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 `3 Hz` to `2 Hz`, by how much does its angular momentum change?

Answer 1

The answer is `=0.523kgms^(-1)`

Explanation:

The angular momentum is `L=Iomega`

and `DeltaL=IDeltaomega`

where `I` is the moment of inertia

For a cylinder, `I=m/2(r_1^2+r_2^2)`

So, `I=8/2*(0.08^2+0.12^2)=4*0.0208=0.0832kgm^2`

`Delta omega=(3-2)*2pi=2pi rads^-1`

`DeltaL=0.0832*2pi=0.523kgms^(-1)`