A cylinder has inner and outer radii of #4 cm# and #8 cm#, respectively, and a mass of #6 kg#. If the cylinder's frequency of counterclockwise rotation about its center changes from #7 Hz# to #8 Hz#, by how much does its angular momentum change?

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Answer 1 · 2017-03-27T06:29:50

The change in angular momentum is #=0.15kgm^2s^-1#

The angular momentum is #L=Iomega#

Mass, #m=6kg#

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

So, #I=6*((0.04^2+0.08^2))/2=0.024kgm^2#

The change in angular momentum is

#DeltaL=IDelta omega#

The change in angular velocity is

#Delta omega=(8-7)*2pi=(2pi)rads^-1#

The change in angular momentum is

#DeltaL=0.024*2pi=0.15kgm^2s^-1#

Answer 2 · 2017-03-27T06:46:08

#L = 0.15kgm^2s^-1#

Angular momentum formula: # L = I * omega #
#L# = Angular momentum
#I# = Moment of Inertia
#omega# = Angular velocity

#I=1/2MR^2#
#M# =Mass
#R# =Distance of Axis
#I# = Moment of Inertia

#I=1/2 * 6kg*((0.04m)^2+(0.08m)^2) = 0.024kgm^2#

#omega = 2pi* F#
#F# = Frequency
#omega# = Angular Vecloicty
#omega = 2pi (8Hz-7hz)= 2piHz or 2pis^-1#

# L= 0.024kgm^2 * 2pis^-1 = 0.15kgm^2s^-1#