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 #2 Hz#, by how much does its angular momentum change?

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Answer 1 · 2017-01-26T14:46:04

The answer is #=0.523kgms^(-1)#

The angular momentum is #L=Iomega#

and #DeltaL=IDeltaomega#

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

So, #I=8/2*(0.08^2+0.12^2)=4*0.0208=0.0832kgm^2#

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

#DeltaL=0.0832*2pi=0.523kgms^(-1)#