Question

A uniform solid brass sphere is rotating with angular speed ${\omega }_0$ about a diameter. If its temperature is now increased by ${100}^{\circ }\text{C}$. What will be its new angular speed? (Given ${\alpha }_B=2.0\times {10}^{-5}{\text{per}}^{\circ }C$)

• A. $1.1{\omega }_0$
• B. $1.01{\omega }_0$
• C. $0.996{\omega }_0$
• D. $0.824{\omega }_0$

Answer 1

The correct option is C $0.996{\omega }_0$

Let $M$ be the mass of the sphere and $R_o$ be its radius before increasing the temperature.
After increasing the temperature,
$R=R_o(1+{\alpha }_B\Delta \theta )$
$\therefore R^2=R_o^2(1+2{\alpha }_B\Delta \theta )$
Then from conservation of angular momentum :
$I\omega =I_0{\omega }_0$
or $\omega =\frac{I_0}{I}.{\omega }_0$

$=\frac{\frac{2}{5}M{R}_o^2{\omega }_0}{\frac{2}{5}M{R}_o^2(1+2{\alpha }_B\Delta \theta )}=\frac{{\omega }_0}{1+2{\alpha }_B\Delta \theta }$
$=\frac{{\omega }_0}{1+2(2.0\times {10}^{-5})\left(100\right)}={\omega }_o(1+4\times {10}^{-3}{)}^{-1}={\omega }_o(1-4\times {10}^{-3})=0.996{\omega }_0$