Question

A top is spinning in clockwise direction about a stationary axis. The angular acceleration of the top is given by $\alpha =-\sqrt{\theta }$, where $\theta$ is angular displacement of the body. At $t=0$, the angular velocity of body was $\sqrt{\frac{32}{3}}{\pi }^{\frac{3}{4}}$ and angular position of the body is zero. The angular velocity of the body when it finishes two revolutions is

• A. $\frac{8}{3}{\pi }^{\frac{3}{2}}$
• B. $\frac{4}{3}{\pi }^{\frac{3}{2}}$
• C. zero
• D. $\frac{2}{3}{\pi }^{\frac{3}{2}}$

Answer 1

The correct option is C zero

Initial angular velocity ${\omega }_0=\sqrt{\frac{32}{3}}rad/s$ and
Initial angular position ${\theta }_0=0rad$
Given that, $\beta =-\sqrt{\theta }$ (negative angular acceleration)
We know that $\beta =\omega \frac{d\omega }{d\theta }$
$\therefore \omega \frac{d\omega }{d\theta }=-\sqrt{\theta }$
$\omega dw=-\sqrt{\theta }d\theta$
Integrating on both sides,
${\int }_{{\omega }_0}^{\omega }\omega d\omega =-{\int }_{{\theta }_0}^{\theta }\sqrt{\theta }d\theta$
$\Rightarrow {\left[\frac{{\omega }^2}{2}\right]}_{{\omega }_0}^{\omega }={[-\frac{2}{3}{\theta }^{\frac{3}{2}}]}_{{\theta }_0}^{\theta }$
$\Rightarrow \frac{{\omega }^2-{\omega }_0^2}{2}=\frac{-2}{3}{\theta }^{\frac{3}{2}}$
$\Rightarrow {\omega }^2={\omega }_0^2-\frac{4}{3}{\theta }^{\frac{3}{2}}$
For two revolutions, $\theta =4\pi$ and given ${\omega }_0=\sqrt{\frac{32}{3}}{\pi }^{\frac{3}{4}}$
$\Rightarrow {\omega }^2=\frac{32}{3}{\pi }^{\frac{3}{2}}-\frac{4}{3}(4\pi {)}^{\frac{3}{2}}$
$\Rightarrow {\omega }^2=\frac{32}{3}{\pi }^{\frac{3}{2}}-\frac{32}{3}{\pi }^{\frac{3}{2}}$
$\Rightarrow \omega =0rad/s$