Question

A solid disk, spinning counter-clockwise, has a mass of `12 kg` and a radius of `5/4 m`. If a point on the edge of the disk is moving at `2/9 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 `=5/3kgm^2s^-1` and the angular velocity is `=8/45rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

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

`omega=v/r`

where,

`v=2/9ms^(-1)`

`r=5/4m`

So,

`omega=(2/9)/(5/4)=8/45rads^-1`

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

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

So, `I=12*(5/4)^2/2=75/8kgm^2`

The angular momentum is

`L=75/8*8/45=5/3kgm^2s^-1`