Physics - Rotation Question with Solution | TestHub

Given, work done, W = 12000 J,
Initial frequency,$f_1=10Hz$
Angular velocity for rotational motion is given by
$\omega =2\pi f$
$\therefore {\omega }_1=2\pi f_1=2\pi \times 10=20\pi \frac{rad}{s}$
$and{\omega }_2=2\pi f_2=2\pi \times 20=\frac{40\pi rad}{s}$
According to work-energy theorem,
Work done in rotation = change in rotational kinetic energy
$\Rightarrow W=\frac{\frac{1}{2}}{{\omega }_2^2}-\frac{\frac{1}{2}}{{\omega }_1^2}\left[\because \frac{K{E}_{rotational}\frac{1}{2}}{{\omega }^2}\right]$
=$\mathrm{}\frac{1}{2}/\left({\omega }_2^2-{\omega }_1^2\right)$…(i)
Where, l = moment of inertia of the flywheel.
Substituting given values in Eq. (i), we get
$\frac{12000\frac{1}{2}}{1600{\pi }^2-400{\pi }^2}$
$\Rightarrow =\frac{1}{1}\left(1200\times 10\right)\left[\because {\pi }^2=10\right]$
$\Rightarrow l=\frac{12000\times 2}{12000}=2kg{m}^2$