Free rotation of a rigid body

Question

So I am currently reading Fowles and Cassidy and there is something I'm confused about in the section about geometric description of free rotation of a rigid body. I will present the stuff first that I am confused about and then I will ask my question.

The book says

..with zero torque the angular momentum of the body, as seen from the outside must remain constant in direction and magnitude according to the general principle of conservation of angular momentum. With respect to rotating axes fixed in the body , however the direction of the angular momentum may change, altough its magnitude must remain constant.

From those statements we derive two equations $I_{1}^2w_{1}^2 + I_{2}^2w_{2}^2 + I_{3}^2w_{3}^2 = L^2 = constant$ (Equation 1)

$I_{1}w_{1}^2 + I_{2}w_{2}^2 + I_{3}w_{3}^2 = 2T_{rot} = constant$. (Equation 2)

These are equations of two ellipsoids whose principal axes coincide with the principal axes of the body. The first ellipsoid has principal diameters in the ratios $I_{1}^{-1} : I_{2}^{-1} : I_{3}^{-1}$.

The second ellipsoid has principal diameters in the ratios $I_{1}^{-1/2} : I_{2}^{-1/2} : I_{3}^{-1/2}$.

So

1. I don't see how come it follows from conservation of angular momentum must remain constant ?
2. I don't understand how come with respect to rotating axes fixed in the body it may experience angular momentum that will change in direction I don't understand and I don't fully visualize it, so it would be perfect if someone can explain this with some also clear visualization of why this happens.

Lastly, I don't see how those two ellipsoids principal axes coincide with the principal axes of the body, also I don't understand what do they mean by "principal diameters" ?

Answer 1

The statement is really about the transformation between inertial co-ordinates and co-ordinates fixed to the body. This is expressed by:

$$D_t = d_t + \omega(t)\times\tag{1}$$

where $D_t$ is the "total" derivative, i.e. the time derivative in the inertial frame, $d_t$ is the time derivative in the frame fixed to the body.

Since there there are no torques on the body, angular momentum is conserved in the inertial frame so that $D_t L=0$. Therefore we get:

$$d_t L = -\omega(t)\times L\tag{2}$$

So, in the frame fixed to the body, the time derivative of $L$ is given by a cross product with $L$, so the time derivative is always at right angles to $L$. Therefore, $|L|$ is constant, but $L$ can certainly change direction by (2): its head is constrained to a sphere.

So this reasoning yields $I_{1}^2w_{1}^2 + I_{2}^2w_{2}^2 + I_{3}^2w_{3}^2 = const.$

The other equation $I_{1}w_{1}^2 + I_{2}w_{2}^2 + I_{3}w_{3}^2 = 2T_{rot}$ does not follow from the paragraph you referenced, but it is correct nonetheless: it is instead a statement of conservation of rotational kinetic energy, not of angular momentum.

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BTW (1) is derived from the Leibnitz rule applied to the matrix equation $X=\exp(H(t))\,Y(t)$ where $H$ is a $3\times 3$ skew-symmetric matrix; $Y$ are the co-ordinates in the frame fixed to the body, $X$ are those in the inertial frame and $\exp(H)$ is the total rotation operator: we think of the co-ordinates written as $3\times 1$ column vectors. At $t=0$ (when our co-ordinate frames are instantaneously aligned) this yields $\dot{X} = \Omega(t)\,Y + \dot{Y}$, where $\Omega(t)$ is derived by a complicated Lie-theoretical formula from the skew symmetric $H$, but $\Omega$ is nonetheless a skew-symmetrix $3\times 3$ matrix, and can thus be represented as a cross product $\omega\times Y$.

Answer 2

I don't have that text, but I can find the table of contents on the internet. Somewhere in that text (most likely chapter 5 on non-inertial reference systems), there should be a derivation that for any vector quantity $\boldsymbol q$, the time derivative of that vector in an inertial frame and a rotating frame that share the same origin are related by $$ \left(\frac{d \boldsymbol q}{dt}\right)_\text{inertial} = \left(\frac{d \boldsymbol q}{dt}\right)_\text{rotating} + \boldsymbol \omega \times \boldsymbol q \tag 1 $$

This is the same expression as the first equation in WetSavannaAnimal's answer, but expressed in terms more consonant with someone reading Fowles & Cassiday.

In the special case of a $\boldsymbol q$ being constant in the inertial frame, the above reduces to $$\left(\frac{d \boldsymbol q}{dt}\right)_\text{rotating} = - \, \boldsymbol \omega \times \boldsymbol q$$ Note that the right hand side is either zero or is normal to $\boldsymbol q$. This raises an important concept: The time derivative of a constant length vector is either zero or is normal to the vector in question. The converse is also true: If the time derivative of a vector is always zero or normal to the vector in question, the magnitude of that vector must necessarily be constant. In the case of torque-free rotation, angular momentum is a conserved quantity in the inertial frame. The above means that from the perspective of a body-fixed frame, the magnitude of the angular momentum is constant in the case of torque-free rotation: $$L^2 = \boldsymbol L \cdot \boldsymbol L = (I_1\omega_1)^2 + (I_2\omega_2)^2 + (I_3 \omega_3)^2 = \text{constant} \tag 2$$

Assuming all of the principal moments of inertial are positive, this is the equation of an ellipsoid. What if the body-fixed frame is not aligned with the principal axes? You'll get a messier quadratic form, but still an ellipsoid if that quadratic form is positive definite. This quadratic form is positive definite iff the principal moments of inertia are positive, and the axes of that ellipsoid are the principal axes.

What about that other ellipsoid? One way to arrive at that result is to assume kinetic energy is conserved. This is not necessarily a valid assumption. In fact, it fails in the case of a non-rigid body. While energy is (momentarily) conserved for a non-rigid body, kinetic energy is not. Some of the rotational energy gets converted into heat in the case of a non-rigid body, and that heat eventually gets radiated out into the universe.

An alternative is to once again use the fact that angular momentum is conserved. I'll look at $\frac{d}{dt}(2T) = \frac{d}{dt}(\boldsymbol \omega \cdot \boldsymbol L)$ from the perspective of the rotating frame. This expands to $\frac{d\boldsymbol \omega}{dt}\cdot \boldsymbol L + \boldsymbol \omega \cdot \frac{d\boldsymbol L}{dt}$. From the above, $\frac{d\boldsymbol L}{dt} = -\boldsymbol \omega \times \boldsymbol L$. This is either zero or orthogonal to $\boldsymbol \omega$, so the second term on the right vanishes. After a bit of algebra (not shown), the first term also vanishes. Thus $\frac{d}{dt}(\boldsymbol \omega \cdot \boldsymbol L)$ is zero; kinetic energy is indeed conserved in the case of a rigid body undergoing torque-free rotation: $$\boldsymbol \omega \cdot \boldsymbol L = I_1{\omega_1}^2 + I_2{\omega_2}^2 + I_3 {\omega_3}^2 = 2T \tag 3$$