PHY6200 W07

Chapter 10: Rotational Motion of Rigid Bodies


In Taylor, read sections 10.3 to 10.5 for today, and 10.5 to 10.6 for Wednesday.


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

I =

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


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.

Angular Momentum for an Arbitrary Angular Velocity

ω31 = ω32 + ω21

Principal Axes of Inertia

Kinetic Energy of a Rotating Body

Finding the Principal Axes; Eigenvalue Equations

Precession of a Top due to a Weak Torque

Euler's Equations

Euler's Equations with Zero Torque

A Body with Three Different Principal Moments

Motion of a Body with Two Equal Moments: Free Precession

Euler Angles

Motion of a Spinning Top

© 2006 Robert Harr