Question

A solid disk, spinning counter-clockwise, has a mass of `6 kg` and a radius of `1/2 m`. If a point on the edge of the disk is moving at `12/5 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 `=3.6kgm^2s^-1`
The angular velocity is `=4.8rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

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

`omega=v/r`

where,

`v=12/5ms^(-1)`

`r=1/2m`

So,

`omega=(12/5)/(1/2)=24/5=4.8rads^-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*(1/2)^2/2=0.75kgm^2`

`L=0.75*4.8=3.6kgm^2s^-1`