Question

A cylinder has inner and outer radii of `12 cm` and `18 cm`, respectively, and a mass of `6 kg`. If the cylinder's frequency of rotation about its center changes from `9 Hz` to `8 Hz`, by how much does its angular momentum change?

Answer 1

The angular momentum changes by `=0.88kgm^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*((0.12^2+0.18^2))/2=0.1404kgm^2`

The change in angular momentum is

`DeltaL=IDelta omega`

The change in angular velocity is

`Delta omega=(9-8)*2pi=(2pi)rads^-1`

The change in angular momentum is

`DeltaL=0.1404*2pi=0.88kgm^2s^-1`