The instantaneous angular position of a point on a rotating wheel is given by the equation θ(t) = 2t3-6t2 The torque on the wheel becomes zero at

The instantaneous angular position of a point on a rotating wheel is given by the equation θ(t) = 2t3-6t2According to question, torque, τ = 0its means that, α = 0α = d2θdt2Given θ(t) = 2t3-6t2So, dθdt = 6t2-12t       d2θdt2= 12t-12                 (α = d2θdt2=0)      12t-12 = 0        t = 1 s