The instantaneous angular position of a point | vedclass

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Question

(A) Given the angular position $	heta(t) = 2t^3 - 6t^2$.First,find the angular velocity $\omega$ by differentiating $	heta(t)$ with respect to time

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$1$

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(A) Given the angular position $	heta(t) = 2t^3 - 6t^2$.First,find the angular velocity $\omega$ by differentiating $	heta(t)$ with respect to time $t$:$\omega = \frac{d	heta}{dt} = \frac{d}{dt}(2t^3 - 6t^2) = 6t^2 - 12t$.Next,find the angular acceleration $\alpha$ by differentiating $\omega$ with respect to time $t$:$\alpha = \frac{d\omega}{dt} = \frac{d}{dt}(6t^2 - 12t) = 12t - 12$.The torque $	au$ is given by $	au = I\alpha$,where $I$ is the moment of inertia.For the torque to be zero,$	au = 0$,which implies $I\alpha = 0$. Since $I 
eq 0$,we must have $\alpha = 0$.Setting $\alpha = 12t - 12 = 0$,we get $12t = 12$,which results in $t = 1 	ext{ sec}$.

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t) = 2t^3 - 6t^2$. The torque on the wheel becomes zero at $t = \dots \text{sec}$.

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