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A particle starting from rest, moves in a circle of radius 'r'. It attains a velocity of $V_0$ m/s in the $n^{th}$ round. Its angular acceleration will be:
Detailed Solution
$\theta=2\pi n$, $\omega_f=\frac{V_0}{r}$, $\omega_i=0$. $\omega_f^2=2\alpha\theta \Rightarrow \alpha=\frac{(V_0/r)^2}{2(2\pi n)}=\frac{V_0^2}{4\pi nr^2}$.