If the equation for the angular displacement | vedclass

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(C) The angular displacement is given by $	heta(t) = 2t^3 + 0.5$.The angular velocity $\omega$ is the rate of change of angular displacement with res

Vedclass · Accepted Answer

$24$

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(C) The angular displacement is given by $	heta(t) = 2t^3 + 0.5$.The angular velocity $\omega$ is the rate of change of angular displacement with respect to time,given by $\omega = \frac{d	heta}{dt}$.Differentiating $	heta(t)$ with respect to $t$: $\omega = \frac{d}{dt}(2t^3 + 0.5) = 6t^2$.To find the angular velocity at $t = 2 \, s$,substitute $t = 2$ into the expression for $\omega$:$\omega = 6 	imes (2)^2 = 6 	imes 4 = 24 \, rad/s$.

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

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