Example: Inertia tensor of a triangular plate
Let's go through the example of studying a simple laminar object. We take a flat, \( 45^{\circ} \) triangular plate of uniform density and total mass \( M \). The density is thus
\[ \begin{aligned} \sigma = M/A = \frac{2M}{a^2}. \end{aligned} \]
We take the \( z \)-axis to be perpendicular to the triangle, so \( \rho = \sigma \delta(z) \). Since this is a laminar object, we know immediately that \( I_{xz} = I_{yz} = 0 \). We can further see that there is symmetry under the exchange \( x \leftrightarrow y \), so we expect \( I_{xx} = I_{yy} \). That leaves us with three integrals to do, using our general formula,
\[ \begin{aligned} I_{ij} = \int\ dV\ \rho(x,y,z) (r^2 \delta_{ij} - r_i r_j). \end{aligned} \]
Let's go through them. To set them up, note that the hypotenuse of the triangle corresponds to the line \( y = a-x \):
\[ \begin{aligned} I_{xx} = \int dV \rho(x,y,z) (y^2 + z^2) \\ = \int_0^a dx \int_0^{a-x} dy \int dz \frac{2M}{a^2} \delta(z) (y^2 + z^2) \\ = \frac{2M}{a^2} \int_0^a dx \left. \frac{1}{3} y^3 \right|_0^{a-x} \\ = \frac{2M}{3a^2} \int_0^a dx (a-x)^3 \\ = \frac{2M}{3a^2} \int_a^0 (-du) u^3 \\ = \frac{2M}{3a^2} \frac{1}{4} a^4 = \frac{1}{6} Ma^2, \end{aligned} \]
where I substituted \( u = a-x \) to make the last integral easy. Skipping some of the details, for the other two integrals we find
\[ \begin{aligned} I_{zz} = \frac{2M}{a^2} \int_0^a dx \int_0^{a-x} dy\ (x^2 + y^2) \\ = \frac{2M}{a^2} \int_0^a dx \left( x^2 (a-x) + \frac{1}{3} (a-x)^3 \right) = \frac{1}{3} Ma^2 \end{aligned} \]
and
\[ \begin{aligned} I_{xy} = \frac{2M}{a^2} \int_0^a dx \int_0^{a-x} dy\ (-xy) \\ = -\frac{2M}{a^2} \int_0^a dx \frac{1}{2} x(a-x)^2 = - \frac{1}{12} Ma^2. \end{aligned} \]
Putting everything together, the inertia tensor takes the form
\[ \begin{aligned} \mathbf{I} = \frac{1}{12} Ma^2 \left( \begin{array}{ccc} 2 & -1 & 0 \\ -1 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right). \end{aligned} \]
Let's keep going and find the principal axes and principal moments. We know before we start that the \( z \)-axis is principal, since there are no off-diagonal \( z \) components. In terms of the eigenvalue problem, this means that the determinant will factorize nicely:
\[ \begin{aligned} \det (\mathbf{I} - \lambda \mathbf{1}) \propto \left| \begin{array}{ccc} 2-\lambda & -1 & 0 \\ -1 & 2-\lambda & 0 \\ 0 & 0 & 4-\lambda \end{array}\right| = 0 \\ (4-\lambda) ((2-\lambda)^2 - 1) = 0 \end{aligned} \]
(the \( \propto \) symbol means "proportional to", and is a reminder to put back in the prefactor \( Ma^2 / 12 \) at the end.) So one solution is \( \lambda = 4 \), the other two come from the quadratic equation \( \lambda^2 - 4\lambda + 3 = 0 \), which gives \( \lambda = 1,3 \). Thus, putting the prefactor back in, the principal moments are
\[ \begin{aligned} \lambda_i = \left( \frac{Ma^2}{12}, \frac{Ma^2}{4}, \frac{Ma^2}{3} \right). \end{aligned} \]
Finally, to find the principal axes we solve \( (\mathbf{I} - \lambda_i \mathbf{1}) \vec{v}_i = 0 \). Let's start with the smallest moment \( \lambda_1 \). Since we're just looking for a direction, now we can ignore the prefactor entirely:
\[ \begin{aligned} \left( \begin{array}{ccc} 1 & -1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 3 \end{array} \right) \left( \begin{array}{c} v_x \\ v_y \\ v_z \end{array} \right) = 0 \end{aligned} \]
We can read off \( v_z = 0 \) and \( v_x = v_y \). Thus, normalizing to get a unit vector, we find the first principal axis corresponding to \( \lambda_1 = 1 \) is
\[ \begin{aligned} \hat{e}_1 = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right). \end{aligned} \]
I leave it as an exercise for you to show that the second axis is
\[ \begin{aligned} \hat{e}_2 = \frac{1}{\sqrt{2}} \left( \begin{array}{c} -1 \\ 1 \\ 0 \end{array} \right), \end{aligned} \]
and the third is just the \( \hat{z} \) axis, i.e. \( \hat{e}_3 = (0,0,1) \).
We could have guessed these principal axes from the sketch, in fact! Notice that there is a reflection symmetry about the \( \vec{v}_1 \) direction, which means that if we take \( \vec{v}_1 \) and \( \hat{z} \) as axes, we must have a diagonal inertia matrix. (The third direction \( \vec{v}_2 \) is then required by orthogonality of the three principal axes.)
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,
\[ \begin{aligned} \vec{\Gamma} = -c \vec{\omega}, \end{aligned} \]
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
\[ \begin{aligned} \mathbf{I} = \left(\begin{array}{ccc} \lambda_1 & 0 & 0 \\ 0 & \lambda_1 & 0 \\ 0 & 0 & \lambda_3 \end{array} \right) \end{aligned} \]
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:
\[ \begin{aligned} \lambda_3 \dot{\omega}_3 = \Gamma_3 = -c \omega_3. \end{aligned} \]
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
\[ \begin{aligned} \omega_3 = \omega_0 e^{-(c/\lambda_3)t}. \end{aligned} \]
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:
\[ \begin{aligned} \lambda_1 \dot{\omega}_1 - (\lambda_1 - \lambda_3) \omega_2 \omega_3 = -c \omega_1 \\ \lambda_1 \dot{\omega}_2 - (\lambda_3 - \lambda_1) \omega_1 \omega_3 = -c \omega_2. \end{aligned} \]
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
\[ \begin{aligned} \tan \alpha = \frac{\sqrt{\omega_1^2 + \omega_2^2}}{\omega_3}. \end{aligned} \]
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
\[ \begin{aligned} \lambda_1 \dot{\omega}_1 \omega_1 - (\lambda_1 - \lambda_3) \omega_1 \omega_2 \omega_3 = -c\omega_1^2 \\ \lambda_1 \dot{\omega}_2 \omega_2 - (\lambda_3 - \lambda_1) \omega_1 \omega_2 \omega_3 = -c\omega_2^2. \end{aligned} \]
Now since this is a symmetric top, if we add the two equations the second term on the left cancels completely, and we have
\[ \begin{aligned} \omega_1 \dot{\omega}_1 + \omega_2 \dot{\omega}_2 = -\frac{c}{\lambda_1} (\omega_1^2 + \omega_2^2) \\ = -\frac{c}{\lambda_1} \omega_3^2 \tan^2 \alpha. \end{aligned} \]
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 \( \tan \alpha \):
\[ \begin{aligned} \frac{d}{dt} (\tan \alpha) = \frac{\omega_1 \dot{\omega}_1 + \omega_2 \dot{\omega}_2}{\omega_3 \sqrt{\omega_1^2 + \omega_2^2}} - \frac{\dot{\omega}_3 \sqrt{\omega_1^2 + \omega_2^2}}{\omega_3^2} \\ = \frac{1}{\omega_3^2 \tan \alpha} \left[ \omega_1 \dot{\omega}_1 + \omega_2 \dot{\omega}_2 - \omega_3 \dot{\omega}_3 \tan^2 \alpha \right], \end{aligned} \]
or
\[ \begin{aligned} \omega_1 \dot{\omega}_1 + \omega_2 \dot{\omega}_2 = \omega_3^2 \tan \alpha \frac{d}{dt} (\tan \alpha) - \frac{c}{\lambda_3} \omega_3^2 \tan^2 \alpha, \end{aligned} \]
using the fact that \( \dot{\omega}_3 = -(c/\lambda_3) \omega_3 \). Plugging back in above, this simplifies very nicely:
\[ \begin{aligned} \frac{d}{dt} (\tan \alpha) = -c \tan \alpha \left[ \frac{1}{\lambda_1} - \frac{1}{\lambda_3} \right]. \end{aligned} \]
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.
Example: the simplest coupled oscillator
Let's return to the simple 3-mass coupled harmonic oscillator and try finding a solution.
We'll take the simplest case where all masses and springs are identical:
\[ \begin{aligned} k_1 = k_2 = k_3 = k \\ m_1 = m_2 = m. \end{aligned} \]
The spring constant matrix then becomes
\[ \begin{aligned} \mathbf{K} = \left( \begin{array}{cc} 2k & -k \\ -k & 2k \end{array} \right) \end{aligned} \]
so the matrix appearing in the eigenvalue equation is
\[ \begin{aligned} \mathbf{K} - \omega^2 \mathbf{M} = \left( \begin{array}{cc} 2k - m\omega^2 & -k \\ -k & 2k - m\omega^2 \end{array} \right). \end{aligned} \]
What are the squared normal frequencies? Taking the determinant, we have
\[ \begin{aligned} \det(\mathbf{K} - \omega^2 \mathbf{M}) = (2k-m\omega^2)^2 - k^2 \\ = 4k^2 + m^2 \omega^4 - 4km\omega^2 - k^2 \\ = 3k^2 - 4km\omega^2 + m^2 \omega^4 \end{aligned} \]
This looks like it will factorize, and the coefficients of \( k \) should be \( 1 \) and \( 3 \). Since the cross-term coefficient is \( -1-3 \), we can see that the factorized product is
\[ \begin{aligned} (k-m\omega^2)(3k-m\omega^2) \end{aligned} \]
so the two solutions are \( \omega^2 = +k/m, +3k/m \).
Let's press on and find the normal modes. For the small frequency \( \omega_a = \sqrt{k/m} \), plugging back in gives us
\[ \begin{aligned} \mathbf{K} - \omega_a^2 \mathbf{M} = \left( \begin{array}{cc} k & -k \\ -k & k \end{array} \right). \end{aligned} \]
so the normal mode must satisfy
\[ \begin{aligned} \left( \begin{array}{cc} k & -k \\ -k & k \end{array} \right) \left( \begin{array}{c} A_1 \\ A_2 \end{array} \right) = 0. \end{aligned} \]
The solution to this equation is fairly obvious: the two amplitudes must be equal, \( A_1 = A_2 = A \). Switching the time-dependenence back on, we can write
\[ \begin{aligned} \vec{x}_a = \left( \begin{array}{c} x_1(t) \\ x_2(t) \end{array} \right) = \left(\begin{array}{c} A \cos (\omega_a t - \delta) \\ A \cos (\omega_a t - \delta) \end{array} \right) = \left( \begin{array}{c} 1 \\ 1 \end{array} \right) A \cos (\omega_a t - \delta)\\ = \vec{v}_a A \cos (\omega_a t - \delta). \end{aligned} \]
where I've introduced the constant vector \( \vec{v}_a \), which defines the "direction" of the normal-mode oscillation. Usually when we use the term "normal mode", we'll be referring to these vectors, which can be separated out from the harmonic oscillator part.
For the other frequency \( \omega_b = \sqrt{3k/m} \), I won't do the derivation on the board, but it's easy to see that the second normal mode is
\[ \begin{aligned} \vec{v}_b = \left( \begin{array}{c} 1 \\ -1 \end{array} \right). \end{aligned} \]
(Notice that unlike the principal axes of rigid bodies, we don't care about the normalization of these vectors, since it can be absorbed into the amplitude \( A \) of our harmonic oscillator solution.)
We can combine the two specific solutions into our general solution:
\[ \begin{aligned} \vec{x}(t) = \vec{x}_a(t) + \vec{x}_b(t) \\ = \left( \begin{array}{c} 1 \\ 1 \end{array} \right) A \cos \left( \sqrt{\frac{k}{m}} t - \delta_a \right) + \left( \begin{array}{c} 1 \\ -1 \end{array} \right) B \cos \left( \sqrt{\frac{3k}{m}} t - \delta_b \right). \end{aligned} \]
Note that we have four unknown quantities \( (A, B, \delta_a, \delta_b) \), which is exactly what we expect: for a second-order differential equation, we need 2 initial conditions (e.g. position and speed at \( t=0 \)) for each of our 2 objects, so 4 initial conditions in all.
Physically, it's fairly easy to see what's going on here. The first normal mode corresponds to the two masses moving together; the distance between them stays fixed. In the other mode, they move at the same frequency and amplitude, but exactly out of phase, stretching and squeezing the middle spring.