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Physics Question on Motion in a plane

A particle starting from rest, moves in a circle of radius 'r'. It attains a velocity of V0 m/s in the nthn^{th} round. Its angular acceleration will be :-

A

∨0nrad/s2\frac{\vee_{0}}{n}rad / s^{2}

B

∨022πnr2rad/s2\frac{\vee^{2}_{0}}{2\pi nr^{2}}rad / s^{2}

C

∨024πnr2rad/s2\frac{\vee^{2}_{0}}{4\pi nr^{2}}rad / s^{2}

D

∨044πnrrad/s2\frac{\vee^{4}_{0}}{4\pi nr}rad / s^{2}

Answer

∨024πnr2rad/s2\frac{\vee^{2}_{0}}{4\pi nr^{2}}rad / s^{2}

Explanation

Solution

θ=(2πn),ω0=0,ω=V0/r\theta = (2\pi n), \omega_0 = 0, \omega = V_0/r
α=ω2−ω022θ=(V0/r)2−02(2πn)\alpha = \frac{\omega^2 - \omega_0^2}{2\theta} = \frac{(V_0/r)^2 - 0}{2(2\pi n)}
=V024πnr2= \frac{V_0^2}{4\pi nr^2}