When I was at high school they teached me this formula $v = \omega × r$, it is the formula for knowing the linear velocity out of an angular velocity. But I've never understand why this works, even the teacher cannot show me why this works, I would have to trust him.

My question is, what is the scientific explanation of why this works?

And I don't know if this is silly, but not even the measuerement units concide. $v = \frac{m}s$, but the velocity obtained from this formula would be $v = \frac{rad × m}s$.

Why this formula works?

  • 2
    $\begingroup$ A radian is a dimensionless quantity. $\endgroup$
    – Bhavay
    Commented Dec 3, 2020 at 21:56
  • $\begingroup$ It might help to think in terms of revolutions per second instead of radians per second. Eg, if r=10 m, and the object is doing 1 revolution per second, what's its speed (in m/s) ? $\endgroup$
    – PM 2Ring
    Commented Dec 3, 2020 at 21:59
  • $\begingroup$ Ok so, the part about the units, is solved, but why this formula works? $\endgroup$ Commented Dec 3, 2020 at 21:59
  • $\begingroup$ Are you familiar with cylindrical polar coordinates and calculus? $\endgroup$
    – Shrey
    Commented Dec 3, 2020 at 22:01
  • $\begingroup$ @Shrey I'm supposed to learn it this year, so nope. Sorry :( $\endgroup$ Commented Dec 3, 2020 at 22:07

6 Answers 6


The circumference of a circle is: $C = 2\pi r $

If the number of revolutions you traveled is n, then the length traveled is:

$L = 2\pi r n$

If you differentiate with respect to time to get velocity, you get:

$v = \frac{dL}{dt}= 2\pi r \times \frac{dn}{dt}$

$\frac{dn}{dt}$ is revolutions per second and $2\pi$ is the radians around a full circle.

This is simple enough you can completely skip the differentiation and do it intuitively. It should be obvious why:

$velocity = circumference \times revolutions.per.second = 2\pi r \times revolutions.per.second$

Continuing on, then $2\pi\frac{dn}{dt}$ (or $2\pi \times revolutions.per.second$ if you prefer) is radians per second $\omega$.

Therefore, $v = 2\pi r \times \frac{dn}{dt} =(2\pi\frac{dn}{dt}) \times r= \omega r$

As pointed out by others, a radian is not a unit. Radians is just a proportional (dimensionless) measure of the arc length around a circle relative to the circumference of ANY circle, of ANY size. Put another way, it is a proportional measure of how far you've gone around an entire circle anglewise or distance wise; same thing.

Start with the circumference of a circle $C = 2\pi r $

Let's say we need to make up a unit; a unit that represents some arc length around the circumference of a circle. It would be awfully convenient if it would be independent of the size of the circle so we should pick something that scales with the circle rather than a fixed number.

There's not many things to pick from: radius, diameter, circumference. Circumference makes no sense since we want a ratio proportional to the circumference which leaves either radius or diameter. Mathematicians in the past picked radius, but they could just have easily as chosen diameter.

So, let's find the ratio of the radius to the circumference of a circle:

$\frac{r}{C} = \frac{r}{2\pi r} = \frac{1}{2\pi}$

That 1 in the numerator is the one radian, and $2 \pi$ in the denominator is the what it takes to get around the circumference. So really, a single radian represents one radius of arc length if you define $2 \pi$ as the circumference. Why do we define things this way? It helps keep the math clean.

There is something similar that exists for a sphere as well, the Stearadian. A single Stearadian is the area $r^2$, $r$ being the radius of the sphere, and $4\pi$ is the entire surface area of the sphere (coming from how $4\pi r^2$ is the surface area of a sphere). It is derived exactly the same way as I just did for a radian, except arc length is replaced by area and circumference is replaced by surface area. Everything moves up one dimension.

I only learned what a radian really was until I had to learn Stearadians and realized that none of it made sense because I did not understand what a radian was.


Depending on the definition of $\vec{r}$ you can state the law as

$$ \vec{v} =\vec{r} \times \vec{\omega}$$ where $\vec{r}$ is location of the rotation axis.

The above is entirely analogous to the definition of torque $$ \vec{\tau} = \vec{r} \times \vec{F}$$ where $\vec{r}$ is the location of the force.

Also the above is entirely analogous to the moment of inertia of particle with momentum $\vec{p}$ which is $$ \vec{L} = \vec{r} \times \vec{p} $$ where $\vec{r}$ is the location of the particle.

So what do all the above have in common? All of the above are a "moment of" calculation, with $\text{(moment)} = \vec{r} \times \text{(quantity)}$. There is velocity = moment of rotation, torque = moment of force and angular momentum = moment of momentum.

The significance of $\vec{r}\times$ is to result in the perpendicular distance to a line. Here is a graphical explanation of $\vec{v} = \vec{r} \times \vec{\omega}$


The vector $\vec{v}$ is perpendicular to the rotation axis $\vec{\omega}$ as well as out of plane where the vector $\vec{r}$ lies with $\vec{\omega}$. More importantly, the magnitude of $\vec{v}$ depends on the perpendicular distance $d$ to the rotation axis.

$$ | \vec{v} | = d \,|\vec{\omega}| $$

So you see the $\vec{r}\times$ results in accounting for the perpendicular distance to the axis of rotation.

So the magic of the cross product is that it results in a perpendicular vector with a magnitude proportional to the perpendicular distance. The result is that you can slide the vectors $\vec{\omega}$, $\vec{F}$ and $\vec{p}$ along their line of action, and it will not change the problem, as the perpendicular distance remains the same.

In this sense $\vec{v}$ describes where the axis of rotation is, just as $\vec{\tau}$ describes where the line of action of the force is and $\vec{L}$ describes where the axis of percussion (line of momentum) is. For all these quantities another cross product can recover the point on the line closest to the reference point.

You can verify for yourself that

$$ \vec{r}_\perp = \frac{1}{| \vec{\omega} |^2} ( \vec{\omega} \times \vec{v} ) $$



$\vec v=\vec \omega\times \vec r$ is always valid , not only in circular motion

Circular motion

the line element is

$$s=\varphi\,r$$ $\Rightarrow$


to see where the cross product come from we have to go to 3D space

3D Space enter image description here

The rotation axis is $\vec{\hat e}_\varphi$ and the rotation angle is $~\varphi~$ thus

$$\vec{s}=S({\hat e}_\varphi\,,\varphi)\,\vec{r}$$

where the components of $\vec r~$ are in body fixed coordinate system, and $S~$ is the rotation matrix.


$$\vec{v}=\vec{\dot s}=\dot{S}({\hat e}_\varphi\,,\varphi)\,\vec{r}$$

with: $$\dot{S}=S\,\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] \\ \vec{\omega }=\hat{e}_{\varphi }\dfrac{d\varphi }{dt}$$ $\Rightarrow$

$$\vec v=S\,\left(\vec{\omega}\times \vec r\right) ~,S^T\,\vec v=\vec{\omega}\times \vec r$$

where $S^T\,\vec v~$ are the components of the vector $v$ in body fixed coordinate system


$$\boxed{\vec v=\vec \omega\times \vec r}$$


Radians are not a proper unit, but more of a book-keeping device, so the units actually do work out. The math to prove this, at least in a way I find satisfying, it somewhat involved. So let's instead consider the example of a particle which is rotating around the $z$-axis with angular frequency $\omega$.

To do this, note that the matrix $$ R(\theta)=\left(\begin{array}{ccc} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0&0&1 \end{array}\right) $$ implements a rotation about the $z$-axis. That is, if for example we took the initial position of our particle to be $$ \boldsymbol{r}=\boldsymbol{\hat x} $$ so it's just sitting one unit away from the origin on the $x$-axis, then the vector $$ \boldsymbol{r}^\prime=R(\theta)\boldsymbol{r} $$ will be the position of our particle after being rotated by an angle $\theta$ about the $z$-axis.

So, if the angular speed is given by $\omega$, then the angular velocity of this particle would be defined to be $\boldsymbol{\omega}=\omega\boldsymbol{\hat z}$. Hence the angular position of the particle will be given by $\theta=\omega t$ and so the position of the particle at time $t$ is given by $$ \boldsymbol{r}(t)=R(\omega t)\boldsymbol{r}=\left(\begin{array}{ccc} \cos\omega t & -\sin\omega t & 0\\ \sin\omega t & \cos\omega t & 0\\ 0&0&1 \end{array}\right)\left(\begin{array}{c} 1\\0\\0\end{array}\right). $$

Now, we know the definition of velocity is $\boldsymbol{v}=\frac{d\boldsymbol{r}}{dt}$, so we need to take a time derivative of the rotation matrix, $$ \frac{d}{dt}\left(\begin{array}{ccc} \cos\omega t & -\sin\omega t & 0\\ \sin\omega t & \cos\omega t & 0\\ 0&0&1 \end{array}\right)=\omega\left(\begin{array}{ccc} -\sin\omega t & -\cos\omega t & 0\\ \cos\omega t & -\sin\omega t & 0\\ 0&0&0 \end{array}\right)=-\omega\left(\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{array}\right)\left(\begin{array}{ccc} \cos\omega t & -\sin\omega t & 0\\ \sin\omega t & \cos\omega t & 0\\ 0&0&1 \end{array}\right). $$ The last step is non-trivial, but you can check yourself that it multiplies out correctly (I may be off by a minus sign). It's very important that we end up with this numeric matrix times the original rotation matrix.

So then, this means that $$ \boldsymbol{v}=-\omega\left(\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{array}\right)\boldsymbol{r}(t). $$ I'll leave it to you to check that the action of this matrix on a vector in the $xy$-plane is equivalent to computing the cross-product with $\boldsymbol{\hat z}$, which when multiplied times that factor of $\omega$ we got from the chain rule, is the angular velocity vector.

While this has only been a very specific example, this fact about the derivative of a rotation matrix being equal to some matrix times the original rotation is actually a generic property of rotations (which can be worked out by writing the rotations as exponentials of the generators of the Lie algebra of SO(3)).

Though the matrix that gets factored out is not always the one written above (in general it will depend on the angles you're rotating by), this fact about it being equivalent to a cross-product {\em is} generic and always occurs. This is the reason behind the formula $\boldsymbol{v}=\boldsymbol{\omega}\times\boldsymbol{r}$. (Again, please forgive any errant minus signs in the above).

There are other approaches to obtaining this same result which you may find online. These typically involve drawing some pictures.

  • 3
    $\begingroup$ Isn't this answer wildly running away with things? Wouldn't it be a lot simpler just to derive it using the circumference of a circle and then differentiate? $\endgroup$
    – DKNguyen
    Commented Dec 3, 2020 at 23:43
  • 1
    $\begingroup$ Oh, is the OP asking about vectors or scalars? Seems to me they're asking about scalars but I guess with the multiplication sign it could be interpreted as vectors. $\endgroup$
    – DKNguyen
    Commented Dec 3, 2020 at 23:46
  • 3
    $\begingroup$ @DKNguyen Ah, that would make sense actually. It didn't occur to me that the $\times$ might be indicating multiplication and not a cross product. Interpreted as a cross product, the equation they wrote is still correct (up to signs I don't remember). $\endgroup$ Commented Dec 4, 2020 at 0:42

The formula you listed is relevant when you have an object moving around a circle of radius $r$.

$\vec{\omega}$ is the amount of the circle that the object traverses in some period of time (normally radians/sec) and has a direction tangent to the plane that the circle lies in in 3D space. $\vec{r}$ is the vector that points from the origin of the circle to object on its outer edge.

The linear velocity is tangent to the circle, so its direction is $\hat{\omega} \times \hat{r}$. You can see this by recalling that the cross product of two vectors has a direction that is normal to both, so it will pick out the other direction in the plane the circle lives in since $\hat{\omega}$ is out of the plane, and $\hat{r}$ is normal to the circle.

The speed of the object is the amount of distance traveled over time, since $|\vec{\omega}|$ is the number of radians the object moves over time, we just need to multiply it by the radius of the circle to get arc length. Recall that the arc length along a circle is $s = r \theta$.

Putting it all together, the speed is $|\vec{\omega}| r$ and the direction is $\hat{\omega} \times \hat{r}$, this is of course $\vec{\omega} \times \vec{r}$.


To add a bit to the earlier answers. A right-handed coordinate system is almost always (implicitly) assumed. $\vec v = \vec \omega \enspace\times \enspace \vec r$ assumes a right-handed coordinate system where angles are measured in a counterclockwise direction. $\vec \omega$ is an axial vector that depends on the "handedness" of the coordinate system, as contrasted with a polar vector that does not; $\vec v$ and $\vec r$ are polar vectors.

In a left-handed coordinate system $\vec v = \vec r \enspace \times \enspace \vec \omega$.

See Symon Mechanics or Goldstein Classical Mechanics for details.

  • $\begingroup$ What's $x$ in those equations? Is it supposed to be a multiplication? If so, you can use \times or a Unicode ×. BTW, I don't think the OP is talking about vectors, just simple 2D planar rotation. $\endgroup$
    – PM 2Ring
    Commented Dec 4, 2020 at 12:28
  • $\begingroup$ @PM 2Ring $x$ is the vector cross product. Maybe my answer was too advanced for the question. $\endgroup$
    – John Darby
    Commented Dec 4, 2020 at 13:59
  • $\begingroup$ @JohnDarby The user PM 2Ring is saying that you should replace $x$ for the cross product with \times; so that it is more clear that you are using a cross product. $\endgroup$
    – Andrew
    Commented Dec 4, 2020 at 14:39
  • $\begingroup$ I replaced x with $\times$. $\endgroup$
    – John Darby
    Commented Dec 4, 2020 at 16:13

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