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: (2011 Pre)
Solution:
Angular velocity $\omega = \frac{d\theta}{dt} = 6t^{2} - 12t$. Angular acceleration $\alpha = \frac{d\omega}{dt} = 12t - 12$. Torque is zero when $\alpha = 0$, which implies $12t - 12 = 0$, or $t = 1\text{ s}$.
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