PHY6200 W07

Chapter 10: Rotational Motion of Rigid Bodies


In Taylor, read sections 10.2 to 10.4 for today, and 10.4 to 10.5 for Wednesday.


Rotation about a Fixed Axis

Angular Momentum about z

Lz = Izω


Iz = ∑a mara⊥²

Angular Momentum about x and y

It may be a bit surprising that there are angular momentum components in the x and y directions, considering that the rotation is fixed about the z axis. These terms exist and are generally non-zero:

Lx = -∑a maxazaω
Ly = -∑a mayazaω

The basic point that the angular velocity vector and angular momentum vectors can, and often will, point in different directions. The basic rule that L = Iω is a special case, not a generally valid rule. A common, everyday, situation where this occurs is when a car has an unbalanced wheel. An unbalanced wheel is one where Lx or Ly is non-zero, where the z axis is in the direction of the axle. Let's assume Lx is the piece that is non-zero. But, the x-axis is fixed to the wheel, so as the wheel rotates, the direction of Lx changes, meaning that Ldot ≠ 0. A torque is required to change L; with your car, the forces needed to create the torque ultimately arise from friction between the tires and the road. To create a torque in the direction of the axle, this friction is along the direction of travel. To create a torque perpendicular to the axle, in the direction of the rotating x-axis, a side-to-side force is needed, causing the car to shake or jitter.

Balanced wheels are important to smooth handling of a car, especially at high speed. Two types of wheel balancing are generally performed: static balancing ensures that mass is distributed evenly around the wheel when laid horizontally; dynamic balancing ensures that mass is distributed evenly in all three planes such that when spun, Lx = Ly = 0. Dynamic balancing is preferred. A wheel can be balanced statically, but have non-zero components of angular momentum in the x or y directions.

The Products of Inertia

We can write the x and y components of the angular momentum as ω times a piece that depends on the mass and geometry of the object.

Lx = Ixz ωz    Ly = Iyz ωz

where I've written a subscript z on ω because the rotation is around the z axis. (The point will be more evident in the next section.) We call Ixz and Iyz "products of inertia" rather than moments of inertia. They are defined as:

Ixz = -∑a maxaza    Iyz = -∑a mayaza

If we now call Iz = Izz, then the angular momentum for an object rotating around the z axis is written

L = (Ixz ωz, Iyz ωz, Izz ωz)

Simple Moments and Products of Inertia

Calculate the moment and products of inertia for rotation about the z axis for the following bodies: (a) A single mass m located at the position (0, y0, z0) (b) A pair of masses, m, located at positions (0, y0, ±z0) (c) A pair of masses, m, located at positions (0, y0, z0) and (0, -y0, -z0). (d) A uniform ring of mass m and radius a centered on the z axis and parallel to the xy plane.

(a) Direct evaluation of the sums yields Ixz = 0, Iyz = -my0z0, and Izz = my0².

(b) Notice that this situation has symmetry in the z direction, if we change every z coordinate to its negative value (a mirror image in the xy plane), the system looks the same. Symmetries such as this play an important role in physics. In this case, it causes some simplification of the result. Evaluation of the sums yields: Ixz = 0, since both masses have x coordinate of zero; Iyz = -my0z0 - my0(-z0) = 0, a result of the z symmetry; and Izz = my0² + my0² = 2my0².

(c) In this case, the symmetry in z is broken, since changing z to -z doesn't yield exactly the same distribution of masses. Evaluation of the sums yields: Ixz = 0, since both masses have x coordinate of zero; Iyz = -my0z0 - m(-y0)(-z0) = 2my0z0, non-zero since the symmetry is broken; and Izz = my0² + m(-y0)² = 2my0².

(d) The ring has symmetry in both x and y. This symmetry means that the products of inertia will be zero, but the moment of inertia won't: Ixz = Iyz = 0, due to the symmetry ; and Izz = (∑ ma) a² = Ma²

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

We can express an arbitrary axis and angular velocity with the notation ω = (ωx, ωy, ωz). The angular momentum is, in the discrete sum notation,

L = ∑a ma ra × va = ∑a ma ra × (ω × ra)

where I've substituted ω × ra for va, a relation we derived for a rotation. The triple cross product is most easily evaluated with the BAC-CAB rule: A × (B × C) = B(AC) - C(AB). The triple cross product becomes

ra × (ω × ra) = ω (rara) - ra (raω)
ra × (ω × ra) = ω ra² - ra (xaωx + yaωy + zaωz)
ra × (ω × ra) = [(y²+z²)ωx - xyωy - xzωz, -xyωx + (x²+z²)ωy - yzωz, -xzωx - yzωy + (x²+y²)ωz]

This can be cast into a more compact form using a notation where the components of r are numbered, ra = (ra1, ra2, ra3) = (xa, ya, za), and we let the indices i and j run from 1 to 3, corresponding to the x, y, and z components. The triple product then can be written as:

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
© 2006 Robert Harr