Find out the increase in moment of inertia $I$ of a uniform rod (coefficient of linear expansion a) about its perpendicular bisector when its temperature is slightly increased by $\Delta T$.
Find out the increase in moment of inertia $I$ of a uniform rod (coefficient of linear expansion a) about its perpendicular bisector when its temperature is slightly increased by $\Delta T$.
Suppose mass of rod is $\mathrm{M}$ and moment of inertia is $\mathrm{I}$. Moment of inertia of rod about perpendicular bisector,
$I=\frac{M l^{2}}{l^{2}}$
Increase in length due to increase in temperature $\Delta \mathrm{T}$, $\Delta l=l \alpha \Delta \mathrm{T}$
$\therefore$ Now, new moment of inertia of rod,
$\mathrm{I}^{\prime} =\frac{\mathrm{M}}{12}(l+\Delta l)^{2}$
$=\frac{\mathrm{M}}{12}\left(l^{2}+2 l \Delta l+\Delta l^{2}\right)$
But, $\Delta l$ is very small, $\Delta l^{2}$ is neglected.
$\mathrm{I}^{\prime}=\frac{\mathrm{M}}{12}\left(l^{2}+2 l \Delta l\right)$
$=\frac{\mathrm{M} l^{2}}{12}+\frac{\mathrm{M} l \Delta l}{6}$
$=\mathrm{I}+\frac{\mathrm{M} l \Delta l}{6}$
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