Question

A ring is rotating with an angular acceleration given by $\alpha =4t^3-\frac{3}{2}{t}^2$, where $t$ is time. Find the equation of angular speed $\omega$ if the ring starts from rest.

• A. $t^4-\frac{t^3}{2}$
• B. $t^3-t$
• C. $\frac{t^4}{2}-t^3$
• D. $12t^2-3t$

Answer 1

The correct option is A $t^4-\frac{t^3}{2}$

We know that angular acceleration $\left(\alpha \right)$ is the time rate of change of angular velocity $\left(\omega \right)$

It is given the angular acceleration is,
$\alpha =4t^3-\frac{3}{2}{t}^2$

and hence for finding angular velocity we integrate $\left(\alpha \right)$

$\alpha =\frac{d\omega }{dt}\Rightarrow \omega =\int \alpha dt$
$\omega =\int (4t^3-\frac{3}{2}{t}^2)dt$
$=\frac{4t^4}{4}-\frac{3}{2}\frac{t^3}{3}+c$
As it starts from rest $t=0$, $\omega =0\Rightarrow c=0$
So,
$\omega =t^4-\frac{t^3}{2}$

Hence option A is the correct answer.