The instantaneous angular position of a rotat | vedclass

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(A) The angular position is given by $	heta (t) = 2t^3 - 6t^2$.The angular velocity $\omega$ is the first derivative of angular position with respect

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(A) The angular position is given by $	heta (t) = 2t^3 - 6t^2$.The angular velocity $\omega$ is the first derivative of angular position with respect to time:$\omega = \frac{d	heta}{dt} = 6t^2 - 12t$.The angular acceleration $\alpha$ is the derivative of angular velocity with respect to time:$\alpha = \frac{d\omega}{dt} = 12t - 12$.According to Newton's second law for rotation,torque $	au = I\alpha$. For the torque to be zero,the angular acceleration $\alpha$ must be zero (assuming the moment of inertia $I 
eq 0$).Setting $\alpha = 0$:$12t - 12 = 0$$12t = 12$$t = 1 	ext{ s}$.

The instantaneous angular position of a rotating wheel is given by the formula $\theta (t) = 2t^3 - 6t^2$. At what time will the torque on this wheel be zero? $t = \dots \text{s}$

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