My teacher gave this as a fact $\vec{v}=\vec{\omega} \times \vec{r}$(cross product) where $\vec{\omega}$ and $\vec{r}$ are angular velocity and position vectors respectively. He also said that this relation held true for any reference point on the body.

But I was confused as I knew the definition of angular velocity only when $\vec{v}$ was perpendicular to $\vec{r}$.

Also I guess that the definition of magnitude of $\vec{\omega}$ is: $$=\frac{\text{component of velocity perpendicular to r }}{\text{magnitude of r}}$$

But how do we define its direction then?

Also my teacher made another claim: He told that the "angular velocity" observed in a frame on any particle on the rigid body is the same and that itself is defined as the angular velocity of the rigid body.
To understand the above claim and the equation $\vec{v}=\vec{\omega} \times \vec{r}$, I needed the definition of angular velocity.

Note: All I need is the definition of angular velocity about any point on the rigid body with magnitude and direction. I looked up related links to questions on Phys.SE itself but was unable to find one with complete relevancy.


3 Answers 3


In terms of magnitude is $$\omega=\frac{d\theta}{dt}=\lim_{\Delta t\rightarrow 0}\frac{\theta(t+\Delta t)-\theta(t)}{\Delta t}=\frac{v}{R}$$

In vectorial terms $$\vec{v}=\vec{\omega} \times \vec{r}$$ $$|\vec{v}|=|\vec{\omega} \times \vec{r}|=\omega r \sin{\alpha}=\omega R$$ the result doesn't depend on the origin of $\vec{r}$ as long as is chosen along the rotation axis.


Then, because the relative velocity between any 2 points in the rigid body is null all the points are rotating around the same axis with the same angular velocity.

Deriving the velocity vector using the Leibniz rule $$\frac{\partial \vec{v}}{\partial t}=\frac{\partial }{\partial t}(|\vec{v}|\hat{v})=\frac{\partial |\vec{v}|}{\partial t}\hat{v}+|\vec{v}|\frac{\partial \hat{v}}{\partial t}$$ You get two terms: the first one is the tangential velocity $\frac{\partial |\vec{v}|}{\partial t}$ directed in the velocity (tangential) direction while the second term is the modulus of the velocity times the derivative of the velocity versor. Then, deriving the versor you get $$\frac{\partial \hat{v}}{\partial t}=\frac{\partial \theta}{\partial t}\hat{n}$$, where $\hat{n}$ is the normal vector (with respect to the tangential one). Thus, collecting the results for the second term you get $|\vec{v}|\frac{\partial \hat{v}}{\partial t}=|\vec{v}|\frac{\partial \theta}{\partial t}\hat{n}=v\omega\hat{n}=\omega^2R\hat{n}$.

Finally $$\vec{a}=\vec{a_t}+\vec{a_n}=\frac{\partial |\vec{v}|}{\partial t}\hat{v}+\omega^2R\hat{n}$$

  • 1
    $\begingroup$ The question is not about acceleration. $\endgroup$
    – nasu
    Commented Sep 19, 2023 at 18:20
  • $\begingroup$ I fixed my answer $\endgroup$
    – Cuntista
    Commented Sep 20, 2023 at 8:49

The magnitude of angular velocity often known as angular frequency is given by this:

Angular velocity is defined as $\frac{\vartriangle \theta}{\vartriangle t}$ , where the rate of change of angular displacement is equal to the angular velocity , and we use this definition to define the linear velocity of the object , now if its a positive change then it points up , negative means it points down which depends on the reference frame.

From the arc of circle equation $\vartriangle S= r\vartriangle \theta$ substituting change in angle in the angular velocity equation

and from this we have $$w=\frac{\vartriangle S}{r\vartriangle t}$$

which yields $V=w \times r$

Edit for the direction:

For the direction of angular velocity , we can define it using the right hand thumb rule where our hands curl up in the direction of rotation

Now since we know the magnitude of V , we can use $w=V/r$ and multiply it as cross product $\frac{\vec r}{r}$ which means that we are just adding a direction to the angular velocity (as its perpendicular to radius and tangential velocity

so $\vec w=\frac{\vec V \times \vec r}{r^{2}}$ which then yields its direction hence we can write $$\vec w=\frac{ V }{r} \hat{n} $$ where $\hat{n}$ is the unit resultant vector in the direction of $\vec V \times \vec r$

  • $\begingroup$ if w is always perpendicular to v, then it means that angular velocity of points in a rigid body depend on the frame of reference on the rigid body. but I know for a fact that it shouldnt depend $\endgroup$ Commented Sep 19, 2023 at 8:49
  • $\begingroup$ Can you say what you mean with "e frame of reference on the rigid body" for example by giving two different frames? The change of an angle should not depend ont the frame? $\endgroup$
    – trula
    Commented Sep 19, 2023 at 9:03

The equation

$$\vec{v}_A = \vec{\omega} \times \vec{r}_A \tag{1}$$

is a solution to the condition that all lengths must be preserved under a rotation. Here $\vec{r}_A$ is the location of an arbitrary point, and $\vec{v}_A$ is the translational velocity of that point.

Take two arbitrary points A and B riding on a rigid body that is rotating about the origin.

The distance between the two points is $\ell_{AB} = \| \vec{r}_B-\vec{r}_A \|$ but it is more convenient to use the square of the distance, and the dot product rule $\| u \|^2 = u \cdot u$

$$ \ell^2_{BA} = ( \vec{r}_A-\vec{r}_B ) \cdot ( \vec{r}_A-\vec{r}_B ) \tag{2} $$

If distances are to be preserved, then the time derivative of (2) must equal to zero. Evaluate $\tfrac{\rm d}{{\rm d}t} \ell_{BA}^2 = 0$ using the product rule and the commutative property of the dot product to get

$$ \tfrac{\rm d}{{\rm d}t} \ell^2_{BA} =2 ( \vec{r}_A-\vec{r}_B ) \cdot ( \vec{v}_A-\vec{v}_B ) = 0 \tag{3} $$

This means that the difference in velocities is always perpendicular to the difference is location. The cross product enforces this perpendicularity such that if there is a constant vector $\vec{\omega}$ such that

$$ ( \vec{v}_A-\vec{v}_B ) = \vec{\omega} \times ( \vec{r}_A-\vec{r}_B ) \tag{4} $$

at all times, and between all locations, then (4) solves (3) also. Remember that $a \cdot (a\times b) = 0$.

Now since the above is true for any arbitrary points, choose B to be the origin and (4) becomes

$$ (\vec{v}_A - 0) = \vec{\omega} \times ( \vec{r}_A - 0 ) $$


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

which is a special case of (4).

Now if you are asking about the meaning of (1), it is exactly the cross-product that provides the perpendicular distance between the origin and moving point


such that speed $ \| \vec{v}_A \| = d_A \| \vec{\omega} \| $ is the magnitude of rotation times the distance.

Also read this related post about why use cross-products in physics.


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