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 `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 `=11.25kgm^2s^-1`
The angular velocity is `=0.45rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

`v=r*((Deltatheta)/(Deltat))=r omega`

`omega=v/r`

where,

`v=9/8ms^(-1)`

`r=5/2m`

So,

`omega=(9/8)/(5/2)=0.45rads^-1`

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

For a solid disc, `I=(mr^2)/2`

So, `I=8*(5/2)^2/2=25kgm^2`

The angular velocity is

`L=25*0.45=11.25kgm^2s^-1`