Physics › Rotational Motion › Angular momentum and Angular impulse
A $20 \mathrm{~kg}$ flywheel in the form of a uniform circular disc, $1 \mathrm{~m}$ in diameter is making…
A $20 \mathrm{~kg}$ flywheel in the form of a uniform circular disc, $1 \mathrm{~m}$ in diameter is making $120 \mathrm{rpm}$. What is its angular momentum?
$3.14 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$ $31.4 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$ $314 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$ $0.314 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$ Solution
Mass of fly wheel,
$
m=20 \mathrm{~kg}
$
Diameter, $D=1 \mathrm{~m}$
$\therefore$ Radius, $R=\frac{1}{2}=0.5 \mathrm{~m}$
Angular velocity, $\omega=2 \pi \times \frac{120}{60}=4 \pi \mathrm{rad} \mathrm{s}^{-1}$
$\therefore$ Moment of inertia of uniform circular disc,
$
\begin{aligned}
I & =\frac{m R^2}{2}=\frac{20 \times(0.5)^2}{2} \\
& =10 \times 0.25 \\
& =2.5 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}
$
$\therefore$ Angular momentum,
$
\begin{aligned}
L & =I \omega \\
& =2.5 \times 4 \pi \\
& =2.5 \times 4 \times 3.14 \\
& =31.4 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}
\end{aligned}
$
Asked in: AP EAMCET 2020 (22 Sep Shift 2)
Practice more Rotational Motion questions on Aicharya