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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^°C and a is the coefficient of linear expansion of the solid.

Answer

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

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Question 21 Mark

A steel wire of cross-sectional area 0.5mm² is held between two fixed supports. If the wire is just taut at 20^°C, determine the tension when the temperature falls to 0^°C. Coefficient of linear expansion of steel is 1.2 × 10^-5 °C^-1and its Young's modulus is 2.0 × 10¹¹N m^-2.

Answer

$\text{A}=0.5\text{mm}^2=0.5\times10^{-6}\text{m}^2$
$\text{T}_1=20^\circ\text{C},$
$\text{T}_2=0^\circ\text{C}$
$\alpha_\text{s}=1.2\times10^{-5}\ /^\circ\text{C},$
$\text{Y}=2\times2\times10^{11}\text{N/m}^2$
Decrease in length due to compression $=\text{L}\alpha\Delta\theta\ ...(1)$
$\text{Y}=\frac{\text{stress}}{\text{strain}}=\frac{\text{F}}{\text{A}}\times\frac{\text{L}}{\Delta\text{L}}$
$\Rightarrow\Delta\text{L}=\frac{\text{FL}}{\text{AY}}\ ...(2)$
Tension is developed due to (1) & (2)
Equating them,
$\text{L}\alpha\Delta\theta=\frac{\text{FL}}{\text{AY}}$
$\Rightarrow\text{F}=\alpha\Delta\theta\text{AY}$
$=1.2\times10^{-5}\times(20-0)\times0.5\times10^{-5}2\times10^{11}=24\text{N}$

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