1
$\begingroup$

I want to proof that the derivative of Rodriguez's Rotation Formula equals the formula for relating the linear velocity of a point on a moving body to the angular velocity (following the footsteps of this article, p. 5; even though the author proofs it there in a simpler manner).
If I have a vector $\vec r$ which is rotated with the Rodriguez's Rotation Formula around the vector $\vec n$ by the angle $\theta$, with the rotation point being the origin, I get: $$ \vec r'=(\vec n \cdotp \vec r)\vec n+\cos(\theta)(\vec r - (\vec n \cdotp \vec r)\vec n)+\sin(\theta)(\vec n \times \vec r) $$ If I considered the equation above as a function which describes the constant vector $\vec r$ rotating around the constant axis $\vec n$ by the changing angle $\theta$ dependent of the time, I could compute the linear velocity of the vector $\vec r'$ by differentiating the equation with respect to the time $t$ to get: $$\dot {\vec r'}=-\sin(\theta)\dot \theta (\vec r - (\vec n \cdotp \vec r)\vec n)+\cos(\theta)\dot \theta(\vec n \times \vec r)=\dot \theta(-\sin(\theta) (\vec r - (\vec n \cdotp \vec r)\vec n)+\cos(\theta)(\vec n \times \vec r))$$ According to that Q&A on physics stack exchange I could also just compute the linear velocity of the vector from the angular velocity: $$\dot{\vec r'} = \vec\omega \times \vec r' $$ where $\vec \omega$ is the angular velocity defined as $\dot {\theta}\vec n$. (Remember, the vector is rotating around the origin and the whole system isn't moving, so I am not in need of subtracting the rotation point from the vector to be rotated and also not in need of adding the velocity of the system in the end.)
Substituting Rodigruez's Rotation Formula into this equation, I get: $$\dot{\vec r'} = \vec\omega \times ((\vec n \cdotp \vec r)\vec n+\cos(\theta)(\vec r - (\vec n \cdotp \vec r)\vec n)+\sin(\theta)(\vec n \times \vec r)) = \dot {\theta}\vec n \times ((\vec n \cdotp \vec r)\vec n+\cos(\theta)(\vec r - (\vec n \cdotp \vec r)\vec n)+\sin(\theta)(\vec n \times \vec r)) $$ Now I can try to proof that the derivative of Rodriguez's rotation formula is the same as the formula for relating the linear velocity of a point to the angular velocity: $$\dot {\theta}(\vec n \times ((\vec n \cdotp \vec r)\vec n+\cos(\theta)(\vec r - (\vec n \cdotp \vec r)\vec n)+\sin(\theta)(\vec n \times \vec r))) = \dot \theta(-\sin(\theta) (\vec r - (\vec n \cdotp \vec r)\vec n)+\cos(\theta)(\vec n \times \vec r)) $$ Or, simplified (if this is done correctly): $$\dot {\theta}(\vec n \times ((\vec n \cdotp \vec r)\vec n+\cos(\theta)(\vec r - (\vec n \cdotp \vec r)\vec n)+\sin(\theta)(\vec n \times \vec r))) = \dot \theta(\vec n \times \cos(\theta)\vec r -\sin(\theta) (\vec r - (\vec n \cdotp \vec r)\vec n))$$ So I get the equation to solve (without $\dot \theta$): $$ \vec n \times ((\vec n \cdotp \vec r)\vec n+\cos(\theta)(\vec r - (\vec n \cdotp \vec r)\vec n)+\sin(\theta)(\vec n \times \vec r)) = \vec n \times \cos(\theta)\vec r -\sin(\theta) (\vec r - (\vec n \cdotp \vec r)\vec n)$$ This includes exactly the following problem involving transformation of vector sums including the vector cross product: I need to rewrite the term $ (\vec a \times \vec b) + \vec c $, where $\vec a$, $\vec b$, and $\vec c$ are some arbitray vectors in $\mathbb R^3$, so that I get $$(\vec a \times \vec b) + \vec c = \vec a \times (\vec b + \vec x)$$ where $\vec x$ is some vector computed from the given vectors.

Edit 1: Including, in addition to the simple mathematical transformation problem involving the vector cross product, the context of the question (trying to proof that the derivative of Rodriguez's Rotation Formula equals the formula for relating the linear velocity of a point on a moving body to the angular velocity).
Edit 2: Changing the title, the tags and the order of the text of the question to properly reflect the new main focus (proof of the derivative of Rodriguez's Rotation Formula --> in accordance with the new answer).

$\endgroup$
  • $\begingroup$ Welcome the Mathematics Stack Exchange! A quick tour of the site (math.stackexchange.com/tour) will help you get the most of your time here. $\endgroup$ – dantopa Dec 27 '18 at 19:18
0
$\begingroup$

Let $\mathbf{K}$ be the matrix such that $\mathbf{K}\vec{v} = \vec{n} \times \vec{v}$ for all vectors $\vec{v}$. The Rodrigues formula can then be stated as $$\vec{r}' = \big(I+(\sin\theta)\mathbf{K} + (1-\cos\theta)\mathbf{K}^2\big)\vec{r}$$

where $\vec{r}$ is constant.

Taking the derivative gives $$\dot{\vec{r}}' = \dot\theta\big((\cos\theta)\mathbf{K} + (\sin\theta)\mathbf{K}^2\big)\vec{r}$$

On the other hand, we have $$\dot{\vec{r}}' = \vec{\omega} \times \vec{r}' = \dot\theta\,\vec{n}\times \vec{r}' = \dot\theta \,\mathbf{K}\vec{r}' = \dot\theta\big(\mathbf{K}+(\sin\theta)\mathbf{K}^2 + (1-\cos\theta)\mathbf{K}^3\big)\vec{r}$$

Now we use the identity $\mathbf{K}^3 = -\mathbf{K}$ to obtain $$\dot{\vec{r}}' = \dot\theta\big((\cos\theta)\mathbf{K} + (\sin\theta)\mathbf{K}^2\big)\vec{r}$$

which reproduces the above result.

$\endgroup$
1
$\begingroup$

For this to be possible, you have $c=a\times x$ so $a$ must be perpendicular to $c$. In which case there are infinitely many possible $x$

$\endgroup$
  • $\begingroup$ Please ignore this answer. The question has been changed beyond all recognition since I answered it. $\endgroup$ – David Quinn Dec 28 '18 at 22:45
  • $\begingroup$ I am sorry, but I needed to include a specific problem, and this problem took the question over. I decided to rewrite the question because of mechanodroid's excellent answer to the specific problem. But thanks for your great help anyway! It was exactly the answer I needed at that time, that I can not proof the specific problem with my approach. $\endgroup$ – alpabrz Jan 3 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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