INERTIA TENSOR

4. INERTIA TENSOR

Inertia tensor If a rigid body is rotating about a fixed axis, then from elemen- tary mechanics we know that τ = dL/dt where τ is the torque and L is the angular momentum about the rotation axis. The angular velocity ω and the angular mo- mentum L are related by the equation L = Iω where I is the moment of inertia of the body about the rotation axis. For rotation about a fixed axis, L and ω are parallel vectors, and I is a scalar. But if the rotation axis is not fixed, the angular velocity and the angular momentum may not be parallel.

506 Tensor Analysis Chapter 10

Example 1. Try the following experiment. Take a small book bound by a rubber band, hold it by one corner and toss it upward giving it a spin. As it falls observe that it tumbles, that is, the angular velocity ω about the center of mass is not fixed in direction. However, by definition of the center of mass, the gravitational torque τ about the center of mass is zero so τ = dL/dt = 0. (We are neglecting air resistance.) Thus L is a constant vector, and a constant L and a changing ω are not parallel. Then if the equation L = Iω is to be true, I cannot be a scalar.

We have seen this situation before; look at the discussion of the quotient rule in Section 3 and the proof of the case we have here in (3.5) to (3.8). Since L and ω are vectors, we see by the quotient rule that (when L and ω are not parallel) the scalar

I must be replaced by a 2 nd -rank tensor with components I jk . Then in component form we have

L j =I jk ω k

Example 2. Next we want to find the components of the inertia tensor. For simplicity, first consider a point mass m at the tip of a vector r with tail at the origin O. From Chapter 6, end of Section 3, the angular momentum of m about the origin is L = mr × (ω × r) where ω is the angular velocity of the mass m about O. (See Chapter 6, Figures 2.6 and 3.8.) We can expand the triple vector product [see Chapter 6, equation (3.8)] to get

− (r · ω)r] = m[r 2 ω − (xω x + yω y + zω z )r]. Next we write the components of L in terms of the components of ω. For example,

(4.2) L = mr × (ω × r) = m[r 2 ω

taking the x component of (4.2), we find (4.3)

L x = m[r 2 ω x − (xω x + yω y + zω z )x] = m[(r 2 −x 2 )ω x − xyω y − xzω z ]. Thus three components of the inertia tensor are (4.4)

I xx = m(r 2 −x 2 ) = m(y 2 +z 2 ), I xy = −mxy, I xz = −mxz. The other 6 components can be found similarly by taking the y and z components

of (4.2) (Problem 1). Example 3. If, instead of a single mass, we have a set of masses or an extended body, then

the expressions for the components of the inertia tensor become sums or integrals.

I xy =− m i x i y i or − xy dm, etc. (Problem 1.)

It is useful to write (4.1) as a matrix equation (see discussion in Section 3 about contraction). Then the inertia tensor components form a square matrix. This matrix is symmetric and so we know from Chapter 3, Section 11, that it can be diagonalized by an orthogonal similarity transformation. The new axes are called the principal axes of inertia and the three eigenvalues are called the principal moments of inertia . We see that the equations of motion are simpler relative to the principal axes.

Section 4 Inertia Tensor 507

Example 4. Find the inertia tensor about the origin for the mass distribution consisting of a mass 1 at (0, 1, 1) and a mass 2 at (1, −1, 0). Find the principal moments of inertia and the principal axes.

Substituting (x 1 ,y 1 ,z 1 ) = (0, 1, 1), m 1 = 1, and (x 2 ,y 2 ,z 2 ) = (1, −1, 0), m 2 =2 into (4.5), we find I xx = (1 2 +1 2 ) + 2(−1) 2 = 4, I xy =I yx = −0 − 2(−1) = 2. Continuing in the same way, we can find the rest of the components (Problem 2) and write them as an inertia matrix

I=  2 3 −1 

Either by hand or by computer we find that the eigenvalues of the matrix I are 6 and √ 3± 3; these are the principal moments of inertia. The corresponding eigenvectors √ √ √ √ are (1, 1, −1), (−1 −

3, 1); these are vectors along the principal axes of inertia.

Example 5. Find the inertia tensor about the origin for a mass of uniform density = 1, inside the part of the unit sphere in the first octant, that is, x > 0, y > 0, z > 0. We will write the integrals for the components of the inertia tensor first in rectangular coordinates and then switch to spherical coordinates [see Chapter 5, equation (4.5)] to evaluate them since the limits are then simpler. Satisfy yourself that in order to cover the required volume, the limits are: r from 0 to 1, θ from 0 to π/2, and φ from 0 to π/2. Then

I xy = (−xy) dV =

15 Similarly, the other integrals can be written and evaluated (Problem 3). Alterna-

tively, it may be clear that by symmetry the three diagonal components are all the same, and all the off-diagonal components are the same. Then the inertia matrix is

As in Example 4, we find (Problem 3):

Principal moments of inertia:

Principal axes of inertia: (1, 1, 1), and any two orthogonal vectors in the plane x + y + z = 0, for example, (1, −1, 0) and (1, 1, −2).

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PROBLEMS, SECTION 4

1. As in (4.3) and (4.4), find the y and z components of (4.2) and the other 6 com- ponents of the inertia tensor. Write the corresponding components of the inertia tensor for a set of masses or an extended body as in (4.5).

2. Complete Example 4 to verify the rest of the components of the inertia tensor and the principal moments of inertia and principal axes. Verify that the three principal axes form an orthogonal triad.

3. As in Problem 2, complete Example 5. 4. Find the inertia tensor about the origin for a mass of uniform density = 1, inside

the part of the unit sphere where x > 0, y > 0, and find the principal moments of inertia and the principal axes. Note that this is similar to Example 5 but the mass is both above and below the (x, y) plane. Warning hint: This time don’t make the assumptions about symmetry that we did in Example 5.

For the mass distributions in Problems 5 to 7, find the inertia tensor about the origin, and find the principal moments of inertia and the principal axes.

5. Point masses 1 at (1, 1, 1) and at (−1, 1, 1). 6. Point masses 1 at (1, 1, −2) and 2 at (1, 1, 1). 7. Mass of uniform density = 1, bounded by the coordinate planes and the plane

x + y + z = 1. 8. For the point mass m we considered in (4.2) to (4.4), the velocity is v = ω × r so the

kinetic energy is T = 1 2 mv 2 = 1 1 T

2 m(ω × r) · (ω × r). Show that T can be written in

matrix notation as T = 2 ω

I ω where I is the inertia matrix, ω is a column matrix,

and ω T is a row matrix with elements equal to the components of ω.