A cylinder has inner and outer radii of #9 cm# and #14 cm#, respectively, and a mass of #9 kg#. If the cylinder's frequency of counterclockwise rotation about its center changes from #1 Hz# to #7 Hz#, by how much does its angular momentum change?

2 Answers
Feb 3, 2018

Change in angular momentum = #(I delta omega)#

now,change in angular velocity= #delta omega = 2pi(7-1) (rad)/s# i.e #12 pi (rad)/s#

#I# = moment of inertia of of a cylindrical shell of given outer and inner radii will be #9/2((9/100)^2 + (14/100)^2) Kg.m^2# i.e #0.124 Kg.m^2# (formula is #M/2((r_1)^2+(r_2)^2)#)

So,change in angular momentum is #(0.124 * 12 pi) N.m.s=4.674 N.m.s#

Feb 3, 2018

Answer:

The change in angular momentum is #=4.7kgm^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=9kg#

The radii of the cylinder are #r_1=0.09m# and #r_2=0.14m#

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

So, #I=9*((0.09^2+0.14^2))/2=0.12465kgm^2#

The change in angular velocity is

#Delta omega=Deltaf*2pi=(7-1) xx2pi=12pirads^-1#

The change in angular momentum is

#DeltaL=IDelta omega=0.12465xx12pi=4.7kgm^2s^-1#