A cylinder has inner and outer radii of #8 cm# and #12 cm#, respectively, and a mass of #8 kg#. If the cylinder's frequency of rotation about its center changes from #3 Hz# to #4 Hz#, by how much does its angular momentum change?

1 Answer
Jun 14, 2017

Answer:

The angular momentum changes by #=0.52kgm^2s^-1#

Explanation:

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

Mass, #m=kg#

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

So, #I=8*((0.08^2+0.12^2))/2=0.0832kgm^2#

The change in angular momentum is

#DeltaL=IDelta omega#

The change in angular velocity is

#Delta omega=(4-3)*2pi=(2pi)rads^-1#

The change in angular momentum is

#DeltaL=0.0832*2pi=0.52kgm^2s^-1#