Question

A cylinder has inner and outer radii of `16 cm` and `18 cm`, respectively, and a mass of `2 kg`. If the cylinder's frequency of rotation about its center changes from `4 Hz` to `9 Hz`, by how much does its angular momentum change?

Answer 1

The change in angular momentum is `=1.82kgm^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=2kg`

The radii of the cylinder are `r_1=0.16m` and `r_2=0.18m`

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

So, `I=2*(0.16^2+0.18^2)/2=0.058kgm^2`

The change in angular velocity is

`Delta omega=Deltaf*2pi=(9-4)*2pi=10pirads^-1`

The change in angular momentum is

`DeltaL=IDelta omega=0.058 xx10pi=1.82kgm^2s^-1`