In Taylor, read sections 10.3 to 10.5 for today, and 10.5 to 10.6 for Wednesday.
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
| Ixx | Ixy | Ixz |
| 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
and likewise for Iyy and Izz. If i=x and j=y
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, 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 = rdr 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)MR². 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.