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Physics Question on System of Particles & Rotational Motion

A solid cylinder of mass 20 kg20\, kg and radius 20 cm20\, cm rotates about its axis with an angular speed of 100 rad s−1100 \,rad\, s ^{-1} . The angular momentum of the cylinder about its axis is

A

40 J s40\,J\,s

B

400 J s400\,J\,s

C

20 J s20\,J\,s

D

200 J s200\,J\,s

Answer

40 J s40\,J\,s

Explanation

Solution

Here, M=20 kgM = 20 \,kg R=20 cm=20×10−2 m,ω=100 rad s−1R = 20 \,cm = 20 \times 10^{-2}\, m, \omega = 100\, rad\, s^{-1} Moment of inertia of the solid cylinder about its axis is I=MR22=(20 kg)(20×10−2m)22=0.4 kg m2I = \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=Iω=(0.4 kg m2)(100 rad s−1)=40 J sL = I\omega = (0.4\, kg \,m^2) (100\, rad\, s^{ -1}) = 40 \,J \,s