A particle moving on a circular path travels first one third part of circumference in 2 sec & next one hand part in 1 sec. Average angular velocity of the particle is (in rad/sec)-

\frac{2 π}{3}

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\frac{π}{3}

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\frac{4 π}{9}

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\frac{5 π}{3}

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Solution

The correct option is A $\frac{2 π}{3}$
Angular velocity = $\frac{a n g u l a r d i s p l a c e m e n t}{t i m e}$

For the first one-third part

$θ_{1} = \frac{s_{1}}{r} = \frac{\frac{2 π r}{3}}{r}$

For the second one third part $θ_{2} = \frac{\frac{2 π r}{3}}{r} = \frac{2 π}{3} r a d$

Total angular displacement $\begin{align}

$θ = θ_{1} + θ_{2} = \frac{4 π}{3}$

Total time = 3sec
$ω_{a v} = \frac{\frac{4 π}{3}}{3} = \frac{2 π}{3} r a d / s$

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A particle moving on a circular path travels first one third part of circumference in 2 sec & next one hand part in 1 sec. Average angular velocity of the particle is (in rad/sec)-

The correct option is A 2π3Angular velocity =angulardisplacementtime For the first one-third part θ1=s1r=2πr3r For the second one third part θ2=2πr3r=2π3rad Total angular displacement $\begin{align} θ=θ1+θ2=4π3 Total time = 3secωav=4π33=2π3rad/s