# Problem of the Week #3: Solution

Cold-Blooded Killer

Hello all and welcome to to another roundup of Problem of the Week. If the time intervals don't seem to be adding up, just remember that "week" is an illusion here at Virtuosi headquarters. "Problem of the Week," doubly so. But enough with the excuses, let's see if Mr. Bond lives to Die Another Day. The situation presented in the problem was one of a glass of water filled all the way to the top with a single ice cube in it. The goal is to see if any water falls to the floor as the ice cube melts. The simplest analysis would be to just consider the displacement of water by the ice cube. By Archimedes' principle we know that the buoyant force on the ice cube is equal to the weight of the water displaced by the ice cube. But since the ice cube is floating, we know that the buoyant force is equal to the weight of the ice cube. Therefore, the ice cube just displaces it's own weight in water. So as the ice cube melts, the water level stays the same and no water is spilled over. Stated a bit more explicitly, we have that the buoyant force is the weight of the displaced water: $$F_b = m_{water}g = {\rho}{water}Vg$$ where we have just expressed the mass of the water as the density of water times the volume of the water displaced. Likewise, the weight of the ice cube is $$W_{ice} = m_{ice}g = {\rho}{ice}Vg$$ But since the ice cube is floating, the buoyant force is equal to the weight. Setting the above equations equal to each other and rearranging a bit, we can solve for the volume of the water displaced by the ice cube: $$V_{water} = \frac{{\rho}{ice}V}{{\rho}{water}}$$ Great, now we have the volume of water that the ice cube displaced, so now we just need to see what volume of water the ice cube will add once it melts. Since we know that the mass of ice cube better be the same as the mass of the water left after it melts, we can write $${\rho}V_{ice} = {\rho}{water}V$$ which gives that the total volume of the water is just $$V_{water} = \frac{{\rho}{ice}V}{{\rho}_{water}}$$ which is exactly the same volume we found before. Therefore, when the ice cube melts it fills up exactly the same amount of volume as the water displaced to hold up the ice cube, so the water level remains the same and James Bond lives! Or does he? Well, he sure does if the system is entirely described by the analysis above. But we need to be exact here since a single drop would set off the Koala death trap. We have considered the biggest contributions to the change of volume of the water (and easiest to calculate!), but we have ignored a large number of much much smaller effects on the volume. Typically these effects would be swept aside and callously labeled "negligible," but since even the smallest change of volume could mean life or death for our hero, they must be accounted for to give a final "exact" answer. Unfortunately, "exact" isn't something science can do very well. We can only do as good as the measurements we make allow. And since I gave you practically no information, this was very tough. However, one could consider the magnitude of the effects given typical values and see which side wins out. But if two competing effects give similar values under reasonable conditions, it becomes very difficult to give an answer. The last two "Problems of the Week" were well posed math problems, each with a clear-cut "correct" answer. This one was a physics problem and physics is a lot messier than math. That may initially seem less satisfying, but it means that there is a whole lot of interesting "mess" to sort out! I really enjoyed reading the comments detailing all the neat small effects to the change of volume that I ignored in my solution. Density variations as a function of temperature, surface tension effects, evaporation and thermal expansion all play some role in determining the final answer. There's a whole lot of interesting physics going on! Without knowing the exact conditions a bit more and doing a whole lot more work, it seems as though I will not be able to provide a final physics solution. But, having seen just about every Bond movie, I can give an answer with a certainty not afforded to me by science: Mr. Bond lives. How do I know this? Well, it is common knowledge that James Bond has tiny smoke grenades in the soles of his shoes that release a potent koala knock-out potion specifically created by Q. So Bond releases these into the koala death pit, rendering the vicious beasts harmless. Then he uses his watch laser to cut any bindings Dr. Cherrycoke may have placed him in, falling into the death pit. The fall leaves Mr. Bond uninjured, as a cuddly mass of sleeping koalas break his fall. By stacking the koalas into a rudimentary series of steps, Mr. Bond ascends out of the death pit and free to save the world's ice supply from the evil Dr. Cherrycoke. Obviously.

# The Law and Large Numbers

Human beings are not equipped for dealing with large numbers. Honestly, 7 thousand, 7 million, 7 billion and 7 trillion all register about the same in my mind, namely 7 big. Unfortunately, there is a world a different between each of these, three whole orders of magnitude, a thousand, the difference between lifting me and a US quarter. This lack of respect for orders of magnitude has really been rearing its head recently with most of the political discussions surrounding the US budget. Turns out the US Budget is really large. In 2010 it weighed in at $3.55 trillion. Thats big. Really big. So big that I can't fathom it. Without getting too political, there has been a site going around recently; the You Cut program, which invites public suggestions for cuts to be made to the budget to try and fix the deficit. Now, personally, I believe we ought to do something about the deficit. To this end, I think it is useful to point out the scales involved. In particular, the link I gave above is to one of the suggested cuts: federal funding of NPR (Disclaimer alert: I love NPR), which weighs in at 7 million dollars. Seven million dollars is a lot of money. A lot of money, more than I can imagine having personally. But to suggest that a 7 million dollar cut is any sort of progress towards solving a$1.2 trillion dollar deficit is a little amusing. As a fraction, this comes out to $$\frac{ 7 \text{ million} }{ 3.55 \text{ trillion} } = 2 \times 10^{-6}$$ Two parts in a million. To give a sense of scale to this, the gravitational influence of the moon on my weight is: $$\frac{ \frac{ G M_{\text{moon}} }{ R_{\text{earth-moon}}^2 } }{ 10 \text{ m/s}^2 } = 3 \times 10^{-6}$$ Three parts in a million. So, suggesting that you have made real gains in reducing the US budget by cutting federal funding for NPR is as silly as suggesting that if I want to lose weight, my first concern should be the current tides. [I want to point out that I don't really mean to get too political, and that I've noticed both parties pulling these kinds of numbers tricks.] So, wanting to get a little better understanding of the numbers at stake, I collected some data (all from the 2010 budget). My goal is to attempt to represent how the US government spends its money. Before I begin I need to plug two great tools towards this end: Here the NYTimes graphically represents government spending, helping to give a sense of scale to different categories. Here the NYTimes lets you try and balance the budget, not only for next year but down the line, letting you choose from a wide array of proposed changes. The Data So, below is a list of some of the relevant numbers I could thing of, and some reference numbers (trillion, billion, million), as well as the US debt, and total deficit. In addition to just reporting the numbers (which you can find anywhere), in the second column I give the fraction of total spending in scientific units.

Name $Fraction US Debt 13.8 trillion 3.9 Total Spending 3.55 trillion 1.0 Budget Deficit 1.17 trillion 3.3E-1 1 trillion 1 trillion 2.8E-1 SS / Def / MM 730 billion 2.1E-1 Education 93 billion 2.6E-2 Taxes (250G+) 54 billion 1.5E-2 Science 31 billion 8.7E-3 1 billion 1 billion 2.8E-4 NPR 7 million 2.0E-6 1 million 1 million 2.8E-7 So here, SS / Def / MM means each of Social Security, Defense spending and Medicare and Medicaid, which each comes in at about the same per program. We spend 730 billion on each of these. To first glance, these three programs are how the US spends money, each of these coming in at 21%. Notice also just how large the deficit is, coming in at 33% of total spending. The rather cryptic "Taxes (250G+)" is how much money would be saved by letting the Bush tax cuts expire for those making more than 250,000 a year, a currently hotly debated topic. Notice that it would only ease the burden by 15%. So this alone would only cut the budget deficit in half. The above table shows the power of scientific notation. (Even though I had to use the ugly "E" notation). A number like "2.8E-4" is really 2.9 x 10^(-4). But honestly, even a list like this doesn't really make an impact for me. So, I thought of a couple other ways to represent the same numbers, scaling them to some 'big' things I can conceive of: #### Barry's Budget Now, this might not be a fair comparison, but let's scale down the US budget to sizes that a person can understand. By scaling by a factor of$100 million, we end up with the story of my friend Barry.

Name $Fraction Barry US Debt 13.8 trillion 3.9$138,495.50 Total Spending 3.55 trillion 1.0 $35,500.00 Budget Deficit 1.17 trillion 3.3E-1$11,690.0 1 trillion 1 trillion 2.8E-1 $10,000.00 SS / Def / MM 730 billion 2.1E-1$7,296.67 Education 93 billion 2.6E-2 $930.00 Taxes (250G+) 54 billion 1.5E-2$540.00 Science 31 billion 8.7E-3 $310.00 1 billion 1 billion 2.8E-4$10.00 NPR 7 million 2.0E-6 $0.07 1 million 1 million 2.8E-7$0.01

As you'll notice, Barry makes $23,810.00 a year (total receipts)[he's a grad student], but spends$35,500 a year. This has created his current debt problem. Barry has $138,495 in credit card debt, and still overspends by$11,690 a year. How does Barry spend his money? Well, every year Barry buys some $7,300 in guns, spends another$14,600 mostly taking care of his grandma. He pays a tuition of $930 a year at school, and$310 on science books. Every year he donates 7 cents to NPR. Every million US dollars is 1 penny in Barry dollars.

#### Work Week

That might not have been fair. This time, lets scale the US government spending to a work week of 40 hours. I think a work week is a large amount of time that I still have a real grasp for.

Name $Work Time Guess at time US Debt 13.8 trillion 156 hours Work month Total Spending 3.55 trillion 40 hours Work week Budget Deficit 1.17 trillion 13 hours All Mon and Tue morn 1 trillion 1 trillion 11.3 hours two days of good work SS / Def / MM 730 billion 8.2 hour true work in a day Education 93 billion 63 minutes lunch break Taxes (250G+) 54 billion 36 minutes cooler chat Science 31 billion 21 minutes Show on Hulu 1 billion 1 billion 40 seconds Stretching at desk NPR 7 million 0.3 seconds mouse click 1 million 1 million 0.04 seconds blink #### Cross Country Now lets scale to a distance. The longest drive I've ever done is from Los Angeles, CA to Orlando, Fl, which was 2511 miles. In terms of this scale, the budget breaks down as: Name$ Dist
US Debt 13.8 trillion 9796 mi NY - Sydney Total Spending 3.55 trillion 2511 mi LA - Orlando Budget Deficit 1.17 trillion 827 mi Texas 1 trillion 1 trillion 707 mi
SS / Def / MM 730 billion 516 mi Orlando - Mobile Education 93 billion 66 mi
Taxes (250G+) 54 billion 38 mi
Science 31 billion 22 mi Cross town 1 billion 1 billion 0.7 mi 14 blocks NPR 7 million 9 yards
1 million 1 million 4 feet

(So I just want to point out again, claiming that cutting NPR funding makes a dent in the US budget is similar to claiming you've moved closer to Orlando (while in LA) by crossing the room.)

### The Lesson

So, without getting too political... I'd just like to point out that if our politicians are serious about solving the budget crisis, they need to stop talking about million dollar programs, and start taking about 100 billion dollar ones. The problem is that it's hard to either slash funding for large programs like defense, or social security, and it's even harder raise taxes (really at all). But if we never consider those options, we're never going to get out of the rut. In my undergraduate physics lab, the instructor had a mantra: "A number without context is meaningless". Now, he originally meant the statement to be a lesson on how important it is to quote errors on your measurements, but I think I can adapt it to apply to giving out numbers like 7 billion without a sense of scale.

# Your Week in Seminars: Short Thanksgiving edition

Good Monday evening all, and welcome to another edition of Your Week in Seminars. Last week was a half week here in Cornell, but we still managed two talks, the general colloquium and the Wednseday-talk-on-Tuesday. Monday we had Charles Marcus of Harvard talk about Building Schrödinger's Chip. This was a quantum computing talk. We don't actually have a quantum computing group in Cornell, but I've taken a couple of courses on it back home, and it's an interesting subject, although I've been a bit disillusion by the notable lack of problems solvable by quantum computers. Marcus started by talking about the wonders and insanities of quantum mechanics - the usual spiel about the two slit experiment, electrons passing through walls, and Schrödinger's cat. He said he dubbed the talk Schrödinger's chip because unlike cats, that appear to break when we put them in the kind of low-temperature vacuum conditions we like to do quantum experiments in, chips keep working pretty well. Next he introduced the concept of entanglement, which is at the basis of the whole concept of quantum computing. Entangled particles are two or more particles that do not have a definite quantum state, but are definitely in the same quantum state. If I take them apart and measure their state - say, spin up or down - then I do not know the answer, but as soon one turns up, the other is up as well, and if one is down the other is immediately down as well. The experimental side of quantum computing is all about making little quantum boxes that contain a small number of states (like up or down) and then making it possible to entangle two of those boxes together. Add up enough boxes, under some criteria that were posited a decade ago but still not achieved, and you have a quantum computer. The majority of the talk was a survey of the various state of the art boxes and the methods used to make them. For those keeping track, 15 is still the largest number factored by a quantum computer. On Tuesday, I came in just in time for the second half of John Terning's talk on Monopoles, Anomalies, and Electroweak Symmetry Breaking. It's not really ideal to go into the second half of a talk in Newman 311, and this one actually sounded like I missed some interesting stuff. The part that I heard was about adding magnetic monopoles to the standard model., Those lovely magnetic equivalents of the electric point charge that we all heard about in our undergraduate E&M course turn out to be surprisingly hard to integrate into basic particle theories, which is perhaps for the best as we have not detected them so far. The gist of what I got from the second half of the talk was that it is not enough to add a magnetic counterpart to the electric part of the standard model, but in fact one needs a magnetic QCD and Weak force as well. Under some conditions, this kind of configuration can work, and predict magnetic monopoles with TeV-scale masses - the kind we might see in the LHC. Terning also talked about how these could be detected in the LHC. It turns out this isn't simple, because at TeV we would be producing monopole-anti-monopole pairs just barely, and so without a lot of kinetic energy they would tend to collapse back on themselves and annihilate, creating what is essentially an omnidirectional shower of photons. He mentioned that one of the big detectors at the LHC - the CMS - was equipped to detect these kind of photon bursts, and so this another prediction or possibility that we can look forward to seeing or not seeing soon. That's it for last week, kind of on the short side due to the holiday and so on. This week is our last normal one, as the semester ends, but I'll have a couple more particle talks the week after that, as well.

# Problem of the Week #2: Solution

Thanks to all who sent in solutions! We are very happy with the vast number of responses, and we will put up a leader board shortly! Solution Let A be the fraction of the rubber band that the ant has traversed. $$A\left(t\right)=\frac{x\left(t\right)}{L+vt}.$$ The ant's velocity relative to any point along the rubber band is equal to the length of the rubber band times the first time derivative of A: $$u=\left(L+vt\right)\frac{dA}{dt}=\left(L+vt\right)\left[\frac{\dot{x}}{L+vt}-\frac{vx}{\left(L+vt\right)^{2}}\right]=\dot{x}-\frac{vx}{L+vt}.$$ This gives us an inseparable, first-order differential equation for x(t). The general differential equation of this type, $$\dot{x}+p\left(t\right)x=q\left(t\right),$$ has the general solution $$x\left(t\right)=e^{-\int_{0}^{t} p\left(t\right)dt}\int_{0}^{t} q\left(t\right)e^{\int_{0}^{t} p\left(t\right)dt}dt.$$ In our case, $$p\left(t\right)=-\frac{v}{L+vt},\quad q\left(t\right)=u.$$ Therefore, $$x\left(t\right)&=&e^{\int\frac{v}{L+vt}dt}\int ue^{-\int\frac{v}{L+vt}dt}dt=\frac{u}{v}\left(L+vt\right)\ln\left(1+\frac{vt}{L}\right).$$ When the ant has reached the other side, x(t) = L + vt, so $$x\left(t\right)=L+vt=\frac{u}{v}\left(L+vt\right)\ln\left(1+\frac{vt}{L}\right).$$ Solving for t, we get $$t=\frac{L}{v}\left(e^{v/u}-1\right).$$ So the ant always makes it to the other side (unless u = 0)! In the limit as v -> 0, we get $$t\approx\frac{L}{u},$$ which is exactly what we would expect. We can show that the ant will reach the other side without doing any calculations. As the ant moves across the rubber band, the velocity of the point on the rubber band that the ant is currently walking on increases from 0 to v. Therefore, the ant is accelerating. Since the other end of the rubber band is moving at a constant velocity, the accelerating ant will always eventually catch up to the far end. Another way to see this is that since the ant is accelerating, if it makes it halfway, it will make it all the way to the other side; if it makes it one-quarter of the way, it will make it halfway, and so on. We know that since the ant is traveling at some nonzero velocity, it must make it some nonzero fraction of the way to the other side. Therefore, it must make it twice that far, and so on, all the way to the other side. Thanks to Frank for pointing this out!

# Your Week in Seminars: One for Two Edition

Hello everyone and welcome to another week of talks here at the physics department. I was out of Ithaca for a bit this past week, so in this very special edition I'm going to present a full week's worth of seminars (one from last week and two from the previous week) in one post covering two weeks. The Colloquium two weeks ago was given by our own Csaba Csaki, who talked about Electroweak Symmetry Breaking and the Physics of the TeV Scale. This was essentially an overview of beyond-the-standard-model physics and the kind of things we expect out of the LHC. Csaba started off by reminding us of the Standard Model, our very successful model of particle physics that's withstood nearly every test over the last thirty or forty years. The Standard Model is a set of three gauge theories - three forces that a relayed by massless particles - along with the theory of electroweak symmetry breaking that explains why one of these powers, the Weak one, is relayed by massive particles. This theory is well-backed by experiment, with the exception of the crucial Higgs boson, the one that gives mass to those previously massless W and Z, which we hope to see soon in the LHC. There are a few problems with the Standard Model, and the big one is the Hierarchy problem. Given what we know of symmetry breaking and how the W and Z get their masses, we expect elementary particles to have masses that correlate with the energy scale of the interaction that gives them this mass. Since that interaction is not one we see at low energies, we expect the elementary particles to be very massive. Since they are not, we conclude that there must be some symmetry that keeps them massless or nearly so. Some solutions to this was mentioned, beginning with current-favorite supersymmetry. This extra symmetry relation bosons to fermions and vice versa works well to solve the original problem, but creates a few of its own, like the Little Hierarchy Problem - if there's all this new physics at energies just a little higher than we've been exploring, why don't we see its effects on the low energy physics? In other words, why does the non-sypersymmetry Standard Model work so well? Csaba went on to mention some ways of solving these problems, such as burying the Higgs by allowing it to decay only in very specific ways. He also talked about a few more, like no-Higgs theories that accomplish electroweak symmetry breaking by different means, and extra-dimensional theories that allow us to give different energy scales to different forces. And the exciting thing about all of this is that we are likely to know a great deal of the answers soon, within the next few years, once the LHC starts giving data. On Wednesday after it we had Sven Krippendorf from Cambridge talk about Particle Physics from local D-brane models at toric singularities. This was a heavy string theory talk and I couldn't follow much of it. The question at hand was how to get the Standard Model, or parts of it, out of string theory models, and the gist of the talk revolved around toric singularities in the spacetime that the string theory lives in. No, I'm not entirely sure what makes a singularity toric. There were a lot of colorful graphs and some explanations. At the end there seemed to be some analogy made between different types of singularities on the manifold in string theory language and different gauge theories in the quantum field theory language, with a way to map them to each other. Possibly exciting, but you'd have to ask a proper string theorist about it. Then just this last Friday, we had Rachel Rosen from Stockholm University talk about Phase Transitions of Charged Scalars and White Dwarf Stars. This was a blackboard talk, which is always exciting and is usually more illustrative than Power Point ones. The subject was the thermodynamics of white dwarf stars - stars that are very dense and old, where fusion has mostly stopped and the only thing preventing the collapse of the star upon itself is the fermionic pressure of the electrons, that cannot fall into the same quantum states. The physical description of these stars is one of relatively free positive ions, specifically helium ions in this case, floating through a background of fermions. They are described by a quantum condensate, which has a good theory explaining it, but with the addition of Coulombic interaction between the ions. This state applies, specifically, to a subgroup of these stars that are mostly made of helium. This kind of ion condensate tends to crystallize depending on the ratio between the kinetic and potential Coulombic energy. Quantum effects, on the other hand, depend on some ratios of mass and charge between the ions. The only material where this applies turns to be helium, but luckily there are plenty of these helium-made white dwarfs. The derivation, as Rosen showed it, starts with a neutral Bose-Einstein condensate, which has a simple phase diagram - uncondensed above some critical temperature, and increasing condensation as the temperature is lowered to zero. The charged condensate introduces photons as it is usually done in field theory and follows the consequences. The result is a more complicated phase diagram. Under the old Tc, the ions still condense, but things change above it. There is now some higher temperature above which there is no condensation, but in between there are two solutions to the equations of motions, a condensate and a non-condensed state, and both a local energy minima. This means that the transition into a condensate is not continuous, and this is a first order phase transition. The nice thing here is that we have such white dwarf stars to observe and we can compare this theory to observations. That's it for these last two weeks. All you Americans out there have a good Thanksgiving, and I'll see you next week with two new seminars.

# Problem of the Week #3

We welcome you to send in solutions, or even any ideas you have about how to solve the problems tothe.physics.virtuosi@gmail.comwith “problem of the week” in the subject line. We will keep track of the top Virtuosi problem solvers. Welcome to the third installment of Problem of the Week! We are very pleased with the number of responses we have gotten so far and we super duper promise to put up some kind of leader board soon. In fact, I super duper promise to relegate that assignment to Alemi. We intend to keep this up as long as we can and give out prizes for high scores maybe...? They will be lame internet prizes...? The solution to the last problem of the week will be up shortly. Adam is the only one who knows the answer and he was busy all weekend taking a magic ring to Mordor. One more housekeeping note. Since we want everyone to have a clean shot at answering the question, we would prefer the solutions to be sent by email instead of posted in the comments. However, we certainly don't want to stop discussion on the problem, so if you don't want any hints you might want to avoid the comments! Now for the problem... ** James Bond has been captured by the evil mastermind Dr. Vontavious Cherrycoke, who is trying to take over the world's ice cube supply. The minions of Dr. Cherrycoke have placed Mr. Bond in an elaborate and unnecessary death contraption that, if triggered, would drop our hero into a pit of ravenous killer koalas. In other words, certain death! Dr. Cherrycoke has constructed a fitting trigger for his death machine / koala feeder. A single ice cube is placed in a glass of water so that the water is completely filling the glass up to the brim. The glass of water is then placed on a very sensitive detector. If even a single drop of water spills out of the glass as the ice cube melts, James Bond will find himself on the wrong end of a murderous marsupial mauling. Vontavious exits the room with his minions, confident that his death trap will be triggered and Bond will be vanquished once and for all. Does Mr. Bond survive? If not, roughly how long does it take until he checks in to the big MI6 in the sky? Be sure to back up your answer!

# Problem of the Week #2

As always, we welcome you to send in solutions, or even any ideas you have about how to solve the problems tothe.physics.virtuosi@gmail.comwith “problem of the week” in the subject line. We will keep track of the top Virtuosi problem solvers.Why did the ant cross the rubber band?** A rubber band is held fixed at one end. The other end is pulled at a velocity v. At time t = 0, the rubber band has a length of L, and an ant starts crawling from one end to the other at velocity u. Does the ant reach the other side? If so, how long does it take to get there? Assume that the rubber band is able to be stretched indefinitely without breaking.

# Pet Projects

You're doing it wrong!

Here at the Virtuosi, we have a very specific way of asking a very specific type of question that sounds anything but specific. These are the "How come [blank]?" questions [1]. These are very simple questions that just about every four year old asks, but likely never get sufficiently answered. To get a feel for what I mean by these questions I provide the following translations of problems we have either considered or will consider: Q: How come trees? Translation: How tall can trees be? Q: How come plants? Translation: Why are plants green? These are my very favorite types of questions because they are completely understandable by everyone and promise to have very interesting physics working behind the scenes. So I've been thrilled to see two such questions considered by scientists lately that have also had a good run in popular media. They are: Q: How come cats? Translation: How do cats drink? and Q: How come dogs? Translation: How do dogs shake? The question of how cats drink was answered recently by a few dudes from MIT, Virginia Tech and Princeton. One morning, one of the guys was just watching his cat drink water and realized he couldn't immediately figure out how it worked, so he decided to do a bit more research and BAM! science happens. It turns out that the cat isn't just scooping up the water with it tongue as one would probably have expected. Instead the cat uses its tongue to effectively punch the water, drawing up a column of fluid. They then bite this column and get a very itty bitty kitty mouthful of water. Here it is in a slow motion video I stole from the Washington Post who stole it from Reuters: (You'll probably have to sit through an ad, apologies) And here's a series of stills that illustrates the same thing: Not a cat person? Well there was also a study done a month or so ago about dog shaking. If you have a dog, you have almost certainly experienced the elegant way in which a wet dog un-wettafies itself. Well, a few students at Georgia Tech were also familiar with this shaking and just happened to have a super high speed camera that could track water droplets. I assume it only took a short period of time to put two and two together and BAM! science happens. In addition to providing the logical extension of the "spherical cow" joke to "dog of radius R", the study also found some fairly surprising results. If we are spinning a wet cylinder (i.e. dog), we would assume that the water is held on by some "sticking force." Then to shake off these droplets, we'd assume that the dog would have to shake fast enough that the centrifugal force would overcome the sticking force. In other equations, $$F_{centrifugal} = m{\omega}^2 R$$ and $$F_{sticking} = C$$ So we would then expect the shaking frequency to be the frequency that gets those two guys to equal each other. We would thus predict $${\omega} \sim R^{-0.5}$$ where the little squiggle just means that the frequency scales as the radius to the -0.5 power, with some constant multipliers out front that we don't know so we just ignore. But this is not what the Ga Tech guys observed! Check out the video below: So there's still something else going on that wasn't in our simple model. But this is what makes science so exciting! Even something dogs and cats figured out long ago can have some really rich and interesting physics. So keep on asking those questions! Notes: [1] Interestingly enough, the first "How come [blank]?" questions we asked did not have the now canonical form. Instead, it was stated as "Why are cows?". After much deliberation, we found that the solution is "Because milkshakes." Fair enough.

# Your Week in Seminars Dark Edition

I thought we could spice things up a bit with a more interactive post on The Virtuosi. Starting this week, a new problem of the week will be posted each week. Solutions will be posted the following week. These problems will be a collection of physics and math problems and riddles, and although hopefully challenging enough to be fun and interesting, they should mostly be solvable using concepts from introductory undergraduate physics and math classes. We welcome you to ponder these problems, and send in solutions, or even any ideas you have about how to solve the problems tothe.physics.virtuosi@gmail.comwith “problem of the week” in the subject line. We will keep track of the top Virtuosi problem solvers. Here it goes…Maximizing Gravity* You are given a blob of Play-Doh (with a fixed mass and uniform density) that you can shape however you choose. How can you shape it to maximize the gravitational force at a given point P* on the surface of the Play-Doh? Solution In cylindrical coordinates, where the point P is taken to be the origin, the z-component of the gravitational field due to any point (s, z) felt at the origin, is proportional to $$\frac{1}{s^{2}+z^{2}}\frac{z}{\sqrt{s^{2}+z^{2}}},$$ where the second factor is necessary to take the z-component. If we have azimuthal symmetry, the magnitude of the gravitational field is given by the sum of all the z-components of the forces due to all points in the planet. For a given contour $$\frac{z}{\left(s^{2}+z^{2}\right)^{3/2}}=C,$$ for some constant C, each point on the interior contributes more to the total gravitational field than each point on the exterior. Thus, the gravitational field is maximized if the surface of the planet takes the shape of one of these contours. Solving for s(z), we get$$s\left(z\right)=\sqrt{\left(\frac{z}{C}\right)^{2/3}-z^{2}}.$$ If the length of the planet in the z-direction is $$z_0,$$ we can replace C in the above expression to get $$s\left(z\right)=\sqrt{\left(z_{0}^{4}z^{2}\right)^{1/3}-z^{2}}.$$ As can be seen by the plot above, this shape looks a lot like a sphere, but slightly smushed toward the point of interest P. We can rigorously compare the field of the maximal gravity solid to that of a sphere with the same volume. The volume of the maximal solid is given by $$V=\pi\int_{0}^{z_{0}}s^{2}\left(z\right)dz=\pi\int_{0}^{z_{0}}\left[\left(z_{0}^{4}z^{2}\right)^{1/3}-z^{2}\right]dz=\frac{4\pi}{15}z_{0}^{3}$$ The volume of a sphere of radius r is of course $$V=\frac{4}{3}\pi r^{3},$$ so in order for the volumes to be the same, the sphere must have a radius of $$r=z_{0}/5^{1/3}.$$ The acceleration due to the maximal solid is proportional to $$a_{max}=\int dVa_{z}=2\pi G\rho\int_{0}^{z_{0}}dz\int_{0}^{s\left(z\right)}sds\frac{z}{\left(s^{2}+z^{2}\right)^{3/2}}=\frac{4\pi G\rho z_{0}}{5},$$ while the acceleration due to the sphere is just $$a_{sphere}=\frac{G\rho\frac{4}{3}\pi r^{3}}{r^{2}}=\frac{4\pi G\rho z_{0}}{3\cdot5^{1/3}}.$$ Thus, we have $$\frac{F_{max}}{F_{sphere}}=\frac{4\pi G\rho z_{0}/5}{4\pi G\rho z_{0}/\left(3\cdot5^{1/3}\right)}=\frac{3}{5^{2/3}}\approx1.026.$$ So the solid that gives the maximum gravitational field at a point is only about 3% better than a sphere. For a more detailed discussion, see Alemi's solution.