I can not wrap my head around the following doubt:

How can we express (or prove the fact that) the position of a material point on a rigid body as the sum of a "traslational component" and a "rotational component"? Namely: $\underline{r_i} = \underline{r_c} + R \cdot \underline{r_c}$, where $R$ is a rotation matrix.

As always, any kind of comment or answer is highly appreciated!


3 Answers 3


First we can define what measuring a point means, and the nomenclature used


The location of a point P as measured from the origin, is $\boldsymbol{r}_P$ and the components are expressed on some common basis vectors (world coordinate system).

You are free at any point to decompose this vector into two vectors, based on another arbitrary point A. I have designated the world basis vectors as {0}.


The vector $\boldsymbol{r}_{P/A}^{\{0\}}$ means the coordinates of P relative to A in the {0} basis vectors.

$$ \boldsymbol{r}_{P/A}^{\{0\}} = \boldsymbol{r}_{P}^{\{0\}} - \boldsymbol{r}_{A}^{\{0\}}$$

Now we can define this relative vector on a body coordinate basis vectors {1}


where $\boldsymbol{r}_{P}^{\{1\}}$ is the location of point P as measured from the {1} coordinate frame. This is a fixed vector. The components of this vector do not change with time.

In order to relate the vectors on two different basis vectors you need a rotation matrix $\bf{R}_{\{0\}}^{\{1\}} $ that contains the unit vectors of {1} as columns.

$$\bf{R}_{\{0\}}^{\{1\}} \boldsymbol{r}_{P}^{\{1\}} = \boldsymbol{r}_{P}^{\{0\}} - \boldsymbol{r}_{A}^{\{0\}}$$

and the re-arranging of the above to get the decomposition

$$\boldsymbol{r}_{P}^{\{0\}} = \boldsymbol{r}_{A}^{\{0\}} + \bf{R}_{\{0\}}^{\{1\}} \boldsymbol{r}_{P}^{\{1\}} \tag{1}$$

So far so good, as it covers the position kinematics as inquired upon in this question.

The next step is to transition to the velocity kinematics to describe the motion P as decomposition of two motions.

The direct time derivative of (1) gives us the motion kinematics of point P. Note that $\frac{\rm d}{{\rm d}t} \bf{R}_{\{0\}}^{\{1\}} = \boldsymbol{\omega}_{\{0\}} \times \bf{R}_{\{0\}}^{\{1\}}$ which is a shortcut notation to a matrix composed with the $\omega\times$ cross product with each column of the rotation matrix.

$$\boldsymbol{v}_{P}^{\{0\}} = \boldsymbol{v}_{A}^{\{0\}} + \boldsymbol{\omega}_{\{0\}} \times \bf{R}_{\{0\}}^{\{1\}} \boldsymbol{r}_{P}^{\{1\}} \tag{2}$$

where $\boldsymbol{v}_{P}^{\{0\}}$ is the rate of change of $\boldsymbol{r}_{P}^{\{0\}}$ and so on for the other quantities.

So now you are asking of why/how can you interpret the above as

$$ \underbrace{ \boldsymbol{v}_{P}^{\{0\}}}_\text{motion of P} = \underbrace{ \boldsymbol{v}_{A}^{\{0\}}}_\text{motion of A} + \underbrace{ \boldsymbol{\omega}_{\{0\}} \times \bf{R}_{\{0\}}^{\{1\}} \boldsymbol{r}_{P}^{\{1\}} }_\text{rotation about A}$$

And the answer the that is that there is nothing special about A. It is an arbitrary point, and any point would work. This is because if you fix A then the remaining allowed motion is always going to be a rotation about A.

And this is Chasle's theorem exactly. That the motion of a point riding on a body is the superposition of the translation of an arbitrary point, and a rotation about this arbitrary point.

For dynamics you often consider the center of mass as the arbitrary point, not because it simplifies the kinematics, but because it simplifies the equations of motion.

In robotics you often consider some joint specific point for decomposition because it does simplify the kinematics, but it makes the dynamics more complex.


Assuming an arbitrary origin, the translational component carries us to the centre of mass of the body while the rotational component takes us to the specific point in the body under consideration. So if the origin coincides with the CoM, the translational component is simply zero. Conversely if the point under consideration is the CoM, the rotational component is zero.


Your question is basically the result of chasles theorem which is Any motion of a rigid body is equivalent to the superpostion of translation, and rotation about any point. There are various proofs scattered over the internet, if you are a potential mathematician, but it should be pretty easy enough to convince yourself.

I see this answer has gotten a down vote, but David morin's book on Introduction to Classical mechanics, has a proof of chasles theorem.


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