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

Question

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

Dr. Abubacker · Accepted Answer

(c) $\omega = \frac{d	heta}{dt} = \frac{d}{dt}\left( 2t^{3} + 0.5 ight) = 6t^{2}$
At $t = 2\ s,\ \omega = 6 	imes (2)^{2} = 24\ rad/s$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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