The angular momentum of a solid cylinder rotating about its geometric axis with angular speed $40…
The angular momentum of a solid cylinder rotating about its geometric axis with angular speed $40 \mathrm{rad} \mathrm{s}^{-1}$ is $2 \mathrm{~kg}$ $\mathrm{m}^2 \mathrm{~s}^{-1}$. If the radius of the cylinder is $10 \mathrm{~cm}$, the mass of the cylinder is
$2 \mathrm{~kg}$
$5 \mathrm{~kg}$
$8 \mathrm{~kg}$
$10 \mathrm{~kg}$
Solution
Angular speed, $\omega=40 \mathrm{rad} / \mathrm{s}$
Angular momentum, $\mathrm{L}=2 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$
Radius of the cylinder, $r=10 \mathrm{~cm}=0.10 \mathrm{~m}$
Moment of inertia of solid cylinder, $I=\frac{\mathrm{mr}^2}{2}$
Angular momentum, $\mathrm{L}=\mathrm{I} \omega$
$\begin{aligned} & 2=\frac{\mathrm{m} \pi^2}{2} \times 40 \\ & 2=\frac{1}{2} \times \mathrm{m} \times(0.10)^2 \times 40 \\ & \mathrm{~m}=\frac{2}{20 \times(0.1)^2}=10 \mathrm{~kg}\end{aligned}$