A cylinder has inner and outer radii of #12 cm# and #18 cm#, respectively, and a mass of #2 kg#. If the cylinder's frequency of rotation about its center changes from #4 Hz# to #9 Hz#, by how much does its angular momentum change?

1 Answer
Mar 22, 2017

Answer:

The change in angular momentum is #=1.47kgm^2s^-1#

Explanation:

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

So, #I=2*((0.12^2+0.18^2))/2=0.0468kgm^2#

The change in angular momentum is

#DeltaL=IDelta omega#

The change in angular velocity is

#Delta omega=(9-4)*2pi=(10pi)rads^-1#

The change in angular momentum is

#DeltaL=0.0468*10pi=1.47kgm^2s^-1#