In Taylor, read sections 10.4 to 10.6 for today, and 10.7 to 10.8 for Wednesday.
In general, the angular momentum of a rotating object is given by
The inertia tensor is
| Ixx | Ixy | Ixz | |
| I = | Iyx | Iyy | Iyz |
| Izx | Izy | Izz |
where the elements are given by
As discussed in the cone example, a diagonal inertia tensor means that a rotation about one of the axes has angular momentum parallel to angular velocity. This is nice, when we happen to have an object that produces a diagonal inertia tensor.
But wait! There's a theorem from linear algebra that says that any symmetric matrix can be rotated to a different basis in which it is diagonal. Every inertia tensor is symmetric, therefore we can always diagonalize the inertia tensor, for any object no matter how oddly shaped.
Physically, this means that every object has directions that they can be spun around such that the angular momentum is parallel to the angular velocity vector. We write this mathematically as saying that the angular momentum equals a constant times the angular velocity or
where the * is to remind us that this angular velocity is in a special direction where this relation holds. The directions where this is true are called the principal axes of the body. For the cone example, the x, y, and z axes defined in the calculation are principal axes. The constants of proportionality are the diagonal elements of the diagonalized inertia tensor (L = λω*), and are called the principal moments.
We can easily calculate rotational kinetic energy by exploiting the principle axes and principal moments.
The most general form for the rotational kinetic energy is
where in the final form the rules for multiplying matrices apply.
The principal axes are directions where the angular momentum is proportional to the angular velocity
The last expression is simply the definition of the inertia tensor. These yield a way to find the principal moments and principal axes, simply solve the equation
Equations of this type appear commonly in physics and are known as
This determinant yields a cubic equation in λ which can have 1, 2, or 3 real roots. The roots are the principal moments, λ1, λ2, and λ3. To find the principal axes, we solve the three equations
for the three corresponding eigenvectors e1, e2, and e3.
Determine the principal moments and axes for a cube rotating about a corner.
The inertia tensor for a cube with sides of length a oriented with its sides parallel to the x, y, and z axes is given in Taylor, and is
| [ | 8 | -3 | -3 | ] | |
| I = (1/12) M a² | [ | -3 | 8 | -3 | ] |
| [ | -3 | -3 | 8 | ] |
First, find the principal moments by solving the equation
| [ | 8 - λ' | -3 | -3 | ] | |
| det(I - λ) = 0 = (1/12) M a² det | [ | -3 | 8 - λ' | -3 | ] |
| [ | -3 | -3 | 8 - λ' | ] |
where I've used λ = (1/12) M a² λ' to simplify the expression somewhat. The constant outside the determinant plays no role in finding the zeros. To take the determinant, I will expand about the entries in the first row:
This can be simplified like:
and expanding the terms in square brackets
Therefore, the principal moments are λ1 = λ2 = (11/12) M a² (a double root), and λ3 = (1/6) M a².
Determine the principal axes by solving ( I - λi ) ei = 0. Beginning with e3 we have
| [ | 6 | -3 | -3 | ] | [ | e3x | ] | ||
| I = (1/12) M a² | [ | -3 | 6 | -3 | ] | [ | e3y | ] | = 0 |
| [ | -3 | -3 | 6 | ] | [ | e3z | ] |
This has the solution e3 = (1/√3)(1, 1, 1), where I've normalized the vector to unit length. This direction corresponds to a body diagonal, a diagonal from the corner the cube is rotating about, to the opposite corner, passing through the middle of the cube.
For e1 and e2 we have
| [ | -3 | -3 | -3 | ] | [ | e1x | ] | ||
| I = (1/12) M a² | [ | -3 | -3 | -3 | ] | [ | e1y | ] | = 0 |
| [ | -3 | -3 | -3 | ] | [ | e1z | ] |
where e1 can also be e2. This has an infinite number of solutions for e1 and e2, which can be any pair of orthogonal directions in a plane perpendicular to e3. For instance, we can take e2 = (1/√2)(1, -1, 0), and then find e1 from e1 = e2 × e3 = (1/√6)(-1, -1, 2).
To get a flavor of what's involved in solving rotation problems, consider the case of a rotating top subject to a weak torque. The top has axial symmetry, and calling e3 the symmetry axis, with e1 and e2 perpendicular to it and each other, we know that the inertia tensor is diagonal with principal moments λ1 = λ2 = λ, and λ3.
Suppose the top begins spinning about its symmetry axis. In the absence of gravity, there is no torque, and it will continue to spin that way with L = λ3 ω = λ3 ω e3. With no torque, the angular momentum will remain constant, equal to this value.
Now turn on gravity, supposing that it causes a weak torque Γ = R × M g. The magnitude of the torque is Γ = M g R sinθ and is directed perpendicular to the plane containing the z and e3 axes. The torque will cause the angular momentum to change, but since it is directed perpendicular to the angular momentum, the direction will change but not the magnitude. The change in L will cause ω1 and ω2 to be small, but non-zero.
where