Question

A solid disk, spinning counter-clockwise, has a mass of `9 kg` and a radius of `3/2 m`. If a point on the edge of the disk is moving at `9/2 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 `=30.38kgm^2s^-1`. The angular velocity is `=3rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

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

`omega=v/r`

where,

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

`r=3/2m`

So,

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

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

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

So, `I=9*(3/2)^2/2=81/8kgm^2`

`L=81/8*3=30.38kgm^2s^-1`