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  • Find out the increase in moment of inertia I of a uniform rod about its perpendicular bisector when its temperature is slightly increased by ∆T.

 Find out the increase in moment of inertia I of a uniform rod about its perpendicular bisector when its temperature is slightly increased by \Delta T.

Answers (1)

Now, we know that moment of inertia (I) of a rod when its axis is along perpendicular bisector it is =

\frac{1}{12}ML^{2}

Now,\Delta L = \alpha .L\Delta T

Thus, I^{'} = \frac{1}{12} M (L + \Delta L)^{2}

            = \frac{1}{12} M (L ^{2}+\Delta L^{2}+2L \Delta L)^{2}

Since \Delta L^{2} is a very small term, we will neglect it,

= \frac{M}{12} (L ^{2}+2L \Delta L)

   = \frac{ML^{2}}{12}+\frac{ML\Delta L}{6}\times \frac{2L}{2L}

= \frac{ML^{2}}{12}+\frac{ML^{2}}{12}. \frac{2\Delta L}{L}

I^{'} = I (1 + 2\alpha \Delta T)

Thus, the new moment of inertia will increase by 2I\alpha \Delta T.

Posted by

infoexpert24

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