Question

A solid disk, spinning counter-clockwise, has a mass of `16 kg` and a radius of `3/7 m`. If a point on the edge of the disk is moving at `8/5 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 `=381.95kgm^2s^-1`
The angular velocity is `=3.73rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

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

`omega=v/r`

where,

`v=8/5ms^(-1)`

`r=3/7m`

So,

`omega=(8/5)/(3/7)=56/15=3.73rads^-1`

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

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

So, `I=16*(8/5)^2/2=512/25=102.4kgm^2`

The angular momentum is

`L=102.4*3.73=381.95kgm^2s^-1`