Question

A disc having mass $1\text{kg}$ and radius $2\text{m}$ is pivoted about its centre. Initially, it is at rest. A force $F=(5t^2+9)\text{N}$ starts acting tangentially on it. Find the angular velocity of the disc after $3\text{seconds}$ in $\text{rad/s}$

Answer 1

Given
Mass of the disc $=1\text{kg}$
Radius of the disc $=2\text{m}$
Force acting on the disc $F=(5t^2+9)$
Moment of inertia of the disc about its centre,
$I=\frac{M{R}^2}{2}$
$I=\frac{1\times 2^2}{2}=2{\text{kg m}}^2$
Initially, disc is at rest, hence initial angular momentum of the disc $L_1=I{\omega }_1=0$
As we know, angular impulse $J=\underset{t_1}{\overset{t_2}{\int }}\tau .dt$
Here,
$\tau =F.R=(5t^2+9)\times 2=(10t^2+18)\text{N-m}$
(because force acts tangentially)
$J=\underset{0}{\overset{3}{\int }}(10t^2+18)dt={[10\frac{t^3}{3}+18t]}_0^3$
$=90+54=144$
But $J=\underset{t_1}{\overset{t_2}{\int }}d\overrightarrow{L}=L_2-L_1$
$\Rightarrow 144=I.{\omega }_2-0$
Here, ${\omega }_2$ is the angular velocity of the disc after 3 seconds
$\Rightarrow 144=2\times {\omega }_2$
$\therefore {\omega }_2=72\text{rad/s}$