Question

A uniform solid brass sphere is rotating with angular speed ${\omega }_0$ about a diameter. If its temperature is now increased by ${100}^{\circ }C.$ What will be its new angular speed? (Given ${\alpha }_B=2\times {10}^{\frac{-5}{{}^{\circ }C}})$

• A. $\frac{{\omega }_0}{1-0.002}$
• B. $\frac{{\omega }_0}{1+0.002}$
• C. $\frac{{\omega }_0}{1+0.004}$
• D. $\frac{{\omega }_0}{1-0.004}$

Answer 1

The correct option is C $\frac{{\omega }_0}{1+0.004}$

$I_0{\omega }_0=I_t{\omega }_t\frac{2}{5}M{r}_0^2{\omega }_0=\frac{2}{5}M{r}_0^2(1+2\alpha \Delta T){\omega }_t{\omega }_t=\frac{{\omega }_0}{1+0.004}$