Question
The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta (t) = 2t^3-6t^2$. The torque on the wheel becomes zero at $t=$ ....... $\sec.$
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Answer
Given $: \theta(t)=2 t^{3}-6 t^{2}$
$\frac{\mathrm{d} \theta}{\mathrm{dt}}=6 \mathrm{t}^{2}-12 \mathrm{t}$
$\frac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}}=12 \mathrm{t}-12$
Angular acceleration, $\alpha=\frac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}}=12 \mathrm{t}-12$
When angular acceleration $(\alpha$ ) is zero, then the torque on the wheel becomes zero
$(\because \tau=\mathrm{I} \alpha)$
$\Rightarrow 12 t-12=0$ or $t=1\,s$
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