Question

A solid disk, spinning counter-clockwise, has a mass of `7 kg` and a radius of `5/2 m`. If a point on the edge of the disk is moving at `6/5 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 `=10.5kgm^2s^-1` and the angular velocity is `=0.48rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

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

`omega=v/r`

where,

`v=6/5ms^(-1)`

`r=5/2m`

So,

The angular velocity is positive (spinning counter clockwise)

`omega=(6/5)/(5/2)=12/25=0.48rads^-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= 7kg`

So, `I=7*(5/2)^2/2=175/8kgm^2`

The angular momentum is

`L=175/8*0.48=10.5kgm^2s^-1`