 2l∆l)

= \(\frac{Ml^2}{12}+\frac{Ml∆l}{6}\) = I + \(\frac{Ml∆l}{6}\) .

Therefore, moment of inertia, increase.

∆I = I₁ − I = \(\frac{M_l∆l}{6}\)

= 2( \(\frac{Ml^2}{12}\) )\(\frac{∆l}{l}\)

∆l = 2l ∝ ∆T.