Consider two people on a disk of radius R, distances $r_1$ > $r_2$ from the center. Suppose the disk rotates at angular velocity $\vec{\omega}$. The velocities of the two people relative to the ground are $\vec{\omega}$ x $\vec{r_1}$ and $\vec{\omega}$ x $\vec{r_2}$ respectively. If you then calculated the velocity of person 2 from the frame of reference of person 1, you would get $\vec{\omega}$ x $\vec{r_2}$ - $\vec{\omega}$ x $\vec{r_1} \ne 0$. But, if two people are on a carousel they should measure each other's velocity as 0.

  • $\begingroup$ The position of the the second person relative to the first person is constantly changing. Why do you say their relative velocity should be zero? $\endgroup$ – M. Enns Sep 3 '16 at 17:03
  • $\begingroup$ I disagree with your statement. The positions are not changing. They are both on the same carousel. I should see person 2 frozen in position. $\endgroup$ – user7348 Sep 3 '16 at 17:06
  • $\begingroup$ Sometimes they are to the north of you, sometimes to the west... $\endgroup$ – M. Enns Sep 3 '16 at 17:07
  • $\begingroup$ @M.Enns That's silly. From my frame of reference by definition I'm always at rest. Suppose initially person 2 is along my radius to the center and I call this North. Now, he's always along my radius to the center which must always be North because I'm at rest. $\endgroup$ – user7348 Sep 3 '16 at 17:12
  • $\begingroup$ OK, if you consider if from a frame of reference that is rotating along with one one of the people then what is $omega$ in that frame of reference? Zero sure $\vec{\omega}$ x $\vec{r_2}$ - $\vec{\omega}$ x $\vec{r_1} = 0$ $\endgroup$ – M. Enns Sep 3 '16 at 17:14

Every point on a rotating rigid body is moving with different tangential velocity according to the rule you stated

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

Yet the distance $\ell_{AB}$ between two arbitrary points A and B remains the same. This can only occur of the difference in velocities $\vec{v}_A = \vec{\omega} \times \vec{r}_A$ and $\vec{v}_B=\vec{\omega} \times \vec{r}_B$ is only along the perpendicular to the line connecting A and B.

It is sufficient to prove that $$(\vec{v}_A - \vec{v}_B) \cdot (\vec{r}_A - \vec{r}_B) =0$$ NOTE: $\cdot$ is the dot product and $\times$ is the cross product


Expand out the velocity terms

$$ ( \vec{\omega} \times \vec{r}_A -\vec{\omega} \times \vec{r}_B) \cdot ( \vec{r}_A-\vec{r}_B) =\left( \vec{\omega} \times ( \vec{r}_A-\vec{r}_B) \right) \cdot ( \vec{r}_A-\vec{r}_B) =0 $$

There is vector identity for triple products that states $\left(\vec{a} \times \vec{b} \right) \cdot \vec{b} =0 $

  • $\begingroup$ Just because the distance between the two is invariant doesn't mean that there is no relative velocity between the two. If I swing a ball around my head in a circle the distance does not change, but it most certainly does have a velocity measured from my frame of reference. $\endgroup$ – user7348 Sep 4 '16 at 4:30
  • $\begingroup$ Unless your answer is intended to mean that the velocity of B from frame A is NOT 0. That would be very interesting and in agreement with the Galilean transformation. Essentially my concern was that this seemed to violate the Galilean transformation even in the non-relativistic limit. OK, if the velocity of B from frame A is non 0, does that mean that from frame A, we see B trace out a path? What would the path look like? I would think that because the relative velocity would always be tangential to the line connecting A, B, the path would be a circle. $\endgroup$ – user7348 Sep 4 '16 at 4:54
  • $\begingroup$ In 3D the instant velocity of a point on rigid body traces a helix. It is rotation plus a parallel translation. $\endgroup$ – ja72 Sep 4 '16 at 15:17
  • $\begingroup$ Then you and the ball are not a rigid body. On a rigid body the relative velocity is zero if the distance is constant. $\endgroup$ – ja72 Sep 4 '16 at 15:19
  • $\begingroup$ BTW Relative velocities are very misleading in mechanics. I go with the rule that since there are no relative forces and no relative momenta, all velocities have to be treated as seen from inertial reference frames. $\endgroup$ – ja72 Sep 4 '16 at 15:20

I want to add another answer that might add clarity to the subject. I am speaking only for rigid bodies (rotating frames) and for classical mechanics.

A velocity measurement $\vec{v}_A$ at a point A, is rather meaningless until you couple it with the vector of rotational motion $\vec{\omega}$ of the body. Together they describe a screw motion in space.

The coupled pair $\boldsymbol{v}_A = (\vec{v}_A,\vec{\omega})$ together with the location of A completely describe the motion of a rigid body at one instant.

If at the same time another pair $\boldsymbol{v}_B = (\vec{v}_B,\vec{\omega})$ is described at a different point B on the same body, then this pair will also describe the exact same motion.

Here by motion I mean the complete vector field of velocities. Refer to this post for more details.

So if you have a moving reference frame (like a disk on a turntable) and measure the velocities at various points A, B, C,.. and construct all the motion pairs $$\begin{aligned} \boldsymbol{v}_A& = (\vec{v}_A,\vec{\omega}) = (\vec{\omega}\times \vec{r}_A,\vec{\omega}) = \boldsymbol{T}_A (0,\vec{\omega}) \\ \boldsymbol{v}_B& = (\vec{v}_B,\vec{\omega}) = (\vec{\omega}\times \vec{r}_B,\vec{\omega}) = \boldsymbol{T}_B (0,\vec{\omega})\\ \boldsymbol{v}_C& = (\vec{v}_C,\vec{\omega}) =(\vec{\omega}\times \vec{r}_C,\vec{\omega}) = \boldsymbol{T}_C (0,\vec{\omega})\\ \ldots \end{aligned}$$

you will notice that all of them a transformation of the same "base" motion of $\boldsymbol{v} =(0,\vec{\omega})$. The 6×6 transformation matrix is well defined as $$\boldsymbol{T}(x,y,z)=\begin{bmatrix}1&0&0&0&z&-y\\0&1&0&-z&0&x\\0&0&1&y&-x&0\\ 0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1 \end{bmatrix}$$

The shift in thinking is that instead of looking at an individual particle as moving with a velocity, the velocity is a measurement of the overall motion of the rigid body (similar to taking a temperature measurement, or a pressure measurement).

This is powerful because it turns out rotational momentum taken at a point also is a measurement of the overall momentum state of the body and together with the motion form a tightly bound description of motion mechanics in 6 dimensions.


  • Linear velocity at a point is a manifestation of a rotation along an axis away from that point.
  • Angular momentum at a point is a manifestation of a linear momentum along an axis away from that point.
  • Torque at a point is a manifestation of a linear force along a line away from the point.
  • That special point with names like instant center of rotation (for motion), percussion axis (for momentum) and line of action (for forces) can be finite or infinite.
  • The math for finding the special location is identical for all the pairs of quantities above.

In the ground frame of reference the person 2 is moving with respect to person 1. Let's put in some simple numbers to make it more concrete. Say our carousel is rotating clockwise (seen from above) at 1 rad/sec and that that at some instant person 1 is 1 m north of the centre and person 2 is 3 m north of the centre. Relative to the ground their velocities will be 1 m/s [east] and 3 m/s [east] respectively. So, the velocity of the second person with respect to the first person will be 3 - 1 = 2 m/s [east] exactly as your equation predicts. If the first person checks the position of the second person a short time later they will find them further east than themselves.

When you say "If two people are on a carousel they should measure each other's velocity as 0." you are thinking of a frame of reference that is rotating with the carousel. In that frame of reference person 1 would always see person 2 in the same position - in this example 2 m ahead of them (assuming they are facing away from the centre). The thing is, in this frame of reference the velocities of the two people are zero because $\omega$ is zero. So you get the expected result 0 - 0 = 0;


As mentioned by you angular velocity is same. So theta1=theta2 that implies r1/l1=r2/l2 now as r1>r2 l1>l2 (here l1 refers to the length of the arc of the particle 1)

If you consider the distance moved by the particle on outer circle is more than the distance moved by the particle in the inner circle u can understand that the speed of outer particle has to be greater than that of inner particle so that the angle remain same. Here the speed is the tangential speed.

The particle 1 sees the particle 2 to have traversed the same angle so it seems that relative to particle 1 particle 2 is stationary on the disk or carousel as u considered here.


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