Question

A solid disk, spinning counter-clockwise, has a mass of `8 kg` and a radius of `3/2 m`. If a point on the edge of the disk is moving at `9/8 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 `=27/4kgms^(-1)`
The angular velocity is `=3/4Hz`

Explanation:

The angular velocity `omega` is
`omega=v/r`

where
`v=velocity=9/8ms^(-1)`
`r=radius=3/2m`

so, `omega=9/8*2/3=3/4 Hz`

The angular momentum is

`L=Iomega`

Where `I`=moment of inertia

In the case of a solid disc, `I=(mr^2)/2`

`m=`mass of the disc, `=8kg`

`r=` radius of the disc, `=3/2 m`

Therefore, `I=1/2*8*9/4=9 kgm^2`

The angular momentum is `L=9*3/4=27/4kgms^(-1)`