Question

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

• A. $8\mspace{6mu} rad/\sec$
• B. $12\mspace{6mu} rad/\sec$
• C. $24\mspace{6mu} rad/\sec$
• D. $36\mspace{6mu} rad/\sec$

Answer 1

Answer: (C)

$24\mspace{6mu} rad/\sec$

Answer 2

$\theta = 2t^{3} + 0.5$ and $\omega = \frac{d\theta}{dt} = 6t^{2}$

at t = 2 sec, $\omega = 6(2)^{2} = 24rad/sec$