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 `7/4 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 `=57kgms^(-1)`
The angular veocity is `=13.7rads^-1`

Explanation:

The angular velocity is

`omega=v/r`

where,

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

`r=4/5m`

So,

`omega=(7/4)/(4/5)*2pi=70/16pi=13.7 rads^-1`

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

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

So, `I=13*(4/5)^2/2=208/50=4.16kgm^2`

`L=13.7*4.16=57kgms^(-1)`