Question

A cylinder has inner and outer radii of `2 cm` and `5 cm`, respectively, and a mass of `5 kg`. If the cylinder's frequency of counterclockwise rotation about its center changes from `12 Hz` to `9 Hz`, by how much does its angular momentum change?

Answer 1

Let's assume this cylinder is uniform (which is very important).

Recall for a solid cylinder,

`I = 1/2mr^2`

Moreover, recall angular momentum,

`L = Iomega`

Now, let's derive the moment of inertia of this circular body,

`I = 1/2mr^2 approx 5"kg"*"m"^2`

which is generally constant in our calculations.

Now, recall,

`omega = 2pif`

Hence,

`I_(1)2pif_1 approx (75"kg"*"m"^2)/"s"`, and

`I_(2)2pif_2 approx (57"kg"*"m"^2)/"s"`

`therefore DeltaL = L_2 - L_1 approx (18"kg"*"m"^2)/"s"`

is the change in angular momentum given your data.

Answer 2

The change in angular momentum is `=0.137kgm^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=5kg`

The radii of the cylinder are `r_1=0.02m` and `r_2=0.05m`

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

So, `I=5*((0.02^2+0.05^2))/2=0.00725kgm^2`

The change in angular velocity is

`Delta omega=Deltaf*2pi=(12-9) xx2pi=6pirads^-1`

The change in angular momentum is

`DeltaL=IDelta omega=0.00725xx6pi=0.137kgm^2s^-1`