Question

A solid disk, spinning counter-clockwise, has a mass of `3 kg` and a radius of `8/9 m`. If a point on the edge of the disk is moving at `11/4 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 `=3.67kgm^2s^-1`
The angular velocity is `=3.09rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

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

`omega=v/r`

where,

`v=11/4ms^(-1)`

`r=8/9m`

So,

The angular velocity is `omega=(11/4)/(8/9)=99/32=3.09rads^-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=3kg`

The radius of the disc is `r=8/9m`

So, `I=3*(8/9)^2/2=32/27kgm^2`

The angular momentum is

`L=32/27*99/32=3.67kgm^2s^-1`