Non-inertial Frames - Duke University

[Pages:5]Physics 181

Non-inertial Frames

Overview

The first of Newton's laws of motion is (in effect) a statement that there are reference frames in which a body will move with constant velocity unless it interacts with some other body. These are the preferred reference frames for the application of Newton's laws. We call them inertial, because in them the law of inertia holds. But even if there are such frames (and it is difficult to say how one might find such a frame) the universe we see from here on earth is mostly observed from non-inertial frames. Our planet is in orbit around the sun, and it spins on its axis, so any frame fixed on the surface of the earth must be non-inertial. The sun rotates around the center of our galaxy, and our galaxy is in motion, probably accelerated, relative to other galaxies. Fortunately for us, the accelerations arising from these phenomena are relatively small by our everyday standards, much smaller than the acceleration with which an object falls to the ground if we drop it. So for many purposes we simply pretend as an approximation that our "laboratory" frame is inertial. But on a larger scale the accelerations resulting from our planet's daily rotation have demonstrable, even large effects, making some wind patterns south of the equator opposite to those north of that line. And on a smaller scale we frequently find ourselves in a reference frame subject to substantial acceleration, as when we turn a corner in our car, or stand in a rapidly rising elevator. It is clumsy at best to analyze the motion of objects in those circumstances by inventing an external inertial frame so that we can use Newton's laws. It is more convenient, as well as more intuitive, to find a way to use the "natural" reference frame, taking account somehow of the fact that it is accelerating. Ways to do that are the subject of these notes.

Effective gravity

We start with simple situations near the surface of the earth, with a particle subject to gravity and perhaps interactions of other types. We divide the total force accordingly: Ftot = mg + Fnon! grav . Here g is the gravitational field of the earth. The symbol m here, which gives the strength of the gravitational force, is called the gravitational mass. The 2nd law tells us that Ftot = ma . The m on the right side of this equation is the inertial mass, which measures how strongly the particle resists being accelerated. There is no a priori reason why these two masses should be related, but careful measurement (beginning perhaps with Galileo and culminating in a famous experiment by E?tv?s in the late 19th century) has shown that they are the same. So it is a fact of nature.

The experiments show only that they are proportional, but one chooses units to make them equal.

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Now suppose that in an inertial frame Ftot = 0 , so the particle moves without acceleration. Consider the situation as seen in a frame moving with acceleration a. In that frame the particle will be observed to move with acceleration !a . Knowing the 2nd law, an observer in that frame will conclude that the total force is Ft!ot = "ma .

Now let the total force in the inertial frame be Ftot = mg + Fnon! grav . Then in the accelerating frame we have Ft!ot = Fnon" grav + m(g " a) . The effect we call gravity has been

altered. We express it as a change in the gravitational field:

g ! geff = g " a .

The quantity geff is the effective gravitational field.

The term !ma that is added to the forces because of the acceleration of the frame is called an inertial force. It is sometimes called "fictitious", but in the non-inertial frame it is as real as any of the forces. And effective gravity is the real gravitational field in that frame. It gives the direction of "down" in the sense of a hanging weight, and any method of measuring the weight of an object will give mgeff .

Example. Shown is a railroad car with a

pendulum attached to its ceiling. The car accelerates to the right as indicated. We ask

!

a

for the angle ! at which the bob will hang at

rest relative to the car, and for the period of

small swings about that direction.

The answer to the first question is easy: the string hangs in the direction of geff . The vector diagram shows the situation. We see that tan! = a/ g . If the bob

has mass m, its weight (the tension in the string) is mgeff = m g2 + a2 .

geff ! g

?a

From introductory physics we know that the period of a simple

pendulum for small swings is given by T = 2! " !/ g , where ! is the length of the

string. So it is easy to substitute in this formula: T = 2! " !/ geff .

Exercise. Solve this problem in an inertial frame at rest on the ground beside the train. In this frame the bob of the pendulum (if it is at rest relative to the car) must have acceleration a. Find the string tension and angle necessary to give it this acceleration.

Ocean tides

The concept of effective gravity is useful to explain one consequence of the dependence on distance in Newton's gravitational law: the ocean tides on earth. The essential point is this: while we easily understand that the moon in its orbit accelerates toward the earth, it is equally true that the earth accelerates toward the moon. Thus -- forgetting

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for the moment the spin of the earth and its orbit around the sun -- a frame fixed on earth is non-inertial because of its (small) acceleration toward the moon.

The amount of this acceleration is given by

a0

=

G

Mm r02

,

where Mm is the mass of the moon and r0 is the distance between the center of the moon and the center of the earth (recall that for spherical bodies gravity from the outside acts effectively at the center). This is the acceleration of the reference frame of the earth, directed toward the moon.

Now we consider the total gravitational force

on two particles of mass m, #1 on the surface of

the earth closest to the moon, #2 directly opposite it on the earth, as shown in the figure. 2

R To moon

1

Of course the gravitational attraction of these particles toward the center of the earth ( mge ) is stronger than the force exerted by the moon on them, so the net force on each will be toward the center of the earth. But we must also include among the forces the inertial force resulting from the earth's acceleration a0 toward the moon.

Consider first #1. The net force is to the left. Its magnitude is

F1

=

mge

+

ma0

!

G

mMm (r0 ! R)2

=

mge

! GmMm

" $ #$ (r0

1 ! R)2

!

1 r02

% '. &'

The quantity in [ ] is positive, so F1 < mge . In other words, the effect of the moon is to produce a small force on #1 toward the moon.

Now consider #2. The net force is to the right, and has magnitude

F2

=

mge

!

ma0

+

G

mMm (r0 + R)2

=

mge

! GmMm

" $ #$

1 r02

!

(r0

1 + R)2

% '. &'

Again the quantity in [ ] is positive, so F2 < mge . The effect of the moon is to produce on #2 a small force away from the moon.

These small forces make the ocean water bulge outward slightly on both sides of the earth, creating the tides. Although the bulge in mid-ocean is tiny in height, it involves a great volume of water, and when this shift in water reaches the shores the effect can be very large.

Since R ................
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