13 Rotational Dynamics
In the previous chapter we built the machinery to compute the inertia tensor and to identify the principal axes of any rigid body. Now we turn to the dynamics: given the inertia tensor, how does the body actually move? We will find that the equations of motion are dramatically simplified if we work in coordinates that rotate with the body, and we will need some new formalism — rotation matrices and Euler angles — to switch between such body-fixed coordinates and the lab frame.
13.1 Space frame and body frame
The fundamental equation governing rotational motion is the torque-angular-momentum relation, \vec{\Gamma} = \dot{\vec{L}}. where I’ll remind you that \vec{\Gamma} = \vec{r} \times \vec{F}. As we saw, in a fixed set of coordinates, in general the right-hand side of this equation splits into two components: \dot{\vec{L}} = \dot{\mathbf{I}} \vec{\omega} + \mathbf{I} \dot{\vec{\omega}}. In principle, we can describe the motion by calculating \mathbf{I}(t). Of course, computing the inertia tensor once for a static object is already a pain, and we really don’t want to evaluate those integrals for a completely arbitrary orientation of the object. A better alternative is to work in a body coordinate system, one which rotates along with the object itself. This forces the inertia tensor to remain fixed.
The body frame is, of course, non-inertial, so we still find that \vec{\omega} can change even with no external torques, because within the rotating frame \vec{L} does change with time (there are fictitious torques, if you like.) As always, these pictures in different frames of reference are identical in terms of the actual motion.
Recall that when changing from a fixed frame (here usually called the space frame) to a rotating frame (the body frame), the time derivative of a vector changes. So for angular momentum, \vec{\Gamma} = \left( \frac{d\vec{L}}{dt} \right)_{\textrm{space}} becomes \vec{\Gamma} = \left( \frac{d\vec{L}}{dt} \right)_{\textrm{body}} + \vec{\omega} \times \vec{L}, or since we’ve specifically chosen the body frame so that \mathbf{I} is constant, \vec{\Gamma} = \mathbf{I} \dot{\vec{\omega}} + \vec{\omega} \times \vec{L}. Warning: it may be confusing to think about expanding the angular velocity vector \vec{\omega} in the body frame. After all, the body frame is co-rotating with the object, so there is no rotating left in the body frame, right?
We have to update our thinking about rotating frames, in which we’ve always set up our coordinates to orient the rotation vector \vec{\omega} with one of the axes. But as we’ll see shortly, the best choice of coordinate axes to use for the body frame is not always aligned with the rotation vector - it will be dictated by the geometry of our object.
There’s nothing wrong with defining a rotating coordinate system that isn’t aligned with \vec{\omega}, but if we set up our coordinates like this, then the components of \vec{\omega} will generally change as the coordinate system spins around. This is clear if \vec{\omega} is fixed in the space frame; it’s a little less clear how to understand what’s happening if \vec{\omega} changes there too, but we’ll see some examples soon.
13.2 Euler’s equations
Now that we know we can always find a set of principal axes and diagonalize \mathbf{I}, let’s see exactly how the equation of motion \vec{\Gamma} = \mathbf{I} \dot{\vec{\omega}} + \vec{\omega} \times \vec{L} simplifies in this case. The cross product on the right becomes \vec{\omega} \times \vec{L} = \vec{\omega} \times \left( \begin{array}{ccc} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{array} \right) \left( \begin{array}{c} \omega_1 \\ \omega_2 \\ \omega_3 \end{array} \right) = \left| \begin{array}{ccc} \hat{e_1} & \hat{e_2} & \hat{e_3} \\ \omega_1 & \omega_2 & \omega_3 \\ \lambda_1 \omega_1 & \lambda_2 \omega_2 & \lambda_3 \omega_3 \end{array} \right|. (where \hat{e_i} are the unit vectors of my three body-frame coordinate directions 1,2,3, pointing along the principal axes of the object.)
Thus, we find in the body frame the following three equations:
\lambda_1 \dot{\omega_1} - (\lambda_2 - \lambda_3) \omega_2 \omega_3 = \Gamma_1 \\ \lambda_2 \dot{\omega_2} - (\lambda_3 - \lambda_1) \omega_3 \omega_1 = \Gamma_2 \\ \lambda_3 \dot{\omega_3} - (\lambda_1 - \lambda_2) \omega_1 \omega_2 = \Gamma_3.
The form of the torque vector \vec{\Gamma} in body-frame coordinates tends to be messy, although we’ll deal with some special cases. First, let’s discuss what happens when there are no torques, so \vec{\Gamma} = 0.
First observation: if \vec{\omega} is aligned with one of the body-frame axes, then two of its components are zero, and all of the equations of motion reduce to simply \dot{\omega}_i = 0. So rotation about any axis is stable in the absence of torques, as long as the inertia tensor is diagonal in our coordinates.
Second observation: for a highly symmetric object like the cube about its CM, all of the differences between moments \lambda_i are zero, and once again we have \dot{\omega}_i = 0. So a cube or a sphere will rotate stably for any rotation about its center of mass.
13.2.1 Torque-free motion and tumbling
Now we’re finally ready to understand the (torque-free) motion of the deck of cards! As you heard in the video, the motion about certain axes is either stable or unstable. Of course, we just noted that if \vec{\omega} is perfectly aligned with one of the body-frame axes, then the equations reduce to \dot{\omega_i} = 0 and the rotation is constant. However, in the real world the alignment is never perfect, so the question of stability is related to what happens when \vec{\omega} is almost aligned with an axis.
We’ll be concrete and use the deck of cards, which is an example of a rectangular prism, a solid with rectangular cross-section in all three directions. The calculation of the moments of inertia for such an object is exactly the same as for the cube about its CM, except that the limits of integration are different for each diagonal component. For example, a standard deck of cards (in the box) has dimensions of about 3.5” x 2.5” x 0.75”, or 90 mm x 60 mm x 20 mm.
If we let \hat{z} point in the longest direction and \hat{x} the shortest, and consider rotation about the CM once again, then the off-diagonal moments are still zero, and the diagonal moments work out to \overset{\leftrightarrow}{I} \approx M(2\ {\rm cm})^2 \left( \begin{array}{ccc} 33 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 11 \end{array} \right). So in our coordinates, the largest moment is for rotation about the \hat{x} axis, i.e. perpendicular to the broadest face of the deck, like so:
The smallest moment is for rotation about the \hat{z} axis, perpendicular to the smallest side.
Finally, the intermediate moment corresponds to the in-between side:
We could have predicted this without doing the full calculation; we know that for simple rotation about a chosen axis, where \vec{L} and \vec{\omega} are aligned, the moment of inertia is given by integrating over \rho r^2, with r the radial distance to that axis. So for an object with constant density, the size of the moment of inertia is given by the size of the cross-sectional area of the object, as viewed from along the rotation axis.
Now, back to Euler’s equations. Let’s pick an arbitrary axis and call it 3, without regard to which \lambda_i is which yet. We suppose that the rotation is nearly in the \hat{e}_3 direction, so \omega_1 and \omega_2 are non-zero but very small. The equation of motion for \omega_3 (in the absence of torque) is \lambda_3 \dot{\omega_3} = (\lambda_1 - \lambda_2) \omega_1 \omega_2 \approx 0. since both \omega_1 and \omega_2 are small. So \omega_3 is approximately constant for this motion. The other two equations are \lambda_1 \dot{\omega}_1 = (\lambda_2 - \lambda_3) \omega_3 \omega_2 \\ \lambda_2 \dot{\omega}_2 = (\lambda_3 - \lambda_1) \omega_3 \omega_1 Let’s differentiate the first equation, and plug in the second: \lambda_1 \ddot{\omega}_1 = (\lambda_2 - \lambda_3) (\dot{\omega}_3 \omega_2 + \omega_3 \dot{\omega_2}) \\ \approx (\lambda_2 - \lambda_3) \omega_3 \left[ \frac{(\lambda_3 - \lambda_1) \omega_3 \omega_1}{\lambda_2} \right] or simplifying, \ddot{\omega}_1 = - \left[ \frac{(\lambda_3 - \lambda_2) (\lambda_3 - \lambda_1)}{\lambda_1 \lambda_2} \omega_3^2 \right] \omega_1 Again, we’re taking \omega_3 to be roughly constant, so as long as the expression in brackets is positive, this looks like the equation for simple harmonic motion! So \omega_1, and similarly \omega_2, will just undergo small oscillations about zero, and the rotation about axis 3 will appear to be stable, as long as the coefficient in brackets is positive!
But that depends on the values of the moments of inertia. If we have three unequal moments of inertia \lambda_1 \neq \lambda_2 \neq \lambda_3, then we see that if \lambda_3 is either smaller or larger than both \lambda_1 and \lambda_2, then the coefficient is negative and we have simple harmonic motion. But if \lambda_1 > \lambda_3 but \lambda_2 < \lambda_3, the coefficient becomes negative, and we find the differential equation \ddot{\omega_1} = +k \omega_1. The solutions to this equation aren’t sines and cosines, but exponentials. So the value of \omega_1 rapidly increases, and \vec{\omega} will tilt away from the third axis. (We don’t know what happens after this point because once \omega_1 gets large, the assumptions we were making to write this equation break down.)
So now we understand what the astronaut was telling us in the video: for an object with three different moments of inertia, the motion about the axis with the largest or smallest moment is stable, but the motion about the middle axis is unstable, and leads to the tumbling behavior that we saw.
13.3 Worked example: cube tipping
A cube of mass m and side length a is balanced perfectly on one edge. At t=0, we nudge it so that it will tip in the +\hat{x} direction. Assuming the initial speed is roughly zero, what is the angular speed of the cube when it hits the ground?
This time, conservation of energy can be applied: the only force acting is gravity, which is conservative. The gravitational potential depends only on the height of the center of mass: U_g = mgh_{\rm CM}, which starts at a height of h_{\rm CM} = a/\sqrt{2}, and ends at h_{\rm CM} = a/2. The kinetic energy is zero beforehand, and after is due to the rotation and due to the rotation of the cube, so we have \frac{mga}{\sqrt{2}} = \frac{mga}{2} + \frac{1}{2} m v_{\rm CM}^2 + \frac{1}{2} I \omega^2. The center of mass is always a/\sqrt{2} from the edge, so its linear speed is related to the angular speed by v_{\rm CM} = \frac{a}{\sqrt{2}} \omega. Finally, we need the moment of inertia for a cube rotating about its CM, which (as we found before) is I = 1/6 ma^2. Plugging back in and rearranging, then, mga (\sqrt{2} - 1) = \frac{ma^2 \omega^2}{2} + \frac{1}{6} ma^2 \omega^2 = \frac{2}{3} ma^2 \omega^2 or inverting, \omega^2 = \frac{3g}{2a} \left( \sqrt{2} - 1 \right). Now, that all went pretty quickly, so let’s stop and think about the details. First of all, I’m cheating a little bit here by going back to the intro physics formula; we know that the full expression for the kinetic energy should be T = \frac{1}{2} \vec{\omega}^T \mathbf{I} \vec{\omega}. However, in this problem we “know” what the motion of the cube is already. If you imagine this as an experiment, once you’ve seen the cube tip over, you know what the motion looks like - it is only rotating about one axis as it falls over. As long as we choose our coordinates properly (which is especially easy for the cube, as a spherical top!), this will be a principal axis, so we just have L_z = I_{zz} \omega_z, or L = I \omega for short. Basically, if there were non-trivial off-diagonal contributions from \mathbf{I} here, we would know about it because the rotation would be more complicated!
Another burning question you might have: the cube is pivoting about its edge, so why aren’t we using the elements of that inertia tensor? The inertia tensor for a cube rotating about its edge is, in fact, different, and in fact it isn’t even diagonal. At the bottom of the fall when the cube is lying flat, it will be equal to \mathbf{I}_{\rm edge} = Ma^2 \left( \begin{array}{ccc} 5/12 & -1/4 & 0 \\ -1/4 & 5/12 & 0 \\ 0 & 0 & 2/3 \end{array} \right). The axis we’re rotating about, \hat{z}, is at least still principal in this way of looking at it. But the value of I here would be 8/3 Ma^2, not 1/6 Ma^2. Why shouldn’t we use that number?
In the equations I wrote above, we’ve silently invoked one of the main results of this chapter, which is separation of kinetic energy between motion of the CM and motion relative to it: T = T_{CM} + T_{\rm rot}. To divide things up like this, the rotation is explicitly taken to be about the center of mass. Hence, we use I for rotation about the CM when we’re accounting for the center of mass’s translational motion separately. Of course, once we have the inertia tensor about the edge, nothing stops us from using that! Now the rotation includes all of the motion of the cube: mga (\sqrt{2} - 1) = \left( \frac{2}{3} ma^2 \right) \omega^2 and we find exactly the same answer for \omega!
You could object one more time at this point: that inertia tensor is only valid at the bottom of the fall. Why aren’t we keeping track of how \mathbf{I} changes with time, in this approach? Really we should be, which tends to make it dangerous to work in the space frame, and was the motivation for developing the formalism to switch to the body frame in the first place. But we now know that we can describe the inertia matrix as it rotates, by using a rotation matrix. The rotation will be about the \hat{z} axis, so the matrix will take the form \mathbf{R} = \left( \begin{array}{ccc} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right), and then at some angle \theta between 0 and \pi/4, the inertia tensor as the cube falls is \mathbf{I}'(\theta) = \mathbf{R} \mathbf{I} \mathbf{R}^T. The bad news is that the expression for \mathbf{I}' is very messy. The good news is that nothing changes in the \hat{z} direction! Since we’re rotating about \hat{z}, which is already principal, the rotation just scrambles the components in \hat{x} and \hat{y}, but has no effect through the motion we’re considering. This is a simple, but important observation: any rotation about a principal axis will leave it principal.
13.4 Free rotation of a symmetric top
Back to Euler’s equations with no torques: \lambda_1 \dot{\omega_1} - (\lambda_2 - \lambda_3) \omega_2 \omega_3 = 0 \\ \lambda_2 \dot{\omega_2} - (\lambda_3 - \lambda_1) \omega_3 \omega_1 = 0 \\ \lambda_3 \dot{\omega_3} - (\lambda_1 - \lambda_2) \omega_1 \omega_2 = 0. We’ve already covered the spherical top — it will stably rotate about any axis at all — and we’ve done what we can with the asymmetric top for now, so let’s move on to the symmetric top.
We’ll choose \lambda_3 to be the unique eigenvalue of our symmetric top, so that \lambda_1 = \lambda_2. Then only the last equation simplifies, \lambda_3 \dot{\omega}_3 = 0. So once again, we find a certain amount of stability if \Gamma_3 = 0: \dot{\omega}_3 = 0. However, \omega_1 and \omega_2 can still evolve in time if they are non-zero. For the asymmetric top, we could only study the case where \omega_1 and \omega_2 were very small; we explored the stability of rotation mostly around the \omega_3 axis. Now we can go a bit further!
We can reorganize the other two equations, plugging in \lambda_1 for \lambda_2: \dot{\omega}_1 = \frac{(\lambda_1 - \lambda_3) \omega_3}{\lambda_1} \omega_2 = \Omega_b \omega_2 \\ \dot{\omega}_2 = -\frac{(\lambda_1 - \lambda_3) \omega_3}{\lambda_1} \omega_1 = -\Omega_b \omega_1,
For a symmetric top with unique moment \lambda_3 undergoing torque-free rotation, \Omega_b = \frac{\lambda_1 - \lambda_3}{\lambda_1} \omega_3.
This is a coupled set of differential equations, which we haven’t learned how to solve in general yet. But we can use the trick you learned last semester, and used on the homework for rotating frames: we define the complex variable \eta \equiv \omega_1 + i \omega_2. Notice that if we add i times the second equation to the first, we have \dot{\omega_1} + i \dot{\omega_2} = \Omega_b (\omega_2 - i\omega_1) \\ = -i \Omega_b (\omega_1 + i \omega_2), so we can combine both differential equations into \dot{\eta} = -i \Omega_b \eta. We know the solution to this is just an exponential, \eta(t) = \eta_0 e^{-i \Omega_b t}. Setting \eta_0 = \omega_0 + 0i to put the initial \omega_2 = 0, we take the real and imaginary part of \eta to get back \omega_1 and \omega_2: \vec{\omega} = (\omega_0 \cos (\Omega_b t), -\omega_0 \sin (\Omega_b t), \omega_3). So \omega_3 remains constant, while \omega_1 and \omega_2 rotate around at frequency \Omega_b.
This motion - constant and periodic rotation of the vector \vec{\omega} about a symmetry axis - is called precession (hence our name for \Omega_b.) Notice that in this frame, \frac{d\vec{\omega}}{dt} = (-\Omega_b \omega_0 \sin (\Omega_b t), -\Omega_b \omega_0 \cos (\Omega_b t), 0) \\ = (0, 0, \Omega_b) \times (\omega_0 \cos (\Omega_b t), -\omega_0 \sin (\Omega_b t), \omega_3) \\ = \vec{\Omega_b} \times \vec{\omega} which is just the familiar vector equation telling us that \vec{\omega} rotates about the axis of \vec{\Omega_b} = \Omega_b \hat{e}_3 at constant angular velocity \Omega_b.
Notice that the sign of \Omega_b can change, depending on the values of the principal moments. If the unique moment \lambda_3 is the smallest moment, then we have a prolate symmetric top; otherwise, the top is said to be oblate if \lambda_3 is larger. The precession direction is opposite in these two cases (counter-clockwise about \hat{e}_3 for a prolate top, from the right-hand rule.)
There’s a more geometrical way to look at this, based on conservation laws. In the body frame, notice that our solution gives a constant magnitude for the angular velocity vector: |\vec{\omega}| = \sqrt{\omega_0^2 \cos^2 (\Omega_b t) + \omega_0^2 \sin^2 (\Omega_b t) + \omega_3^2} = \textrm{const} If we sketch the motion of the vector, it traces out a cone, sometimes called the body cone:
The same is true for the angular momentum \vec{L}, although it points in a slightly different direction from \vec{\omega} so it won’t trace out the same cone.
What about in the space frame? We know that because there are no torques \vec{L} is constant, and the total energy T = \frac{1}{2} \vec{\omega} \cdot \vec{L} is constant too. So the component of \vec{\omega} pointing along \vec{L} is constant. With the magnitude of \vec{\omega} fixed, the only possible motion is precession once again, this time tracing out the space cone:
If we imagine the motion of \vec{\omega} and \hat{e}_3 in the space frame, then \vec{\omega} is the point of contact between the body and space cones, and we can picture the imaginary body cone rolling around the (also imaginary) space cone without slipping.
It’s much more informative to see this in motion instead of my still sketches:
YouTube - free precession of a simulated satellite
If you watch very carefully and compare the angular momentum vector to the solar panels on the satellite, you should be able to see that the direction of \vec{\omega} is changing in the body frame (obviously it’s changing in the space frame, which is our point of view.)
This leads us to yet another interesting effect involving the rotation of the Earth. Because the Earth does have a slight bulge at the center, it doesn’t quite spin as a spherical top, but is better described as a symmetric top! This means that independent of any torques, it will undergo the free precession that we just described here. As a reminder, the precession frequency is \Omega_b = \frac{\lambda_1 - \lambda_3}{\lambda_1} \omega_3 and the Earth’s moment about the pole \lambda_3 is about 1/300 larger than the other moment, \lambda_1. This gives a precession frequency of \Omega_b = \omega_3 / 300 or about 300 days for a full cycle. The amplitude is much harder to predict, of course, since it depends on the misalignment between the Earth’s rotational and polar symmetry axes. In fact, this small wobble in the Earth’s rotation was predicted by both Newton and Euler over 200 years ago, but wasn’t discovered until 1891, by American astronomer Seth Chandler; it is called the Chandler wobble after him.
It’s an interesting lesson in physics, because part of the reason that Chandler finally found the wobble is because he didn’t trust the theory completely! Most previous searches had been narrowly focused on fluctuations with period of about 300 days, but Chandler looked more broadly and finally found the predicted rotation with a period of over 400 days. The difference is mainly explained by the fact that the Earth is not quite a rigid body.
Even more interesting, the Chandler wobble is not precisely fixed; its amplitude fluctuates somewhat over time. Oddly, the phase of the Chandler wobble has occasionally changed quickly and dramatically, definitively in 1925 and (according to recent observations) again around 1850 and 2005. It remains an open problem in geophysics to explain what effect could cause the wobble to suddenly change in such a way!
13.5 Rotation matrices
To make progress into more complicated and interesting problems, we need to understand how to describe rotations of coordinates in general. First, we need to talk in general about coordinate transformations.
If you and I come across a ladder leaning against the wall, let’s say I use the floor and wall as my x-y coordinates, and I measure the distances under the ladder on the floor and wall in yards. On the other hand, you decide the direction of the ladder is a better coordinate, and just measure its length directly, in meters. Our numbers will be different, we will have different ideas about the size of the components of the vector describing the ladder; but once we account for units, we must agree on its length, and on the angle it makes with the wall and the floor, and so forth.
Here we are taking the point of view here that the vector \vec{v}, or the ladder, is fixed in space, and we’re rotating our coordinate system around it. This is called a passive transformation. On the other hand, we could sit in a fixed coordinate system and actually change the orientation of the vector, for example if I start with the ladder oriented straight up and then lean it against the wall; this is an active transformation.
In physics terms, the difference is in whether the object is moving, or the observer looking at it is.
The mathematics of active and passive rotations is very similar but with crucial minus signs: a vector rotating clockwise looks just like a coordinate rotation counter-clockwise. Be careful with formulas you get from other sources like Wikipedia! You can always just ask what happens with the formula in front of you if you rotate by something like \pi/2 or \pi, and check that the result is what you expect.
Let’s consider a passive rotation from coordinates (x,y,z) to (x',y',z'), as pictured. In the original (un-primed) frame, we can divide the vector \vec{v} into components, \vec{v} = {v}_x \hat{x} + {v}_y \hat{y} + {v}_z \hat{z}. Of course, we can do the same thing in the primed coordinates, \vec{v} = {v}_{x'} \hat{x}' + {v}_{y'} \hat{y}' + {v}_{z'} \hat{z}'. Any differences between the components of \vec{v} in the two coordinate systems is because the unit vectors themselves are different. We can write this out in terms of dot products between the unit vectors, v_{x'} = \vec{v} \cdot \hat{x}' \\ = (v_x \hat{x} + v_y \hat{y} + v_z \hat{z}) \cdot \hat{x}' \\ = v_x (\hat{x} \cdot \hat{x}') + v_y (\hat{y} \cdot \hat{x}') + v_z (\hat{z} \cdot \hat{x}'). Note that it doesn’t matter whether we think of \hat{x}' as a vector in the unprimed coordinates, or vice-versa; the dot products are all the same. If we expand all three components of \vec{v} in the same way, we find a matrix equation \left(\begin{array}{c} v_{x'} \\ v_{y'} \\ v_{z'} \end{array} \right) = \left( \begin{array}{ccc} \hat{x} \cdot \hat{x}' & \hat{y} \cdot \hat{x}' & \hat{z} \cdot \hat{x}' \\ \hat{x} \cdot \hat{y}' & \hat{y} \cdot \hat{y}' & \hat{z} \cdot \hat{y}' \\ \hat{x} \cdot \hat{z}' & \hat{y} \cdot \hat{z}' & \hat{z} \cdot \hat{z}' \end{array} \right) \left( \begin{array}{c} v_x \\ v_y \\ v_z \end{array} \right) or more compactly
\vec{v}' = \overset{\leftrightarrow}{R} \vec{v}.
\overset{\leftrightarrow}{R} is a rotation matrix, which transforms the vector \vec{v} from the unprimed to the primed coordinates. (Remember that we’re taking the attitude that this is a passive transformation, so \vec{v} is the same vector; I’m just writing \vec{v}' as shorthand for “the vector \vec{v} expressed in the primed coordinates.”)
What if we wanted to go back the other way, and find the components of a vector in the unprimed coordinates from the primed ones? We can write that as a matrix of dot products again, for example the \hat{x} component will be \vec{v} \cdot \hat{x} = (v_{x'} \hat{x}' + v_{y'} \hat{y}' + v_{z'} \hat{z}') \cdot \hat{x} and we end up with another rotation matrix, which I can write as \vec{v} = \overset{\leftrightarrow}{R'} \vec{v}'. How is \overset{\leftrightarrow}{R'} related to \overset{\leftrightarrow}{R}? Since the order in dot-products doesn’t matter, you can check that it’s clearly the transpose matrix: \overset{\leftrightarrow}{R'} = \overset{\leftrightarrow}{R}^T. Of course, since applying \overset{\leftrightarrow}{R} and then \overset{\leftrightarrow}{R}' takes us from unprimed to primed coordinates and then back again, we also know that \mathbf{R'} \mathbf{R} = \mathbf{R}^T \mathbf{R} =\mathbf{1}. A matrix whose transpose is also its inverse is known as an orthogonal matrix; all rotation matrices are orthogonal, as we can now see.
As I’ve repeatedly warned, the inertia tensor \mathbf{I} is intimately connected to the choice of coordinates, which certainly means that a rotation will change its form. But how is \mathbf{I'} related to \mathbf{I}? We know that the relationship between \vec{L} and \vec{\omega} should still be the same after rotation: \vec{L'} = \mathbf{I'} \vec{\omega'}. But we can write the primed vectors using the rotation matrix: \mathbf{R} \vec{L} = \mathbf{I'} \mathbf{R} \vec{\omega}. Multiply on the left by \mathbf{R}^T: \mathbf{R}^T \mathbf{R} \vec{L} = \mathbf{R}^T \mathbf{I'} \mathbf{R} \vec{\omega} \\ \vec{L} = (\mathbf{R}^T \mathbf{I'} \mathbf{R}) \vec{\omega}, since as we just saw, \mathbf{R}^T is the inverse of \mathbf{R}. But now, the matrix in the middle is just the original \mathbf{I}! Multiplying by \mathbf{R} and \mathbf{R}^T again to move things around, we see that
\mathbf{I'} = \mathbf{R} \mathbf{I} \mathbf{R}^T.
Now we can see in detail what I meant when I said that “a tensor is a matrix that transforms in a certain way under coordinate changes.” In fact, this equation is sometimes presented as the definition of a tensor. So we’ve seen three kinds of transformations under a rotation: \vec{v'} \cdot \vec{w'} = \vec{v} \cdot \vec{w} \\ \vec{v'} = \mathbf{R} \vec{v} \\ \mathbf{I'} = \mathbf{R} \mathbf{I} \mathbf{R}^T describing, in turn, a scalar, a vector, and a tensor. It’s instructive to write these out again in index notation: \sum_i v'_i w'_i = \sum_i v_i w_i \\ v'_i = \sum_j R_{ij} v_j \\ I'_{ij} = \sum_{k,l} R_{ik}^T I_{kl} R_{lj} \\ = \sum_{kl} I_{kl} R_{ki} R_{lj} There is, in fact, a very simple pattern here. For any object with any number of indices that aren’t already being summed over, under a rotation, that index is “contracted” (summed over with a common index) with one copy of the rotation matrix. In other words, we can imagine an arbitrarily complicated, higher-dimensional tensor, and we know how it transforms under rotation: T_{ijkl...} = \sum_{abcd} T_{abcd...} R_{ai} R_{bj} R_{ck} R_{dl} ... In classical mechanics, we’ll never go any further than a matrix, so you don’t have to worry about these more abstract tensor structures. If you go on to study general relativity, you will encounter them there!
13.6 Rotation about an axis
Let’s see how our rotation matrix formalism works for a specific coordinate change. Consider a rotation by +\theta about the \hat{z} axis. (As always, the right-hand rule tells us that +\theta corresponds to a counter-clockwise rotation about \hat{z}.)
We can read off the dot products from the geometry of the rotation. For example, \hat{x} \cdot \hat{x}' is the \hat{x} component of \hat{x'}, which is just \cos \theta (since everything here is a length-1 unit vector.) Similarly, the \hat{y} component of \hat{y}' is \cos \theta. Since this rotation only moves the x,y axes, the z coordinates don’t change at all; so \hat{z} \cdot \hat{z}' = 1. Putting everything together, the rotation matrix takes the form \overset{\leftrightarrow}{R}_z(\theta) = \left( \begin{array}{ccc} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right). You may have seen this before, or at least the 2\times 2 submatrix.
Let’s move on to rotation about the \hat{y} axis. We can work this out, as before, by drawing our coordinates out and evaluating the components. But we have to notice that if we draw \hat{x} and \hat{z} to the right and up, then \hat{y} points into the board, because our coordinates are right-handed!
As a result, the rotation by +\theta about \hat{y} is clockwise in the x-z plane. So the x and z components (the corners of the matrix) rotate with an extra minus sign, compared to how x-y rotated. \overset{\leftrightarrow}{R}_y(\theta) = \left( \begin{array}{ccc} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{array} \right). We can do the same exercise for rotation about \hat{x}. Here the orientation of the coordinates is back to normal, so the minus sign is on the bottom left, and we find: \overset{\leftrightarrow}{R}_x(\theta) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & -\sin \theta & \cos \theta \end{array} \right). Now that we understand rotations about fixed axes, what happens if we combine them? Any combination of rotation matrices will just give us another rotation; if \vec{v}' = \overset{\leftrightarrow}{R}_1 \vec{v} and \vec{v}'' =\overset{\leftrightarrow}{R}_2 \vec{v}', then \vec{v}'' = \overset{\leftrightarrow}{R}_2 \overset{\leftrightarrow}{R}_1 \vec{v} = \overset{\leftrightarrow}{R}_3 \vec{v}.
A combined rotation depends on the order in which the individual rotations are applied — equivalently, rotation matrices do not commute, \mathbf{R}_1 \mathbf{R}_2 \neq \mathbf{R}_2 \mathbf{R}_1 in general.
Consider two rotations, number 1 being a rotation about the \hat{z} axis by +90^\circ, and then number 2 a rotation about the (rotated!) \hat{x} axis by +90^\circ.
Intuitively, this makes sense; the first rotation changes the axes around, which changes how the second rotation acts. Mathematically, this makes perfect sense too. We now know how to write the two rotations out as matrices, \overset{\leftrightarrow}{R}_1 = \overset{\leftrightarrow}{R}_z (90^\circ) = \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) \\ \overset{\leftrightarrow}{R}_2 = \overset{\leftrightarrow}{R}_x (90^\circ) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array} \right). Doing the rotations in different orders is equivalent to multiplying the matrices in different orders (which is which?): \overset{\leftrightarrow}{R}_2 \overset{\leftrightarrow}{R}_1 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array} \right) \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right) \\ \overset{\leftrightarrow}{R}_1 \overset{\leftrightarrow}{R}_2 = \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array} \right) = \left( \begin{array}{ccc} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \end{array} \right) ## Euler angles
How many fixed-axis rotations do we need to describe any possible orientation of a rigid body in three dimensions? If we consider, say, the \hat{x} vector, it’s clear that two rotations are enough: one rotation about the \hat{z} axis puts it anywhere in the xy plane, and then one more rotation will let it point anywhere on the sphere. The same is true for \hat{y} or \hat{z}.
However, when we have a collection of vectors, as in a rigid body, to describe an arbitrary orientation of the collection we actually need three rotations. For example, if we attach a \hat{z} vector to the \hat{x} vector we considered and put it through the same rotations, the first rotation does nothing, so it’s clear that we can’t reach an arbitrary orientation of both vectors.
Although we always need three angles, there are multiple conventions by which we can choose the axes and the order of rotation. For example, you may know that airplane pilots use a system of three angles called pitch, yaw and roll:
Once again, the order of rotation matters! A 90-degree roll to the right followed by a 45-degree pitch up gives a completely different orientation from a 45-degree pitch up and then a 90-degree roll to the right. So we can specify the airplane’s rotation with three angles, as long as we have a fixed convention for which order we apply the rotations in (i.e., pitch always comes before roll.)
For this class, we’ll use a different set of conventions invented by Euler in the 18th century. The Euler angles are sometimes also known as the z-y’-z’’ or just zyz convention, for the choices and order of axes we rotate around.
As the name implies, here’s how the Euler angles are defined:
First angle, \phi: rotate around the \hat{z} axis.
Second angle, \theta: rotate around the new \hat{y} axis.
Third angle, \psi: rotate around the new \hat{z} axis.
The disk is just to help you visualize the plane of rotation; you’ll notice that for a disk, the first and third rotations seem completely redundant. This is because a disk is more symmetric than a more general object, for which we would need the third rotation.
Be aware that this is one particular convention: not even all mechanics textbooks agree on exactly what the Euler angles are! But this is the choice we’ll use; it’s Taylor’s choice, and it’s a nice convention because the first two angles match on to the usual \phi and \theta of a spherical coordinate system for the orientation of \hat{e}_3.
One thing you’ll always see for Euler angles in any book is that the first and third axes are the same; this is distinct from the conventions for an airplane, where all three rotation axes are distinct.
As we’ve just learned, the combined effect of these three rotations can be written as the product of three rotation matrices: \overset{\leftrightarrow}{R}_1(\phi) = \left( \begin{array}{ccc} \cos \phi & \sin \phi & 0 \\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{array} \right) \\ \overset{\leftrightarrow}{R}_2(\theta) = \left( \begin{array}{ccc} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{array} \right) \\ \overset{\leftrightarrow}{R}_3(\psi) = \left( \begin{array}{ccc} \cos \psi & \sin \psi & 0 \\ - \sin \psi & \cos \psi & 0 \\ 0 & 0 & 1 \end{array} \right) If you multiply these out, you will arrive at one large and complicated rotation matrix, which can describe any spatial orientation of a rigid body. (It’s not very illuminating to multiply it out here, but the combined form is used very often in e.g. computer games or other simulations to keep track of rotating objects.)
13.7 The Lagrangian for rotational motion
There’s one immediate application of the Euler angles for us: they will serve as the perfect generalized coordinates for a Lagrangian solution. They are time-dependent, but remember that time dependence in GCs is fine - you’ve done lots of Lagrangian problems involving elevators and boxcars.
We start with the kinetic energy, focusing only on rotation and ignoring any other motion of the CM. The general expression for rotational kinetic energy is: T = \frac{1}{2} \vec{L} \cdot \vec{\omega}. Since relative angular velocities about the same pivot can be added as vectors, we can split \vec{\omega} into parts around each Euler axis: \vec{\omega} = \dot{\phi} \hat{z} + \dot{\theta} \hat{e}'_2 + \dot{\psi} \hat{e}_3. These are three vectors in three different coordinate systems, so we’ll have to unpack them in some common coordinates. We’ll expand in the body frame, since there \overset{\leftrightarrow}{I} is diagonal and constant, so it will be much easier to obtain \vec{L} from \vec{\omega}.
We can work backwards through the Euler rotations, starting at the end. For the last rotation, we rotated around \hat{e}_3 = \hat{e}'_3 by \psi, and what was originally a unit vector \hat{e}_2' will now generally have components in both \hat{e}_1 and \hat{e}_2. It’s helpful to look back at the diagram again here:
The Euler angle \psi describes a passive transformation, changing the unit vectors in the plane perpendicular to \hat{e}_3. Before the rotation, \hat{e}'_2 is a unit vector. In the rotated coordinates (in terms of \hat{e}_1 and \hat{e}_2), it will thus be a rotation matrix applied to a unit vector: \hat{e}'_2 = \mathbf{R}_{e_3}(\psi) \hat{e}_2 \\ = \left( \begin{array}{ccc} \cos \psi & \sin \psi & 0 \\ -\sin \psi & \cos \psi & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} 0\\ 1 \\ 0 \end{array} \right) \\ = \sin \psi\ \hat{e}_1 + \cos \psi\ \hat{e}_2. We can see from the diagram that in terms of the \hat{e}_1 and \hat{e}_2 axes, the vector \hat{e}_2' is in the upper-right quadrant, i.e. both its components should be positive. So the answer we have above checks out.
We can immediately see similarly that \hat{e}'_1 = \cos \psi\ \hat{e}_1 - \sin \psi\ \hat{e}_2. On to the next angle. \hat{e}_3 is obtained by rotating \hat{z} by \theta about \hat{e}_2': \hat{z} = \overset{\leftrightarrow}{R}_{e_2'} (\theta) \hat{e}_3 \\ = \left( \begin{array}{ccc} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{array} \right) \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right) \\ = -\sin \theta\ \hat{e}_1' + \cos \theta\ \hat{e}_3 \\ = -\sin \theta \cos \psi\ \hat{e}_1 + \sin \theta \sin \psi\ \hat{e}_2 + \cos \theta\ \hat{e}_3. Putting everything together, we have that
\vec{\omega} = (\dot{\theta} \sin \psi - \dot{\phi} \sin \theta \cos \psi) \hat{e}_1 + (\dot{\theta} \cos \psi + \dot{\phi} \sin \theta \sin \psi) \hat{e}_2 \\ + (\dot{\phi} \cos \theta + \dot{\psi}) \hat{e}_3.
This is completely general — the unique formula for the angular velocity vector in the body frame of an object described by the Euler angles. For the kinetic energy we need the angular momentum too, but since we’re working in the body frame where \mathbf{I} is diagonal and constant, we immediately have \vec{L} = (\lambda_1 \omega_1, \lambda_2 \omega_2, \lambda_3 \omega_3), so the total kinetic energy is T = \frac{1}{2} \vec{L} \cdot \vec{\omega} = \frac{1}{2} \left(\lambda_1 \omega_1^2 + \lambda_2 \omega_2^2 + \lambda_3 \omega_3^2 \right). For free motion, we’re done: if we have any forces that depend on the Euler angles, we’ll have to go on to figure out what the function U(\theta, \phi, \psi) is. From that point forward, we just apply the Euler-Lagrange equations as usual.
In the most general case, this is a fairly complicated expression, as you’ll notice if you try to write out T in terms of the Euler angles explicitly. But if we have a symmetric top, \lambda_1 = \lambda_2, then we find that \omega_1^2 + \omega_2^2 = \dot{\theta}^2 (\sin^2 \psi + \cos^2 \psi) + \dot{\phi}^2 \sin^2 \theta (\cos^2 \psi + \sin^2 \psi) + 2 \dot{\theta} \dot{\psi} (0) \\ = \dot{\theta}^2 + \dot{\phi}^2 \sin^2 \theta, which gives us a greatly simplified kinetic energy: T = \frac{1}{2} \lambda_1 (\dot{\theta}^2 + \dot{\phi}^2 \sin^2 \theta) + \frac{1}{2} \lambda_3 (\dot{\psi} + \dot{\phi} \cos\theta)^2. Notice that the explicit dependence on the angle \psi has vanished completely from the Lagrangian! This is what we expect physically for a symmetric top: because the first two moments are equal, the principal axes \hat{e}_1 and \hat{e}_2 are not unique; they can lie anywhere in the plane perpendicular to \hat{e}_3. So nothing should change when we rotate by \psi, as we have found. (Of course, the top can still carry kinetic energy by spinning in that direction, so T still depends on \dot{\psi}.)
13.8 The symmetric top with torque
Here we pick up directly where we left off in deriving the Lagrangian in terms of Euler angles, and consider an example of symmetric top motion including a torque. We’ll consider a simple but very interesting example, which is a top in a gravitational field:
If we assume that the top is spinning about a point at its bottom, \vec{R} points in the \hat{e}_3 direction, with the CM along the axis. So only the angle \theta - the tilt of \hat{e}_3 relative to \hat{z} - is relevant for gravity. The height of the CM above the point of contact is, doing some simple trig, MgR \cos \theta; the potential should be maximum when the top is standing straight up, at \theta = 0, so U should be proportional to \cos \theta and it should be positive: U = MgR \cos \theta The Lagrangian is then \mathcal{L} = T - U as always. We have three coordinates, which means three Lagrange equations; ordinarily, this would be a tremendous pain to solve. But notice that if we write \mathcal{L} out: \mathcal{L} = \frac{1}{2} \lambda_1 (\dot{\phi}^2 \sin^2 \theta + \dot{\theta}^2) + \frac{1}{2} \lambda_3 (\dot{\psi} + \dot{\phi} \cos \theta)^2 - MgR \cos \theta, that is doesn’t depend on either \psi or \phi! So we have two conserved generalized momenta, and the corresponding Lagrange equations give us the corresponding expressions in terms of our variables: p_\psi = \mbox{const} = \frac{\partial \mathcal{L}}{\partial \dot{\psi}} \\ = \lambda_3 (\dot{\psi} + \dot{\phi} \cos \theta). Glancing back up at our expression for \vec{\omega}, this is exactly L_3, the component of \vec{L} along the \hat{e}_3 axis. This is what we should expect; the torque due to gravity on the symmetric top is always perpendicular to \hat{e}_3, so that component of the angular momentum is exactly conserved.
The other conserved generalized momentum is with respect to \phi: p_\phi = \mbox{const} = \frac{\partial \mathcal{L}}{\partial \dot{\phi}} \\ = \lambda_1 \dot{\phi} \sin^2 \theta + \lambda_3 (\dot{\phi} \cos \theta + \dot{\psi}) \cos \theta. It’s a little less obvious what this corresponds to, but you can show (see Taylor 10.10) that it is in fact the component of \vec{L} along the \hat{z} axis, in the space frame. This makes sense too; the torque \vec{\Gamma} = \vec{r} \times \vec{F} is always perpendicular to the \hat{z} direction for a gravitational torque.
Finally, for \theta we have a more complicated Euler-Lagrange equation to write down. Let’s do it: \frac{\partial \mathcal{L}}{\partial \theta} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\theta}} \\ \lambda_1 \dot{\phi}^2 \sin \theta \cos \theta - \lambda_3 \dot{\phi} \sin \theta (\dot{\psi} + \dot{\phi} \cos \theta) + MgR \sin \theta = \lambda_1 \ddot{\theta}. ### Steady precession
What can we do with these equations? First, notice that the momentum p_\psi appears in the p_\phi equation: p_\phi = \lambda_1 \dot{\phi} \sin^2 \theta + p_\psi \cos \theta. If \theta is constant, this equation immediately tells us that \dot{\phi} has to be constant too! Since \phi is the polar angle of the top’s axis, we find constant precession, as we’ve seen before. The motion is relatively simple: \dot{\phi} and \dot{\psi} (check the equation for p_\psi!) are both constant, and \theta is fixed. If we take \dot{\phi} = \Omega to be the precession frequency, then we can plug in to the \theta equation of motion: 0 = \lambda_1 \Omega^2 \cos \theta - \lambda_3 \Omega \omega_3 + MgR where I’ve recognized that \dot{\psi} + \dot{\phi} \cos \theta = L_3 / \lambda_3 = \omega_3.
This is a quadratic equation for \Omega, which means for any given \omega_3 (spin of the top) and \theta (initial tilt with respect to the axis), there will be up to two values of \dot{\phi} which will give us totally stable precession. Of course, since it’s a quadratic equation, we’re not guaranteed that any solutions work! We have to check the discriminant (for the equation ax^2 + bx + c, the discriminant is D = b^2 - 4ac.) D = \lambda_3^2 \omega_3^2 - 4\lambda_1 \cos \theta MgR > 0 \\ \Rightarrow \omega_3 > \frac{2}{\lambda_3} \sqrt{MgR \lambda_1 \cos \theta}. For \omega_3 too small, then, there are no solutions for stable rotation; the top just falls over after we let it go. The solutions, when they exist, are given by the formula \Omega = \frac{1}{2\lambda_1 \cos \theta} \left( \lambda_3 \omega_3 \pm \sqrt{\lambda_3^2 \omega_3^2 - 4\lambda_1 MgR \cos \theta} \right) In the limit that \omega_3 is very large, it’s easy to see that the plus-sign solution is just \Omega_1 \approx \frac{\lambda_3 \omega_3}{\lambda_1 \cos \theta}. This term is present even if g = 0; in fact, it should look familiar, because it’s the free precession frequency that we found using Euler’s equations!
The minus-sign solution is a little trickier, since the \lambda_3 \omega_3 terms cancel out: but remembering that \sqrt{1+x} \approx 1-x/2 for small x, we see that \Omega_2 \approx \frac{1}{2\lambda_1 \cos \theta} \left( \lambda_3 \omega_3 - \left[\lambda_3 \omega_3 - \frac{2\lambda_1 MgR \cos \theta}{\lambda_3 \omega_3}\right] \right) \\ = \frac{MgR}{\lambda_3 \omega_3}. This second frequency can also be found by treating gravity as an applied torque, and solving for the motion assuming \omega_3 is very large; Taylor does this as an example, and you can verify that our answer matches his.
13.9 Precession and nutation of the symmetric top
Outside of these very special sets of initial conditions, in general \theta will evolve with time as well. Motion in \theta is called nutation, which is Latin for “nodding repeatedly” - exactly what the top appears to do if it undergoes periodic motion in \theta as well.
To solve for the motion in \theta, in principle we just need to eliminate \dot{\phi} and \dot{\psi} in favor of their associated conserved momenta. But the resulting differential equation is very messy, and we can’t in general solve for the motion in terms of elementary functions. However, we can use them to reduce to an equivalent one-dimensional problem: we write E = T + U = \frac{1}{2} \lambda_1 \dot{\theta}^2 + U_{\textrm{eff}}(\theta). We have to do some algebra to express U_{\textrm{eff}} in terms of the conserved momenta; I’ll skip it here, but the result is U_{\textrm{eff}}(\theta) = \frac{(p_\theta - p_\psi \cos \theta)^2}{2\lambda_1 \sin^2 \theta} + \frac{p_\psi^2}{2\lambda_3} + MgR \cos \theta. What does this potential look like? The first two terms are positive, while the last can be negative for \theta > \pi/2. But at \theta = 0 or \theta = \pi, the first (positive) term goes to infinity! So there must be a potential barrier at both these values of \theta. The exact shape of U_{\textrm{eff}} in the middle is harder to see, but it’s clear that the potential will be U-shaped overall.
We can think of this like a central-force motion problem: we know that E = T + U so E \leq U_{\textrm{eff}}. So based on this sketch, we know that the top will undergo bound nutation; in other words, \theta will oscillate between some minimum and maximum values.
What does the overall motion of the top look like? Remember that once we know \theta, the radial velocity of the top about the z axis, \dot{\phi} is fixed; if we take the equation for p_\phi and solve for \dot{\phi}, we arrive at \dot{\phi} = \frac{p_\phi - p_\psi \cos \theta}{\lambda_1 \sin^2 \theta}. There are three possibilities here. If p_\psi < p_\phi, then \dot{\phi} is always greater than zero, and the precession is always in the same direction:
The same thing occurs if p_\psi is much larger than p_\phi, except that the precession is clockwise instead. However, if the two momenta are similar in magnitude, then things start to get interesting. We know that \theta is between 0 and \pi, so if \theta_1 is the smaller value, then \cos \theta_1 > \cos \theta_2. Thus it is possible that p_\phi < p_\psi \cos \theta_1, but p_\phi > p_\psi \cos \theta_2. If this happens, then \dot{\phi} changes direction as the top nutates; if we trace the path of the top’s axis, it will trace out a spiraling path:
Finally, if p_\phi - p_\psi \cos \theta = 0 at one of the boundaries, say \theta_1, then we find a cusp-like motion:
This isn’t as odd as it might look; if we spin a top about its axis, and then hold it fixed at an angle to the ground and release it, it will begin by falling due to gravity, so the motion will look exactly like this. To get it to move in one of the other two paths, we have to give the top a bit of a push one way or the other when releasing.
Here’s some videos in order (note that their ordering of “first, second, third” type is completely different):
Nutation of a gyroscope, first type Nutation of a gyroscope, second type Nutation of a gyroscope, third type
13.9.1 The tippe top
(We didn’t get to this in lecture, but I’m leaving it here for your entertainment.) With just the methods we have so far, it’s hard to get much further on pen and paper. For a number of spinning objects subject to torques, we can only make qualitative arguments about the motion unless we’re willing to turn to numerical solutions. A good example is the “tippe top” or “flip top”:
The tippe top. The tippe top, in slow motion.
This bizarre motion is caused partly by the geometry of the top, but mostly by torque due to friction. Not a large torque; angular momentum is still mostly conserved, which is why the top “flips over” and continues to spin about its stem.
Like the loaded die from the homework, the geometric center of the tippe top does not match its center of mass. As a result, the point of contact on release tends to be slightly off of the rotational axis. Thus, the friction supplied by the table at the point of contact provides a torque, which tilts the stem of the top over. You should also notice that the rotation gets slower once the top inverts; this is slightly due to friction, but mostly related to the fact that the CM of the top is higher once it inverts. So some of the kinetic energy is transferred to potential energy.
(By the way, you can get something like this to happen by spinning a football on the floor; if you spin it the long way, the friction will cause a torque which makes it “stand up” briefly.)
13.10 Worked example: the flying aerobie
Let’s study an example of rotational motion with torque: the movement of an aerobie.
We’ll study a typical physical situation for this object: assume it has been thrown with \vec{\omega} almost perpendicular to the object, but not quite. We ignore gravity (since that won’t provide a torque), but we will include a torque due to air resistance, \vec{\Gamma} = -c \vec{\omega}, where c>0. (As with linear motion, the air resistance gives a torque which opposes the current direction of motion.) How does the presence of this torque affect the rotation of the aerobie?
Based on symmetry, we know this is a symmetric top, i.e. the inertia tensor takes the form \mathbf{I} = \left(\begin{array}{ccc} \lambda_1 & 0 & 0 \\ 0 & \lambda_1 & 0 \\ 0 & 0 & \lambda_3 \end{array} \right) in the body frame. The actual values of \lambda_1 and \lambda_3 aren’t too hard to calculate if you know the inner and outer radius and the mass, but we’ll just leave them as symbols.
As always, for a symmetric top we take the \hat{e}_3 axis to be the unique one; as always, for a flat object like this that is the axis perpendicular to it. Taking the space-frame rotation vector \vec{\omega} in the vertical \hat{z} direction, the situation looks like this:
I’ve noted the projections of \vec{\omega} along and perpendicular to \hat{e}_3, which will be very helpful later on.
For now, let’s focus on the evolution of \omega_3. Since this is a symmetric top, the Euler equation for \omega_3 is particularly simple: \lambda_3 \dot{\omega}_3 = \Gamma_3 = -c \omega_3. We recognize this differential equation right away, or if not, splitting \dot{\omega}_3 = d\omega_3 / dt and integrating both sides quickly gives the result \omega_3 = \omega_0 e^{-(c/\lambda_3)t}. This is a decaying exponential, which again, makes perfect sense: if we throw the aerobie so that \vec{\omega} is perfectly aligned with \hat{e}_3, it will stay perfectly aligned since it’s a principal axis, but the rotation will slow down over time due to the air resistance.
But we said that \vec{\omega} isn’t perfectly aligned with \hat{e}_3. So what happens on the other two axes? Let’s write the Euler equations: \lambda_1 \dot{\omega}_1 - (\lambda_1 - \lambda_3) \omega_2 \omega_3 = -c \omega_1 \\ \lambda_1 \dot{\omega}_2 - (\lambda_3 - \lambda_1) \omega_1 \omega_3 = -c \omega_2. This is a coupled differential equation, and in fact it’s a much trickier one than most similar equations we’ve seen so far. As usual, there’s a clever way to simplify it, one which becomes obvious if we go back to stare at the diagram above: we know that the individual components of \omega_1 and \omega_2 will have some periodic time dependence, just from the fact that they’re spinning around on the surface of the aerobie (from the point of view of the space frame.) But what we’re actually interested in isn’t that part of the motion, but the tilt between \hat{e}_3 and \vec{\omega}, which is parameterized by the angle \alpha. From the diagram, we notice that \tan \alpha = \frac{\sqrt{\omega_1^2 + \omega_2^2}}{\omega_3}. Can we reduce our two equations above into a single differential equation for \alpha? The simplest thing we can try to do is just add the two equations together; notice that if we multiply by \omega_1 in the first and \omega_2 in the second, then \lambda_1 \dot{\omega}_1 \omega_1 - (\lambda_1 - \lambda_3) \omega_1 \omega_2 \omega_3 = -c\omega_1^2 \\ \lambda_1 \dot{\omega}_2 \omega_2 - (\lambda_3 - \lambda_1) \omega_1 \omega_2 \omega_3 = -c\omega_2^2. Now since this is a symmetric top, if we add the two equations the second term on the left cancels completely, and we have \omega_1 \dot{\omega}_1 + \omega_2 \dot{\omega}_2 = -\frac{c}{\lambda_1} (\omega_1^2 + \omega_2^2) \\ = -\frac{c}{\lambda_1} \omega_3^2 \tan^2 \alpha. This is looking promising! Moreover, we notice that the combination of objects on the left-hand side here should pop out when we take the time derivative of \tan \alpha: \frac{d}{dt} (\tan \alpha) = \frac{\omega_1 \dot{\omega}_1 + \omega_2 \dot{\omega}_2}{\omega_3 \sqrt{\omega_1^2 + \omega_2^2}} - \frac{\dot{\omega}_3 \sqrt{\omega_1^2 + \omega_2^2}}{\omega_3^2} \\ = \frac{1}{\omega_3^2 \tan \alpha} \left[ \omega_1 \dot{\omega}_1 + \omega_2 \dot{\omega}_2 - \omega_3 \dot{\omega}_3 \tan^2 \alpha \right], or \omega_1 \dot{\omega}_1 + \omega_2 \dot{\omega}_2 = \omega_3^2 \tan \alpha \frac{d}{dt} (\tan \alpha) - \frac{c}{\lambda_3} \omega_3^2 \tan^2 \alpha, using the fact that \dot{\omega}_3 = -(c/\lambda_3) \omega_3. Plugging back in above, this simplifies very nicely: \frac{d}{dt} (\tan \alpha) = -c \tan \alpha \left[ \frac{1}{\lambda_1} - \frac{1}{\lambda_3} \right]. Once again, we find exponential solutions for the motion! However, this time the sign depends on the relative size of \lambda_1 and \lambda_3. The aerobie is an example of an oblate symmetric top, i.e. whatever the precise values are, it will have \lambda_3 > \lambda_1 just due to its shape. Thus, the effect of air resistance is to stabilize the rotation of the aerobie, keeping it flying smoothly even if the initial orientation is a bit off.