Physics - Circular Motion Question with Solution | TestHub

From third equation of motion for circular motion $${\omega }^2-{\omega }_0^2=2\alpha \theta$$ where, $\omega =$ final angular velocity of particle ${\omega }_0=$ initial angular velocity $\alpha =$ angular acceleration and $\theta =$ angular displacement Here, $\omega =\frac{{\mathrm{v}}_0}{\mathrm{r}}\mathrm{rad}/\mathrm{s}$ (where, $\mathrm{r}$ radius of the circle) ${\omega }_0=0$ (initially particle is at rest) $\theta =2\pi n$ (for $n$ rounds) Substituting these values in Eq. (i), we get $${\left(\frac{{\mathrm{v}}_0}{\mathrm{r}}\right)}^2-0=2\alpha (2\pi \mathrm{n})\Rightarrow \alpha =\frac{{\mathrm{v}}_0^2}{4\pi {\mathrm{nr}}^2}\mathrm{rad}/{\mathrm{s}}^2$$