# PHY5210 W08

## Chapter 10: Rotational Motion of Rigid Bodies

In Taylor, read sections 10.3 to 10.4 for today, and review for Monday.

### Rotation about Any Axis; the Inertia Tensor

#### Angular Momentum for an Arbitrary Angular Velocity

L = ∑a ma ra × (ω × ra)
ra × (ω × ra) = &sumij ωiij ra² - rairaj)

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

δij = 1 if i=j
δij = 0 if i≠j

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

 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

Iij = &suma maij ra² - rairaj)

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

Ixx = &suma ma(ya² + za²)

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

Ixy = -&suma maxaya

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

Ixy = -&suma maxaya = -&suma mayaxa = 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 - rirj)dm = ∫ V ρ(r) (δij r² - rirj)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 = 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

M = ∫ 0 dφ ∫ 0hdz ∫ 0zR/h rdr ρ = (πρ) (R/h)² ∫ 0hdz z² = πR² h ρ/3

or

ρ = 3M / (πR² h).

The integral Izz is

Izz = ∫ 0 dφ ∫ 0hdz ∫ 0zR/h rdr ρr² = (½πρ) (R/h)40hdz z4
Izz = (πρhR4)/10 = (3/10)MR².

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 ρy² + ∫ cone dV ρz²

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

0 dφ ∫ 0hdz ∫ 0zR/h rdr ρz² = (πρ) (R/h)² ∫ 0hdz z4 = πρh³ R²/5 = (3/5)Mh²

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.

## Chapter 10: Rotational Motion of Rigid Bodies

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

### Rotation about Any Axis; the Inertia Tensor

L = I ω

where L and ω are (1×3) column vectors, and I is the (3×3) matrix

I =
 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

Iij = &suma maij ra² - rairaj)

#### Example: Inertia Tensor for 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 = &suma maya²
Iyy = &suma maxa²
Izz = &suma ma(xa² + ya²) = Ixx + Iyy

The remaining element, Ixy = &suma maxaya

ω31 = ω32 + ω21

### Principal Axes of Inertia

As discussed in the cone example, a diagonal inertia tensor means that a rotation about one of the axes has angular momentum parallel to angular velocity. This is nice, when we happen to have an object that produces a diagonal inertia tensor.

But wait! There's a theorem from linear algebra that says that any symmetric matrix can be rotated to a different basis in which it is diagonal. Every inertia tensor is symmetric, therefore we can always diagonalize the inertia tensor, for any object no matter how oddly shaped.

Physically, this means that every object has directions that they can be spun around such that the angular momentum is parallel to the angular velocity vector. We write this mathematically as saying that the angular momentum equals a constant times the angular velocity or

L = Iω = λω

The directions where this is true are called the principal axes of the body. For the cone example, the x, y, and z axes defined in the calculation are principal axes. The constants of proportionality are the diagonal elements of the diagonalized inertia tensor (L = λω), and are called the principal moments.

#### Kinetic Energy of a Rotating Body

We can easily calculate rotational kinetic energy by exploiting the principle axes and principal moments.

T = ½∑i λiωi²