If equation for displacement of a particle moving on a circular path is given by (θ) = 2t^3 + 0.5 ,

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3-2.Motion in Plane

medium

If the equation for the displacement of a particle moving on a circular path is given by $(\theta) = 2t^3 + 0.5$, 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$36$

Solution

$\omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(2 \mathrm{t}^{3}+0.5\right)=6 \mathrm{t}^{2}$
at $t=2 s, \omega=6 \times(2)^{2}=24 r a d / s$

Standard 11

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If the equation for the displacement of a particle moving on a circular path is given by $(\theta) = 2t^3 + 0.5$, where $\theta$ is in radians and $t$ in seconds, then the angular velocity of the particle after $2\, sec$ from its start is ......... $rad/sec$

Solution

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