Question

A cylinder has inner and outer radii of `12 cm` and `15 cm`, respectively, and a mass of `4 kg`. If the cylinder's frequency of rotation about its center changes from `2 Hz` to `8 Hz`, by how much does its angular momentum change?

Answer 1

The change in angular momentum is `=76.8kgm^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=4kg`

The radii of the cylinder are `r_1=0.12m` and `r_2=0.15m`

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

So, `I=4*((0.12^2+0.15^2))/2=2.0369kgm^2`

The change in angular velocity is

`Delta omega=Deltaf*2pi=(8-2) xx2pi=12pirads^-1`

The change in angular momentum is

`DeltaL=IDelta omega=2.0369xx12pi=76.8kgm^2s^-1`