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

• A. t = 1 s
• B. t = 0.5 s
• C. t = 0.25 s
• D. t = 2 s

Answer 1

Answer: (A)

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