In Taylor, read sections 10.2 for today, and 10.3 for Friday.
For the remainder of this chapter we will concentrate on the rotation of a body about its CM (projectile or satellite for example) or about a fixed point (a spinning top). We'll begin with the special, but common, case of an object rotating on a fixed axis. We can imagine this to be a rotating wheel, disc, or fan. For simplicity, we'll take the z axis as the axis of rotation, and we can write the angular velocity vector ω = ωzhat = (0, 0, ω)
The angular momentum of the body is the sum of the angular momenta of its parts:
In chapter 9 we learned that for a rotating object, va = ω × ra. For the choice of angular velocity and an arbitrary position ra = (xa, ya, za), this yields va = (-ω ya, ω xa, 0). The angular momenta terms become la = ra×mava = ma ω (-za xa, -za ya, xa² + ya²). Notice that there are three components to the angular momentum despite the fixed axis of rotation.
Summing the angular momenta terms yields the total angular momentum. The total angular momentum has three non-zero components, L = (Lx, Ly, Lz), and I'll treat them each below.
The quantity xa² + ya² is just the perpendicular distance from the axis of rotation, squared, (ra⊥)². So we can write
where
is the usual expression for the moment of inertia about the z axis. This is the standard, elementary physics equations for angular momentum.
If the system of particles are part of a rigid body, then all the relative velocities, v'a, are due to the object's rotation about the CM and can be related to the angular velocity and distance from the axis of rotation (passing through the CM), v'i = ω ra⊥:
which is the familiar result.
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:
The basic point is that the angular velocity vector and angular momentum vectors can, and often will, point in different directions! The basic rule that L = I ω 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.
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.
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:
If we now call Iz = Izz, then the angular momentum for an object rotating around the z axis is written
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 = -m y0z0, and Izz = m y0².
(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 = -m y0 z0 - m y0 (-z0) = 0, a result of the z symmetry; and Izz = m y0² + m y0² = 2m y0².
(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 = -m y0 z0 - m (-y0) (-z0) = 2m y0 z0, non-zero since the symmetry is broken; and Izz = m y0² + m (-y0)² = 2m y0².
(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 = M a²