Question

A solid disk, spinning counter-clockwise, has a mass of `9 kg` and a radius of `3/2 m`. If a point on the edge of the disk is moving at `5/2 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 `=16.875kgm^2s^-1` and the angular velocity is `=1.67rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

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

`omega=v/r`

where,

`v=5/2ms^(-1)`

`r=3/2m`

So,

The angular velocity is positive (spinning counter clockwise)

`omega=(5/2)/(3/2)=5/3=1.67rads^-1`

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

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

The mass is `m= 9kg`

So, `I=9*(3/2)^2/2=81/8kgm^2`

The angular momentum is

`L=81/8*5/3=16.875kgm^2s^-1`