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 #1 Hz# to #5 Hz#, by how much does its angular momentum change?

1 Answer
Dec 16, 2017

Answer:

The change in angular momentum is #=2.09kgm^2s^-1#

Explanation:

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

and #omega# is the angular velocity

The mass of the cylinder is #m=8kg#

The radii of the cylinder are #r_1=0.08m# and #r_2=0.12m#

For the cylinder, the moment of inertia is #I=m((r_1^2+r_2^2))/2#

So, #I=8*((0.08^2+0.12^2))/2=0.08322kgm^2#

The change in angular velocity is

#Delta omega=Deltaf*2pi=(5-1) xx2pi=8pirads^-1#

The change in angular momentum is

#DeltaL=IDelta omega=0.0832 xx8pi=2.09kgm^2s^-1#