Question

A solid body rotates with deceleration about a stationary axis with an angular deceleration $\beta \propto \sqrt{\omega }$, where $\omega$ is its angular velocity. If the mean angular velocity of the body averaged over the whole time of rotation is $⟨\omega ⟩=\frac{{\omega }_0}{x}$, (at the initial moment of time its angular velocity was equal to ${\omega }_0$). Find the value of $x$.

Answer 1

In accordance with the problem, ${\beta }_z<0$
Thus, $-\frac{d\omega }{dt}=k\sqrt{\omega }$, where $k$ is proportionality constant
$\Rightarrow -{\int }_{\sqrt{{\omega }_0}}^{\sqrt{\omega }}\frac{d\omega }{\sqrt{\omega }}=k{\int }_0^{t}dt$ or, $\sqrt{\omega }=\sqrt{{\omega }_0}-\frac{kt}{2}$
when, $\omega =0$, total time of rotation, $t=\tau =\frac{2\sqrt{{\omega }_0}}{k}$
Average angular velocity $⟨\omega ⟩=\frac{\int \omega dt}{\int dt}=\frac{{\int }_0^{\frac{2\sqrt{{\omega }_0}}{k}}({\omega }_0+\frac{k^2{t}^2}{4}-kt\sqrt{{\omega }_0})dt}{\frac{2\sqrt{{\omega }_0}}{k}}$
Hence, $⟨\omega ⟩=\frac{{[{\omega }_{0}t+\frac{k^2{t}^3}{12}-\frac{k}{2}\sqrt{{\omega }_0}{t}^2]}_0^{\frac{2\sqrt{{\omega }_0}}{k}}}{2\frac{\sqrt{{\omega }_0}}{k}}=\frac{{\omega }_0}{3}$

Therefore, $x=3$