Euler Equations for Dynamics of rigid body

[Abhishek11235]

I have been studying the dynamics of free top from Morin's book. In his book when describing the dynamics,he writes down the equation of motion as shown in screenshot. However,I am not able to understand which term refers to which coordinate system. For eg: Here $\omega$ refers to angular velocity in the frame of centre of mass(body frame) or in fixed frame(space frame)? As Morin tells,the torque on L.H.S are calculated in frame of fixed point or CoM so that torque vanishes. So $\omega$ should be in body frame? In short, can you answer in brief which term in Euler Equation refers to which coordinate?

 

[vanhees71]

The Euler equations all refer to the body-fixed reference frame. The general equation in this case reads
$$\mathrm{D}_t \vec{J}'=\dot{\vec{J}}'+\vec{\omega}' \times \vec{J}'=\vec{M}',$$
where I've put a prime on the vector symbols to indicate that these are components wrt. the non-inertial body-fixed system. $\vec{M}'$ are the corresponding components of the torque. If the only force acting on the body is the homogeneous gravitational field of the Earth i.e., $\vec{F}=m \vec{g}$ for a point particle, then $\vec{M}$ is given as $m \vec{s} \times \vec{g}$, where $\vec{s}$ is the vector from the body-fixed point of rotation to the center of mass. If the body is freely falling or if the body-fixed point of rotation is chosen to be the center of mass, then $\vec{M}'=0$ and this gives the equation of motion (9.46) in the book, since
$$\vec{J}'=\hat{\Theta}' \vec{\omega}' \; \Rightarrow \; \hat{\Theta}' \dot{\vec{\omega}}'+\vec{\omega}' \times \hat{\Theta}' \vec{\omega}'.$$
Since $\hat{\Theta}'$ are the components of the tensor of inertia with respect to the body-fixed reference basis around the body-fixed point of rotation you can always choose this basis such to make $\hat{\Theta}'=\mathrm{diag}(I_1,I_2,I_3)$, and then writing out the equation in components, you get (9.46), which however is for the special case of a symmetric top, where (at least) two of the principle moments of inertia are equal, i.e., as stated in the book $I_1=I_2=I$.

The free symmetric top is not too difficult to solve. For a full understanding, you should introduce Euler angles between the body- and space-fixed reference bases, choosing the space-fixed (inertial) basis such that the 3-axis points into the direction of the conserved angular momentum.

 

[Abhishek11235]

Ok. Now here is the next page. On this page,he has chosen com as Origin. My question is what is $\omega_{i}$ in this equation? Are they angular velocity seen in CoM frame. If yes then I have aligned my axis in Principal axis direction. And due to this there should be no $\omega_{1,2}$ only $\omega_{3}$. Can you help me to visualise this?

 

[vanhees71]

It's not so clear to me since I don't know the specific definitions of that book. The little picture is clearly in the body frame as stated in the caption to the figure. In the body frame of the free top usually the 3-axis is chosen as the symmetry axis of the top, i.e., the principle momenta of inertia $I_1=I_2=I$. That's why this symmetry axis is at rest in the body frame, and the angular momentum as well as the angular velocity precesses around this symmetry axis with the same precession frequency.

I have a complete analysis in my recent mechanics manuscript for teacher students. Unfortunately it's in German. So I guess I can only refer to the figures for the point of view in the space-fixed inertial frame (i.e., the frame we are in as experimenters playing with spinning tops :-)):

https://th.physik.uni-frankfurt.de/~hees/publ/theo1-l3.pdf

You find the figure on p. 126 for the two cases of a prolate and an oblate body. In your notation $A=I_1=B=I_2$ and $C=I_3$. Seen from the body-fixed system the angular momentum is at rest due to angular-momentum conservation for the free top, i.e., there's no torque, because the top is fixed in its center of mass. Of course in general both the symmetry axis of the body (the red vector $\vec{f}$) as well as the angular velocity (the blue vector $\vec{\omega}$) precess around this fixed angular-momentum axis. At the same time, as seen from the body-fixed frame the angular velocity precesses around the axis of symmetry. As the analysis with help of the Euler angles shows, the red "Polkegel" (English: "body cone") rolls on the "Spurkegel" (English: "space cone"). As seen in the space-fixed frame, at the same time the symmetry axis precesses around the fixed angular-momentum axis, making up the black "Präzessionskegel" (English: "precession cone").

 

[Abhishek11235]

That is what I wanted to ask. Does $\omega$ precesses around symmetry axis? I think it defies my intuition. Am I right?

 

[vanhees71]

Yes, $\omega$ precesses around the symmetry axis. To get the intuition think yourself as sitting at the top (i.e., being an observer in the body-fixed non-inertial reference frame).