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 #4 Hz# to #9 Hz#, by how much does its angular momentum change?

Question

1180 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-08T07:05:37

The answer is #=1.47kgm^2s^(-1)#

The angular momentum is #L=Iomega#

The change in angular momentum is

#DeltaL=IDelta omega#

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

So, #I=6*(0.12^2+0.18^2)/2=0.0468kgm^2#

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

#Delta L=0.0468*10pi=1.47kgm^2s^(-1)#