Question

A solid disk, spinning counter-clockwise, has a mass of `2 kg` and a radius of `7/4 m`. If a point on the edge of the disk is moving at `5/9 m/s` in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Answer 1

The angular velocity is `=20/63 Hz`
The angular momentum is `=35/36kgm^2s^(-1)`

Explanation:

The angular velocity is `omega=v/r`

where `v= velocity =(5/9)ms^(-1)`

and `r=radius=7/4 m`

`omega=(5/9)/(7/4)=5/9*4/7=20/63 Hz`

The angular momentum is `L=Iomega`

Where `omega=`angular velocity

and `I=` moment of inertia

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

`I=2*(7/4)^2*(1/2)=49/16 kgm^2`

So, `L=(49/16)*20/63=35/36kgm^2s^(-1)`