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