Question

A solid disk, spinning counter-clockwise, has a mass of `9 kg` and a radius of `3/8 m`. If a point on the edge of the disk is moving at `9/2 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 `=7.59kgm^2s^-1` and the angular velocity is `=12rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

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

`omega=v/r`

where,

`v=9/2ms^(-1)`

`r=3/8m`

So,

The angular velocity is

`omega=(9/2)/(3/8)=12rads^-1`

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

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

The mass of the disc is `m=9 kg`

So, `I=9*(3/8)^2/2=81/128kgm^2`

The angular momentum is

`L=81/128*12=7.59kgm^2s^-1`