In Taylor, read section 10.3 for today, and 10.4 to 10.6 for Monday.
To treat the most general rotation problems, we need to study the case of an arbitrary and variable axis of rotation. There are two main classes of problems we wish to consider:
The first class of problems the fixed point is the natural choice of origin. In the second class of problems, we know that the general motion can be decomposed into the motion of the CM, and rotational motion about the CM, so the CM will be the natural choice of origin. In the following we will work with coordinates and velocities measured relative to these origins.
To simplify this, it is helpful to use the BAC-CAB rule
a helpful relation for simplifying a vector triple product. Notice that the result has components parallel to B and C, but must be perpendicular to A due to the nature of the cross product. With this rule, the vector triple product becomes
where δij is the "kronecker delta", equal to 1 if i=j, and 0 otherwise.
Multiply by ma and sum to give the components of L. It is useful to have a convenient way to write this in a yet more compact form. The required form is a matrix expression:
where L and ω are (1 × 3) column vectors, and I is the (3 × 3) matrix of moments and products of inertia.
| Ixx | Ixy | Ixz | |
| I = | Iyx | Iyy | Iyz |
| Izx | Izy | Izz |
An individual element of the matrix, ij, where i and j = x, y, or z, is
This form reduces to the same expressions appearing in the text, namely, if i = j = x, then it yields
and likewise for Iyy and Izz. If i = x and j = y then it yields
and likewise for the other combinations.
The matrix I is called the inertia tensor (a tensor is a matrix with certain transformation properties that we'll discuss), and has some important properties. Possibly the most important is that it is symmetric I = IT, or Iij = Iji for i,j = x, y, or z. This is readily seen by comparing the expression for Ixy with that for Iyx
Of course to calculate the inertia tensor for a continuous solid, we take the sum to an integral. This is done by imagining dividing the solid object up into many small cubes, the cubes enumerated by the index a. But we can also identify the cube by its position in the object, (x, y, z). Each cube has a small mass, dm = ρdV where ρ is the density (at that point if the density varies) and dV = dxdydz is the volume of the cube. The sum becomes the integral
Let's look at some examples.
Find the inertia tensor I for a solid cone of mass M, height h, and base radius R, that spins on its tip. With the z axis chosen along the axis of symmetry of the cone, find the cone's angular momentum L for an arbitrary angular velocity ω.
It is convenient to do this integral in cylindrical coordinates -- the cone does have cylindrical symmetry. The volume element in cylindrical coordinates is dV = r dr dφ dz. The radius of the cone depends on the z coordinate like Rz = zR/h. First, find the density of the cone by computing the mass
or
The integral Izz is
The remaining diagonal terms are seen to be equal, by symmetry; Ixx = Iyy. They are calculated easily with some tricks
The first integral is just like the integral for Izz, but half as large, yielding (3/20) M R². The second integral is straightforward to evaluate
Summing these up we find
Due to the symmetry in x and y, we can deduce that the products of inertia vanish, Ixz = Iyz = Ixy = 0.
Derive relations among the elements of the inertia tensor for a lamina.
A lamina is a planar object. Being flat, we can orient it to lie in the x-y plane so that all points have z = 0. Then we see immediately that Ixz = Iyz = 0. Additionally there's a relation between the diagonal elements. To see this, notice that, since z = 0 for all points,
The remaining nonzero element is Ixy = ∑a ma xa ya