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Week 2: Newton’s Laws: Continued |
a)The force added or subtracted to a real force (i.e. – mg, or a normal force) in a frame accelerating uniformly. The elevator and boxcar examples below illustrate this nearly ubiquitous experience. This is the “force” that pushes you back in your seat when riding in a jet as it takes o , or a car that is speeding up. Note that it isn’t a force at all – all that is really happening is that the seat of the car is exerting a normal force on you so that you accelerate at the same rate as the car, but this feels like gravity has changed to you, with a new component added to mg straight down.
b)Rotating frames account for lots of pseudoforces, in part because we live on one (the Earth). The “centrifugal” force mv2/r that apparently acts on an object in a rotating frame is a pseudoforce. Note that this is just minus the real centripetal force that pushes the object toward the center. The centrifugal force is the normal force that a scale might read as it provides the centripetal push. It is not uniform, however – its value depends on your distance r from the axis of rotation!
c)Because of this r dependence, there are slightly di erent pseudoforces acting on objects falling towards or away from a rotating sphere (such as the earth). These forces describe the apparent deflection of a particle as its straight-line trajectory falls ahead of or behind the rotating frame (in which the ”rest” velocity is a function of ω and r).
d)Finally, objects moving north or south along the surface of a rotating sphere also experience a similar deflection, for similar reasons. As a particle moves towards the equator, it is suddenly travelling too slowly for its new radius (and constant ω) and is apparently “deflected” west. As it travels away from the equator it is suddenly traveling too fast for its new radius and is deflected east. These e ects combine to produce clockwise rotation of large air masses in the northern hemisphere and anticlockwise rotations in the southern hemisphere.
Note Well: Hurricanes rotate counterclockwise in the northern hemisphere because the counterclockwise winds meet to circulate the other way around a defect at the center. This defect is called the “eye”. Winds flowing into a center have to go somewhere. At the defect they must go up or down. In a hurricane the ocean warms air that rushes toward the center and rises. This warm wet air dumps (warm) moisture and cools. The cool air circulates far out and gets pulled back along the ocean surface, warming as it comes in. A hurricane is a heat engine! There is an optional section on hurricanes down below “just for fun”. I live in North Carolina and teach physics in the summers at the Duke Marine Lab at Beaufort, NC, which is more or less one end of the “bowling alley” where hurricanes spawned o of the coast of Africa eventually come to shore. For me, then, hurricanes are a bit personal – every now and then they come roaring overhead and do a few billion dollars worth of damage and kill people. It’s interesting to understand at least a bit about them and how the rotation of the earth is key to their formation and structure.
The two forces just mentioned (pseudoforces in a rotating spherical frame) are commonly called coriolis forces and are a major driving factor in the time evolution of weather patterns in general, not just hurricanes. They also complicate naval artillery trajectories, missile launches, and other long range ballistic trajectories in the rotating frame, as the coriolis forces combine with drag forces to produce very real and somewhat unpredictable deflections compared to firing right at a target in a presumed cartesian inertial frame. One day, pseudoforces will one day make pouring a drink in a space station that is being rotated to produce a kind of ‘pseudogravity’ an interesting process (hold the cup just a bit antispinward, as things will not – apparently – fall in a straight line!).
2.4.1: Identifying Inertial Frames
We are now finally prepared to tackle a very di cult concept. All of our dynamics so far is based on the notion that we can formulate it in an inertial frame. It’s right there in Newton’s Laws – valid
Week 2: Newton’s Laws: Continued |
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only in inertial frames, and we can now clearly see that if we are not in such a frame we have to account for pseudoforces before we can solve Newtonian problems in that frame.
This is not a trivial question. The Universe doesn’t come with a frame attached – frames are something we imagine, a part of the conceptual map we are trying to build in our minds in an accurate correspondence with our experience of that Universe. When we look out of our window, the world appears flat so we invent a Cartesian flat Earth. Later, further experience on longer length scales reveals that the world is really a curved, approximately spherical object that is only locally flat – a manifold 67 in fact.
Similarly we do simple experiments – suspending masses from strings, observing blocks sliding down inclined planes, firing simple projectiles and observing their trajectories – under the assumption that our experiential coordinates associated with the Earth’s surface form an inertial frame, and Newton’s Laws appear to work pretty well in them at first. But then comes the day when we fire a naval cannon at a target twenty kilometers to the north of us and all of our shots consistently miss to the east of it because of that pesky coriolis force – the pseudoforce in the rotating frame of the earth and we learn of our mistake.
We learn to be cautious in our system of beliefs. We are always doing our experiments and making our observations in a sort of “box”, a box of limited range and resolution. We have to accept the fact that any set of coordinates we might choose might or might not be inertial, might or might not be “flat”, that at best they might be locally flat and inertial within the box we can reach so far.
In this latter, highly conservative point of view, how do we determine that the coordinates we are using are truly inertial? To put it another way, is there a rest frame for the Universe, a Universal inertial frame S from which we can transform to all other frames S′, inertial or not?
The results of the last section provide us with one possible way. If we systematically determine the force laws of nature, Newton tells us that all of those laws involve two objects (at least). I cannot be pushed unless I push back on something. In more appropriate language (although not so conceptually profound) all of the force laws are laws of interaction. I interact with the Earth by means of gravity, and with a knowledge of the force law I can compute the force I exert on the Earth and the force the Earth exerts on me given only a knowledge of our mutual relative coordinates in any coordinate system. Later we will learn that the same is more or less true for the electromagnetic interaction – it is a lot more complicated, in the end, than gravity, but it is still true that a knowledge of the trajectories of two charged objects su ces to determine their electromagnetic interaction, and there is a famous paper by Wheeler and Feynman that suggests that even “radiation reaction” (something that locally appears as a one-body self-force) is really a consequence of interaction with a Universe of charge pairs.
The point is that in the end, the operational definition of an inertial frame is that it is a frame where Newton’s Laws are true for a closed, finite set of (force) Law of Nature that all involve welldefined interaction in the coordinates of the inertial frame. In that case we can add up all of the actual forces acting on any mass. If the observed movement of that mass is in correspondence with Newton’s Laws given that total force, the frame must be inertial! Otherwise, there must be at least one “force” that we cannot identify in terms of any interaction pair, and examined closely, it will have a structure that suggests some sort of acceleration of the frame rather than interaction per se with a (perhaps still undiscovered) interaction law.
There is little more that we can do, and even this will not generally su ce to prove that any given frame is truly inertial. For example, consider the “rest frame” of the visible Universe, which can be thought of as a sphere some 13.7 billion Light Years68 in radius surrounding the Earth. To
67Wikipedia: http://www.wikipedia.org/wiki/manifold. A word that students of physics or mathematics would do well to start learning...
68Wikipedia: http://www.wikipedia.org/wiki/Light Year. The distance light travels in a year.
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Week 2: Newton’s Laws: Continued |
the best of our ability to tell, there is no compelling asymmetry of velocity or relative acceleration within that sphere – all motion is reasonably well accounted for by means of the known forces plus an as yet unknown force, the force associated with “dark matter” and “dark energy”, that still appears to be a local interaction but one we do not yet understand.
How could we tell if the entire sphere were uniformly accelerating in some direction, however? Note well that we can only observe near-Earth gravity by its opposition – in a freely falling box all motion in box coordinates is precisely what one would expect if the box were not falling! The pseudoforce associated with the motion only appears when relating the box coordinates back to the actual unknown inertial frame.
If all of this gives you a headache, well, it gives me a bit of a headache too. The point is once again that an inertial frame is practically speaking a frame where Newton’s Laws hold, and that while the coordinate frame of the visible Universe is probably the best that we can do for a Universal rest frame, we cannot be certain that it is truly inertial on a much larger length scale – we might be able to detect it if it wasn’t, but then, we might not.
Einstein extended these general meditations upon the invariance of frames to invent first special relativity (frame transformations that leave Maxwell’s Equations form invariant and hence preserve the speed of light in all inertial frames) and then general relativity, which is discussed a bit further below.
Example 2.4.1: Weight in an Elevator
a
Figure 27: An elevator accelerates up with a net acceleration of a. The normal force exerted by the (scale on the floor) of the elevator overcomes the force of gravity to provide this acceleration.
Let’s compute our apparent weight in an elevator that is accelerating up (or down, but say up) at some net rate a. If you are riding in the elevator, you must be accelerating up with the same acceleration. Therefore the net force on you must be
X
F = ma |
(207) |
Week 2: Newton’s Laws: Continued |
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where the coordinate direction of these forces can be whatever you like, x, y or z, because the problem is really one dimensional and you can name the dimension whatever seems most natural to you.
That net force is made up of two “real” forces: The force of gravity which pulls you “down” (in whatever coordinate frame you choose), and the (normal) force exerted by all the molecules in the scale upon the soles of your feet. This latter force is what the scale indicates as your ”weight”69.
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Thus: |
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or |
X F = ma = N − mg |
(208) |
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N = W = mg + ma |
(209) |
where W is your weight.
Note well that we could also write this as:
W = m(g + a) = mg′ |
(210) |
The elevator is an example of a non-inertial reference frame and its acceleration causes you to experiences something that feels like an additional force of gravity, as if g → g′. Similarly, if the elevator accelerates down, gravity g′ = g − a feels weaker and you feel lighter.
Example 2.4.2: Pendulum in a Boxcar
Tx
m |
Ty |
+y’ |
+y |
T |
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a |
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mg’ θ mg |
θ |
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+x |
−ma |
+x’ |
Figure 28: A plumb bob or pendulum hangs “at rest” at an angle θ in the frame of a boxcar that is uniformly accelerating to the right.
In figure 28, we see a railroad boxcar that is (we imagine) being uniformly and continuously accelerated to the right at some constant acceleration ~a = axxˆ in the (ground, inertial) coordinate frame shown. A pendulum of mass m has been suspended “at rest” (in the accelerating frame of the boxcar) at a stationary angle θ relative to the inertial frame y axis as shown
We would like to be able to answer questions such as:
a)What is the tension T in the string suspending the mass m?
b)What is the angle θ in terms of the givens and knows?
We can solve this problem and answer these questions two ways (in two distinct frames). The first, and I would argue “right” way, is to solve the Newton’s Second Law dynamics problem in the inertial coordinate system corresponding to the ground. This solution (as we will see) is simple
69Mechanically, a non-digital bathroom scale reads the net force applied to/by its top surface as that force e.g. compresses a spring, which in turn causes a little geared needle to spin around a dial. This will make more sense later, as we come study springs in more detail.
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Week 2: Newton’s Laws: Continued |
enough to obtain, but it does make it relatively di cult to relate the answer in ground coordinates (that isn’t going to be terribly simple) to the extremely simple solution in the primed coordinate system of the accelerating boxcar shown in figure 28. Alternatively, we can solve and answer it directly in the primed accelerating frame – the coordinates you would naturally use if you were riding in the boxcar – by means of a pseudoforce.
Let’s proceed the first way. In this approach, we as usual decompose the tension in the string in terms of the ground coordinate system:
X Fx |
= |
Tx = T sin(θ) = max |
(211) |
X Fy |
= |
Ty − mg = T cos(θ) − mg = may = 0 |
(212) |
where we see that ay is 0 because the mass is “at rest” in y as the whole boxcar frame moves only in the x-direction and hence has no y velocity or net acceleration.
From the second equation we get: |
mg |
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T = |
(213) |
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cos(θ) |
and if we substitute this for T into the first equation (eliminating T ) we get:
mg tan(θ) = max |
(214) |
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or |
ax |
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tan(θ) = |
(215) |
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g |
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We thus know that θ = tan−1(ax/g) and we’ve answered the second question above. To answer the first, we look at the right triangle that makes up the vector force of the tension (also from Newton’s Laws written componentwise above):
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Tx |
= |
max |
(216) |
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Ty |
= |
mg |
(217) |
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and find: |
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T = q |
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= mp |
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= mg′ |
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Tx2 + Ty2 |
ax2 + g2 |
(218) |
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p
where g′ = a2x + g2 is the e ective gravitation that determines the tension in the string, an idea that won’t be completely clear yet. At any rate, we’ve answered both questions.
To make it clear, let’s answer them both again, this time using a pseudoforce in the accelerating frame of the boxcar. In the boxcar, according to the work we did above, we expect to have a total e ective force:
~ ′ |
~ |
(219) |
F |
= F − m~aframe |
~
where F is the sum of the actual force laws and rules in the inertial/ground frame and −m~aframe is the pseudoforce associated with the acceleration of the frame of the boxcar. In this particular problem this becomes:
− mg′yˆ′ = −mgyˆ − maxxˆ |
(220) |
or the magnitude of the e ective gravity in the boxcar is mg′, and it points “down” in the boxcar frame in the yˆ′ direction. Finding g′ from its components is now straightforward:
g′ = p |
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ax2 + g2 |
(221) |
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as before and the direction of yˆ′ is now inclined at the angle |
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θ = tan−1 µ gx ¶ |
(222) |
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a |
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