If equation for displacement of a particle moving on a circular path is given by (θ ) = 2{t^3} + 0.

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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 by $(\theta ) = 2{t^3} + 0.5$, where $\theta $ is in radians and $t$ in seconds, then the angular velocity of the particle after $2 \,sec$ from its start is ......... $rad/sec$

A$8$

B$12$

C$24$

D$36$

(AIIMS-1998)

Solution

(c) $\omega = \frac{{d\theta }}{{dt}} = \frac{d}{{dt}}(2{t^3} + 0.5) = 6{t^2}$
at $t =2 \,s$, $\omega = 6 \times {(2)^2} = 24\,rad/s$

Standard 11

Physics

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