Question

A solid disk, spinning counter-clockwise, has a mass of `13 kg` and a radius of `4/5 m`. If a point on the edge of the disk is moving at `7/3 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 `=12.15kgm^2s^-1` and the angular velocity is `=2.92rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

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

`omega=v/r`

where,

`v=7/3ms^(-1)`

`r=4/5m`

So,

The angular velocity is

`omega=(7/3)/(4/5)=35/12=2.92rads^-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=13 kg`

So, `I=13*(4/5)^2/2=4.16kgm^2`

The angular momentum is

`L=4.16*2.92=12.15kgm^2s^-1`