Question

The angular position of a point on a rotating wheel varies with time $t$ as $\theta =2t^3-6t^2$, where $\theta$ is in radian and $t$ is in $s$. The angular velocity at the momment when torque on the wheel becomes zero, is

• A. $-3\text{rad/s}$
• B. $-6\text{rad/s}$
• C. $4\text{rad/s}$
• D. $12\text{rad/s}$

Answer 1

The correct option is B $-6\text{rad/s}$

$\theta =2t^3-6t^2$
$\Rightarrow \frac{d\theta }{dt}=\omega =6y^2-12t$
$\Rightarrow \frac{d\omega }{dt}=\alpha =12t-12=0$
$\Rightarrow t=1s$
$\therefore \omega =6\times (1{)}^2-12\times 1=-6$
$\omega =-6\text{rad/s}$