A uniform solid brass sphere is rotating with angular speed ω₀ about a diameter. If its temperature is now increased by 100⁰C. What will be its new angular speed? (Given ${\mathrm{\alpha }}_{\mathrm{B}}=2.0\times {10}^{-5} \mathrm{per} {}^0\mathrm{C}$)

Let M be the mass of the sphere and R its radius before increasing the temperature.

$R\text{'}=R\left(1+\alpha \Delta \theta \right)$

$\Rightarrow R{\text{'}}^2=R^2{\left(1+\alpha \Delta \theta \right)}^2$

$\Rightarrow R{\text{'}}^2=R^2\left(1+2\alpha \Delta \theta \right)$

 Then from conservation of angular momentum 
$\begin{array}{l}\text{I}\omega {\text{=I}}_{\text{0}}{\omega }_0\\ or\omega =\frac{{\text{I}}_{\text{0}}}{\text{I}}{\omega }_0\\ \Rightarrow \omega =\frac{\frac{2}{5}M{R}^2{\omega }_0}{\frac{2}{5}M{R}^2(1+2\alpha \Delta \theta )}=\frac{{\omega }_0}{(1+2\alpha \Delta \theta )}\\ =\frac{{\omega }_0}{1+2(2.0\times {10}^{-5})\left(100\right)}=0.996{\omega }_0\end{array}$