Question

A solid disk, spinning counter-clockwise, has a mass of `4 kg` and a radius of `2 m`. If a point on the edge of the disk is moving at `6/7 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 `=3.43kgm^2s^-1` and the angular velocity is `=0.43rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

The velocity is `v=r*((Deltatheta)/(Deltat))=r omega`

Therefore, `omega=v/r`

where,

`v=6/7ms^(-1)`

`r=2m`

So,

`omega=(6/7)/(2)=3/7=0.43rads^-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*(2)^2/2=8kgm^2`

The angular momentum is

`L=8*3/7=24/7=3.43kgm^2s^-1`