## Posts Tagged ‘physics’

### Why is kinetic energy proportional to v^2?

November 23, 2013

I answered this question here, channeling this. That seems like the best answer – kinetic energy is proportional to v^2 because of Galilean relativity.

The more common answer, though, is to say that work is force times distance. Then because f=ma, we can show that the work done is proportional to the change in v^2. Ron, at least, is not satisfied with this approach, saying

The previous answers all restate the problem as “Work is force dot/times distance”. But this is not really satisfying, because you could then ask “Why is work force dot distance?” and the mystery is the same.

I’m not sure if this is the case. Suppose we assume that acceleration is some function of position, so

$\mathbf{\ddot{x}} = a(\mathbf{x})$

Then we have

$\mathbf{\dot{x}}\mathbf{\ddot{x}} = \mathbf{\dot{x}} a(\mathbf{x})$

$\frac{1}{2}\frac{d}{dt}(\mathbf{\dot{x}}^2) = \frac{d}{dt} \int_{x_0}^{\mathbf{x}} a(\mathbf{x'}) ds$

This shows that we can create a conserved quantity, namely $\frac{1}{2}\dot{x}^2 - \int_{x_0}^{\mathbf{x}} a(\mathbf{x'}) ds$, using just the knowledge f=ma, which doesn’t seem to be begging the question at all. It does, however, assume that the line integral is independent of path. So if we’re told f=ma and told that f is a gradient, we can find that kinetic energy is proportional to v^2. I suppose we ought to be able to, since from f=ma it is manifest that physics has Galilean invariance, and the thought experiment linked up top shows that energy depends on v^2.

So both explanations seem correct, but the one based on invariance certainly feels better.

### Simple experiments with beats and hearing

September 29, 2013

In the class I’m TAing, we are discussing beats. Beats are when you play two tones that are close to each other, but at slightly different frequencies. They will slowly drift from in phase to out of phase, like when your turn signal has slightly different timing from that of the car in front of you at a stop light. In sound, this results in the “wa-wa-wa” sound you hear when tuning two instruments to each other.

Beats depend on space as well as time. If two speakers play a tone, they can be in phase at one place and out of phase at another, since the distances to the speakers is different.  I had never actually played around with this with sound before, so I set up two speakers in a room where I could move both them and myself around pretty freely.

Making the speakers play the same pure tone (440 Hz), I found it was quite easy to observe the interference pattern of the speakers by moving my head around. And when the speakers were placed far enough apart, the phase difference could be quite different at opposite ears, so the sound would appear only in my left or only in my right ear. Also, by putting the speakers the right distance apart, I was able to observe turning the volume on the closer speaker up from zero and hearing the volume of the sound go down as the front speaker interferes with the back one. (It’s not practicable to make the sound seem to disappear entirely.)

Next, playing notes separated by 0.5 Hz, I found I could stand in one spot and hear the sound move from my left ear to my right ear and back again once every two seconds. This never sounded as if the source itself was moving around. Instead, the sound felt as if it was being played right next to my ear, even though the speakers were a couple meters in front of me.

These are just simple things, but still striking, since we rarely encounter coherent sound sources set up this way in daily life. Some time ago I did one other experiment accidentally that had an interesting result. I was listening to an app that creates beats by outputting different frequencies to the left and right audio channels, but I didn’t hear the beats. After a bit of confusion, I realized it was because I was wearing earbuds – there was no interference because nobody was getting both signals at once. Playing the tones through speakers instead, I heard the beats. However, when I turned the frequency down, I could hear the beats even with the earbuds. Somehow, your brain collects phase information about sound, but only if the sound is slow enough. The transition occurred somewhere around 600 Hz, or roughly a millisecond. This corresponds fairly well to the width of a nerve impulse, so a likely hypothesis is that you can only determine the arrival of the peak or trough of a wave to a time comparable to the width of a nerve impulse, which means you only have meaningful phase information down to around the 600 Hz I heard. (In a discrete Fourier transform, you have information about frequencies up to half the inverse of your sampling rate, so this checks out pretty well.)

The app I used was similar to this one, but I think my cheap headphones don’t process audio channels properly, as I can hear the beats even with one earbud in, both with that app and with WIkipedia’s file.

### Weights on a pendulum

September 23, 2013

A friend of mine is one of the people in charge of winding a clock tower. He was showing my physics class how the tower works, and we were looking at the pendulum.

Here is a photo of the pendulum in question, from this blog.

We asked the class:

Why are there weights on the top of the pendulum?

This may sound like a bit of a trick question; one can come up with many possible reasons. But I’ll tell you that the weights are put there by humans deliberately as they maintain the clock. Sometimes they add a few and sometimes they take a few off.

A student then asked why the weights don’t slip off.

For a given pendulum length and swing amplitude, what is the minimum coefficient of friction to keep the weights from sliding off?

### Dropping a Slinky (calculation)

July 30, 2012

Let’s do a quick bit of math related to Dropping a Slinky. Last time, I estimated that it takes about 0.3 seconds for the slinky to collapse. To get a more precise answer, note that however the slinky falls, its center of mass must accelerate downwards at gravitational acceleration.

Where is the slinky’s center of mass? When it’s just hanging, the slinky is in equilibrium, so the derivative of the tension is proportional to the density. Also, if we assume an ideal spring with zero rest length, the tension is inversely proportional to the density (why?). Therefore, we write

$\frac{\mathrm{d}T}{\mathrm{d}x} = g \rho$

$T = \frac{\alpha}{\rho}$

This can be solved to show that the density follows

$\rho \propto \frac{1}{\sqrt{x}}$

Integrating, we find that the center of mass is one third the way up the slinky. The time for the slinky to collapse is the same as the time for the center of mass to fall to the bottom, or

$t = \sqrt{\frac{2 (1/3 l)}{g}}$

This is the same answer, but modified by a factor of 0.81. Notice that this only depends on the “slinkiness” – the zero rest length ideal spring. We expect thick and thin slinkies of different stiffnesses to act in essentially the same way.

### 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:

### 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.

### 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.

### Bounce, Part 6

January 11, 2010

Last time, we looked at what Galileo had to say about free fall. This time, we’ll take one more example from his dialog and try to squeeze a little moral out of it.

Galileo presents his ideas through the character Salviati, who explains them to his companions Sagredo and Simplicio. Salviati’s interlocutors raise all manner of objection to his theories, but Salviati answers them and convinces everyone of his point all the more surely in the process. One such objection is given by Sagredo, who doesn’t believe that the velocity of a falling object increases evenly with each second of falling:

So far as I see at present, the definition might have been put a little more clearly perhaps without changing the fundamental idea, namely, uniformly accelerated motion is such that its speed increases in proportion to the space traversed; so that, for example, the speed acquired by a body in falling four cubits would be double that acquired in falling two cubits and this latter speed would be double that acquired in the first cubit.

Sagredo is suggesting that rather than Galileo’s law

$v \propto t$,

that the velocity of a falling body increases the same amount each second, we should instead have

$v \propto x$,

the the velocity increases the same amount each meter the body falls. These are different hypotheses, and so we need to distinguish between them. Given that Salviati states he is not interested in examining the fundamental cause of gravity, and only in characterizing its behavior, there is only one way to do this – experiment.

Instead, Salviati offers the following retort:

…that motion should be completed instantaneously; and here is a very clear demonstration of it. If the velocities are in proportion to the spaces traversed, or to be traversed, then these spaces are traversed in equal intervals of time; if, therefore, the velocity with which the falling body traverses a space of eight feet were double that with which it covered the first four feet (just as the one distance is double the other) then the time-intervals required for these passages would be equal. But for one and the same body to fall eight feet and four feet in the same time is possible only in the case of instantaneous [discontinuous] motion;

What a strange counterargument! It makes absolutely no sense. Gaining an even increment of speed for each unit of time is a perfectly consistent mathematical law, and does not at all imply instantaneous motion. We can write this law as

$\frac{dx}{dt} = c(x - x_0)$,

which implies that the distance fallen increases exponentially with time. This is completely contrary to observation, and it would be hard to build a unified mechanics like Newton’s that respects this law, but it isn’t logically impossible for the reasons Salviati cites.

And how do Salviati’s friends respond to this argument? Do they rip it apart, or restate their objection more clearly, or request further detail?

Sagredo replies,

You present these recondite matters with too much evidence and ease; this great facility makes them less appreciated than they would be had they been presented in a more abstruse manner. For, in my opinion, people esteem more lightly that knowledge which they acquire with so little labor than that acquired through long and obscure discussion.

I guess it’s easier to convince the people you’re arguing with when they’re fictional characters you invented yourself!

Everyone makes mistakes, and they hadn’t gotten around to inventing peer review in the sixteenth century, so let’s forgive Galileo, and take a further look at this hypothesis.

Suppose we have a projectile with constant horizontal velocity and vertical velocity that changes according to the distance traveled in the vertical direction.

What happens if we shoot it up out of a cannon? We know from experience that the cannonball slows down, so it must be losing a constant amount of velocity for each unit height it gains.

The cannonball slows down its vertical velocity, but as it does so, its vertical position changes less. Since its vertical position changes less, the change in its vertical velocity slows down more. In fact, the cannonball asymptotically reaches a certain height above ground, and then stays there!

If the vertical height isn’t changing, then according to this law the vertical distance traveled is zero, and because vertical velocity only changes when vertical height changes, the vertical velocity stays zero. This cannonball would never come down.

On the other hand, if it were pushed down just a little bit, it would gain speed very rapidly, falling exponentially back toward Earth. The motion under Sagredo’s law is absurd, but I wonder why Galileo brought it up at all, only to miss the point.

Too much philosophizing can be dangerous, but this sort of philosophy – extracting results from speculative physical laws – is exactly what theoretical physicists do. The name of the game for a theoretical cosmologist, for example, is to come up with some crazy ideas about how the universe might work, the way Sagredo came up with an idea about falling bodies. Then, the cosmologist tries to work out the consequences of the theory, for example that cannonballs ought to hang in midair until a slight breeze comes along and gives them a downward tap, and they come plummeting back to Earth extremely quickly. If the theory doesn’t agree with observation, it’s wrong.

One difference between the Renaissance and Internet Age versions is that Sagredo’s and Salviati’s theories about falling are easy to test. The experiment doesn’t require any equipment. You just drop something. If you don’t have a thing, you can try jumping instead. But with advanced theoretical ideas, it can be very difficult to make the required observations. That’s why we need giant particle accelerators and kilometer-long interferometers and thirty-meter telescopes and ridiculously-good gyroscopes.

But another problem is that working out the consequences of modern theories is hard. We saw an example of Galileo failing to work out the consequences of a theory, but that was a simple mistake, and if someone had brought his attention to it, he’d have been able to fix it. Some of today’s new ideas about physics are so complicated that even if we can state the law (the equivalent of Sagredo’s idea about velocity being proportional to distance fallen), we may not know how to get to the conclusion (cannonballs hanging over our heads).

We’ve come long way since Galileo, figuring out lots of ways to check ourselves and test our ideas. But there’s a much longer way left to go.

### Bounce, part 3

December 25, 2009

Today we’ll introduce the principle of Galilean relativity and use it to continue out examination of the tennis ball/basketball experiment. I don’t want to talk about relativity much, because I’m too stupid to say anything new or interesting. Instead, I’ll just start using it.

We’ll think about the experiment as a series of steps. First, the tennis ball and basketball are dropped, one on top the other, with a tiny gap in between them. Next, the basketball hits the ground, bounces, and changes direction. Then, the tennis ball, coming down, hits the basketball, now coming up, and bounces off. The question we will try to answer is, “Once the tennis ball bounces off the basketball in our experiment, was is the maximum speed it could have going back up?” That will be enough for this post.

We begin by dropping the tennis ball and basketball from some distance above the ground. Just as we drop them, they aren't moving.

From our previous investigations, we already understand the first step of the basketball bouncing by itself. Ideally, it can bounce back up with exactly the same speed it had coming down. Let’s call that speed $v$.

The tennis ball and basketball fall together, and pick up some speed v just before reaching the ground.

In the next step, the basketball is coming up at a speed $v$, and the tennis ball coming down at speed $v$ when they collide. To understand this case, we’ll begin with a simpler one.

The basketball bounces off the ground, changing direction.

The tennis ball bounces off the moving basketball, shooting back up at an unknown speed.

If we were in an elevator moving up at speed $v$ right at the moment of the collision, we would have a different opinion on the speeds of the basketball and tennis ball. The basketball is going up the same speed we are, so from our perspective it isn’t moving at all.

We go back to just before the tennis ball bounces off the basketball, and imagine we're riding up in an elevator at the same speed as the basketball.

On the other hand, by looking at things from the point of view of the ground, we see that the distance between us on the elevator and the falling ball is shrinking at a rate $2v$. So, if we believe that we in the elevator aren’t moving, then the tennis ball must be falling towards us at speed $2v$, to keep the gap between us and the tennis ball shrinking at the same rate.

To look at things from the elevator's point of view, we add a downward velocity v to everything in the scene, including the ground.

Now the tennis ball bounces off the basketball. If the basketball is much larger than the tennis ball, it is essentially like bouncing off a brick wall, or the ground, and the tennis ball reverses direction keeping the same speed. So from your point of view in the elevator, the tennis ball is going up at speed $2v$.

The tennis ball bounces off the stationary basketball, reversing its direction and keeping the same speed, all viewed from the elevator's reference frame.

Finally, we return to the point of view of the ground. We know that the distance between you and the tennis ball is increasing at a speed $2v$. Since you’re going up at $v$, the tennis ball must be going up at $3v$. So in the ideal case, where the basketball is so much larger than the tennis ball that it isn’t deflected at all, and the tennis ball’s collisions don’t lose any energy, the tennis ball can shoot upward with three times the velocity it picks up by falling. This shows us why it can bounce higher than it came from. It bounces up going faster, and so reaches a greater height. But it also tells us that there’s still a maximum height. By making the basketball bigger and bigger, and pumping it up better, we’ll still only approach a certain limit where the tennis ball bounces back at $3v$, so we can’t launch the tennis ball into outer space this way.

To go back to the ground's reference frame, we add a speed v in the upward direction to everything, and see that the tennis ball goes up at speed 3v in the ideal case.

Before continuing, I’d like to look at where the principle of relativity came into this discussion. Most of what I’ve said, I hope, seems obviously true. It is based on the argument “if the distance between A and B is changing at a certain rate, it will change at that rate even if you begin moving”. For example, if you are playing catch, and you throw a ball away from you at 20 mph, then someone driving past in a car still thinks the difference in speeds between you and the ball is 20mph. This isn’t relativity – it’s simple kinematics. The only place we needed relativity, the idea that physics is the same in different reference frames, was in saying that in the elevator frame, the tennis ball bounces off the basketball reversing its direction with the same speed, just as it would if bouncing off a stationary basketball on Earth.

In fact, in the theory of special relativity, it’s this physics principle that holds, and not the kinematics of switching between reference frames (but that’s a different story).

In the next post, we’ll look at what today’s conclusion means in terms of how high the tennis ball goes.

### Bounce, part 2

December 6, 2009

In the previous post, I arrogantly announced I could explain the classic experiment in which you drop a tennis ball and basketball together, and the tennis ball goes flying. Then, I got as far as examining a single bounce. Part way. This might take a while.

Today I’m going to keep talking about that single bounce. When you drop a basketball and see how high it bounces (assume it goes straight up-and-down), there are three possibilities:

1. It bounces higher than you dropped it.
2. It bounces back to the same height as you dropped it.
3. It bounces lower than you dropped it (or does not bounce).

We’ve already said that the first one is impossible given that we can’t build a perpetual motion machine, and that a ball that bounces higher than it started would let us make a successful one. When we make the experiment, it turns out the basketball does not bounce back as high as it was dropped from. Today we’ll see why this is true.

On the face of it, bouncing back to exactly the same height seems a reasonable thing to do. After all, if the basketball is going to bounce to, say, 72% of the height it was dropped from, where does that number come from? Why not 71% instead? Bouncing back to exactly the same height is a sort of natural choice if the problem is very simple, because it’s a very simple answer. The problem, however, is not very simple.

Watch this remarkable video of a golf ball colliding with a wall at high speed.

This is clearly not simple! After bouncing off the wall, the motion of the golf ball is very different than before bouncing – all that vibrating and oscillating got added in. So it makes sense that the trip back up should be different than the trip down.

The oscillation of the ball is a form of motion, and if we had a ball oscillating the way it does in that video, we could find some way to use that to drive a cuckoo clock, if we were clever. So the ball can’t bounce to its original height because if it did, we could catch it there, damp the oscillations while driving the cuckoo clock, and drop it again.

Further, when the ball hits the ground, it makes some noise. The noise is motion of the air, and this motion could be used to drive a cuckoo clock. The wall the ball bounces off is shoved back during the collision – this too is a motion that could drive the clock. Additionally, the ball and wall both heat up a little in the collision. If you’ve ever driven a nail with a hammer and touched the nailhead immediately afterward, you’ve probably noticed this phenomenon. The difference in heat between the part of the wall that the ball hit and the rest of the wall could, in principle, drive the cuckoo clock as well.

I’m not interested in the details and mechanics of how the various types of motion I just mentioned could be converted into driving the cuckoo clock. The goal is to understand a bounce, and for now we simply need to know that when the bounce occurs, there’s a lot more going on than simply a ball changing direction. All that other stuff makes the sequence distinctly irreversible. We go from having the motion concentrated in once place – the overall movement of the ball – to many places – the oscillations of the ball, noise, heat, the movement of the wall, etc. As various other bits of the environment pick up motion (which, in general, we call “energy” in physics), the ball loses it, and can’t bounce to the same height again.

This brings up an interesting question. We know the ball doesn’t bounce as high as it was dropped because its motion gets spread out to other places. Does that mean that the process could, in principle, work the other way? Could various bits of the environment give motion to the ball, so that in fact it does bounce higher than it was dropped in that special case? For example, in the video, the ball oscillates after hitting the wall, but not before. What if we struck the ball to set it oscillating, then dropped it? Could it then work the other way, losing most of its oscillation when it bounces off the wall, but actually picking up speed and going higher than it was dropped?

That’s possible. It would be a difficult trick to pull. But notice it doesn’t violate our principle of no perpetual motion, because in order to make that scheme work you need to strike the ball and set up oscillations, which is against the “no outside influences” rule.

Another way to make the ball bounce back faster would be to move the wall it crashes into. If we push the wall forward to meet the ball, then the wall might give some of its energy to the ball, and the ball could bounce away quickly and go higher than dropped. This is what happens when a baseball player hits a pitch. The batted ball can travel much further than the pitcher could have thrown it, since the bat adds energy to the ball. This is what’s happening with the basketball and tennis ball. We’ll get more into it next time.

Before I go, check out this additional video of a much tamer golf ball collision. In the slower collision, the golf ball is still deformed, but not so severely. You can guess that if you want something to bounce up to nearly its original height, it’s better to drop it from a low height than a high one – and that’s true (try it).