PHY6200 W07

Chapter 9: Mechanics in Noninertial Frames


In Taylor, read sections 10.1 to 10.2 for today, and 10.2 to 10.4 for Friday.


The Foucault Pendulum

The Foucault pendulum is a striking demonstration of the earth's rotation (or technically that the surface is a non-inertial frame) that can be seen at a numerous museums around the world. The basic idea behind the Foucault pendulum is easy to visualize if one imagines locating the pendulum at the north or south pole. The basic idea is to decouple the pendulum from the rotation of the earth by decoupling it from the rotation of the structure from which it is suspended, and to have a pendulum that will swing sufficiently long to see the effect of the earth's rotation (at least several hours, preferably more than a day. Imagine the pendulum at the south pole, set in motion, and decoupled from the earth's rotation. As seen by an observer in space, the pendulum swings always in the same plane while the earth rotates beneath it. As seen by an observer on the ground, the plane of the pendulum's swing moves (precesses), changing by 90° in 6 hours, 180° in 12 hours, 270° in 18 hours, and a full 360° in 24 hours.

There are technical details in how to make such a pendulum: it needs a massive weight hanging from the end of a long wire, possibly with a gimbled mount. We'll ignore these details and look at the motion.

Let's begin with the forces. There are two regular forces on the bob, gravity and the tension in the wire. There are two non-inertial forces, the centrifugal force and the Coriolis force. Choose a coordinate system with z pointing up, y pointing north, x pointing east, and the origin at the equilibrium location of the bob. The equation of motion is

ma = T + mg - mΩ×(Ω×r) + 2mv×Ω

As before, the gravitational force and the centrifugal force combine to give local gravity. We'll restrict the motion to small oscillations -- this is generally the case for real Foucault pendula -- so if the angle of the bob from its equilibrium location is β, then sinβ≈β, and cosβ≈1. This means that mg ≈ Tz = Tcosβ ≈ T. We must allow the pendulum to swing in any x-y direction. Therefore, the bob can be at any x-y location, and the tension has x and y components, Tx = -mgx/L and Ty = -mgy/L. The angular velocity &Omega = Ω(sinθj + cosθk) and Coriolis force has x and y components of v×Ω = Ω[(vy cosθ - vz sinθ)i - vx cosθj + vx sinθk. In the small angle approximation, we can neglect vz, yielding the x and y equations of motion

xddot = -gx/L + 2vyΩ cosθ
yddot = -gy/L 2 2vxΩ cosθ

For a simple pendulum, g/L = ω0⊃2. And Ω cosθ = Ωz, the z component of the earth's angular velocity.

xddot - 2vyΩz + ω0² x = 0
yddot + 2vxΩz + ω0² y = 0

We can decouple these equations by introducing the complex variable η = x + iy. Then ηdot = xdot + iydot, and ηddot = xddot + iyddot. The equation of motion becomes

ηddot + 2iΩzηdot + ω0² η = 0

This is a second order, linear, homogeneous differential equation. To find the solutions, try the solution η = e-iαt where α is a yet-to-be-determined constant. Substitute and cancel out the common exponential factor to find

α² - 2Ωzα - ω0² = 0

which is satisfied for

α = Ωz ± (Ωz² + ω0²) ≈ Ωz ± ω0.

The approximation assumes that the angular velocity of the pendulum is much greater than the earth's angular velocity. We have the required pair of independent solutions, so the general solution cna be written:

η(t) = e-iΩzt (C1e0t + C2e-iω0t).

If the pendulum is started from rest with x=A and y=0, then the solution becomes, as you should find in your homework problem:

η(t) = x(t) + iy(t) = Ae-iΩzt cos(ω0t)

This solution says that the amplitude doesn't change. But since Ωz<<ω0, we can break the motion down into a fast part and a slow part. The fast motion is the swinging of the pendulum, goverened by the cos(ω0t) term. During one period of the pendulum, the slow part of the motion, the e-iΩzt term, barely changes. To see an appreciable change in the exponential piece, we must wait many minutes, or hours, while the period of the pendulum is typically a few seconds. So we can view the motion as the pendulum swinging in a plane, with a slow change to the plane of the swing. The plane of the swing is determined by the complex exponential, e-iΩzt = cos(Ωzt) - i sin(Ωzt). The angle of the plane to the x-axis is Ωzt = (Ω cosθ)t, and the change of the angle is called the precession.

Let's verify that the precession is slow, everywhere on earth. At the north or sout pole, cosθ = ±1, so to change the angle by 2π takes 24 hours -- recall that Ω ≈ 2π/24 hrs. At the equator, cosθ = 0; the pendulum doesn't precess! (It has an infinite period of precession.) At a location inbetween, like Detroit where θ ≈ 47°, the pendulum precesses, but it takes more than a day to precess 360°. In particular, it takes time T = 24hrs/cosθ which is about 35 hours at the latitude of Detroit.

Coriolis Force and Coriolis Acceleration

Please read this section on your own. Taylor compares the centrifugal and Coriolis forces to the terms that arise when working with spherical coordinates.

Chapter 10: Rotational Motion of Rigid Bodies

We covered much of the first two sections last semester. They will be briefly reviewed before continuing with new material.

Properties of the Center of Mass

For a system of n particles, we define the center of mass (CM) of the system as

RCM = (m1r1 + m2r2 + ... + mnrn) / (m1 + m2 + ... + mn) = (1/M) ∑i mi ri

where mi and ri are the mass and position vector of the ith particle, and M = ∑i mi is the total mass of the system. For a continuous object the sum can be taken to an integral over the entire object

RCM = (1/M) &intvol r dm = (1/M) ∫vol rρ(r)dV

where the integrals are taken over the volume of the object, dm is an element of mass, ρ(r) is the density of the object (possibly a function of position), and dV is a volume element (dxdydz in rectangular coordinates). The velocity of the center of mass is found by differentiating with respect to time:

VCM = dRCM / dt = (1/M) ∑i mi dri / dt = (1/M) ∑i mi vi

The Total Momentum and the CM

Logically we define the center of mass momentum as the total mass times of the system times the center of mass velocity. The result is:

PCM = M VCM = ∑i mi vi = ∑i pi

To summarize, the momentum of the center of mass of a system of particles is equal to the vector sum of the momenta of the particles.

Fext = dPCM / dt = M(RCMddot)

where it's straightforward to see that the time derivative of the CM is the total mass of the system times the second derivative of the position vector of the CM. Therefore, the center of mass of a system moves like a point particle located at the CM, and having a mass equal to the mass of the system, subject to the external force applied to the system.

© 2006 Robert Harr