Question

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

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

Answer 1

The correct option is A $t=1s$

Given, $\theta \left(t\right)=2t^3-6t^2$

We, know $T=\propto \frac{I{d}^2\theta }{d{t}^2}=I\frac{d}{dt}\left(\frac{d\theta }{dt}\right)=I\frac{d}{dt}(6t^2-12t)=I(12t-12)=0$

Then, $12t-12=0\Rightarrow t=1s$