Formula: 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
\[ \begin{aligned} \mathbf{R}_z(\theta) = \left( \begin{array}{ccc} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right). \end{aligned} \]
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.
\[ \begin{aligned} \mathbf{R}_y(\theta) = \left( \begin{array}{ccc} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{array} \right). \end{aligned} \]
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:
\[ \begin{aligned} \mathbf{R}_x(\theta) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & -\sin \theta & \cos \theta \end{array} \right). \end{aligned} \]
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 \( \vec{v}' = \mathbf{R}_1 \vec{v} \) and \( \vec{v}'' =\mathbf{R}_2 \vec{v}' \), then
\[ \begin{aligned} \vec{v}'' = \mathbf{R}_2 \mathbf{R}_1 \vec{v} = \mathbf{R}_3 \vec{v}. \end{aligned} \]
As you know, the order of rotation matters! 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,
\[ \begin{aligned} \mathbf{R}_1 = \mathbf{R}_z (90^\circ) = \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) \\ \mathbf{R}_2 = \mathbf{R}_x (90^\circ) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array} \right). \end{aligned} \]
Doing the rotations in different orders is equivalent to multiplying the matrices in different orders (which is which?):
\[ \begin{aligned} \mathbf{R}_2 \mathbf{R}_1 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array} \right) \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right) \\ \mathbf{R}_1 \mathbf{R}_2 = \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array} \right) = \left( \begin{array}{ccc} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \end{array} \right) \end{aligned} \]
Concept: 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:
\[ \begin{aligned} \mathbf{R}_1(\phi) = \left( \begin{array}{ccc} \cos \phi & \sin \phi & 0 \\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{array} \right) \\ \mathbf{R}_2(\theta) = \left( \begin{array}{ccc} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{array} \right) \\ \mathbf{R}_3(\psi) = \left( \begin{array}{ccc} \cos \psi & \sin \psi & 0 \\ - \sin \psi & \cos \psi & 0 \\ 0 & 0 & 1 \end{array} \right) \end{aligned} \]
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.)
Formula: 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:
\[ \begin{aligned} T = \frac{1}{2} \vec{L} \cdot \vec{\omega}. \end{aligned} \]
Since relative angular velocities about the same pivot can be added as vectors, we can split \( \vec{\omega} \) into parts around each Euler axis:
\[ \begin{aligned} \vec{\omega} = \dot{\phi} \hat{z} + \dot{\theta} \hat{e}'_2 + \dot{\psi} \hat{e}_3. \end{aligned} \]
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 \( \mathbf{I} \) 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:
\[ \begin{aligned} \hat{e}'_2 = \mathbf{R}_{e_3}(\psi) \hat{e}_2 \\ = \left( \begin{array}{ccc} \cos \psi & \sin \psi & 0 \\ -\sin \psi & \cos \psi & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} 0\\ 1 \\ 0 \end{array} \right) \\ = \sin \psi\ \hat{e}_1 + \cos \psi\ \hat{e}_2. \end{aligned} \]
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
\[ \begin{aligned} \hat{e}'_1 = \cos \psi\ \hat{e}_1 - \sin \psi\ \hat{e}_2. \end{aligned} \]
On to the next angle. \( \hat{e}_3 \) is obtained by rotating \( \hat{z} \) by \( \theta \) about \( \hat{e}_2' \):
\[ \begin{aligned} \hat{z} = \mathbf{R}_{e_2'} (\theta) \hat{e}_3 \\ = \left( \begin{array}{ccc} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{array} \right) \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right) \\ = -\sin \theta\ \hat{e}_1' + \cos \theta\ \hat{e}_3 \\ = -\sin \theta \cos \psi\ \hat{e}_1 + \sin \theta \sin \psi\ \hat{e}_2 + \cos \theta\ \hat{e}_3. \end{aligned} \]
Putting everything together, we have that
$$ \begin{aligned} \vec{\omega} = (\dot{\theta} \sin \psi - \dot{\phi} \sin \theta \cos \psi) \hat{e}_1 + (\dot{\theta} \cos \psi + \dot{\phi} \sin \theta \sin \psi) \hat{e}_2 \
- (\dot{\phi} \cos \theta + \dot{\psi}) \hat{e}_3. \end{aligned} $$
So far, this is completely general; this is 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
\[ \begin{aligned} \vec{L} = (\lambda_1 \omega_1, \lambda_2 \omega_2, \lambda_3 \omega_3), \end{aligned} \]
so the total kinetic energy is
\[ \begin{aligned} T = \frac{1}{2} \vec{L} \cdot \vec{\omega} = \frac{1}{2} \left(\lambda_1 \omega_1^2 + \lambda_2 \omega_2^2 + \lambda_3 \omega_3^2 \right). \end{aligned} \]
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
\[ \begin{aligned} \omega_1^2 + \omega_2^2 = \dot{\theta}^2 (\sin^2 \psi + \cos^2 \psi) + \dot{\phi}^2 \sin^2 \theta (\cos^2 \psi + \sin^2 \psi) + 2 \dot{\theta} \dot{\psi} (0) \\ = \dot{\theta}^2 + \dot{\phi}^2 \sin^2 \theta, \end{aligned} \]
which gives us a greatly simplified kinetic energy:
\[ \begin{aligned} T = \frac{1}{2} \lambda_1 (\dot{\theta}^2 + \dot{\phi}^2 \sin^2 \theta) + \frac{1}{2} \lambda_3 (\dot{\psi} + \dot{\phi} \cos\theta)^2. \end{aligned} \]
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 \( \dot{\psi} \).)
Application: 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:
\[ \begin{aligned} U = MgR \cos \theta \end{aligned} \]
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:
\[ \begin{aligned} \mathcal{L} = \frac{1}{2} \lambda_1 (\dot{\phi}^2 \sin^2 \theta + \dot{\theta}^2) + \frac{1}{2} \lambda_3 (\dot{\psi} + \dot{\phi} \cos \theta)^2 - MgR \cos \theta, \end{aligned} \]
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:
\[ \begin{aligned} p_\psi = \mbox{const} = \frac{\partial \mathcal{L}}{\partial \dot{\psi}} \\ = \lambda_3 (\dot{\psi} + \dot{\phi} \cos \theta). \end{aligned} \]
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 \( \phi \):
\[ \begin{aligned} p_\phi = \mbox{const} = \frac{\partial \mathcal{L}}{\partial \dot{\phi}} \\ = \lambda_1 \dot{\phi} \sin^2 \theta + \lambda_3 (\dot{\phi} \cos \theta + \dot{\psi}) \cos \theta. \end{aligned} \]
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:
\[ \begin{aligned} \frac{\partial \mathcal{L}}{\partial \theta} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\theta}} \\ \lambda_1 \dot{\phi}^2 \sin \theta \cos \theta - \lambda_3 \dot{\phi} \sin \theta (\dot{\psi} + \dot{\phi} \cos \theta) + MgR \sin \theta = \lambda_1 \ddot{\theta}. \end{aligned} \]
Steady precession
What can we do with these equations? First, notice that the momentum \( p_\psi \) appears in the \( p_\phi \) equation:
\[ \begin{aligned} p_\phi = \lambda_1 \dot{\phi} \sin^2 \theta + p_\psi \cos \theta. \end{aligned} \]
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:
\[ \begin{aligned} 0 = \lambda_1 \Omega^2 \cos \theta - \lambda_3 \Omega \omega_3 + MgR \end{aligned} \]
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 \( D = b^2 - 4ac \).)
\[ \begin{aligned} D = \lambda_3^2 \omega_3^2 - 4\lambda_1 \cos \theta MgR > 0 \\ \Rightarrow \omega_3 > \frac{2}{\lambda_3} \sqrt{MgR \lambda_1 \cos \theta}. \end{aligned} \]
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
\[ \begin{aligned} \Omega = \frac{1}{2\lambda_1 \cos \theta} \left( \lambda_3 \omega_3 \pm \sqrt{\lambda_3^2 \omega_3^2 - 4\lambda_1 MgR \cos \theta} \right) \end{aligned} \]
In the limit that \( \omega_3 \) is very large, it's easy to see that the plus-sign solution is just
\[ \begin{aligned} \Omega_1 \approx \frac{\lambda_3 \omega_3}{\lambda_1 \cos \theta}. \end{aligned} \]
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
\[ \begin{aligned} \Omega_2 \approx \frac{1}{2\lambda_1 \cos \theta} \left( \lambda_3 \omega_3 - \left[\lambda_3 \omega_3 - \frac{2\lambda_1 MgR \cos \theta}{\lambda_3 \omega_3}\right] \right) \\ = \frac{MgR}{\lambda_3 \omega_3}. \end{aligned} \]
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.
Application: 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
\[ \begin{aligned} E = T + U = \frac{1}{2} \lambda_1 \dot{\theta}^2 + U_{\textrm{eff}}(\theta). \end{aligned} \]
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
\[ \begin{aligned} U_{\textrm{eff}}(\theta) = \frac{(p_\theta - p_\psi \cos \theta)^2}{2\lambda_1 \sin^2 \theta} + \frac{p_\psi^2}{2\lambda_3} + MgR \cos \theta. \end{aligned} \]
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
\[ \begin{aligned} \dot{\phi} = \frac{p_\phi - p_\psi \cos \theta}{\lambda_1 \sin^2 \theta}. \end{aligned} \]
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 \( \cos \theta_1 > \cos \theta2 \). Thus it is possible that \( p\phi < p_\psi \cos \theta1 \), 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
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.)
Review: one mass on a spring
(This is an optional review - you covered this last semester, but it might be a useful reminder since we're about to start dealing with the more complicated version of this problem.)
Let's start with some important reminders of how a single mass on a spring behaves.
The spring exerts a force on the mass, proportional to the spring constant \( k \), and to the displacement \( (x-x_0) \) of the mass from its equilibrium position, \( x_0 \):
\[ \begin{aligned} |F_{k}| = -k(x-x_0). \end{aligned} \]
This is a restoring force; it always pulls the mass back to equilibrium. It's also a conservative force, which means we can write it as \( F = -\nabla U \) from a potential energy
\[ \begin{aligned} U_k = \frac{1}{2}k(x-x_0)^2. \end{aligned} \]
We can use Newton's laws to solve for the motion, or the Lagrangian. For this simple system, we have
\[ \begin{aligned} \mathcal{L} = \frac{1}{2} m \dot{x}^2 - \frac{1}{2} k (x-x_0)^2. \end{aligned} \]
This is a one-dimensional problem, with no constraints, so we have one degree of freedom, which is described by the generalized coordinate \( x \). Its equation of motion is:
\[ \begin{aligned} \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \right) = \frac{\partial \mathcal{L}}{\partial x} \\ \frac{d}{dt} (m\dot{x}) = -k(x-x_0) \\ \ddot{x} = -\frac{k}{m} (x-x_0). \end{aligned} \]
This is the familiar equation for a simple harmonic oscillator; the general solution is a sine or cosine, since we get back the same function with a minus sign after two time derivatives. We can write the solution as
\[ \begin{aligned} x(t) = A \cos (\omega t - \delta) \end{aligned} \]
where \( A \) and \( \delta \) are unknown coefficients, determined by the initial conditions, and \( \omega \) is the angular frequency
\[ \begin{aligned} \omega = \sqrt{k/m}. \end{aligned} \]
There are some complications possible, as you studied last semester; we can introduce friction to damp the oscillations, or add a driving force, and the motion will be more interesting. But we'll start with the simplest case: no damping, no driving.
Concept: Intro to coupled oscillators
We turn now to systems of coupled oscillators; in terms of classical mechanics, when you see the word "oscillator" you should think "mass on a spring". The simplest example of a coupled oscillator is then a system of masses, which are connected by springs.
This seems like much more of a special case than rigid body rotation, and it is; any everyday object in this room is a rigid body, but a coupled oscillator isn't nearly as common, and probably makes you picture some contraption with air carts and springs which was dreamed up by a physics professor to confuse you in lab. But the motion of coupled oscillators is much more important once we start to look at the world on an atomic scale.
In fact, I misled you just now because the room is full of coupled oscillators, in the air, within the objects around us. Molecular bonds behave an awful lot like springs when we consider small displacements, which are the only kind important for determining, say, the vibrational energy states of a \( CO_2 \) molecule, or of a crystalline solid:
Physical applications aside, our study of coupled oscillators will also be an introduction to some important mathematical ideas, involving the solution of systems of differential equations, which will apply beyond the cases where we can describe everything in terms of springs and Hooke's law.
Now let's move on to our first example of a coupled oscillator: two masses, connected by three springs.
To make things easier, let's start by defining generalized coordinates \( x_1 \) and \( x_2 \) as pictured, so that equilibrium is at \( x_1 = x_2 = 0 \). We could still solve this in terms of forces, but let's use the Lagrangian again. We will have two kinetic terms (two masses),
\[ \begin{aligned} T = \frac{1}{2} m_1 \dot{x_1}^2 + \frac{1}{2} m_2 \dot{x_2}^2 \end{aligned} \]
The potential energy has three terms, one for each spring: from the diagram, we see that
\[ \begin{aligned} U = \frac{1}{2} k_1 x_1^2 + \frac{1}{2} k_2 (x_1 - x_2)^2 + \frac{1}{2} k_3 x_2^2 \end{aligned} \]
Putting it together, our Lagrangian takes the form
\[ \begin{aligned} \mathcal{L} = \frac{1}{2} m_1 \dot{x_1}^2 + \frac{1}{2} m_2 \dot{x_2}^2 - \frac{1}{2} k_1 x_1^2 - \frac{1}{2} k_2 (x_1 - x_2)^2 - \frac{1}{2} k_3 x_2^2 \\ = \frac{1}{2} m_1 \dot{x_1}^2 + \frac{1}{2} m_2 \dot{x_2}^2 - \frac{1}{2} (k_1 + k_2) x_1^2 - \frac{1}{2} (k_2 + k_3) x_2^2 + k_2 x_1 x_2. \end{aligned} \]
Writing the Euler-Lagrange equations as usual gives us
\[ \begin{aligned} m_1 \ddot{x_1} = -(k_1 + k_2) x_1 + k_2 x_2 \\ m_2 \ddot{x_2} = k_2 x_1 - (k_2 + k_3) x_2. \end{aligned} \]
If we take away the middle spring (\( k_2 = 0 \)), then the oscillators decouple, and this reduces to a pair of simple harmonic oscillators as you should expect. But otherwise, this is a pair of coupled differential equations. Since the derivative of \( x_1 \) depends on \( x_2 \) and vice-versa, we can't just solve one equation and plug in to the other; we have to solve both at once.
One approach would be to just assume that the solution is still given by sines and cosines, and just plug in to see what we get. If you try that, you will end up with a mess! For the simple harmonic oscillator, we can cancel the cosine out after plugging in to the equation, just giving us the simple relation \( \omega^2 = k/m \). But here, we have two cosines with (possibly) two different frequencies; we can't simplify easily!
Before we go on to the general solution, notice that there's one other special case we've seen before; if we set \( k_1 = k_3 = 0 \), then only the middle spring remains, and we have
\[ \begin{aligned} \ddot{x_1} = \frac{k_2}{m_1} (x_2 - x_1) \\ \ddot{x_2} = \frac{k_2}{m_2} (x_1 - x_2) \end{aligned} \]
You solved this on one of the homework sets; we can decouple the equations by switching coordinates to the center-of-mass position \( X = \frac{1}{m_1 + m_2} (m_1 x_1 + m_2 x_2) \) and the relative displacement \( d = x_1 - x_2 \). The CM equations end up being trivial (no external forces), and for the displacement we found
\[ \begin{aligned} \ddot{d} = \ddot{x_1} - \ddot{x_2} = -\frac{k_2}{m_1} d - \frac{k_2}{m_2} d = -\frac{k_2}{\mu} d \end{aligned} \]
where \( \mu \) is the usual reduced mass, \( \mu = m_1 m_2 / (m_1 + m_2) \).
Note that this particular change of coordinates won't help us with all three springs present! You can try, but you won't find the same simplification, basically because the equation for the CM's motion is not trivial; the springs attached to the walls provide "external forces". But that doesn't mean that there isn't some clever choice of coordinates which will let us decompose this into two separate equations. (As you probably expect, finding these coordinates will involve an eigenvalue equation.)
Let's go back to the equations of motion with all three springs present:
\[ \begin{aligned} m_1 \ddot{x_1} = -(k_1 + k_2) x_1 + k_2 x_2 \\ m_2 \ddot{x_2} = k_2 x_1 - (k_2 + k_3) x_2. \end{aligned} \]
We can write this more compactly as a matrix equation, starting by gathering the two coordinates into a vector:
\[ \begin{aligned} \vec{x} \equiv \left( \begin{array}{c} x_1 \\ x_2 \end{array} \right). \end{aligned} \]
In this notation, both the left- and right-hand sides of the equation of motion can be written as matrix-vector products. By inspection,
\[ \begin{aligned} \left( \begin{array}{cc} m_1 & 0 \\ 0 & m_2 \end{array} \right) \left( \begin{array}{c} \ddot{x_1} \\ \ddot{x_2} \end{array} \right) = -\left( \begin{array}{cc} k_1 + k_2 & -k_2 \\ -k_2 & k_2 + k_3 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \end{array} \right). \end{aligned} \]
or more compactly,
\[ \begin{aligned} \mathbf{M} \ddot{\vec{x}} = -\mathbf{K} \vec{x}. \end{aligned} \]
This is basically the usual equation for a simple harmonic oscillator, in matrix form. Notice that both matrices involved are symmetric; for the mass matrix, this is automatic because it is diagonal. For the matrix of spring constants, the symmetry arises from the fact that the middle spring is only sensitive to changes in its length, not which side is being pushed or pulled. (In terms of forces, the symmetry is sort of a manifestation of Newton's third law.)
Concept: Normal modes and normal frequencies
The simple structure of the matrix equation above motivates us to look for solutions in terms of sines and cosines. In particular, let's try a solution where the entire vector \( \vec{x} \) oscillates at a single frequency:
\[ \begin{aligned} \vec{x} = \left( \begin{array}{c} A_1 \\ A_2 \end{array} \right) \cos \omega t \end{aligned} \]
This might seem like a big oversimplification to you, but suppose we find any such solutions \( \vec{x}_1(t) \) and \( \vec{x}_2(t) \) that work. We can then see that if we take \( \vec{x}_3 = a_1 \vec{x}_1 + a_2 \vec{x}_2 \), then
\[ \begin{aligned} \mathbf{M} \ddot{\vec{x}}_3 = \mathbf{M} (a_1 \ddot{\vec{x}}_1 + a_2 \ddot{\vec{x}}_2) \\ = a_1 (-\mathbf{K} \vec{x}_1) + a_2 (-\mathbf{K} \vec{x}_2) \\ = -\mathbf{K} (a_1 \vec{x}_1 + a_2 \vec{x}_2) \\ = -\mathbf{K} \vec{x}_3. \end{aligned} \]
In other words, this differential equation obeys the superposition principle; linear combinations of solutions give us another solution. I'll invoke a fundamental result at this point from the mathematics of differential equations, and state that we only need two such solutions here to have the completely general solution. (The two is not because our matrices are two by two here! It's because the differential equation is second order. However, this is allowing the vector in front of the cosine to have \( N \) independent components. If we fix the vector in front, as we're about to, then we need \( 2N \) such solutions.)
With a solution of this form, we have \( \ddot{\vec{x}} = -\omega^2 \vec{x} \), which immediately lets us rewrite the vector equation above as
\[ \begin{aligned} \mathbf{M} (-\omega^2 \vec{x}) = - \mathbf{K} \vec{x} \\ (\mathbf{K} - \omega^2 \mathbf{M}) \vec{x} = 0. \end{aligned} \]
This is a big, important equation! Let's notice a few things about it:
- There is, as usual, a trivial solution to this equation: \( \vec{x} = 0 \). I've emphasized that trivial solutions are usually boring from the point of view of a physicist, and here that's obviously true; the trivial solution corresponds to no motion at all.
- If the masses are equal, then \( \mathbf{M} = m \mathbf{1} \). In that case, this is the eigenvector equation for the matrix \( \mathbf{K} \), and we identify the eigenvalues as \( \lambda = \omega^2 m \).
- If the masses aren't equal, this is called a generalized eigenvalue equation. The solutions are not really eigenvectors of \( \mathbf{K} \), nor are they eigenvectors of the combined matrix \( \mathbf{K} - \omega^2 \mathbf{M} \) (they are actually vectors which live in the null space of the combined matrix, to use the proper terminology.)
Much like the ordinary eigenvalue equation, the condition for non-trivial solutions to exist is that the determinant of the combined matrix must vanish:
\[ \begin{aligned} \det(\mathbf{K} - \omega^2 \mathbf{M}) = 0. \end{aligned} \]
As in the case of rigid bodies, the fact that the matrices involved here are symmetric (and thus so is their combination) guarantees that we can always find \( n \) solutions for an \( n \times n \) matrix, that is, \( n \) independent vectors \( \vec{x} \) which satisfy the generalized eigenvalue equation. (The symmetry of the inertia tensor guaranteed that any rigid body always has three principal axes, you'll recall.)
Terminology: the solutions \( \vec{x}_i \) are called normal modes, and the corresponding eigenvalues \( \omega_i \) are the normal frequencies. In general there will not be two normal frequencies, but \( N \) for \( N \) objects (which makes the mass and spring constant matrices \( N \times N \).) For each normal mode we can write two solutions to the differential equation - sine and cosine - which gives us the required \( 2N \) solutions to construct the general solution.
Like the inertia tensor, there is of course no guarantee that the normal frequencies are unique, or even non-zero; we only know that we can always write down \( N \) independent normal modes. Unlike the inertia tensor, we don't have to make the normal modes orthogonal, just linearly independent, although orthogonal will sometimes be more convenient.
The best way to understand this is just to see it, so here's a nice PhET simulation: