Given: Initial angular velocity, $\omega_0 = 0$ (since the particle starts from rest) Final linear velocity, $v = v_0$ Radius of the circle $= r$ Final angular velocity, $\omega = \frac{v}{r} = \frac{v_0}{r}$ Angular displacement, $\theta = 2\pi n$ (since it completes $n$ rounds)

Using the third equation of rotational kinematics: $\omega^2 = \omega_0^2 + 2\alpha\theta$

Substituting the values: $\left(\frac{v_0}{r}\right)^2 = 0 + 2\alpha(2\pi n)$ $\frac{v_0^2}{r^2} = 4\pi n \alpha$ $\alpha = \frac{v_0^2}{4\pi n r^2} \text{ rad/s}^2$