The instantaneous angular position of a point | vedclass

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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the angular position: $	heta(t) = 2t^3 - 6t^2$. First,find the angular velocity $\omega$ by differentiating $	heta$ with respect to time $t$: $\omega = \frac{d	heta}{dt} = 6t^2 - 12t$. Next,find the angular acceleration $\alpha$ by differentiating $\omega$ with respect to time $t$: $\alpha = \frac{d\omega}{dt} = 12t - 12$. The torque $	au$ is related to angular acceleration by the equation $	au = I\alpha$,where $I$ is the moment of inertia. For the torque to be zero,the angular acceleration $\alpha$ must be zero (assuming $I 
eq 0$). Setting $\alpha = 0$: $12t - 12 = 0$. Solving for $t$: $12t = 12$,which gives $t = 1 	ext{ s}$.

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 = $ ....... $\text{sec}$.

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