If the equation for the displacement of a particle moving on a circular path is given by $ heta = 2t^3 + 0.5$

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:

Accepted Answer

1.  **Definition of Angular Velocity:** Instantaneous angular velocity ($\omega$) is defined as the time rate of change of angular displacement ($	heta$). Mathematically, $\omega = \frac{d	heta}{dt}$ .
2.  **Differentiate the Equation:** Given $	heta = 2t^3 + 0.5$. Differentiating with respect to time $t$:
    $\omega = \frac{d}{dt}(2t^3 + 0.5) = 2(3t^2) + 0 = 6t^2$.
3.  **Calculate at Specific Time:** Substitute $t = 2$ s into the expression for $\omega$:
    $\omega = 6(2)^2 = 6(4) = 24 	ext{ rad/sec}$.

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