A cylinder has inner and outer radii of 9 cm9cm and 14 cm14cm, respectively, and a mass of 9 kg9kg. If the cylinder's frequency of counterclockwise rotation about its center changes from 1 Hz1Hz to 7 Hz7Hz, by how much does its angular momentum change?

2 Answers
Feb 3, 2018

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

now,change in angular velocity= delta omega = 2pi(7-1) (rad)/sδω=2π(71)rads i.e 12 pi (rad)/s12πrads

II = 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^292((9100)2+(14100)2)Kg.m2 i.e 0.124 Kg.m^20.124Kg.m2 (formula is M/2((r_1)^2+(r_2)^2)M2((r1)2+(r2)2))

So,change in angular momentum is (0.124 * 12 pi) N.m.s=4.674 N.m.s(0.12412π)N.m.s=4.674N.m.s

Feb 3, 2018

The change in angular momentum is =4.7kgm^2s^-1=4.7kgm2s1

Explanation:

The angular momentum is L=IomegaL=Iω

where II is the moment of inertia

and omegaω is the angular velocity

The mass of the cylinder is m=9kgm=9kg

The radii of the cylinder are r_1=0.09mr1=0.09m and r_2=0.14mr2=0.14m

For the cylinder, the moment of inertia is I=m((r_1^2+r_2^2))/2I=m(r21+r22)2

So, I=9*((0.09^2+0.14^2))/2=0.12465kgm^2I=9(0.092+0.142)2=0.12465kgm2

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