Question

A cylinder has inner and outer radii of `8 cm` and `12 cm`, respectively, and a mass of `8 kg`. If the cylinder's frequency of rotation about its center changes from `1 Hz` to `5 Hz`, by how much does its angular momentum change?

Answer 1

The change in angular momentum is `=2.09kgm^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=8kg`

The radii of the cylinder are `r_1=0.08m` and `r_2=0.12m`

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

So, `I=8*((0.08^2+0.12^2))/2=0.08322kgm^2`

The change in angular velocity is

`Delta omega=Deltaf*2pi=(5-1) xx2pi=8pirads^-1`

The change in angular momentum is

`DeltaL=IDelta omega=0.0832 xx8pi=2.09kgm^2s^-1`