Question

A solid disk, spinning counter-clockwise, has a mass of `4 kg` and a radius of `3/2 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 `=0.68kgm^2s^-1` and the angular velocity is `=0.15rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

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

`omega=v/r`

where,

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

`r=3/2m`

So,

`omega=(2/9)/(3/2)=4/27=0.15rads^-1`

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

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

So, `I=4*(3/2)^2/2=4.5kgm^2`

The angular momentum is

`L=4.5*0.15=0.68kgm^2s^-1`