Question

A solid cylinder of mass $20\, kg$ and radius $20\, cm$ rotates about its axis with an angular speed of $100 \,rad\, s ^{-1}$ . The angular momentum of the cylinder about its axis is

• A. $40\,J\,s$
• B. $400\,J\,s$
• C. $20\,J\,s$
• D. $200\,J\,s$

Answer 1

Answer: (A)

Here, $M = 20 \,kg$ $R = 20 \,cm = 20 \times 10^{-2}\, m, \omega = 100\, rad\, s^{-1}$
Moment of inertia of the solid cylinder about its axis is
$I = \frac{MR^2} {2} = \frac{(20\,kg)(20\times 10^{-2} m)^2}{2} = 0.4\,kg\,m^2$
Angular momentum of the cylinder about its axis is
$L = I\omega = (0.4\, kg \,m^2) (100\, rad\, s^{ -1}) = 40 \,J \,s$