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

Answer 1

The angular momentum changes by `=2.09kgm^2s^-1`

Explanation:

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

Mass, `m=8kg`

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

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

The change in angular momentum is

`DeltaL=IDelta omega`

The change in angular velocity is

`Delta omega=(7-3)*2pi=(8pi)rads^-1`

The change in angular momentum is

`DeltaL=0.0832*8pi=2.09kgm^2s^-1`