Bounce, part 3

December 25, 2009 by Mark Eichenlaub

part 1
part 2

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.

Sine Error comic 24: “Snowmunculus”

December 24, 2009 by Mark Eichenlaub

I was going to make these out of real snow until I realized that would mean going outside.

The Stroop Test, Revisited

December 24, 2009 by Mark Eichenlaub

Making these videos is way easier than doing my chores.

Blindfold

December 21, 2009 by Mark Eichenlaub

A directorial/acting debut, filmed on the shortest day of the year, with Shelley (who is actually solving that Rubik’s cube blindfolded).

Things To Do Before I Die

December 17, 2009 by Mark Eichenlaub
  • Play Lady Macbeth
  • Travel to the distant stars… only to discover that what I was truly looking for was inside me all along.
  • (hint: it’s trace amounts of selenium).
  • Be where the people are. See them dancing. Walking around on those feet.
  • Climb Mount Ever. Then climb Mount Everer. Finally, climb Mount Everest.
  • Shoot fish in a barrel. The recoil of the gun breaks my collarbone and shoves me violently backwards just as the barrel explodes and I am dowsed me in a hundred gallons of water and fish guts, which knock me over the edge of a large cliff I was standing near for absolutely no reason. Tell people it was pretty easy.
  • Kill my grandfather first, then go back in time. If someone tells me I got the order wrong, say “What are you talking about? I can fucking time travel.”
  • Have a man light a cigar, blow smoke in my face, and say, “You know, we’re not so different, you and I.”
  • Write the book, “Introduction to Sexual Intercourse for Scientists and Engineers”
  • … after years of exhaustive research
  • urinate on a jellyfish
  • go to a bar with a priest and a rabbi and have absolutely nothing funny happen
  • skin about a hundred different cats, all in exactly the same way
  • tell someone a piece of information is on a need-to-know basis, and have them say some back besides “and I need to know!”
  • perform the Krebs cycle
  • make a self-referential entry on a list
  • get killed by an electric eel (this is not completely “before” I die, but I’ll allow an exception)
  • convince the internet that bacon is over-rated
  • and teach it how to use the word “comprise”
  • actually throw some shit at a fan, just so when anyone uses a certain expression, I can interrupt, ”well, let me tell you about this one time…”
  • pee openly in public when I’m out with my grandmother, and use the excuse that she doesn’t know where she is
  • make peace with my jigsaw puzzles

Bounce, part 2

December 6, 2009 by Mark Eichenlaub

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).

Factorials, Primes, and Long Life

December 3, 2009 by Mark Eichenlaub

Today my age rolls over from a factorial to a square, and this will happen again on December 3, 2105. How old am I?

It occurred to me a little while ago to try to find all the times this could ever happen. A little calculation shows that 6!+1 is not a square, but that 7!+1 = 71^2.

After that, I couldn’t find any more factorials that were one less than a square. I set up a quick computer program to search, and didn’t find any looking up to n = 200. So I began to wonder whether there were any more, and if not, how could I prove it?

One of the most famous proofs in number theory states that there are infinitely many primes, because if not, then there must be a largest prime p. But then p!+1 has no divisors other than one, since all of the primes have remainder one when they divide it. This is nonsense – of course p!+1 has a divisor – itself! So by contradiction there are not a finite number of primes. This suggests that perhaps we’d expect n!+1 to be prime much of the time. Or, since we could have used p!-1 in the above argument, that n!-1 may frequently be prime. And, since n!+1 and n!-1 differ by two, maybe there are infinitely-many twin primes?

No one knows how to answer these simple questions. The numbers n! \pm 1 are called factorial primes, and we’ve found 42 of them so far. That’s a long way short of infinity. On the other hand, who knows? We might already have them all.

The factorials one less than a square have also been studied, but the catalog here is much less extensive. In fact, I unwittingly discovered the entire thing in my first investigation. These are called Brown Numbers, and they are the solution to Brocard’s Problem. There are no more up to n=10^9, and people who professionally think about these things are strongly suspicious that there are no more at all. It has been proven that there are finitely-many. That doesn’t mean there couldn’t be a fourth sitting way, way out there, waiting for us to find it. And if we did, that doesn’t mean there couldn’t be a fifth…

It looks like if I want an answer to this question, I’m either going to have to invent some new mathematics, or wait for someone else to, or else live a very, very long time.

Bounce, part 1

November 7, 2009 by Mark Eichenlaub

When I was a kid, my dad sat my sisters and I down one day to inquire why the ceiling in our garage was peppered with blurry brown splotches. I speculated that perhaps it was camouflage – the ceiling needed to blend into the herd. He didn’t buy it, and I accused him of not paying much attention to what the signs at the zoo had said about the ungulates when we visited two weeks before.

Juvenile creativity aside, the reason for the disfigured ceiling was that someone at school had told me about that thing with the basketball and the tennis ball. You know, the one where you hold a tennis ball directly over a basketball, drop them both straight down, and the tennis ball goes flying away crazily. You have seen that, right? It looks like this:

That experiment is why the ceiling was messed up: we had dirty tennis balls.

At that time, I didn’t know why the experiment should work. Why does the tennis ball bounce higher than it was dropped from? Not even a super-bouncer ball does that. Evidently it has something do with the basketball being so much larger than the tennis ball. So, what if I got something even bigger than a basketball? If I had a ball ten times as big as the basketball, and dropped the tennis ball on top of that, then how high would it blast up? Could I send something up to outer space this way? What if I use three balls – say a ping pong ball on top the tennis ball? What if I use four, or five, or infinity? What if I use two basketballs rather than a basketball and tennis ball? Can I still make one shoot up super-high then? What if I reverse the order, dropping the basketball on top the tennis ball? Will it bounce higher or lower than it was dropped from? Will it go higher than it would if the tennis ball wasn’t even there? Is there a way to make the basketball and tennis ball trick work even better? Does it matter that they’re ball-shaped, rather than bouncy cubes, for example?

The remarkable thing is that although I didn’t understand my experiment at the time, I could have, in principle, if I’d had someone around to explain it. That’s what I’m going to try to do now – answer the questions from the last paragraph with a series of thought experiments, rather than with forces, potential and kinetic energy, and a bevy of equations. It’ll take a few posts to do it, but the fundamental ideas are simple: that a ball bounces off a wall as fast as it came in, that you won’t get carsick if the ride is smooth enough, and that gravity pulls evenly.

Let’s begin by thinking about bouncing in the simplest context: bounce the basketball all by itself. When we do this in real life, it turns out that the basketball bounces back up, but not quite all the way. If we drop it from waist height, it might rise back to thigh-level, then bounce again, going to knee level, and so on.

Why lower than the original height? Why can’t it bounce higher? If I pump it up more, it bounces better. So if I pump it up high enough, could it bounce up higher than it began? The answer is no. You might get it to bounce better by pumping it up, but each bit of pumping would give you less and less benefit, so that you’d never quite get the ball to come all the way back up.

Suppose the ball bounced a little higher than it was dropped from. Then if it bounced a little higher on the next bounce, and a little higher the bounce after that, it would bounce to arbitrarily great heights, all by itself. Eventually it’d go to the moon. This is absurd, so the basketball can’t keep bouncing higher. But that doesn’t mean it can’t bounce higher for some particular height. Why can’t it, for example, bounce from 1 meter to 1.1 meters, then to .8 meters, then to .9 meters, then to .6 meters, etc. in some strange up-and-down pattern? Or why not bounce higher than dropped for all heights under one meter, and lower than dropped for all heights above one meter? Or higher than dropped on the first bounce, and lower than dropped on all subsequent bounces?

All of these scenarios are contrived, intuitively wrong, and even silly. Intuition can fail though, as it did for me when seeing the tennis ball bounce away so high. So it’s interesting that what intuition says about the single bouncing basketball can be derived from a simple principle.

Here is the first bit of physics: We assume that perpetual motion is impossible. By this, I mean that it is impossible to build a cuckoo clock that keeps on going forever without drawing in energy from the outside (for example, a solar-powered cuckoo clock with a battery to store energy for the night might run indefinitely, but if we block out the sun with a large wall of Coppertone bottles, the clock will wind down quickly and halt.)

There’s nothing special about my example of a cuckoo clock. If you could build something else, like a train, that ran forever, you could use its motion to power a cuckoo clock. And if you could build a cuckoo clock that ran forever, you could make a million copies of it and use their combined motion to power a train. So the cuckoo clock is just a convenient example for a general idea that you can’t build any machine that goes on forever.

One way to make a cuckoo clock go is to let a basketball fall, and use this to drive the clock. What you do is tie a string to the basketball, then tie the other side of the string to the gears of the clock. As the clock ticks, the basketball pulls on the gears, forcefully turning them a little bit. The gears deliver a tiny push to the pendulum each swing to keep it going. This way, the clock can run for a long time. Eventually it wears down though, because the basketball falls all the way to the floor. Then there’s nothing pulling the gears around, so the gears can’t power the pendulum, and friction slows everything to a stop. (For more on how a pendulum clock like a cuckoo clock works, see the How Stuff Works article. The important point for this argument is that you can power the clock by carefully and slowly lowering the basketball.)

This leads into why the basketball can’t bounce higher than we dropped it from. Say we drop the basketball from 1 meter, and it bounces back up to 1.1 meters. Then we can concoct a scheme to run a cuckoo clock forever. Start by lifting the ball up to 1.1m. Tie the string to it and use it to make the cuckoo clock run. When the ball falls to just 1 meter, take off the string and drop the ball. It bounces up to 1.1 meters again. Catch it there and re-attach the string. Let the ball fall back to 1 meter, running the clock… This creates an infinite process that’s self-contained. The system is the clock, the basketball, the floor it’s bouncing off, and the Earth, which creates gravity. Nothing outside this system is supplying any energy, and yet the cuckoo clock can run forever. If we assume that such perpetual motion is impossible, then it’s also impossible for a basketball to bounce higher than it was dropped from, no matter what the height.

In the next post, we’ll look at why the basketball doesn’t bounce back up to exactly the same height, which will lay the foundation for understanding the interaction of the basketball and tennis ball.

Freezing in Warm Air

October 25, 2009 by Mark Eichenlaub

I took a bottle of soda out of the freezer today, and looked at it. Not frozen. I opened it, and was surprised to find that a minute later, there was a chunk of soda-ice floating on top. Somehow, the soda had frozen after I took it out of its cold environment and into a warm one.

This didn’t happen because of the temperature of the room. Instead, it happened because I opened the bottle. When sealed, the bottle is under higher pressure. If you put ice under pressure it will try to take up less space, and since water is more dense than ice (as you can test by observing that ice floats), high pressure melts ice. Once I released the pressure, the soda, which was below the normal freezing point, froze sitting out on my desk.

Falling Bullets and the Density of Air

October 19, 2009 by Mark Eichenlaub

Matt discusses a Mythbusters episode where they drop a bullet and shoot one horizontally from a gun, and see which falls first. I want to take this up where he left it off

They find that to within their experimental error, the two bullets hit the ground at the same time. It doesn’t matter how fast you move in the x-direction; gravity still gives you the same acceleration in the y-direction.

It turns out that with standard assumptions about air resistance, this is not true. What I hope to do here is, by assuming that bullets falling and shot horizontally fall at very nearly the same rate, put a bound on the density of air.

First we need a model for air resistance. If the bullet is a non-rotating sphere and air is non-viscous, then the bullet slows down because air smashes into it. We can imagine the bullet leaving a cylindrical trail behind it as it flies. Let’s say that as the bullet flies through its path, all the air it encounters gets brought from rest up to some set fraction x of the bullet’s speed, and that x does not depend on the bullet’s speed v.

Then in one incremental unit of time \Delta t, the bullet intersects a mass of air that is the density of air times the volume of its path over that time, and the volume of the path is given by the velocity of the bullet times is cross-sectional area. This is \rho \pi r^2 v dt grams of air, where \rho is the density of air, v is the speed of the bullet, and r is its radius.

The momentum imparted to the bullet is the same as that imparted to the air. The momentum imparted to the air is the mass of air encountered times the speed it’s being sped up to. This is (\rho \pi r^2 v dt) x*v = dp.

The force on the bullet is the time derivative of its momentum, so dp/dt = F = x \rho \pi r^2 v^2. For the time being, set k = x \rho \pi r^2 so the force on the bullet is just kv^2.

Now let’s make some simplifying assumptions. We imagine that air resistance plays a minor role, because otherwise the Mythbusters would not have found so close to a tie as they did. This means the bullets have very nearly the same y-height at any time, and that it’s the same as the normal, physics 1a parabola falling under gravity. Let the bullet fired out of the gun have the same horizontal velocity throughout its flight. Also assume the bullet goes much faster in the horizontal than in the vertical direction. Using these assumptions, we’ll try to find the relative acceleration of the two bullets, rather than finding both of their paths and then subtracting.

For the bullet falling straight down, all motion is in the y-direction, and the upward force is kv_y^2. For the bullet fired out of the gun, the total force is k(v_x^2 + v_y^2), but only part of this is in the y-direction, so the force in the y-direction is k v_y\sqrt{v_x^2 + v_y^2}. That can be simplified to k v_y*v_x because velocity in the y-direction is small.

Because air resistance plays a minor role, assume the bullets fall for almost the same amount of time, t_f = \sqrt{\frac{2g}{h}}. Over that time, there is a difference in the force on the bullets given by F_{shot} - F_{drop} = k(v_xv_y - v_y^2) = kv_y(v_x - v_y) = (kv_x)v_y. Notice that this is equivalent to assuming the dropped bullet doesn’t get slowed down at all. The shot bullet experiences so much more air resistance that we are essentially assuming only air resistance on the shot bullet matters.

The bullets are in free fall, so v_y = gt and the total distance between the two bullets when they fall comes from integrating their relative acceleration once for their relative velocity, and again for their relative displacement. When the mass of the ball is m, we have

v_{rel}(t) = \int_0^t accel_{rel}(t')dt' = \int_0^t kv_x v_y(t')/m dt' = \frac{k v_x}{m} \int_0^t gt' dt' = \frac{1}{2m}kv_x g t^2

y_{rel}(t) = \int_0^t v_{rel}(t') = \frac{1}{6m} k v_x g t^3.

Plugging in t_f, the falling time, for t gives the distance between the bullets as they land. Dividing by their speed at that time gt_f, gives the difference in their falling times. This comes to

\Delta t = \frac{x \rho \pi r^2 v_x h}{3mg}

To check, the dimensions work out to time. The difference increases with the density of air, with the radius of the ball (for fixed mass), with the horizontal firing speed, with the height dropped, and with the x factor of air resistance. It decreases with increasing gravity and mass of ball (for fixed radius). All that seems plausible enough.

We can solve this for \rho, the density of air. Since \pi = 3 I can cancel those.

\rho = \frac{\Delta t m g}{x r^2 v_x h} .

Let’s make some guesses. They fired the bullet from about 1m off the floor. It traveled some 300m down the hall in the 1/3 s it had to fall, so v_x = 1000m/s. g = 10 m/s^2, m = 100g and r = .01m. x = .5.

That gives

\rho = \Delta t \frac{20 kg}{s m^3} .

The Mythbusters think \Delta t < .1s (they said 39 milliseconds), which implies

\rho < \frac{2 kg}{ m^3} .

And it turns out the density of air is indeed less than this – about 1 kg/m^3.

I should admit, though, that the first time I crunched through these numbers, I got that the density of air should be less than 0.1 kg/m^3. That was plugging in 5g for the bullet weight, thinking of a 1cm^3 bullet about as dense as a typical rock. But the bullet I hypothetically used had radius 1cm, not diameter, and also bullets are made of metals that are pretty dense. I googled "bullet weight" and learned that google automatically interprets that to mean the weight of rugby player Tom James, nicknamed "The Bullet". (His weight was 15 st 8 lbs).

Incidentally, I think my calculation worked by accident. The aerodynamics of a bullet are probably much more complicated than I've assumed here, since the bullet spins and probably entrains air and makes eddy currents and does all kinds of ungodly things. v_x isn't constant, but according to my model it should drop by less than 50m/s in the time it takes the bullet to fall. I guessed kind of wildly on things like the height of the drop and the factor by which air is sped up when the ball passes through. Also, the experiment has error in terms of simultaneous dropping and how level the gun is. But things seem to have worked out to order of magnitude.