I know the written derivation of $v=rω$, we just differentiate $∆s$ and $r∆\theta$ but can someone please provide an intuitive derivation.

  • 1
    $\begingroup$ Have you tried drawing a diagram? $\endgroup$
    – DanDan面
    Commented Feb 23, 2021 at 6:36
  • 2
    $\begingroup$ If you run around a circle with the same frequency you better run faster and faster if the circle gets larger and larger! $\endgroup$ Commented Feb 23, 2021 at 8:00
  • 1
    $\begingroup$ But what does it have to do with linear velocity $\endgroup$
    – Shiv
    Commented Feb 23, 2021 at 8:54
  • 4
    $\begingroup$ I suggest you read the through the answers by John Alexiou, it is probably one of the most simplest detailed explanation on this topic you can find $\endgroup$ Commented Feb 23, 2021 at 11:58
  • 1
    $\begingroup$ Does this answer to another question address your question? $\endgroup$ Commented Feb 23, 2021 at 14:22

4 Answers 4


I think it might help to consider a case of uniform circular motion. The OP is already aware about its general derivation through differentiation, but uniform circular motion may help develop some intuition.
Consider that an object undergoes circular motion with constant angular velocity $\omega$ and thus constant speed $v$. A complete angle is $2 \pi$, so we may state:
$$\omega = \frac{2 \pi}{T} \ \ \ \ (1)$$
where $T$ is the time period for one revolution.
Circumference of circle is $2 \pi R$, so speed may be represented as: $$v= \frac{2 \pi R}{T} \ \ \ \ (2)$$ Substituting $(2)$ in $(1)$, we get $$\omega = \frac{v}{R}$$

Now that we have derived this equation, we notice that while $\omega$ tells us about the rate at which angle is covered in radians, multiplying it by $R$ leads us to the rate at which the circumference is traced.

image showing uniform circular motion and basic formulae


You can write down the velocity in cm/s, or m/s, or km/h... for example, the physical velocity $v$ may be expressed as

$$ v = v_{km/h} \, \frac{km}{h}= v_{m/s} \, \frac{m}{s}= v_{L/T} \, \frac{L}{T} $$

where $L$ and $T$ are generic lenght and time quantities, while $v_{L/T} $ is the "value" of $v$ when you take $L$ and $T$ as units.

If you are on a circle of radius $R$, then you have a natural unit of lenght, $R$, and a natural unit of time, the period $P$. Take $T=s$ to be the usual "second", then

$$ v = \frac{2 \pi R}{P} = v_{R/P} \frac{R}{P} = v_{R/s} \frac{R}{s}$$

Here $v_{R/s} $ is the angular velocity (the velocity measured in "natural" units is always $v_{R/P} =2 \pi$).

For non-uniform circular motion it's the exactly the same, but you thave to work with time derivatives. The same stuff works also for non-circular motion, but you have approximate the trajectory locally with a circle (by Taylor expanding the trajectory), or to choose a "pole" wrt measure the angular velocity.


Simple answer

From geometry, you know the length of an arc is $s = r\, \theta$ where $r$ is the radius of the arc, and $\theta$ the angle it spans.

Take the time derivative of the above as the arc describes the motion of a particle riding on a rotating body and you have $$v = r \, \omega $$ where $\omega = \frac{\rm d}{{\rm d}t} \theta$. Also $r$ is the moment arm of motion as instantaneously it is constant (remains fixed in value over small time frames).

Deep Understanding

I hope you are familiar with vectors because at this level kinematics (the study of motion) becomes more intuitive rather than at the component level. Specifically, you should be familiar with the concept of the cross product and the dot product between two vectors. More importantly that an expression of the form $\text{(position)}\times\text{(something)}$ introduces us to the concept of the lever arm.

Together with the lever arm, comes the concept of moment of. Where torque is the moment of force $\vec{\tau} = \vec{r} \times \vec{F}$, and translational velocity is the moment of rotation $\vec{v} = \vec{r} \times \vec{\omega}$.

Let's talk about the rotational velocity (vector) $\vec{\omega}$

I will use three different aspects of motion to talk about rotational velocity.

  • [Kinematics] $\vec{\omega}$ is a mathematical tool to simplify the description of the motion of a rigid body. Instead of specifying velocity vectors at every point on a body, rotational velocity allows us a convenient calculation using the velocity of one point to estimate the velocity of all other points $$ \vec{v}_{\rm point} = \vec{v}_{\rm ref} + \vec{\omega} \times \vec{r} $$

  • [Vector Field] $\vec{\omega}$ is difficult to visualize because it is directly measurable from the observed motion, unlike translational velocity $\vec{v}$ which can be directly measured at every point. Rotational sensing is a fascinating topic. But understand that the true general motion of a rigid body is described by a velocity vector field with both a rotating component and a parallel translating component. This is called Chasles' Theorem.

  • [Geometry] $\vec{\omega}$ is key to understanding the underlying geometry of motion. By geometry, I mean key points and lines in space that provide insight into the motion. The concept of axis of rotation is what I am talking about here.

So how to define $\vec{\omega}$? Each of the three aspects above has its own definition of rotational motion, and each is talking about the same thing.

  • Using kinematics you would describe the derivative of a vector riding on a rotating frame to establish the general rule that $$ \frac{{\rm d} \vec{A} }{{\rm d}t} = \frac{ \partial \vec{A}}{\partial t} + \vec{\omega} \times \vec{A} $$ which is used to find the motions of particles on bodies and derive the kinematic equation seen above. This is kind of a circular argument as the concept of rotational velocity of a frame is just thrown in there and because it all works out it means our guess was correct.

  • Using the vector field described by points that move, but keep their distances fixed with each other also leads to a definition of $\vec{\omega}$. The math is bit more complex, but taking the distance between two points A and B as $\ell_{AB} = \sqrt{ (\vec{r}_A - \vec{r}_B) \cdot ( \vec{r}_A - \vec{r}_B)}$ and differentiating over time with the chain rule yields something like $$(\vec{r}_A - \vec{r}_B) \cdot ( \vec{v}_A - \vec{v}_B) = 0$$ which is solved by either rotational relative motion, or purely translational motion. The absolute motion the combination of the two and hence Chasles' theorem.

  • Using geometry you can show that there is a locus of points in space that describes an infinite line. On this line, the motion of particles of the body is only parallel to the line. This is the definition of the axis of rotation. This velocity on the axis is decomposed as a scalar multiple of the rotation vector $\vec{v}_{axis} = h\, \vec{\omega}$ where $h$ is called the pitch and it is the ratio of translational speed to rotational speed. The motion of all particles on a body is described by the location of the rotation axis $\vec{r}$, the pitch $h$ and the magnitude and direction of the rotational velocity $\vec{\omega}$ $$ \vec{v} = h\,\vec{\omega} + \vec{r} \times \vec{\omega} $$ You can go back and forth between the 3 velocity components and the geometric decomposition of $\vec{r}$, $h$ and $\vec{\omega}$ freely.

  • $\begingroup$ As you can see this question is a can of worms and you can go as deep as you want. Please comment here if you want more specific information about any of the above, or more high level comments/intuition here. $\endgroup$ Commented Feb 23, 2021 at 15:25
  • $\begingroup$ In $𝑣⃗_{point}=𝑣⃗_{ref}+𝜔⃗×𝑟⃗$, does the reference point have to be the axis of rotation? $\endgroup$
    – Dodo
    Commented Mar 5, 2021 at 11:11
  • $\begingroup$ @Dodo It can but it does not have to be. The equation is valid between any two points riding on a rigid body, and it depends on the definition of $\vec{r}$. If the reference point is the axis of rotation, then $\vec{r}$ is the position of the target point relative to the rotation axis. In general $$ \vec{r} = \vec{r}_{\rm point} - \vec{r}_{\rm ref} $$ $\endgroup$ Commented Mar 5, 2021 at 13:11

What you call linear velocity is just the classical velocity of the test mass in m/s. The angular velocity tells you how much "angle" you cover per second when moving in a circle (so rotating).

Assume you do one full rotation per second. The bigger the radius of the revolution is the faster you have to move to cover the full circumference. That is why the linear velocity grows with the radius if the angular velocity is supposed to stay constant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.