Question

A solid disk, spinning counter-clockwise, has a mass of `2 kg` and a radius of `7/2 m`. If a point on the edge of the disk is moving at `7/4 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 `=38.5kgm^2s^(-1)`
The angular velocity is `=3.14rads^-1`

Explanation:

The angular velocity is

`omega=v/r`

where,

`v=7/4ms^(-1)`

`r=7/2m`

So,

`omega=(7/4)/(7/2)*2pi=pi=3.14rads^-1`

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

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

So, `I=2*(7/2)^2/2=49/4kgm^2`

`L=3.14*49/4=38.5kgm^2s^(-1)`