PHY5210 W15

Chapter 10: Rotational Motion of Rigid Bodies

Reading:

In Taylor, read section 10.3 for today, and 10.4 to 10.6 for Monday.

Rotation about Any Axis; the Inertia Tensor

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:

  1. an object with a fixed point that constrains or limits the possible motion, things like a spinning top with its tip constrained to a cup, or a pendulum with a fixed point; and
  2. an object that moves in space, and is free to rotate.

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.

Angular Momentum for an Arbitrary Angular Velocity

L = ∑a ma ra × (ω × ra)

To simplify this, it is helpful to use the BAC-CAB rule

A × (B × C) = B (AC) - C (AB)

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

ra × (ω × ra) = ∑i ωiij ra² - rai raj)

where δij is the "kronecker delta", equal to 1 if i=j, and 0 otherwise.

δij = 1 if i = j
δij = 0 if ij

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:

L = I ω

where L and ω are (1 × 3) column vectors, and I is the (3 × 3) matrix of moments and products of inertia.

IxxIxyIxz
I = IyxIyyIyz
IzxIzyIzz

An individual element of the matrix, ij, where i and j = x, y, or z, is

Iij = ∑a maij ra² - rai raj)

This form reduces to the same expressions appearing in the text, namely, if i = j = x, then it yields

Ixx = ∑a ma (ya² + za²)

and likewise for Iyy and Izz. If i = x and j = y then it yields

Ixy = -∑a ma xa ya

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

Ixy = -∑a ma xa ya = -∑a ma ya xa = 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

Iij = ∫ Vij r² - ri rj) dm = ∫ V ρ(r) (δij r² - ri rj)dV

Let's look at some examples.

Inertia Tensor for a Solid Cone

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.

Izz = ∫ cone dV ρ (x² + y²)

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

M = ∫ 0 dφ ∫ 0h dz0zR/h r dr ρ = (π ρ) (R/h)² ∫ 0h dz z² = π R² h ρ/3

or

ρ = 3M / (π R² h).

The integral Izz is

Izz = ∫ 0 dφ ∫ 0h dz0zR/h r dr ρ r² = (½ π ρ) (R/h)40h dz z4
Izz = (π ρ h R4)/10 = (3/10) M R².

The remaining diagonal terms are seen to be equal, by symmetry; Ixx = Iyy. They are calculated easily with some tricks

Ixx = ∫ cone dV ρ (y² + z²) = ∫ cone dV ρ y² + ∫ cone dV ρ z²

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

0 dφ ∫0h dz0zR/h r dr ρ z² = (π ρ) (R/h)² ∫0h dz z4 = π ρ h³ R²/5 = (3/5) M h²

Summing these up we find

Ixx = Iyy = (3/20) M (R² + 4h²)

Due to the symmetry in x and y, we can deduce that the products of inertia vanish, Ixz = Iyz = Ixy = 0.

Example: Inertia Tensor for a Lamina

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,

Ixx = ∑a ma ya²
Iyy = ∑a ma xa²
Izz = ∑a ma (xa² + ya²) = Ixx + Iyy

The remaining nonzero element is Ixy = ∑a ma xa ya

© 2015 Robert Harr