Question

A solid disk, spinning counter-clockwise, has a mass of `6 kg` and a radius of `3/5 m`. If a point on the edge of the disk is moving at `5/6 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 `=1.5kgm^2s^-1`
The angular velocity is `=1.39rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

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

`omega=v/r`

where,

`v=5/6ms^(-1)`

`r=3/5m`

So,

`omega=(5/6)/(3/5)=25/18=1.39rads^-1`

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

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

So, `I=6*(3/5)^2/2=27/25kgm^2`

The angular momentum is

`L=27/25*25/18=1.5kgm^2s^-1`