The instantaneous angular position of a point on a rotating wheel is given by the equation $ heta(t

Question

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

Professor PC Thomas Classes · Answer

(a) : Given : $	heta(t)=2 t^{3}-6 t^{2}$\ $	herefore \frac{d 	heta}{d t}=6 t^{2}-12 t \Rightarrow \frac{d^{2} 	heta}{d t^{2}}=12 t-12$\ Angular acceleration, $\alpha=\frac{d^{2} 	heta}{d t^{2}}=12 t-12$\ When angular acceleration $(\alpha)$ is zero, then the torque on the wheel becomes zero\ $(\because 	au=I \alpha)$\ $\Rightarrow 12 t-12=0$ or $t=1 \mathrm{~s}$