Question

A solid disk, spinning counter-clockwise, has a mass of `2 kg` and a radius of `7/4 m`. If a point on the edge of the disk is moving at `7/9 m/s` in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Answer 1

The angular momentum is `=49/36 kgm^2s^(-1)`
The angular velocity is `=4/9 Hz`

Explanation:

The angular velocity is `omega=v/r`

Here `v=7/9 ms^(-1)` and `r=7/4 m`

therefore `omega = 7/9*4/7=4/9 Hz`

The angular momentum is `L=I*omega`
Where `I` is the moment of inertia
For a solid disc `I=(mr^2)/2=2*(7/4)^2/2=49/16 kgm^2`

So, the angular momentum is `49/16*4/9=49/36 kgm^2s^(-1)`