The instantaneous angular position of a point on a rotating wheel is given by the equation $Q(t)=2 t^3-6 t^2$ The torque on the wheel becomes zero at

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Aicharya · Accepted Answer

According to question, torque, $	au=0$ Its means that, $\alpha=0$ $\alpha=\frac{d^2 	heta}{d t^2}$ Given So $	heta(t)=2 t^3-6 t^2$ $\begin{aligned} \frac{d 	heta}{d t} & =6 t^2-12 t \ \frac{d^2 	heta}{d t^2} & =12 t-12 \quad\left(\because \alpha=\frac{d^2 	heta}{d t^2}=0ight) \ 12 t-12 & =0 \ t & =1 \mathrm{~s} \end{aligned}$

The instantaneous angular position of a point on a rotating wheel is given by the equation $Q(t)=2 t^3-6…

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