A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s–1.
Question:

A solid cylinder of mass $20 \mathrm{~kg}$ rotates about its axis with angular speed $100 \mathrm{rad} \mathrm{s}^{-1}$. The radius of the cylinder is $0.25 \mathrm{~m}$. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Solution:

Mass of the cylinder, m = 20 kg

Angular speed, $\omega=100 \mathrm{rad} \mathrm{s}^{-1}$

Radius of the cylinder, $r=0.25 \mathrm{~m}$

The moment of inertia of the solid cylinder:

$I=\frac{m r^{2}}{2}$

$=\frac{1}{2} \times 20 \times(0.25)^{2}$

$=0.625 \mathrm{~kg} \mathrm{~m}^{2}$

$\therefore$ Kinetic energy $=\frac{1}{2} I \omega^{2}$

$=\frac{1}{2} \times 6.25 \times(100)^{2}=3125 \mathrm{~J}$$=\frac{1}{2} \times 6.25 \times(100)^{2}=3125 \mathrm{~J}$

$\therefore$ Angular momentum, $L=1 \omega$

$=6.25 \times 100$

$=62.5 \mathrm{JS}$