Question

A solid disk, spinning counter-clockwise, has a mass of `8 kg` and a radius of `5/2 m`. If a point on the edge of the disk is moving at `5/8 m/s` in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Answer 1

`....`

Explanation:

`r=5/2m`, The circumference of the disk is `C=2\pi r`

Because a point on the outside of the disk moves along the circumference with `v=5/8 m/s`, one rotation is completed in

`T=C/v={2\pi r}/v`

The angular velocity, `w`, has units of radians per second. One rotation has `2\pi` radians so

`w={2\pi}/T=v/r= 0.25` radians per second

As a side note, you could obtain this immediately if you are familiar with the expression `v_\text{perpendicular}=w\times r`

The angular momentum of the disk is `L=Iw`, where `I` is the inertia of the disk. You could obtain the inertia by integration, or just look it up in a table.

For a uniform disk, `I=1/2mR^2= 25 \quad kg\cdot m^2`.

Therefore,

`L=Iw=6.25 \quad kg\cdot m^2/s`

Done, but if you wanted to solve for the inertia on your own without looking it up, you would have to integrate.

Assume the disk has uniform density `\sigma={mass}/{area}=m/{\pi R^2}`

`I=\int r^2 dm=\int r^2 (\sigma dA)`

where `dm=\sigma dA`. You just need to substitute `(dA=r\quad d\phi\quad dr)` and integrate to get the answer, `I=1/2 mR^2`.