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

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t = 0.25s

t = 2s

t =1s

t = 0.5s

answer is A.

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Detailed Solution

$\begin{array}{l} θ = 2 t^{3} - 6 t^{2} \\ \Rightarrow ω = \frac{dθ}{dt} = 6 t^{2} - 12 t \\ \Rightarrow α = \frac{dω}{dt} = 12 t - 12 \end{array}$
$Torque τ = Iα = 0 \Rightarrow α = 0 so t = 1 s$

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