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°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°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°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°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, $\text{I}=\text{I}_0(1+2\alpha\theta)$
