PHY6200 W07

Chapter 9: Mechanics in Noninertial Frames


In Taylor, read sections 9.7 to 9.8 for today, and 9.9 to 9.10 for Monday.


Newton's Second Law in a Rotating Frame

ma' = Freal - mΩ[dot] × r' - 2mΩ × v' - mΩ × (Ω × r')

Example: Bead on a Rod

A smooth rod of length l rotates in a plane with a constant angular velocity Ω about an axis fixed at the end of the rod and perpendicular to the plane of rotation. A bead of mass m can slide along the rod without friction is initially positioned at the stationary end of the rod and given a slight push such that its initial speed directed down the rod is ε = Ωl (draw figure). Calculate how long it takes for the bead to reach the other end of the rod.

Choose a coordinate system that rotates with the rod. Let xhat be along the rod and yhat be perpendicular to the rod. Since the bead is constrained to the rod, the problem becomes one dimensional. The only regular force is the one constraining the bead to the rod, and since this force is perpendicular to the rod, it points in the y direction. There is no regular foce in the x direction.

We need to find the fictitious forces on the bead. To do this, assume the bead is a distance x from the origin, that is, it is located at position (x, 0, 0) in the rotating frame, and assume it has velocity (xdot, 0, 0). (The y and z components of the velocity must be zero in this frame, because the bead can only move along the rod, in x.) Using Ω = Ωkhat = (0, 0, Ω), the fictitious forces are

Fcor = 2mv × Ω = -2m(xdot)Ωyhat.
Fcent = -mΩ × (Ω × r) = -mΩ²x(khat × (khat × xhat)) = mΩ²xxhat
Ftrans = 0

The Coriolis force points in the y direction and will be canceled by the force that constains the bead to the rod. The centrifugal force will contribute to the motion of the bead along the wire. The equation of motion in x is

m(xddot) = mΩ²x
xddot = Ω²x.

This is a second order linear equation in x, and the solution is

x(t) = AeΩt + Be-Ωt.
xdot(t) = ΩAeΩt - ΩBe-Ωt.

The initial conditions are that x(0) = 0, and xdot(0) = ε = Ωl. Applying these yields A + B = 0, and ε = Ω(A - B), or A = -B = ε/2Ω. Therefore, x(t) is

x(t) = (ε/2Ω)(eΩt - e-Ωt) = (ε/Ω)sinh(Ωt).

The time, T, for the bead to reach the end of the rod is found by solving x(T) = (ε/Ω)sinh(ΩT) = l. Recall that ε = Ωl, therefore sinh(ΩT) = 1, or T = (1/Ω)sinh-1(1) = 0.88/Ω.

Effects of Earth's Rotation

Some interesting results can be derived by taking account of earth's rotation and treating it as a non-inertial reference frame. The earth's rotation is small, 2π radians in a sidereal day, 23 hrs 56 mins (the time it takes the earth to rotate once with respect to the stars).

Ωearth = [2π/(23×60 + 56)mins](1min/60sec) = 7.29×10-5rad/s

This rate is rather small, and the effects are generally small. Nevertheless, they can be important at times.

Static Effects: The Plumb Line

On the earth, the direction that we call vertical is normally defined locally as the direction that a plumb bob hangs. (A plumb bob is simply a mass hanging from the end of a string, often used in construction to set walls vertically.) It is safe to assume that (neglecting hills, bumps, and such) the ground (or water) lies perpendicular to vertical, so that on average, the shape of the earth can be related to the verticals at different locations around the globe. Due to the rotation of the earth, local vertical doesn't point back directly to the center of the earth. It will at the equator and the poles, but at other locations, it points slightly away from the center. By integrating the effect of the deviations of the local vertical, we can, iteratively, determine the shape of the earth.

We can begin by approximating the earth as a perfect sphere, so that the gravitational force of the earth on a plumb bob points precisely along a radius to the center. Now choose a coordinate system fixed to the earth and centered on the plumb bob. In this system, the plumb bob is at rest, so a = 0. Call g0 the gravitational acceleration purely due to the earth's mass, and therefore directed to the center of the earth. The Coriolis and transverse forces are zero (since v=0 and Ωdot=0), and since the coordinate system is centered on the bob, r=0, and the centrifugal force is also zero. But the origin of the coordinate system is rotating around a circle of radius R0 = Re cosλ where λ is the latitude on earth and Re is the radius of the earth. The coordinate system is undergoing an acceleration, A0 = Ω²R0 = Ω²Re cosλ directed inward toward the axis of rotation [diagram].

So the plumb bob experiences three forces: tension in the string, T; gravity, mg0, and the fictitious force -mA0. Since the bob is not accelerating, these must sum to zero. Local vertical is parallel to the tension in the string,

T = mA0 - mg0 = -m(g0 - A0)

and the effective local gravity is

g = g0 - A0

The distribution of local verticals tells us that the shape is not exactly sphereical (the starting assumption), but slightly oblate (flattened at the poles, wider at the equator). This could be used as the starting point for a second iteration, and the whole procedure repeated to get a slightly more accurate result, and so on. But for now, let's continue treating the earth as a sphere and evaluate the difference in effective gravity at the equator and the poles.

At the poles, λ = ±90° therefore A0 = 0 and g = g0. At the equator, λ = 0° therefore g = g0 - Ω²Re, a bit smaller than at the poles. The difference amounts to Δg = Ω²Re = (7.3×10-5 rad/s)²(6.38×106m) = 0.034 m/s² or a relative difference of Δg / g = 0.35%. Variations of this order already occur due to terrain, and local mineral deposits.

Another interesting calculation is to determine by what amount (angle) the direction of gravity deviates from a radial line. The deviation is 0 at the poles and the equator, and maximum for latitudes of ±45°. Read the derivation in the text. But note that locally, vertical is always the direction of gravity and is perpendicular to the ground (horizontal).

Dynamic Effects: Motion of a Projectile

Let's now look at the effects on a projectile, launched from the earth's surface at latitude λ. It turns out to be very convenient to choose a coordinate system with origin at the launch point, z axis pointing vertically upward (opposite local gravity), y axis pointing north along a great circle passing through the poles, and x axis pointing east. This yields a right handed coordinate system. This coordinate system moves in a circle or radius Re cosλ and must experience an acceleration towards the axis of rotation, A0, as in the previous example. The general equation of motion is

ma = Fnon g + mg0 - mA0 + 2mv×Ω - mΩ×(Ω×r)

where Fnon g are any non-gravitational forces.

We can simplify this a bit. First, g0 - A0 is value of local gravity, g = -g khat (see discussion of static effects, above). Second, Ω is a small number, so we can expand the corrections in a power series in Ω. For our purposes, we'll keep just the terms that are first order in Ω, so we can neglect the centrifugal force term that is second order in Ω. Finally, we'll neglect air resistance. The equation of motion becomes:

ma = -mg khat + 2mv×Ω .

We will break this up into component equations, but first we need to evaluate the cross product. In the corrdiate system Ω = Ω(0, cosλ, sinλ), and we'll take the velocity to be arbitrary v = (xdot, ydot, zdot) giving

v×Ω = Ω[(zdot cosλ - ydot sinλ)ihat + xdot sinλ jhat - xdot cosλ khat].

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 + vy0
zdot = -gt + 2Ω x cosλ + vz0

where the velocity at t=0 is v0 = (vx0, vy0, vz0). The x equation becomes

xddot = 2Ω(gt cosλ - vz0 cosλ + vy0 sinλ).

Integrate this equation twice to get the result

x(t) = (1/3)Ωgt³ cosλ - Ωt²(vz0 cosλ - vy0 sinλ) + vx0t + x0.

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) = vy0t - Ωvx0t² sinλ + y0
z(t) = -½gt² + vz0t + Ωvx0t² cosλ + z0

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.

Example: Falling Body

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 (vx0, vy0, vz0) = (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)Ωh3/2 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)3/2cos 43° / (9.8 m/s²)½ = 0.0009m = 0.9mm. For a drop of 100m the deflection increases to about 1.6cm.

Example: Deflection of a Rifle Bullet

A bullet is fired horizontally east with initial speed v0. 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 (vx0, vy0, vz0) = (v0, 0, 0) at t=0. Our interest is only in the equation for y to determine the north-south deflection

y(t) = -Ωv0t²sinλ

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/v0, so the deflection is

y(t) = -ΩH²sinλ / v0

The size of the deflection grows with H².

© 2006 Robert Harr