A cylinder has inner and outer radii of #12 cm# and #15 cm#, respectively, and a mass of #4 kg#. If the cylinder's frequency of rotation about its center changes from #5 Hz# to #3 Hz#, by how much does its angular momentum change?

1 Answer
Jun 29, 2017

Answer:

The change in angular momentum is #=0.93kgm^2s^-1#

Explanation:

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

The mass, #m=4kg#

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

So, #I=4*((0.12^2+0.15^2))/2=0.0738kgm^2#

The change in angular momentum is

#DeltaL=IDelta omega#

The change in angular velocity is

#Delta omega=(5-3)*2pi=(4pi)rads^-1#

The change in angular momentum is

#DeltaL=0.0738*4pi=0.93kgm^2s^-1#