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A particle starting from rest moves in a circle of radius rr. It attains a velocity of v0 m/sv_0 \text{ m/s} on completion of nn rounds. Its angular acceleration will be:

A

v0n rad/s2\frac{v_0}{n} \text{ rad/s}^2

B

v022πnr2 rad/s2\frac{v_0^2}{2\pi n r^2} \text{ rad/s}^2

C

v024πnr2 rad/s2\frac{v_0^2}{4\pi n r^2} \text{ rad/s}^2

D

v024πnr rad/s2\frac{v_0^2}{4\pi n r} \text{ rad/s}^2

Step-by-Step Solution

Given: Initial angular velocity, ω0=0\omega_0 = 0 (since the particle starts from rest) Final linear velocity, v=v0v = v_0 Radius of the circle =r= r Final angular velocity, ω=vr=v0r\omega = \frac{v}{r} = \frac{v_0}{r} Angular displacement, θ=2πn\theta = 2\pi n (since it completes nn rounds)

Using the third equation of rotational kinematics: ω2=ω02+2αθ\omega^2 = \omega_0^2 + 2\alpha\theta

Substituting the values: (v0r)2=0+2α(2πn)\left(\frac{v_0}{r}\right)^2 = 0 + 2\alpha(2\pi n) v02r2=4πnα\frac{v_0^2}{r^2} = 4\pi n \alpha α=v024πnr2 rad/s2\alpha = \frac{v_0^2}{4\pi n r^2} \text{ rad/s}^2

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