I was studying about Euler's equations in rotational dynamics

if $t$ denotes torque then

$$t_1 = I_{11}\frac{d\omega_1}{dt} + \omega_2\omega_3(I_{33}-I_{22})$$

Consider a situation where a disk's CM rotates about a vertical axis with angular velocity $Q$ and the disk itself rolls on the ground, rotating about the CM with angular velocity $\omega_s$. $\hat{1}$, $\hat{2}$, and $\hat{3}$ denote the directions of the instantaneous principal axes of the disk.

Consider problem 8.4 of Kleppner https://bayanbox.ir/view/7764531208313247331/Kleppner-D.-Kolenkow-R.J.-Introduction-to-Mechanics-2014.pdf for the figure (pg 337).

$\hat{1}$ would point slightly out of the page towards us.(consider rotating the figure so that the axis is in the horizontal plane. I know the problem can be solved without Euler's equations but I was using them anyway.

Here my professor said that on calculating $\dfrac{d\omega_1}{dt}$, we can write it as the component of $\mathbf{Q}\times\vec{\omega}_\mathrm{net}$ in the $\hat{1}$ direction.

Doing this seems correct since I get the right answer when I solve all such problems, but I don't see why this should be true. First of all $\omega_1$ is the component of $\omega$ and it is a scalar, whose magnitude does not change, so I feel like $\dfrac{d\omega_1}{dt}$ should be zero since the $\omega_1$ we are talking about in the Euler equation is not a vector quantity (right?) But this obviously gives the wrong answer.

And if it is a vector quantity the derivation of Euler's equations does not show it at all since we use the product rule to take separate derivatives of $L_1$ and $\hat{1}$ in the derivation.

Where am I incorrect?

  • 1
    $\begingroup$ When we are considering problems like these, we need to be careful about the frame of reference from which we are analyzing the motion. If a vector is constant in a rotating frame, it is not necessarily constant in an inertial frame. The Omega Theorems are useful in this context. $\endgroup$ Oct 20, 2017 at 14:35
  • $\begingroup$ @SayanMandal: Omega Theorems? Can you provide a reference? $\endgroup$
    – NickD
    Oct 20, 2017 at 15:05
  • $\begingroup$ @Nick: Here is what I am talking about. Turns out that "Omega Theorem" is not a very popular name, but this is exactly it. ocw.mit.edu/courses/aeronautics-and-astronautics/… $\endgroup$ Oct 20, 2017 at 15:36
  • $\begingroup$ Assuming the edit is approved, I changed most w's to omega's - I left the expression that @SayanMandal had a question about in previous comment alone: it should be edited but I'm not sure to what. Maybe the OP would care to answer his question? $\endgroup$
    – NickD
    Oct 20, 2017 at 17:39
  • $\begingroup$ @SayanMandal: yes I understand that w1 won't be constant in an inertial frame as the principal axes themselves are changing but won't it's magnitude be constant? And we are taking the derivative of the magnitude right? $\endgroup$
    – Stormlight
    Oct 20, 2017 at 17:51

1 Answer 1


Consider the vector $\vec r'=U\vec r$, where prime denotes quantities in the lab (or inertial) frame, and unprimed in the body-fixed (or rotating) frame.

Taking the time-derivative: $$ \frac{d\vec r'}{dt}= \dot U\vec r+ U\frac{d\vec r}{dt} \tag{1} $$ The velocity in the lab frame is the result of two changes: the first is captured by $\dot U\vec{r}$, and is the change in the relative positions of the frames. The second term $d\vec{r}/dt$ is the velocity of a point in the body-fixed system. In the case of a rigid body, this term is $0$ since no particle changes its position in that frame.

The matrix $U$ is a rotation matrix, and its inverse is just its transpose, denoted by $\tilde U$, so that $$ U\cdot \tilde U=\tilde U\cdot U=\hat 1 $$ Now insert the unit and use $\frac{d\vec r}{dt}=0$ in (1) to get $$ \frac{d\vec{r}\,'}{dt} = \dot{U}\cdot \tilde U\cdot U\vec{r}= \dot{U}\cdot \tilde U\,\vec r'\, . $$ Note that this relates the time derivative of $\vec r'$ (a vector in the lab frame) and $\vec r'$ itself.

The key is to examine $\dot U\cdot \tilde U$. Since $U\cdot\tilde U=\hat 1$, it follows that $$ \frac{d}{dt}\hat 1= 0= \frac{d}{dt}\left(U\cdot\tilde U\right)= \dot U\cdot\tilde U + U\dot{\tilde U}\, . $$ Call $A'=\dot U\cdot\tilde U$. Note that $\tilde A'= \widetilde{(\dot U\cdot\tilde U)}= U\cdot \dot{\tilde U}$ so we really have $$ 0= A'+\tilde A'\qquad \Rightarrow\qquad \tilde A'=-A'\, $$ i.e. $A'$ is an antisymmetric matrix. The most general form of the matrix $A'$ that satisfies this is \begin{align} A'=\left(\begin{array}{ccc} 0&-\omega'_{12}&\omega'_{13}\\ \omega'_{12}&0&-\omega'_{23}\\ -\omega'_{13}&\omega'_{23}&0\end{array}\right):= \left(\begin{array}{ccc} 0&-\omega'_{3}&\omega'_{2}\\ \omega'_{3}&0&-\omega'_{1}\\ -\omega'_{2}&\omega'_{1}&0\end{array}\right)\, . \end{align} Moreover, note that $$ \left(\begin{array}{ccc} 0&-\omega'_{3}&\omega'_{2}\\ \omega'_{3}&0&-\omega'_{1}\\ -\omega'_{2}&\omega'_{1}&0\end{array}\right)\left(\begin{array}{c} r'_1\\ r'_2\\r'_3\end{array}\right)= \left(\begin{array}{c} \omega'_2r'_3-\omega'_3r'_2\\ \omega'_3r'_1-\omega'_1r'_3\\ \omega'_1r'_2-\omega'_2r'_1 \end{array}\right):=\vec\omega\,'\times \vec r\,'\, . \tag{2} $$ where the formal vector $\vec \omega'$ has been introduced so as to write the matrix action on $\vec r'$ in (2) as a cross-product. Thus $$ \frac{d\vec r'}{dt}=\vec\omega'\times \vec r'\, . $$

To answer your question explicitly, defining $\vec\omega'$ as a vector is seen to be just to trick to encapsulate all the algebra above. The components $(\omega'_1,\omega'_2,\omega'_3)$ are of course not vectors, just components.

The use of the cross-product notation is provides the following geometrical insight.

enter image description here

We imagine a vector $d\vec{\phi}'$ of magnitude $d\phi'$ and pointing in the direction $\vec{n}$, which is the axis of rotation. As a result of this rotation, a vector $\vec{r}'$ is transformed to $\vec{r}'+d\vec{r}'$. Evidently, $d\vec{r}'$ is perpendicular to $\vec{n}$ and to $\vec{r}'$. The length of $d\vec{r}$ is geometrically $r'\,\sin\theta\,d\phi$ where $\theta$ is the angle between $\vec{n}$ and $\vec{r}$, since the area of the shaded triangle is clearly $\frac{1}{2}r'\sin\theta\,dr'$, which is just one half of the area $\vert d\vec{r}'\times \vec{r}'\vert $ of the parallelogram with sides of length $r'\sin\theta$ and $dr'$. If this rotation occurs during a time interval $dt$, then $\vec{\omega}=\frac{d\vec{\phi}}{dt}$ is just the rate at which $\vec{r}$ rotates.

[Picture credit: from the textbook by L.N. Hand and J.D. Finch, Analytical Mechanics, Cambridge University Press]


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.