If equation for displacement of a particle moving on a circular path is given as θ = 2{t^3} +

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3-2.Motion in Plane

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If the equation for the displacement of a particle moving on a circular path is given as

$\theta = 2{t^3} + 0.5,$

where $\theta $ is in radians and $t$ is in second. Then the angular velocity of the particle after two second will be ....... $rad/sec$

A

$36$

B

$8$

C

$48$

D

$24$

Solution

$\theta = 2{t^3} + 0.5$

$\omega = \frac{{d\theta }}{{dt}} = 6{t^2} = 6 \times {2^2}$

$ = 6 \times 4 = 24\,radian/\sec $

Standard 11

Physics

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