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

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Answer 1 · 2017-01-29T08:25:17

The answer is #=1.57kgms^(-1)#

The angular momentum is #L=Iomega#

The change of angular momentum is

#DeltaL=I Delta omega#

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#

#Delta omega=(7-4)*2pi =6pirads^-1#

#L=0.0832*6pi=1.57kgms^(-1)#