Question

A solid disk, spinning counter-clockwise, has a mass of `2 kg` and a radius of `3 m`. If a point on the edge of the disk is moving at `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 `=27kgm^2s^-1` and 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=9ms^(-1)`

`r=3m`

So,

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

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

The mass of the disc is `m=2kg`

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

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

The angular momentum is

`L=9*3=27kgm^2s^-1`