Question

A point on the rim of a disc starts circular motion from rest and after time $t$, it gains an angular acceleration given by $\alpha =3t-t^2$. The angular velocity after $2\text{sec}$ is:

• A. $\frac{10}{3}\text{rad/s}$
• B. $\frac{5}{2}\text{rad/s}$
• C. $\frac{5}{4}\text{rad/s}$
• D. $\frac{5}{3}\text{rad/s}$

Answer 1

The correct option is A $\frac{10}{3}\text{rad/s}$

Angular acceleration is the rate of change of angular velocity w.r.t time.

$\therefore \alpha =\frac{d\omega }{dt}$

$\Rightarrow d\omega =\alpha dt$

$\Rightarrow d\omega =(3t-t^2)dt$

Putting the limits of time from $t=0$ to $t=2\text{sec}$.

$\omega =\int d\omega =\underset{0}{\overset{2}{\int }}(3t-t^2)dt$

$\Rightarrow \omega ={\left[\frac{3t^2}{2}\right]}_0^2-{\left[\frac{t^3}{3}\right]}_0^2$

$\Rightarrow \omega =3[2-0]-\left[\frac{8}{3}\right]$

$\therefore \omega =6-\frac{8}{3}=\frac{10}{3}\text{rad/s}$

Hence, option $\left(b\right)$ is correct.

Why this Question? Caution: The kinetic equation can be used only when $\alpha$ is constant, in problems involving variable $\omega$ or $\alpha$, we have to use the basic definition of $\alpha$ or $\omega$ and solve it using differentiation or integration (as applicable).