MCQ
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
    $12$
  • $24$
  • D
    $36$

Answer

Correct option: C.
$24$
c
$\omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(2 \mathrm{t}^{3}+0.5\right)=6 \mathrm{t}^{2}$

at $t=2 s, \omega=6 \times(2)^{2}=24 r a d / s$

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