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?

2 Answers
Mar 27, 2017

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

Explanation:

The angular momentum is L=Iomega

where I is the moment of inertia

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

Mar 27, 2017

L = 0.15kgm^2s^-1

Explanation:

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