In Taylor, read sections 10.5 to 10.6 for today, and 10.7 to 10.8 for Friday.
Since the inertia tensor is symmetric, we can find an orthogonal set of axes where it is diagonal. These axes are known as principal axes and the diagonal values of the inertia tensor are the principal moments, λi.
We can easily calculate rotational kinetic energy by exploiting the principle axes and principal moments.
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) Ma² | [ | -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) Ma² det | [ | -3 | 8 - λ' | -3 | ] |
| [ | -3 | -3 | 8 - λ' | ] |
where I've used λ = (1/12)Ma² 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) Ma² (a double root), and λ3 = (1/6) Ma².
Determine the principal axes by solving ( I - λi ) ei = 0. Beginning with e3 we have
| [ | 6 | -3 | -3 | ] | [ | e3x | ] | ||
| I = (1/12) Ma² | [ | -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) Ma² | [ | -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 = λ3ω = λ3ωe3. 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×Mg. The magnitude of the torque is Γ = RMg 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.