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

Answer 1

The change in angular momentum is `=0.93kgm^2s^-1`

Explanation:

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

The mass, `m=4kg`

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

So, `I=4*((0.12^2+0.15^2))/2=0.0738kgm^2`

The change in angular momentum is

`DeltaL=IDelta omega`

The change in angular velocity is

`Delta omega=(5-3)*2pi=(4pi)rads^-1`

The change in angular momentum is

`DeltaL=0.0738*4pi=0.93kgm^2s^-1`