Angular velocity or frequency in an uniform circular motion? (for students of an high school withouts the calculus)

Question

In an uniform circular motion, we know that the angular velocity is:

$$\omega=\frac{2\pi}{T}$$ and your measure is:

$$[\omega]=\frac{\text{rad}}{\text{s}}$$ i.e. revolutions per second (in Italian language giri al secondo).

Now the frequence $\nu$ is the number of cycles per second (in Italian language cicli al secondo),

$$\nu=\frac1T \iff [\nu]=\frac{1\, \text{cycles}}{1\, \text s}$$ with the unit of measure of $\nu$ is $1/\text{s}=\text{Hz}$.

But into my textbook the instantaneous angular velocity $\omega$, where in the SI it is measured in radians per second ($\text{rad/s}= \text{s}^{-1}$), because the $\text{rad}$ it is not a true unit of measure. There is written also this:

In the unit SI it is measures in $\text{rad/s}= \text{s}^{-1}$.

My question is:

Doesn't this create confusion between angular velocity and frequency in terms of units of measurement? When revolutions per minute are given in an exercise, are we considering an angular velocity or a frequency?

In this exercise, for example, it is used

the frequence $\nu=f$ and not $\omega$. But in another exercise it is taken, instead like the $\omega$ and not as the frequence $\nu=f$. See the image,

Being true that $$\omega=2\pi\nu\equiv 2\pi f$$

but...

is there a trick or expedient to tell if it's angular velocity or frequency?

Answer 1

Your book uses the following relation

$$\omega = \frac{d \theta}{dt}$$

One can write ds element as $ds = Rd\theta$ where this becomes

$$\omega = \frac{1}{R}\frac{ds}{dt}$$

So it's in $s^{-1}$ units. It's because radian is a dimensionless unit in SI thus rad/s is equivalent to $1/s$. Frequency $f$ or $\nu$ are in units of $1/s$ and period $T$ is in second. In practice $rad/s$ is used for angular frequency and $1/s$ is for normal frequencies not for angular frequencies. I think rad term was omitted with the formulation of quantum mechanics. For instance $E = hf$, energy is in eV or in J and frequency is in Hertz thus planck constant is in $J\cdot s$ but h-bar (reduced planck constant) is expressed in units of $J\cdot s /rad$

Answer 2

It should not create confusion because a magnitude has some units associated to it, but the inverse is not always true, so it can be dangerous to think it that way.

In other words, the units can give you clues of what it is, but it is not the only thing you should look at.

For example, energy is measured in "joules", $1\text J=1\text N\cdot 1\text m$, but on the other hand, torque has also units of $\text N\, \text m$. No one would ever think of adding energy and torque; in fact, torque is a vector and energy is a scalar...

So units must not be our only consideration.

It's true that the fact that radians are a "ghost unit" is misleading. Then, there's this truly bad habit of the physicists being lazy (me included, of course). Physicists usually work with $\omega$ quite more often than with the "actual" frequency... so they've ended up calling it frequency as well.

Nevertheless, the unit for "real" frequency has a special name: Hertzs (Hz). So in sum,

Normally, $\text s^{-1}$ refers to $\omega$, while $\text{Hz}$ refers to $\nu$.

This, however, is not an absolute truth, each author can have his own criteria.

As I see it, if they just say "frequency", you should stick to the literal meaning and they could not punish you. Especially, when it comes to official exams and all that.

...Unless there's something in the further development that reveals it was actually $\omega$, you know, it's usually relying on the logic and the context...

Answer 3

The circumference of a circle radius $r$ is $2\pi r$.

So if the circle rotates at $f$ revolution per second, a point on the cirumference moves a distance $2\pi r f$ in one second, so its velocity is $2\pi r f$.

If we measure the rotation speed as $\omega$ radians per second, a point on the circumference moves a distance $r\omega$ in one second.

Since other formulas for circular motion are also simpler when measuring angular velocity in radians/sec, a good way to avoid confusion is to convert other units (for example revolutions per minute) into radians/sec at the start of the calculations.

When using calculus, all the formulas use angles measured in radians, not revolutions or degrees, so it is better to use radians right from the start.

Answer 4

I think it should also be pointed out that angular velocity is what is known as a vector which means it has both magnitude and direction. Your equation for instantaneous angular velocity, should really be expressed with a a unit vector so that

$$\vec \omega = \frac{ \Delta \theta}{ \Delta t} \hat u$$

where the unit vector $\hat u$ points in the direction of the axis of rotation. A lot of other physicists use velocity in non-vector form too. That is technically incorrect.

You can compare angular frequency and angular velocity in the same way you compare speed and velocity. Angular frequency is the magnitude of angular velocity which is a vector.

Angular velocity is a measure of how fast the angular position of an object changes with time, relative to some point in space.

Angular frequency or angular speed is a scalar measure of this rotation rate without reference to a particular point in space or orientation.

Doesn't this create confusion between angular velocity and frequency in terms of units of measurement? When revolutions per minute are given in an exercise, are we considering an angular velocity or a frequency?

Angular velocity and angular frequency can be used to both explain the same thing. Remember that the prior is a vector and the latter is a scalar.

Frequency will be given with units $s^{-1}$ or $Hz$ "Hertz" or cycles $s^{-1}$ or per minute etc., and whenever you are dealing with angular velocity/angular frequency you will see $rad \ s^{-1}$. i.e., look for "rad".

is there a trick or expedient to tell if it's angular velocity or frequency?

One is a vector and the other is a scalar. Apart from that, see the other answers and your intuition and experience in the future will make you more comfortable with such terms.