# Arcsecond

## Posts Tagged ‘mechanics’

### Unwinding: Physics of a spool of string

June 1, 2012

It’s been a long day. Let’s unwind with a physics problem.

This problem was on the pre-entrance exam I took before arriving at Caltech for my freshman year. I’ve seen it from time to time since, and here I hope to find an intuitive solution.

You have a spool of thread, already partially unwound. You pull on the thread. What happens?

Here it is in side view. The dashed circle is the inside of the spool and the green line is the thread. Take a minute to see if you can tell how it works. Does the spool go right or left?

The usual method is to work it out with torques. The forces you must account for are the force of tension from the string and the force of friction from the table.

Torques are actually a pretty easy way to solve this problem, especially if you calculate the torque around the point of contact between the spool and the table (since in that case friction has no moment arm and exerts no torque).

This method is direct, but it’s useful to find another viewpoint if you can.

Let’s first examine a different case where the string is pulled up rather than sideways.

In this case, even if the first situation was unclear, you probably know that the spool will roll off to the left. To see why, let’s imagine that the thread isn’t being pulled by your hand, but by a weight connected to a pulley.

I put a red dot on the string to help visualize its motion.

The physics idea is simply that the weight must fall, so the red dot must come closer to the pulley. Which way can the spool roll so the red dot moves upward?

When the spool rolls (we assume without slipping), the point at the very bottom, where it touches the table, is stationary. The spool’s motion can be described, at least instantaneously, as rotation around that contact point.

Googling, I found a nice description of this by Sunil Kumar Singh at Connexions. This image summarizes the point:

If the spool rolls to the right, as above, the point where the string leaves the spool (near point B), will have a somewhat downward motion. This will pull the red dot down and raise the weight. That’s the opposite of what we want, so what really happens is that the spool rolls to the left, the string rises, and the weight falls.

With this scenario wrapped-up (or unwrapped, I suppose), let’s return to the horizontal string segment.

Again, the weight must fall and so the red dot must go towards the pulley.

If we check out Mr. Singh’s graphic, we’re now concerned with the motion of a point somewhere near the bottom-middle, between points A and C. As the spool rolls to the right, this point also moves to the right. This is indeed what happens as the weight falls.

Notice that the red point actually moves more slowly than the spool as a whole. This means the spool catches up to the string as we move along – the spool winds itself up. If the inside of the spool is 3/4 as large as the outside (like it is in my picture), the spool rolls 4 times as fast the string moves, and so for every centimeter the weight falls, the spool rolls four centimeters.

Here’s a short video demonstration:

Tags:mechanics, physics, problems and solutions

### Leakier, Slower, and No Rain

December 7, 2010

A while ago, I asked a standard freshman physics problem about a cart that has rain fall into it, then opens a hole and rain leaks out. Then I gave an answer saying that as rain falls vertically into an open cart running on a frictionless track, the cart slows down, but as rain leaks out it shows no change in speed.

That was mostly correct, given the picture I drew of the hole:

Water leaks out a hole in the bottom of the cart as it slides to the right.

The key is that the hole is in the center. Yesterday, Martin Gales posed a question on Physics.StackExchange pointing out that this makes a difference, because if we imagine a stationary cart with a hole all the way to the left, then as it drains, the water moves left, and so the cart will have to move right a little to conserve momentum. But then once the cart starts moving the water leaking out of the cart is moving…

I spent three hours last night trying to solve this seemingly-trivial problem. (My answer is at the original question.) It’s simple enough to pose to first-term freshmen, and yet I went through dozens of slightly-wrong ideas and calculations before hitting on the surprising answer. Further, once I knew what happens, it didn’t seem very complicated any more, leaving me to wonder what the hell is wrong with my overclocked simian brain. The feeling you get when thinking about such a problem is an asymmetric oscillation of healthy frustration and premature joy unparalleled in other pursuits. I want to be mind-fucked like this every night.

Tags:mechanics, physics, physics problems, problems

### Three-Way

October 21, 2010

I’ve been told that when you and your sweetheart get accepted to different grad schools, you’ve encountered the “Two-Body Problem”. It’s an inapt analogy, because the two-body problem can be solved elegantly. What’s really difficult is the three-body problem.

If we have three massive bodies interacting through Newton’s gravitational law, with arbitrary initial positions and velocities, it turns out that exact analytic solutions are extremely difficult to find. That’s not to say there is no solution – the massive bodies have no trouble finding it. But if we want to predict their motion we’ll need to do some numerical integration.

However, we might wonder if there are some particularly simple or intriguing solutions, perhaps in very symmetrical situations. There are. Quite a few of them are cited in the Scholarpedia article. Here are some:

I’m not sure how these solutions were discovered. But if we consider the problem a moment it’s clear the center of mass cannot accelerate, and that energy and angular momentum must be preserved. An especially simple way to do this is to have the angular momentum and energy of each individual mass be constant by making them all rotate around the center of mass at a fixed frequency.

To see if this works, measure the positions of the three masses by vectors $\vec{r_1}, \vec{r_2}, \vec{r_3}$ from the axis of rotation, which is through the center of mass and perpendicular to the plane of the bodies. Let the masses of the bodies be $m_1, m_2, m_3$. Then the equation for the center of mass states

$\vec{r_1}m_1 + \vec{r_2}m_2 + \vec{r_3}m_3 = 0$.

If we look at a reference frame comoving with the center of mass and corotating with the masses, the masses will be stationary, so there must be no force on them. Let’s take $m_1$ as an example. The centrifugal force on it is

$F_c = \omega^2 m_1 \vec{r_1}$.

The gravitational force on it is

$F_g = G m_1 \left(m_2\frac{\vec{r_2} - \vec{r_1}}{\left|\vec{r_1} - \vec{r_2}\right|^3} + m_3\frac{\vec{r_3} - \vec{r_1}}{\left|\vec{r_3} - \vec{r_1}\right|^3}\right)$

If those forces are going to add to zero to give $m_1$ zero acceleration, they better at least point the same direction. It’s not clear that they do. The centrifugal force points only in the direction of $\vec{r_1}$, while the gravitational force has all three position vectors in there. So we need the amounts of $\vec{r_2}$ and $\vec{r_3}$ to be such that they add up to point towards $\vec{r_1}$.

From the equation for the center of mass we get,

$-m_1\vec{r_1} = m_2\vec{r_2} + m_3\vec{r_3}$

whence we know the proportions in which the $\vec{r_2}$ and $\vec{r_3}$ must appear in the gravitational force. We conclude

$m_2\vec{r_2} + m_3\vec{r_3} \propto \left(\frac{m_2\vec{r_2}}{\left|\vec{r_1} - \vec{r_2}\right|^3} + \frac{m_3 \vec{r_3}}{\left|\vec{r_3} - \vec{r_1}\right|^3}\right)$.

This will work perfectly iff the denominators on the right hand side of that equation are equal, meaning mass $m_1$ must be equidistant from masses $m_2$ and $m_3$. Repeating the argument while writing out forces on $m_2$ this time, the masses must be in an equilateral triangle.

We still don’t know if such a solution exists, only that it’s possible to get the centrifugal force to point opposite the gravitational force for arbitrary masses, as long as we put them in an equilateral triangle. Let’s continue, setting the centrifugal and gravitational forces equal for $m_1$ and see what the resulting rotation rate is. Also let’s let the sides of the triangle be length $l$

$F_c = \omega^2 m_1 \vec{r_1} = F_g = \frac{G m_1}{l^3} \left( m_2(\vec{r_2} - \vec{r_1}) + m_3 (\vec{r_3} - \vec{r_1}) \right)$

Applying the center of mass equation

$-m_1\vec{r_1} = m_2\vec{r_2} + m_3\vec{r_3}$

one more time and setting the total mass equal to $M$, this simplifies to

$\omega^2 = \frac{G M}{l^3}$.

A clean result that is the same for all three masses, meaning these orbits indeed solve the equations of motion.

Tags:gravity, mechanics, three-body problem

### Shaking The Right Way

April 30, 2010

The other day I was working at a job – a real one, like grownups have. I used to think this job was boring. But things aren’t boring; they’re just what they are. People can be bored, but that’s their business. Don’t blame the thing!

I work at a place where they make widgets. One of the pieces for the widgets is a little, flat, plastic rod. There’s an assembly line with a robot that puts the rods in the partially-assembled widgets. But before that, the rods have to be fed into the robot one by one. The rods come from the rod factory in a big pile, though, and the robot can’t reach in a pick one out. We need a way to take the rods from a jumbled mess and feed them one-at-a-time into the robot.

To do that we dump the plastic rods in a bowl about a meter across. There’s a ramp winding up along the edge of it. We vibrate the bowl, and the rods march up the ramp and into the rod-eating widget-making robot. The bowl looks a little like this:

There was a library nearby when I was a kid that had a ramp like this in it. I always wanted to go to that library. I never read any of its books.

This is supposed to be a cut-away view of the side of the bowl. The ramp is on the inside of the bowl, angled up slightly to keep the rods from falling off.

When we turn it on, the bowl starts vibrating fast enough that it’s just a blur (60 Hz seems plausible). And the rods walk their way up the ramp at a very even pace. The rods will go up the ramp whether there are one thousand or just one of them in the bowl (although if there are lots of them some will get pushed off the ramp, fall to the bottom, and start over).

This is really crazy! How do the rods go up the ramp, not down? Things are supposed to go down, generally speaking. It’s pretty freaky to watch a single rod climb right up this long winding ramp as the bowl vibrates. The bowl doesn’t spin. It doesn’t visibly tilt. It works for different sizes of rods and even for a penny (I think. I haven’t found an opportunity to slip one in yet, but I’m pretty sure it’ll work for a penny. Not a marble, though.)

I asked the technician who works with the bowl, and he said he wasn’t sure, but that the rods don’t always go up. The people who make the bowl can make some adjustments to it, and then the rod will go up slower, then stop, then march their way back down as the workers keep adjusting.

I couldn’t figure it out, so I asked my boss, who’s mechanically-minded, and we worked out a theory. In short, it goes like this: the bowl can vibrate in two different ways. It can bob up and down, and it can rotate back and forth. It can move a couple of millimeters in each direction. The bowl makes the rods go up the ramp by doing both of these at the same time, and we can adjust the speed of the rods up the bowl’s ramp by adjusting the phase lag between the two oscillation modes.

Draw a little dot on the side of the bowl. Since a few millimeters (the amplitude of the bowl’s vibrations) is small compared to a meter (the bowl’s diameter), for any given point vibrating the bowl by rotation is the same as shaking back a forth horizontally in the direction tangent to the edge of the bowl. So we’ve got a sine wave motion horizontally.

Close-up of one part of the ramp. The red arrows show the motion of the the red dot, although any other point on the ramp would move the same way.

If the bowl instead bobs up and down, the motion of the dot is like this:

Same as last time, but the bowl bobs up and down.

Now make the bowl rotate and bob simultaneously. There are different ways we could do this. Assuming we give both modes the same frequency (easier to engineer, I’d think), then we can describe the way these modes are combined with a single parameter – the phase lag between them. When the phase lag is zero the bowl moves all the way forward at the same time it’s all the way up, and all the way back at the same time it’s all the way down. The red dot traces out a diagonal line, like this:

Moving forward/backward and up/down in sync.

If the phase lag is 180 degrees, the diagonal line switches directions, like this:

But if the phase lag is 90 degrees, the dot instead traces out a circle. Make it -90 degrees and the circle goes the other direction.

Combining back/forth and up/down just so, we get a circle.

We can also make the circle go the other way, if we want.

This is just right for moving a rod up a ramp. Remember that when something moves in circle, its acceleration points towards the center of the circle. So imagine the ramp shaking this way when it’s at the bottom of the circle. The ramp is accelerating up, pushing into the rod. That means there’s a big normal force on the rod, and a big normal force means more friction. To take advantage of this high friction, the ramp should move forward, pushing the rod ahead in space.

The CCW circle moves forward at the bottom of its loop (green arrow). The ramp is accelerating up (blue arrow), jamming into the rod and carrying it forward along with it.

Now the red dot has reached the high point of the circle. It’s accelerating down. That means it’s falling away from beneath the rod, so the normal force goes down. If the acceleration is high enough, the ramp may even pull away from the bottom of the rod completely (I’m not sure whether it actually does this). Now the friction is low, and the ramp can slide backwards, leaving the rod where it is. The net result is that the rod is further along the ramp at the end of the cycle than at the beginning.

Same circle, but now the ramp is at the top of its motion. The rod will get left behind as the ramp pulls away from underneath.

This really works, even without a fancy industrial bowl. I did it just now. I found the longest hardback book in my room (The Feynman lectures) and an appropriate rod substitute (a Rubik’s Cube), and I easily sent the cube uphill on the tilted book by rotating circularly one direction with my hands, and brought it back down by rotating circularly the other direction.

The book was still pretty short, but I have a longer flat, mobile surface – a white board. I’m a bit too clumsy to shake it in a circle with my hands, though. But I have a bicycle…

I duct taped my white board to the pedal of my bike in an attempt to recreate the circular motion I posited for the ramp. I propped the other end of the white board up against a table,

This didn’t work well. The board was tilting during the cycle because the far end, leaning on the table, wasn’t going up and down the same way as the near end. I tried propping the table up at an angle in hopes of keeping the white board close to flat, but that didn’t help much.

Time to recruit the neighbors. It was 11:20 PM so my neighbors, who are college students, were awake and drunk enough to agree to anything that sounded weird or stupid. With a human at the other end of the white board, I could keep the board pretty much flat, or pretty close to tilted at a constant angle as it went around.

The results were that as long as I kept the white board flat, my Rubik’s Cube would go forward and backward the way I predicted, but I couldn’t get it to climb any significant hill. I tried switching out the cube for a high-friction rock-climbing shoe, but that still didn’t make much progress. I conclude the bowl at my job relies on high-speed, high frequency vibrations, the opposite of my bicycle.

Finally, in order to be a good scientist I needed a more objective test of my theory. I was able to get my conveyor belt to work roughly, but I already knew the result I wanted. Maybe I was subconsciously tipping the book/board the direction that I wanted things to go?

To test this, I called a friend and asked him to replicate my experiment with a book and a penny, without telling him what I expected to happen. He reported the same result! We can indeed make things move forward or backward, even up slight inclines, just by shaking the right way.

Tags:boredom, friction, mechanics

### New Problem: Leaky, Rainy, and Slow

February 22, 2010

Here’s a classic physics problem I asked my MCAT students yesterday. They unanimously chose the wrong answer.

A cart runs along a frictionless track on a rainy day. The rain falls straight down, and some of it lands in the open cart. As the cart accumulates rain, does it slow down, speed up, or keep going the same rate? (Do not worry about the cart running into raindrops ahead of it. We imagine that the raindrops fall in such a way that they either land in the cart or don’t hit it. Also, there’s no wind resistance.)

Rain falls straight down into the cart, which is coasting to the right.

Next, the rain stops, but the cart gets a leak. Water pours out a hole in the bottom of the cart. Does the cart get faster, slower, or stay the same speed? How does its final speed, when all the water has leaked out, compare to its original speed before the rain? (Again, ignore friction.)

Water leaks out a hole in the bottom of the cart as it slides to the right.

The answer is now up here.

Tags:cart, mechanics, physics, physics problems, water
Posted in problems and solutions | 7 Comments »

### Conservation of Momentum

February 20, 2009

Here’s something that’s in textbooks, but they tend to leave out lots of little bits and pieces, the way I used to when I made sandwiches for Arby’s one summer. Not that you’ll get the full story here, either. But you’ll get a more satisfying hunk of disgusting, gray, dampish meat clumps, and a little piece of metaphorical lettuce, too.

When two particles interact, Newton’s third law postulates

$F_{12} = - F_{21}$,

where $F_{12}$ means, “the force particle ’1′ exerts on particle ’2′.” This is useless knowledge unless you have some sort of interpretation of force. Force is defined by the second law

$F = ma$.

Someone once tried to tell me that Newton’s second law is not just a definition of force, but has some deeper meaning. I think they were lying because they wanted to seduce me. (No luck there, Grandpa!)

So Newton’s second law defines force, and is meaningless without some rules about what force should do. For example, if you say that a particle with absolutely nothing around to interact with must have no force on it, you’ve said something about force and now Newton’s second law can step in. In this case it says

$0 = ma$

so that a free particle does not accelerate. (That’s Newton’s first law. However, there are philosophical problems with such a conclusion. If there is nothing around for the particle to interact with, then how could you tell whether or not it’s accelerating?)

The third law, a rule about force, is lame without a definition of force. The second law, a definition of force, is lame without any rules. They were made for each other, like rabbits and lawn mowers (but with less of those annoying screaming sounds). By combining Newton’s second and third laws for two interacting particles, we get

$m_1a_1 = F_{21} = -F_{12} = -m_2a_2$

By the transitive law

$m_1a_1 = -m_2a_2$

or

$m_1a_1 + m_2a_2 = 0$

and assuming that mass is constant

$\int_{t_a}^{t_b}dt \left(m_1a_1 + m_2a_2\right) = \int_{t_a}^{t_b} dt*0 = 0$

for arbitrary times $t_a$ and $t_b$. Using the fundamental theorem of calculus and the definition

$a = \frac{dv}{dt}$

yields

$\left(m_1v_1(t_b) + m_2v_2(t_b)\right) - \left(m_1v_1(t_a) + m_2v_2(t_a)\right) = 0$

again for arbitrary times $t_a, t_b$. What we’ve discovered is that if you take measure the quantity

$m_1v_1 + m_2v_2$

at any two times, you will always get the same answer. That quantity is called “momentum”, and the fact that it doesn’t change is called “conservation of momentum.”

We haven’t proved it to be true. Science doesn’t prove anything to be true. What we’ve proved is that it follows from certain assumptions. If we make some measurements and find that the “law” of momentum conservation doesn’t hold, there are a few possibilities that I can think of:

1. We made a mistake with the measurements. Our apparatus is broken, or we did something dumb like converting units wrong, etc.
2. Newton’s laws are wrong. They do not accurately represent the interaction of particles.
3. We were not doing an experiment with exactly two particles. (That is the only situation for which we did the proof. Maybe the theorem failed because there was some third particle around that we didn’t see, or maybe the objects in our experiment were not particles, but instead more complicated composite things that are not bound by Newton’s laws.)
4. The mass of the particles is not constant. (Remember that this was an assumption used in the proof).

Maybe you can think of other explanations. I can’t at the moment. But it turns out that these explanations can account for a lot of situations. Item (1) comes up frequently enough – it’s just a fact that people make misteaks.

Explanation (2) is sometimes correct as well; Newton’s laws aren’t true. Special relativity modifies them. General relativity pretty much scraps them (er, don’t quote me on that). In quantum mechanics, momentum is important, but no longer has an interpretation as mass*velocity. In fact it (mathematically) no longer has any “interpretation” – instead it is its own primary quantity, equally as fundamental to the theory as the concept of “position”. It even steals “position”‘s claim to the letter ‘p’. Momentum is nobody’s bitch.

Complication (3), that we aren’t using two isolated particles, arises in practice as well. There are obvious examples, such as everything. When I drop my spoon, it starts gaining momentum until it hits the floor, when it loses momentum. Then I pick it up and lick it clean, and its momentum bounces all around as I lick more and more violently. All this occurs because a spoon is not a system of two particles.

There are more interesting (but less tasty) examples where the “not-two-particles” explanation manifests. Take two charged nonrelativistic, non-quantum particles and let them interact. They won’t conserve momentum. The reason is charged bodies generate electromagnetic fields, and our assumption that the only things around are the charged bodies fails. The electromagnetic field can carry its own momentum, although technically in order to break the proof all it would have to do is exist. In another example, Wolfgang Pauli was thinking about another case in which momentum is not conserved – beta decay. He decided options (1), (2), and (4) were not for him, and instead guessed that beta decay must involve some previously-unseen stuff. That stuff is the neutrino.

Finally, explanation (4), that the mass is variable, is not something that occurs in practice to my knowledge, but it could. Of course, if a meteor shooting through space hurls off some of its rock-junk when it get near the sun and heats up, then the meteor’s mass decreases. But that doesn’t count because it’s not two particles, and also momentum actually is conserved in that situation if you consider the momentum of the space junk, the meteor, and the sun altogether. What I mean is that I’m not aware of any evidence that fundamental particles can have variable mass.

What if there are three particles? Can we prove that momentum is still conserved if we define momentum to be

$p = m_1v_1 + m_2v_2 + m_3v_3?$

No. We can’t because we could only prove anything by getting some knowledge about force from Newton’s third law. But Newton’s third law is only telling us the story for two lone particles. When there’s a third, all bets are off. However, there is another assumption that we usually take along with Newton’s laws, often implicitly. This is that forces add linearly.

Imagine conducting an experiment with particles 1 and 2, and no particle 3 around. Measure the force on particle 1. Now conduct a new experiment where particle 1 does the same thing it did before, but particle 2 is absent, and particle 3 is around doing whatever it wants. Again, measure the force on particle 1.

We assume that if we conduct a third experiment with particles 1, 2, and 3 all together, the force on particle 1 will be the sum of the forces in the first two experiments.

With this law that forces add linearly, we can prove momentum conservation for three particles. And if we assume forces continue to add in the simple manner for any number of particles, then momentum conservation also holds for any number of particles.

tomorrow: energy

Tags:conservation, conservation law, mechanics, momentum, Newon's Laws, Newton, second law, Third Law
Posted in physics | 2 Comments »

### A Brief Illustration of One Forms and Tensors in Mechanics

December 26, 2008

note: Don’t worry too much about dimensions and constants. I drop them like I would a new born child.

other note: I originally identified the eigenvalues incorrectly. Changing it didn’t alter the math significantly. 12/27/08

In this post I will examine the mechanics of a classical particle in the 2-D potential

$U(x,y) = \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2$

This is a harmonic oscillator potential, but $\omega$ is different in the $x$ and $y$ directions.

A simple harmonic oscillator in one dimension is canonically the mass on a spring:

or a grandfather clock pendulum

These correspond to the potential

$U(x) = \frac{k x^2}{2}$.

The fastest way to get the equation of motion is Newton’s second law, which yields

$\begin{array}{rcl} -\frac{d U}{d x} & = & \frac{d^2 x}{dt^2} = k x \\ x(t) & = & A\sin\left(\omega t + \phi\right)\end{array}$

with $\omega$ some function of $k$.

The easiest way to make the hop to two dimensions is to keep the apparatus the same, but give the mass another degree of freedom. For the mass on a spring, this might correspond to sitting the mass on a table and tacking down one side of the spring, while leaving it free to spin as well as expand/contract. You have to think of this as a spring with no rest length, and collapsible, so that it can pass through itself. Combine the properties of a slinky and a rubber band and you’re pretty much there. The 2-D pendulum is just a pendulum free to swing to and fro in whatever direction it wants.

This scheme would set

$\begin{array}{rcl} U(x,y) &=& \left( \frac{ k x^2}{2}\right)+\left(\frac{k y^2}{2}\right) \\ {} & = & \left(\frac{k (x^2 + y^2)}{2}\right) \\ {} & = & \left(\frac{k r^2}{2}\right) \end{array}$

with $r$ the displacement distance from equilibrium. Then $r$ and $\theta$ are decoupled, and we just get simple harmonic motion of $r$, while $\theta$ increases linearly with time. The mass traces out an ellipse in space, with an eccentricity and orientation depending on the initial conditions, and with the same period as before.

It is a bit more interesting to break the symmetry between the two orthogonal directions and treat them distinctly, as you would people with different ethnicities. In the example of the spring, we could set up a situation like this:

displacement and force on a mass with springs on either side

Before examining the dynamics of this configuration, let’s first hash out the language we’ll be using by describing the kinematics.

The displacement of the mass from equilibrium is a two-dimensional vector, and the set of all possible displacements constitutes a vector space.

It can be easier to manipulate this vector when we set up a coordinate system, so we’ll set up standard Cartesian coordinates with x the left/right direction and y the up/down direction.

Now we are ready to define a (1,1) tensor that characterizes this system. You might be worried because a (1,1) tensor takes a vector and a one form as its arguments, but we haven’t yet identified any one forms! You’ll see in a moment that we can do quite a bit without knowing exactly what the one forms are, but don’t worry, we’ll return and identify them at the end.

Remember that a (0,1) tensor is a vector. That means that a (1,1) tensor can also be considered a function that takes in a vector and gives a (0,1) tensor – another vector. (1,1) tensors are the set of linear operators on the vector space.

On to the tensor. The spring constant of the system is a (1,1) tensor, $K$, which takes in the displacement of the mass and returns the force on the mass.

Notice that the force does not necessarily point in the same direction as the displacement. If you push the mass straight up/down, the springs will pull it back towards the center, as they will with a pure left/right displacement. But suppose you push the mass at a 45 degree angle. Then the springs push left/right much more than they pull up/down. Let’s quantify this by working out the tensor. We need only work out its action on unit displacements in the x and y directions, and then linearity will take care of the rest.

If we displace the mass a small amount to the right, we pull the left spring out and push the right spring in. They each create a restoring force of 2k .

$K(\hat{\mathbf{x}},\_\_) = -2k\hat{\mathbf{x}}$.

I have used the underscore in the second argument of $K$ to indicate this is still an open slot. $K$ must still take on a one form before it can spit out a scalar. Right now, it’s a vector – a (0,1) tensor.

If we push the mass up a small amount in the y direction, we don’t make the springs any longer to first order, but the forces they exert on the mass are no longer balanced. Each spring creates a restoring force that is the equilibrium tension in the spring times the sine of the angle the springs make with the horizontal.

$K(\hat{\mathbf{y}},\_\_) = -2\frac{\textrm{T}_0}{L}\hat{\mathbf{y}}$

with $\textrm{T}_0$ the equilibrium tension of the springs and $L$ their length. If the springs had zero rest length, this would evaluate to the same thing as in the x direction, but let’s just pretend these springs don’t have that.

This defines the tensor spring constant, and allows us to write the force on the mass by the equation

$\mathbf{F} = K(\mathbf{x},\_\_)$.

We can again use Newton’s second law to find the equations of motion.

$K(\mathbf{x},\_\_) = m \frac{d^2 \mathbf{x}}{d t^2}$.

Note that this is a vector equation, not a coordinate equation. Before we solve this equation, we’ll return to the discussion of kinematics left incomplete earlier.

We have a vector space of displacements, so we know there exists a dual space of one forms. This is the space of linear, scalar-valued functions of the displacement. It also has dimension two. We can find a basis for this space by finding two independent one-forms. Call them $p_x$ and $p_y$, and define them by

$p_x(\hat{\mathbf{x}}) = 1$
$p_x(\hat{\mathbf{y}}) = 0$

$p_y(\hat{\mathbf{x}}) = 0$
$p_y(\hat{\mathbf{y}}) = 1$

A general one form can be expressed in this basis as follows:

$\begin{array}{rcl} p(\mathbf{x} &=& p(x\hat{\mathbf{x}} + y\hat{\mathbf{y}}) \\ {} &=& x*p(\hat{\mathbf{x}}) + y*p(\hat{\mathbf{y}}) \\ {} &=& x*\alpha + y*\beta \\ {} &=& \alpha*p_x(\mathbf{x}) + \beta*p_y(\mathbf{x}) \\ p &=& \alpha p_x + \beta p_y \end{array}$

You may have noticed that the basis one forms have a simple interpretation as projection operators. From the above discussion, we see that

$\left[\alpha p_x + \beta p_y\right](\mathbf{x}) = \alpha*x + \beta*y$

So that the one forms are acting in a way similar to the dot product in elementary physics. However, since I haven’t defined the dot product in this little series of posts, I won’t pursue that notion.

Now we can return to the equations of motion, and put each of the basis one forms into the vector equations.

$\mathbf{F}(p_x) = K(\mathbf{x},p_x) = -2 k x$,
$\mathbf{F}(p_y) = K(\mathbf{x},p_y) = -2 \frac{T_0}{L} y$.

You might verify that these forces are consistent with the potential I quoted at the beginning of the post. You probably won’t, though. Nobody ever does shit like that. Anyway, the solutions are of the form

$\mathbf{x}(t) = A\sin(\omega_1 t)\hat{\mathbf{x}} + B\sin\left(\omega_2 t+\phi\right)\hat{\mathbf{y}}$

where $\omega_1$ is a function of $k$ and $\omega_2$ is a function of $T_0$ and $L$. In general they are different, and their ratio will determine the orbit of the mass in 2-D space. The motion of the mass traces out a Lissajous figure, one example of which is this:

A Lissajous figure with frequency ratio 1.8

It might not seem like the tensor accomplished anything, since we could easily have found the same system of equations without knowing anything about tensors. The payoff here is theoretical. The equation

$\mathbf{F} = K(\mathbf{x},\_\_)$

involves no coordinates. It is simply a statement about vectors and linear functions of vectors. The coordinate system was an easy way to do some calculations, but in writing the tensor equation we’ve mathematically depicted the system geometrically instead of algebraically.

In the x-y basis, $K$ can be written as a diagonal matrix with eigenvalues $-2k$, $-2\frac{T_0}{L}$. If we were to change coordinate systems, for example rotating x and y by some angle, a coordinate-based approach would have to recalculate all the forces. On the other hand, our tensor equation would still hold, and we could evaluate it just as before – by plugging in the one forms corresponding to projection operator on the axis of the springs and orthogonal to that axis. The only work would be in finding new coordinate representation of the one forms.

Tags:force, Lissajous, Lissajous figures, mass, mechanics, one form, oscillator, pendulum, projection operator, spring, spring constant, tensor, vector
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