A cylinder has inner and outer radii of #12 cm# and #18 cm#, respectively, and a mass of #6 kg#. If the cylinder's frequency of rotation about its center changes from #9 Hz# to #6 Hz#, by how much does its angular momentum change?

1 Answer
Jul 10, 2017

Answer:

The angular momentum changes by #=2.65kgm^2s^-1#

Explanation:

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

The mass, #m=6kg#

For a cylinder, #I=m((r_1^2+r_2^2))/2#

So, #I=6*((0.12^2+0.18^2))/2=0.1404kgm^2#

The change in angular momentum is

#DeltaL=IDelta omega#

The change in angular velocity is

#Delta omega=(9-6)*2pi=(6pi)rads^-1#

The change in angular momentum is

#DeltaL=0.1404*6pi=2.65kgm^2s^-1#