Question 11 Mark
Show that moment of inertia of a solid body of any shape changes with temperature as $\text{I}=\text{I}_0(1+2\alpha\theta),$ where $\text{I}_0$ is the moment of inertia at $0^\circ C$ and a is the coefficient of linear expansion of the solid.
Answer
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$\text{I}_0$ = Moment of Inertia at $0^\circ C$
$\alpha =$ Coefficient of linear expansion
To prove, $\text{I}=\text{I}_0=(1+2\alpha\theta)$
Let the temp. change to $\theta$ from $0^\circ C$
$\Delta\text{T}=\theta$
Let $‘R\ ’$ be the radius of Gyration,
Now, $\text{R}'=\text{R}(1+\alpha\theta),$
$\text I_0=\text{MR}^2$ where $M$ is the mass.
Now, $\text{I}'=\text{MR}'^2=\text{MR}^2(1+\alpha\theta)^2\approx=\text{MR}^2(1+2\alpha\theta)$
$[$By binomial expansion or neglecting $\alpha^2\theta^2$ which given a very small value.$]$
so, $\text{I}=\text{I}_0(1+2\alpha\theta)$
$\text{I}_0$ = Moment of Inertia at $0^\circ C$
$\alpha =$ Coefficient of linear expansion
To prove, $\text{I}=\text{I}_0=(1+2\alpha\theta)$
Let the temp. change to $\theta$ from $0^\circ C$
$\Delta\text{T}=\theta$
Let $‘R\ ’$ be the radius of Gyration,
Now, $\text{R}'=\text{R}(1+\alpha\theta),$
$\text I_0=\text{MR}^2$ where $M$ is the mass.
Now, $\text{I}'=\text{MR}'^2=\text{MR}^2(1+\alpha\theta)^2\approx=\text{MR}^2(1+2\alpha\theta)$
$[$By binomial expansion or neglecting $\alpha^2\theta^2$ which given a very small value.$]$
so, $\text{I}=\text{I}_0(1+2\alpha\theta)$
