Question
The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t)=2 t^{3}-6 t^{2}$. The torque on the wheel becomes zero at(a) $t=15$(b) $t=0.5 \mathrm{~s}$(c) $t=0.25 \mathrm{~s}$(d) $t=2 s$
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Step 1: We are given the instantaneous angular position of a point on the rotating wheel as $\theta(t)=2 t^{3}-6 t^{2}$. Show more…
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