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

Solution :
According to question, torque $$\tau = 0$$
It means that, $$\alpha = 0$$
$$\eqalign{ & \alpha = \frac{{{d^2}\theta }}{{d{t^2}}} \cr & {\text{Given,}}\,\,\theta \left( t \right) = 2{t^3} - 6{t^2} \cr & {\text{So,}}\,\,\frac{{d\theta }}{{dt}} = 6{t^2} - 12t \cr & \alpha = \frac{{{d^2}\theta }}{{d{t^2}}} = 12t - 12 \cr & 12t - 12 = 0 \Rightarrow t = 1\,s \cr} $$