Question

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta \left(t\right)=3t^3-9t^2$ (Here $t$ is in seconds). The angular acceleration of the wheel becomes zero at

• A. $t=1\text{s}$
• B. $t=0.5\text{s}$
• C. $t=0.25\text{s}$
• D. $t=2\text{s}$

Answer 1

The correct option is A $t=1\text{s}$

We know that angular velocity is the rate of change of angular position of a rotating body and angular acceleration refers to the rate of change of angular velocity

It is given that $\theta \left(t\right)=3t^3-9t^2$

hence,
$\omega =\frac{d\theta }{dt}=9t^2-18t$
$\alpha =\frac{d^2\theta }{d{t}^2}=\frac{d\omega }{dt}=18t-18$
It is said that angular acceleration becomes $0$, hence
$\alpha =18t-18=0$
$t=1\text{s}$