Question

What is the angular momentum of an object with a mass of `5 kg` that moves along a circular path of radius `9 m` at a frequency of `2/3 Hz`?

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Answer 1

We will first convert `Hz` into `{rad}/sec` using the formula below:
`\omega=2\pi f`
`\omega= 2\times \pi\times2/3`
`\omega=4.188{rad}/sec`

We then use the formula for Angular momentum
`L=\vec{r}\times \vec{p}`= `L=\vec{r}\times m \vec{v}`
Since we are calculating Angular momentum wrt to the center of the circle, and during circular motion `\vec{v}` is perpendicular to `\vec{r}` hence the cross product simply becomes a simple multiplication, with the angular momentum pointing along Z-axis( assuming motion is happening in the X-Y plane)

Velocity in a circular motion, given its angular velocity is given by:
`v=\omega r`, where `r` is the radius of circle

Therefore,
`L=m\omegar^2 \hat{z}`
`L= 5\times 4.188\times 9^2`= `1696.46 Kg` `m^2/sec` `\hat{z}`

Answer 2

`L= 1696.7" ""kg.m"^2"s"^(-1)`

Explanation:

The expression for angular momentum is:

`L=mxxvxxr`

`m` = mass

`v` = velocity

`r` = radius

We can use the information given to find the velocity `v`:

The frequency of rotation `f` is `2/3"Hz"`

This means that the time period `T` is `1/f` which equals `3/2=1.5"s"`

This means that it takes `1.5"s"` to make one complete revolution.

We know the circumference of the circle is `2pir`.

This is the distance travelled in `1.5"s"`

So the velocity is given by:

`v=(2pi9)/1.5=12pi"m/s"`

So we can now get `L`:

`L=5xx12pixx9`

`:.L= 1696.7" ""kg.m"^2"s"^(-1)`