In Taylor, read sections 9.9 to 9.10 for today, and 10.1 to 10.2 for Wednesday.

Divide through by the common mass term and the component equations are

xddot = -2Ω(zdot cosλ - ydot sinλ)

yddot = -2Ω xdot sinλ

zddot = -g + 2Ω xdot cosλ

These are coupled linear diferential equations. They cannot be separated, but it is possible to integrate each of them once. In particular, we can integrate the y and z equations to obtain expressions for ydot and zdot that depend only on x, and these can be inserted into the x equation.

ydot = -2Ω x sinλ + v_{y0}

zdot = -gt + 2Ω x cosλ + v_{z0}

where the velocity at t=0 is **v**_{0} = (v_{x0}, v_{y0}, v_{z0}).
The x equation becomes

xddot = 2Ω(gt cosλ - v_{z0} cosλ + v_{y0} sinλ).

Integrate this equation twice to get the result

x(t) = (1/3)Ωgt³ cosλ - Ωt²(v_{z0} cosλ - v_{y0} sinλ) + v_{x0}t + x_{0}.

Insert this result for x(t) into the equations for ydot and zdot, drop terms involving Ω² or higher, and integrate to find y and z

y(t) = v_{y0}t - Ωv_{x0}t² sinλ + y_{0}

z(t) = -½gt² + v_{z0}t + Ωv_{x0}t² cosλ + z_{0}

Note: This problem can be exactly solved in an inertial frame, by treating the motion of the projectile as a Keplerian orbit about the earth. This may be exact, but it is certainly not easy, and solving the problem in the non-inertial frame of the earth allows more easily for corrections due to local variations of gravity, air resistance, or wind.

Where will an object dropped from a height h above the ground land, relative to a point directly below according to gravity. Estimate the deflection for a drop down the stairwell of the Physics building.

This is a classic problem; I was asked this on my oral qualifying exam.
Choose a coordinate system attached to the surface of the earth, with the z axis vertically up, the y axis pointing north, and the x axis pointing east.
The initial position of the object is then (x,y,z)_{0} = (0,0,h), and the initial velocity is (v_{x0}, v_{y0}, v_{z0}) = (0, 0, 0) at t=0.
Putting these initial conditions into the above results yields

x(t) = (1/3)Ωgt³ cosλ

y(t) = 0

z(t) = -½gt² + h

We see that in both the northern and southern hemispheres, the object will vall to the east -- x(t) is always positive. We can express the displacement in terms of the height by finding t for z=0 giving

x(t) = (2/3)sqrt(2)Ωh^{3/2} cosλ/sqrt(g)

Let's extimate the size of the deflection for a drop down the physics stairwell, from the third floor to the basement.
The height is approximately 15m, and the latitude of Detroit is about 43°, giving x = (2/3)sqrt(2)(7.3×10^{-5} rad/s)(15m)^{3/2}cos 43° / (9.8 m/s²)^{½} = 0.0009m = 0.9mm.
For a drop of 100m the deflection increases to about 1.6cm.

A bullet is fired horizontally east with initial speed v_{0}.
Assume that the bullet travels a distance H.
What is the deflection?

Since we are given the eastward distance of motion, the deflection we are looking for is north or south of the intended target (determined with an accurate laser pointer for instance).
Again, choose a coordinate system fixed to the surface of the earth, with z pointing up, y pointing north, x pointing east, and the origin at the point where the bullet is fired.
The initial conditions are (x,y,z)_{0} = (0,0,0) and (v_{x0}, v_{y0}, v_{z0}) = (v_{0}, 0, 0) at t=0.
Our interest is only in the equation for y to determine the north-south deflection

y(t) = -Ωv_{0}t²sinλ

All quantities are positive except sinλ which is positive in the northern hemisphere and negative in the southern hemishphere.
So the bullet veers south in the northern hemisphere and north in the southern hemisphere (always towards the equator).
The time for the bullet to travel a distance H is t = H/v_{0}, so the deflection is

y(t) = -ΩH²sinλ / v_{0}

The size of the deflection grows with H².

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 ≈ T_{z} = 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 = ω_{0}⊃2.
And Ω cosθ = Ω_{z}, the z component of the earth's angular velocity.

xddot - 2vyΩ_{z} + ω_{0}⊃2 x = 0

yddot + 2vxΩ_{z} + ω_{0}⊃2 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 + 2iΩ_{z}ηdot + ω_{0}⊃2 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 - 2Ω_{z}α - ω_{0}² = 0

which is satisfied for

α = Ω_{z} ± (Ω_{z}² + ω_{0}²) ≈ Ω_{z} ± ω_{0}.

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} (C_{1}e^{iω0t} + C_{2}e^{-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(ω_{0}t)

This solution says that the amplitude doesn't change.
But since Ω_{z}<<ω_{0}, 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(ω_{0}t) 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(Ω_{z}t) - i sin(Ω_{z}t).
The angle of the plane to the x-axis is Ω_{z}t = (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.

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

© 2006 Robert Harr