Question

A solid disk, spinning counter-clockwise, has a mass of `1 kg` and a radius of `7 m`. If a point on the edge of the disk is moving at `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 momentum is `=31.5kgm^2s^-1`
The angular velocity is `=(1.29)rads^-1`

Explanation:

The angular velocity is

`omega=(Deltatheta)/(Deltat)`

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

`omega=v/r`

where,

`v=9ms^(-1)`

`r=7m`

So,

`omega=(9)/(7)=(1.29)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=1*(7)^2/2=24.5kgm^2`

The angular momentum is

`L=24.5*1.29=31.5kgm^2s^-1`

Answer 2

Angular momentum: `31.5` `("kgm"^2)/"s"`
Angular velocity: `1.28571` `"rad"/"sec"` rounded to 5 decimals

Explanation:

In this problem, we are given the mass of the disk, the radius of the disk, and the tangential velocity, therefore, we can write out the given information as (in SI units):
`m=1`
`r=7`
`v=9`

Note that angular momentum is `L=Iomega`, where `L` is angular momentum, `I` is the moment of inertia and `omega` is the angular velocity.

Also note that angular velocity is `omega=v/r`, where `omega` is angular velocity, `v` is tangential velocity, and `r` is the radius of the disk.

First, we need find the angular velocity by plugging the given values into the formula `omega=v/r`:
`omega=9/7~~1.28571` `"rad"/"sec"` rounded to 5 decimals

Now, since we know the moment of inertia for a solid disk is `I_(disk)=1/2mr^2`, we can find the angular momentum by plugging the given values into the formula `L=Iomega`:
`L=(1/2(1)(7^2))(9/7)=63/2=31.5` `("kgm"^2)/"s"`

Therefore, the angular momentum is `31.5` `("kgm"^2)/"s"` and the angular velocity is `1.28571` `"rad"/"sec"` rounded to 5 decimals.