Top 10 Martin Gardner Physics Stumpers
November 27, 2014 7:09 AM Subscribe
The list to follow is offered purely in a spirit of fun and education, and is not intended to be definitive. It concerns only the most basic physics concepts, and nothing electronic.
No answers are offered.
Remember the wise words of Bob Crease in Physics World (Oct 2014):
Remember the wise words of Bob Crease in Physics World (Oct 2014):
"Googling is not the Gardner way. The Gardner way is to ignite your fascination
so that you experience the pleasure of finding the answer yourself."
Oh good, something to fight about over thanksgiving dinner, I needed that.
posted by backseatpilot at 7:20 AM on November 27, 2014 [6 favorites]
posted by backseatpilot at 7:20 AM on November 27, 2014 [6 favorites]
Poor pigeons.
posted by Flashman at 7:23 AM on November 27, 2014 [3 favorites]
posted by Flashman at 7:23 AM on November 27, 2014 [3 favorites]
That's typical of physics. In mathematics, we give them cozy little pigeonholes to nest in, all safe and sound.
posted by Wolfdog at 7:24 AM on November 27, 2014 [7 favorites]
posted by Wolfdog at 7:24 AM on November 27, 2014 [7 favorites]
Yay, I love this kind of thing. Now I'm going to sit back and enjoy reading through all the explanations of how I drew the most completely, backwardly wrong conclusion to each question. Don't let me down, MetaFilter.
posted by heyho at 7:25 AM on November 27, 2014 [2 favorites]
posted by heyho at 7:25 AM on November 27, 2014 [2 favorites]
Better alert the moderators now. Things are liable to get ugly in here.
posted by tommyD at 7:26 AM on November 27, 2014 [2 favorites]
posted by tommyD at 7:26 AM on November 27, 2014 [2 favorites]
Okay, here are my answers to some of them.
SPOILERS (or maybe not)
Why does a mirror appear to reflect left right and not up down? A mirror will reflect up/down if you lie on your side.
A piece of solid iron in the form of a doughnut is heated. Will the diameter of its hole get larger or smaller? Trick question. The iron doughnut hole is sold separately.
Two missiles speed directly toward each other, one at 9,000 miles per hour and the other at 21,000 miles per hour. They start 1,317 miles apart. Without using pencil and paper, calculate how far apart they are one minute before they collide.
250 miles. (21000/60) + (9000/60) / 2. (I did this with a computer, no paper)
A cube of ice floats in a beaker of water, the entire system at 0 degrees centigrade. Just enough heat is supplied to melt the cube without altering the system's temperature. Does the water level in the beaker rise, fall, or stay the same?
A small boy is sailing a plastic boat in the bathtub. It is loaded with nuts and bolts. If he dumps all this cargo into the water, allowing the boat to float empty, will the water level in the tub rise or fall? Rise.
Assuming that the truck's compartment is airtight, can anything be said for the driver's line of reasoning? The airtight compartment means all the pigeons are dead.
posted by dances_with_sneetches at 7:35 AM on November 27, 2014 [4 favorites]
SPOILERS (or maybe not)
Why does a mirror appear to reflect left right and not up down? A mirror will reflect up/down if you lie on your side.
A piece of solid iron in the form of a doughnut is heated. Will the diameter of its hole get larger or smaller? Trick question. The iron doughnut hole is sold separately.
Two missiles speed directly toward each other, one at 9,000 miles per hour and the other at 21,000 miles per hour. They start 1,317 miles apart. Without using pencil and paper, calculate how far apart they are one minute before they collide.
250 miles. (21000/60) + (9000/60) / 2. (I did this with a computer, no paper)
A cube of ice floats in a beaker of water, the entire system at 0 degrees centigrade. Just enough heat is supplied to melt the cube without altering the system's temperature. Does the water level in the beaker rise, fall, or stay the same?
A small boy is sailing a plastic boat in the bathtub. It is loaded with nuts and bolts. If he dumps all this cargo into the water, allowing the boat to float empty, will the water level in the tub rise or fall? Rise.
Assuming that the truck's compartment is airtight, can anything be said for the driver's line of reasoning? The airtight compartment means all the pigeons are dead.
posted by dances_with_sneetches at 7:35 AM on November 27, 2014 [4 favorites]
I love the questions, but I can do all the thought experiments I want and still be wrong (indeed I'm pretty sure I am.) What would amuse me would be to read the actual answers with explanations of why.
posted by cccorlew at 7:47 AM on November 27, 2014
posted by cccorlew at 7:47 AM on November 27, 2014
dances_with_sneeches, why are you dividing by two in the missiles question? Unless I'm misreading the question, you shouldn't. Maybe you're dividing by two because there are two missiles?
Check your answer against what would happen if it were correct: The 21000 mph missile by itself would cover that 250 miles in less than a minute.
posted by Flunkie at 7:53 AM on November 27, 2014 [1 favorite]
Check your answer against what would happen if it were correct: The 21000 mph missile by itself would cover that 250 miles in less than a minute.
posted by Flunkie at 7:53 AM on November 27, 2014 [1 favorite]
Two missiles speed directly toward each other, one at 9,000 miles per hour and the other at 21,000 les per hour.
Assuming "21,000 les per hours" is a typo for "lis per hour" rather than "miles per hour," I think this question is trying to normalize the idea of a Sino-Western missile-based conflict, which would only be slightly worse than a missal-based conflict.
posted by GenjiandProust at 7:55 AM on November 27, 2014 [7 favorites]
Assuming "21,000 les per hours" is a typo for "lis per hour" rather than "miles per hour," I think this question is trying to normalize the idea of a Sino-Western missile-based conflict, which would only be slightly worse than a missal-based conflict.
posted by GenjiandProust at 7:55 AM on November 27, 2014 [7 favorites]
Two missiles speed directly toward each other, one at 9,000 miles per hour and the other at 21,000 miles per hour. They start 1,317 miles apart. Without using pencil and paper, calculate how far apart they are one minute before they collide.
250 miles. (21000/60) + (9000/60) / 2. (I did this with a computer, no paper)
Don't divide by two. The two combined travel at 30,000 miles per hour, therefore they are (30000/60) miles apart one minute before collision.
posted by Etrigan at 7:56 AM on November 27, 2014 [2 favorites]
250 miles. (21000/60) + (9000/60) / 2. (I did this with a computer, no paper)
Don't divide by two. The two combined travel at 30,000 miles per hour, therefore they are (30000/60) miles apart one minute before collision.
posted by Etrigan at 7:56 AM on November 27, 2014 [2 favorites]
Woah, wait, I didn't look at your other answers; the missiles one just stood out to me upon a glance.
posted by Flunkie at 7:56 AM on November 27, 2014 [1 favorite]
posted by Flunkie at 7:56 AM on November 27, 2014 [1 favorite]
In the instance of the nuts and bolts, the water level would fall if the cargo was dumped.
When the cargo is in the boat, floating on the surface of the water, they're displacing a volume of water that is equivalent to their mass. When they go in the water, they'll sink and only displace a volume of water equivalent to their size.
posted by The Notorious SRD at 7:57 AM on November 27, 2014 [3 favorites]
When the cargo is in the boat, floating on the surface of the water, they're displacing a volume of water that is equivalent to their mass. When they go in the water, they'll sink and only displace a volume of water equivalent to their size.
posted by The Notorious SRD at 7:57 AM on November 27, 2014 [3 favorites]
I divided by two because I was doing it wrong.
posted by dances_with_sneetches at 8:01 AM on November 27, 2014 [10 favorites]
posted by dances_with_sneetches at 8:01 AM on November 27, 2014 [10 favorites]
When the cargo is in the boat, floating on the surface of the water, they're displacing a volume of water that is equivalent to their mass. When they go in the water, they'll sink and only displace a volume of water equivalent to their size.
Isn't this assuming that the bathtub is not moving near the speed of light?
posted by GenjiandProust at 8:01 AM on November 27, 2014 [6 favorites]
Isn't this assuming that the bathtub is not moving near the speed of light?
posted by GenjiandProust at 8:01 AM on November 27, 2014 [6 favorites]
Isn't this assuming that the bathtub is not moving near the speed of light?
posted by GenjiandProust
Get out of my tub, McConaughey.
posted by The Notorious SRD at 8:03 AM on November 27, 2014 [3 favorites]
posted by GenjiandProust
Get out of my tub, McConaughey.
posted by The Notorious SRD at 8:03 AM on November 27, 2014 [3 favorites]
If the bathtub were moving near the speed of light, all the water would splash out.
posted by tommyD at 8:04 AM on November 27, 2014 [1 favorite]
posted by tommyD at 8:04 AM on November 27, 2014 [1 favorite]
Get out of my tub, McConaughey.
You know what I love about black holes? High school girls get older, I stay the same age. Alright alright alright.
posted by officer_fred at 8:09 AM on November 27, 2014 [29 favorites]
You know what I love about black holes? High school girls get older, I stay the same age. Alright alright alright.
posted by officer_fred at 8:09 AM on November 27, 2014 [29 favorites]
Zeno proved the missiles never collide. This is why real scientists know that Missile Defense is an expensive unworkable boondoggle.
posted by humanfont at 8:11 AM on November 27, 2014 [3 favorites]
posted by humanfont at 8:11 AM on November 27, 2014 [3 favorites]
1: it doesn't really reflect left and right, you just expect to face people with their right on your left, and their left on your right. It just reflects everything back directly.
2: If it increases uniformly in volume, smaller, I think.
3: 500
5: rises
6: should stay the same.
7: no because equal and opposite. They exert as much force downward as they would if they were sitting.
8: it goes forward. Acceleration is the same as gravity, heavy air moves back, helium goes forward.
9: If I remember the math properly, all the gravity inside a perfect sphere cancels out, so it would float freely.
10: too many numbers, my head hurts.
posted by empath at 8:12 AM on November 27, 2014
2: If it increases uniformly in volume, smaller, I think.
3: 500
5: rises
6: should stay the same.
7: no because equal and opposite. They exert as much force downward as they would if they were sitting.
8: it goes forward. Acceleration is the same as gravity, heavy air moves back, helium goes forward.
9: If I remember the math properly, all the gravity inside a perfect sphere cancels out, so it would float freely.
10: too many numbers, my head hurts.
posted by empath at 8:12 AM on November 27, 2014
Thanks to Smarter Every Day for answering #8 for me.
posted by obfuscation at 8:15 AM on November 27, 2014
posted by obfuscation at 8:15 AM on November 27, 2014
For the can of soda, assume the can has height 1, and the soda in it has depth y (where y is between 0 and 1).
The can by itself has mass m1 = 1.5 ounces, and its center of mass is at height h1 = 1/2. The soda by itself has mass m2 = 12*y ounces, and its center of mass is at height h2 = y/2.
Therefore the center of mass of the system is at height
H = (m1*h1 + m2*h2) / (m1 + m2) = (.75 + 6*y2) / (1.5 + 12*y).
Take the derivative with respect to y and set that equal to 0, to see that the center of mass is lowest when y = 1/4. Using calculus seems like cheating, though!
posted by zeptoweasel at 8:16 AM on November 27, 2014
The can by itself has mass m1 = 1.5 ounces, and its center of mass is at height h1 = 1/2. The soda by itself has mass m2 = 12*y ounces, and its center of mass is at height h2 = y/2.
Therefore the center of mass of the system is at height
H = (m1*h1 + m2*h2) / (m1 + m2) = (.75 + 6*y2) / (1.5 + 12*y).
Take the derivative with respect to y and set that equal to 0, to see that the center of mass is lowest when y = 1/4. Using calculus seems like cheating, though!
posted by zeptoweasel at 8:16 AM on November 27, 2014
SPOILERS
Mirror: It doesn't flip left/right, it flips in/out! The reason it looks that way is that if you are looking at a person, you're right eye would be on the same side of the midline as their left eye (they're rotated 180 degrees, compared to you.) But in the mirror image, the images right eye is on the same side as your right eye. The easiest way to see this is to hold an arrow and look in the mirror. When you point the arrow right, the image will be pointing right. When you point the arrow up, the image will be pointing up. But when you point the arrow directly at the mirror, the image will be pointing right back at you. You perceive it as a left/right flip because your mind thinks that the object is rotated 180 degrees, and the image isn't.
Iron Torus: The iron torus' hole will get larger. Railroad wheels have steel "tires" on them, which are, for all intents, iron-carbon alloy toruses. They are removed and installed by heating the tires, which expand.
Missiles: Math, go do it. (Hint: The answer requires knowing the speed per minute, which means dividing miles per hour by 60, which is hard, or by 10, then 3, then 2, which is easy.)
Melting ice cube -- water level stays the same (assuming the atmosphere is at 100% humidity, otherwise, it drops a tiny bit as water evaporates.) This is easy proven -- fill a glass with icewater, mark the level, and let the ice melt. The ice floats because it is less dense, but the total amount of H2O is the same. A floating object displaces it's weight in water, but as it melts, it displaces less, which would make the level fall, but as it melts, it happens to replace the missing displacement with water, which means it stays just the same. Funny enough, displacement is going to show up a couple of more times here.
Bathtub boat: The water level falls. A floating object displaces its weight in water, a sunken object displaces its volume in water. Since the nuts and bolts sunk, they are more dense than water, which means they would displace more water while in the boat, since they weigh more than an equivalent amount of water, but will only displace their volume of water when completely underwater.
Helium balloon: Goes forward. Helium is less dense than air. When you accelerate, the air wants to stay behind (inertia) and in effect moves backwards a bit. (This is also why the car rocks backwards, lowering the rear suspensions and raising the front suspension when you hit the gas pedal.) Since helium is less dense than air, it's displace by the air and goes forward.
Pigeons. Assuming the truck is sealed, when they're flying, they're forcing air down, that force hits the truck. So, mass on the truck remains the same. If you open the top, then things change. Usually, you see this posed as birds in an airliner, which you can assumed is (mostly) sealed from the outside air.
Hollow moon: Inside such a sphere (assuming it's a spherical shell of uniform density) the gravity will be 0g throughout from the moon mass. However, it's a moon, which means there's a planet nearby. Thus, you'll drift to the point inside the shell closest to the planet. This force can be surprisingly strong -- the moon's gravity, after all, is strong enough to cause the tides on Earth.
Soda Can: This...this I don't know. I know the trick to finding the point (tilt the can so that the top edge of the closed end is the same as the bottom edge of the open end, fill with water. Hmm. That means it's half full. Well, hell, that's easy. 6 oz. of soda, or with the can weighing 1.5oz, you need a total weight of 7.5 oz.
posted by eriko at 8:20 AM on November 27, 2014 [6 favorites]
How many people's lifes here were changed by Gardner? In sixth grade I discovered Gardner in my friend's dad's library.
We were already kind of antisocial, a perfect afternoon consisted on taking turns to use his Amiga, while the other one read Gardner books or Mad and Scientific American magazines (remember when those were still good). There were also some Heavy Metal magazines, but those were for private reading.
I can trace a very straight line between discovering Gardner that rainy afternoon in Mexico and becoming a Google engineer (before Googler became a slur).
I wish Gardner was still alive and writing his columns. I am sure he would come up with games and riddles that would be more interesting with the use of Google and everything else www.
Donut hole becomes larger (imagine that you heat a disk of iron, then punch out the hole, then watch it contract as it cools), balloon in car moves in the same direction as car (the air is heavier than the balloon, when the car accelerates all the heave air rushes to the back of the car, and the balloon 'floats' to the front), water level lowers, truck driver is wasting his time.
posted by Doroteo Arango II at 8:24 AM on November 27, 2014 [2 favorites]
We were already kind of antisocial, a perfect afternoon consisted on taking turns to use his Amiga, while the other one read Gardner books or Mad and Scientific American magazines (remember when those were still good). There were also some Heavy Metal magazines, but those were for private reading.
I can trace a very straight line between discovering Gardner that rainy afternoon in Mexico and becoming a Google engineer (before Googler became a slur).
I wish Gardner was still alive and writing his columns. I am sure he would come up with games and riddles that would be more interesting with the use of Google and everything else www.
Donut hole becomes larger (imagine that you heat a disk of iron, then punch out the hole, then watch it contract as it cools), balloon in car moves in the same direction as car (the air is heavier than the balloon, when the car accelerates all the heave air rushes to the back of the car, and the balloon 'floats' to the front), water level lowers, truck driver is wasting his time.
posted by Doroteo Arango II at 8:24 AM on November 27, 2014 [2 favorites]
If the truck is airtight, all the pigeons are dead.
posted by 445supermag at 8:27 AM on November 27, 2014 [3 favorites]
posted by 445supermag at 8:27 AM on November 27, 2014 [3 favorites]
Isn't there more of the moon orthogonal to the section of it you're closest to, therefore pulling you towards the center of the sphere (ignoring any nearby planets)? Where is the force coming from to cancel out the greater mass of the moon far away from you?
posted by cthuljew at 8:28 AM on November 27, 2014
posted by cthuljew at 8:28 AM on November 27, 2014
The clue to most of them is in what information is not given. For instance, he doesn't specify the thickness of the moon's shell.
posted by 445supermag at 8:32 AM on November 27, 2014
posted by 445supermag at 8:32 AM on November 27, 2014
The hollow moon problem isn't hard. It is super easy. Its just that we are taught gravity in the wrong way, using Newton's laws instead of Gauss.
I can trace a very straight line between discovering Gardner that rainy afternoon in Mexico and becoming a Google engineer (before Googler became a slur).
I can also draw a straight line between discovering Gardner in elementary school and majoring in physics in college a decade later.
posted by vacapinta at 8:32 AM on November 27, 2014
I can trace a very straight line between discovering Gardner that rainy afternoon in Mexico and becoming a Google engineer (before Googler became a slur).
I can also draw a straight line between discovering Gardner in elementary school and majoring in physics in college a decade later.
posted by vacapinta at 8:32 AM on November 27, 2014
Since the force of gravity is dependent on distance, an object inside the hollow moon near the shell would fall toward the nearest part of the shell, would it not? Only if it were in the exact center of the sphere would it stay there, and a nudge in any direction would cause it to fall in that direction toward the shell.
posted by tommyD at 8:34 AM on November 27, 2014
posted by tommyD at 8:34 AM on November 27, 2014
Gardner of course influenced so many people, many of them now well-known mathematicians and physicists, that there is Gathering for Gardner, "a conference, a foundation, and a community of people who have a strong personal connection to Martin Gardner."
posted by vacapinta at 8:38 AM on November 27, 2014
posted by vacapinta at 8:38 AM on November 27, 2014
We solved the soda can problem a long time ago on askme .
posted by mbd1mbd1 at 8:39 AM on November 27, 2014 [1 favorite]
posted by mbd1mbd1 at 8:39 AM on November 27, 2014 [1 favorite]
The magnet testing one is easy: the magnetic field on the bar magnet is stronger at the ends than the middle. Hold the end of the magnet to the middle of the non-magnet and it sticks. Other way round and the forces are smaller.
posted by YAMWAK at 8:39 AM on November 27, 2014 [4 favorites]
posted by YAMWAK at 8:39 AM on November 27, 2014 [4 favorites]
The easiest way to see this is to hold an arrow and look in the mirror.
IIRC, in the Gardner book I had as a kid, someone reports testing this by putting a bag on his foot and standing on his head, with the mirror on the floor. After seeing his own face in the mirror, he immediately abandoned all further experimentation.
posted by effbot at 9:05 AM on November 27, 2014
IIRC, in the Gardner book I had as a kid, someone reports testing this by putting a bag on his foot and standing on his head, with the mirror on the floor. After seeing his own face in the mirror, he immediately abandoned all further experimentation.
posted by effbot at 9:05 AM on November 27, 2014
The pigeon one doesn't have a concrete answer. If the pigeons took off just as the truck started over the bridge, then followed a ballistic trajectory without flapping, before hitting the floor again as the truck reached the other side of the bridge, then they would have added nothing to the weight of the truck.
posted by CaseyB at 9:11 AM on November 27, 2014 [1 favorite]
posted by CaseyB at 9:11 AM on November 27, 2014 [1 favorite]
The simplest answer to the mirror question is that mirrors don't swap left and right, they swap front and back.
(which is what eriko already said, I guess.)
Eriko's answer to the soda can question can't be right, though, because it doesn't depend on the density of the soda. If the soda were replaced with, say, neutron-star material, the tiniest amount of it would plummet the center of gravity near the bottom of the can.
CaseyB: That would have to be a really short bridge or a really tall truck.
posted by narain at 9:14 AM on November 27, 2014
(which is what eriko already said, I guess.)
Eriko's answer to the soda can question can't be right, though, because it doesn't depend on the density of the soda. If the soda were replaced with, say, neutron-star material, the tiniest amount of it would plummet the center of gravity near the bottom of the can.
CaseyB: That would have to be a really short bridge or a really tall truck.
posted by narain at 9:14 AM on November 27, 2014
I love the mirror one. Thinking about mirrors and left and right is always a fun pasttime. Of course a mirror flips up and down. If the mirror is on the ceiling. Why you have a mirror on the ceiling is your business. A mirror on the wall flips front and back.
So mirrors flip left and right, regardless of how the mirror is oriented. I mean really mean a mirror flips "left-handed" and "right-handed," which are distinct from "the left side of the room" and "the right side of the room." So both eriko and I can both be right as to whether a mirror flips left and right, it just depends what concept you really mean by those word. The direction the arrow points is an absolute direction (maybe call it east or west, rather than left or right), handedness is a relative direction. (Also, if you hold a sign pointing to your left and look into a mirror, the image sees it point to their right. Both arrows of course point to the same absolute direction.)
You can see that a mirror flips handedness because if you hold up your left hand, your reflection holds up their hand, but you cannot rotate the mirror image hand into your left hand, because the relative orientation of the fingers and thumbs are switched. So your image has lifted their right hand. This would be true regardless of whether the mirror is in front of you, on the ceiling, or to one side.
Mirrors invert one spatial direction (whichever one is perpendicular to the mirror surface). And inverting one spatial direction is equivalent to inverting any odd number of directions, but inequivalent to switching two directions. Handedness is the relative orientation of two directions (which way "east" is relative to "north," for example), so a mirror flips handedness, regardless of how the mirror is oriented.
That does mean that if you look at a reflection from two mirrors, you get the "real" image back: that double reflection's left hand is equivalent to your left hand: you can image rotating that image in space and having it line up with the original. So if you have two mirrors without edges, put them together like a book at 90 degrees to each other, and look directly in the center. You'll see your true image, how other people see you. Not the one you see in the mirror. It'll look slightly off compared to what you usually recognize.
The reason I like thinking about mirrors is that mirrors, by flipping handedness, also flip spin. So particles spinning one way (let's call it "left-handed") become particles spinning the other way ("right-handed") in the mirror. Now, strangely, left and right handed particles are not the same. A left-spinning electron feels different forces than a right-handed one. It turns out that if you take all the particles, turn them into their antiparticles, and look at them in a mirror, the laws of Nature are nearly the same as in the un-mirrored, anti-antimattered version. But not quite. However, the laws of Nature for an antimatter mirror running backwards in time are identical to ours. (Ignore the 2nd law of thermodynamics here)
Also, while I'm discussing deep facts about the Universe, it is very important for you to realize that jalapeno cranberry sauce is no joke the most important part of your Thanksgiving meal. Seriously people, make some right now.
posted by physicsmatt at 9:19 AM on November 27, 2014 [7 favorites]
So mirrors flip left and right, regardless of how the mirror is oriented. I mean really mean a mirror flips "left-handed" and "right-handed," which are distinct from "the left side of the room" and "the right side of the room." So both eriko and I can both be right as to whether a mirror flips left and right, it just depends what concept you really mean by those word. The direction the arrow points is an absolute direction (maybe call it east or west, rather than left or right), handedness is a relative direction. (Also, if you hold a sign pointing to your left and look into a mirror, the image sees it point to their right. Both arrows of course point to the same absolute direction.)
You can see that a mirror flips handedness because if you hold up your left hand, your reflection holds up their hand, but you cannot rotate the mirror image hand into your left hand, because the relative orientation of the fingers and thumbs are switched. So your image has lifted their right hand. This would be true regardless of whether the mirror is in front of you, on the ceiling, or to one side.
Mirrors invert one spatial direction (whichever one is perpendicular to the mirror surface). And inverting one spatial direction is equivalent to inverting any odd number of directions, but inequivalent to switching two directions. Handedness is the relative orientation of two directions (which way "east" is relative to "north," for example), so a mirror flips handedness, regardless of how the mirror is oriented.
That does mean that if you look at a reflection from two mirrors, you get the "real" image back: that double reflection's left hand is equivalent to your left hand: you can image rotating that image in space and having it line up with the original. So if you have two mirrors without edges, put them together like a book at 90 degrees to each other, and look directly in the center. You'll see your true image, how other people see you. Not the one you see in the mirror. It'll look slightly off compared to what you usually recognize.
The reason I like thinking about mirrors is that mirrors, by flipping handedness, also flip spin. So particles spinning one way (let's call it "left-handed") become particles spinning the other way ("right-handed") in the mirror. Now, strangely, left and right handed particles are not the same. A left-spinning electron feels different forces than a right-handed one. It turns out that if you take all the particles, turn them into their antiparticles, and look at them in a mirror, the laws of Nature are nearly the same as in the un-mirrored, anti-antimattered version. But not quite. However, the laws of Nature for an antimatter mirror running backwards in time are identical to ours. (Ignore the 2nd law of thermodynamics here)
Also, while I'm discussing deep facts about the Universe, it is very important for you to realize that jalapeno cranberry sauce is no joke the most important part of your Thanksgiving meal. Seriously people, make some right now.
posted by physicsmatt at 9:19 AM on November 27, 2014 [7 favorites]
it is very important for you to realize that jalapeno cranberry sauce is no joke the most important part of your Thanksgiving meal. Seriously people, make some right now.
posted by physicsmatt at 9:19 AM
eponyhysterical.
posted by chavenet at 9:59 AM on November 27, 2014
posted by physicsmatt at 9:19 AM
eponyhysterical.
posted by chavenet at 9:59 AM on November 27, 2014
When explaining the mirror question to people, it's a good idea to use "-wards" words, i.e.
upwards / downwards,
backwards / forwards, and
leftwards / rightwards.
Then when some says a mirror switches left and right, ask them whether they mean "leftwards" or "left-oriented". English is lazy and uses "left" for both concepts, but one is a direction and the other is an orientation. Then, like physicsmatt says, mirrors flip one of the three direction pairs, leaves two alone, and always changes the orientation.
posted by benito.strauss at 10:04 AM on November 27, 2014 [2 favorites]
upwards / downwards,
backwards / forwards, and
leftwards / rightwards.
Then when some says a mirror switches left and right, ask them whether they mean "leftwards" or "left-oriented". English is lazy and uses "left" for both concepts, but one is a direction and the other is an orientation. Then, like physicsmatt says, mirrors flip one of the three direction pairs, leaves two alone, and always changes the orientation.
posted by benito.strauss at 10:04 AM on November 27, 2014 [2 favorites]
I want to know if the pigeons are spherical.
The easiest way to think about the expanding donut is to consider that heat expansion is in proportion to the amount of metal being heated. The inside half of the ring has the least metal and will gain the least volume; the outside has the most and will be dragging the inside of the torus with it as it expands. Overall the outside is going to win and the hole will be larger.
posted by Tell Me No Lies at 10:09 AM on November 27, 2014
The easiest way to think about the expanding donut is to consider that heat expansion is in proportion to the amount of metal being heated. The inside half of the ring has the least metal and will gain the least volume; the outside has the most and will be dragging the inside of the torus with it as it expands. Overall the outside is going to win and the hole will be larger.
posted by Tell Me No Lies at 10:09 AM on November 27, 2014
Iron Torus: Haven't you ever put a jar under hot water when the lid is stuck? Practical physics.
posted by jeather at 10:14 AM on November 27, 2014
posted by jeather at 10:14 AM on November 27, 2014
> And inverting one spatial direction is equivalent to inverting any odd number of directions, but inequivalent to switching two directions.
Maybe it's math vs. physics, but you might not want to say "switching". To me, if you start with an ordered triple of vectors (a,b,c),
"Inverting" is
(a, b, c) --> (-a, b, c)
and "switching two" is
(a, b, c) --> (b, a, c)
both of which are orientation reversing.
In fact, "switching two" is what happens if the plane of your mirror is between two axes instead of along one. For instance, if your mirror is the plane {x = y} in 3-space, then it leaves z^ alone and switches x^ and y^.
I'm guessing you just thought "I'm using the word 'invert' an awful lot. Maybe use a different word to make the writing more interesting."
posted by benito.strauss at 10:14 AM on November 27, 2014
Maybe it's math vs. physics, but you might not want to say "switching". To me, if you start with an ordered triple of vectors (a,b,c),
"Inverting" is
(a, b, c) --> (-a, b, c)
and "switching two" is
(a, b, c) --> (b, a, c)
both of which are orientation reversing.
In fact, "switching two" is what happens if the plane of your mirror is between two axes instead of along one. For instance, if your mirror is the plane {x = y} in 3-space, then it leaves z^ alone and switches x^ and y^.
I'm guessing you just thought "I'm using the word 'invert' an awful lot. Maybe use a different word to make the writing more interesting."
posted by benito.strauss at 10:14 AM on November 27, 2014
From Richard Feynman's "Fun to Imagine" interview videos:
Mirrors
Then, "A harder one...What keeps a train on the track?" Trains
posted by jjj606 at 10:35 AM on November 27, 2014 [3 favorites]
Mirrors
Then, "A harder one...What keeps a train on the track?" Trains
posted by jjj606 at 10:35 AM on November 27, 2014 [3 favorites]
So is the answer to the pigeon one, "nothing changes"? I mean, there's still the same amount of air and pigeons and truck there. If it was a flatbed, that would be different; for one thing the pigeons would escape.
posted by Acey at 11:13 AM on November 27, 2014
posted by Acey at 11:13 AM on November 27, 2014
I remember that Feynman video about train wheels. It's great, one of those "I've seen 'em a thousand times but never noticed that thing."
posted by benito.strauss at 11:19 AM on November 27, 2014
posted by benito.strauss at 11:19 AM on November 27, 2014
Take the derivative with respect to y and set that equal to 0, to see that the center of mass is lowest when y = 1/4. Using calculus seems like cheating, though!
The only qualitative approach I can think of involves calculus-like thinking, so I'm not sure what's gained ...
but as we remove liquid from the can the c. o. g. falls until a certain point and then rises, so if we remove a tiny amount right around that point in the process, the c. o. g. will stay in the same place.
Yet in a center of gravity situation, the only single place (as opposed to multiple separated places) you can remove mass w/o changing the location of the c. o. g. is right at the spatial location of the c. o. g. itself, which can only happen in this situation when the surface of the liquid is right at the c. o. g. .
In other words, the c. o. g. is at its lowest point as it passes through the falling surface of the liquid, and the rest is an algebraic calculation involving the density of the liquid and the mass of the can and volumes and heights and stuff.
posted by jamjam at 11:41 AM on November 27, 2014 [2 favorites]
The only qualitative approach I can think of involves calculus-like thinking, so I'm not sure what's gained ...
but as we remove liquid from the can the c. o. g. falls until a certain point and then rises, so if we remove a tiny amount right around that point in the process, the c. o. g. will stay in the same place.
Yet in a center of gravity situation, the only single place (as opposed to multiple separated places) you can remove mass w/o changing the location of the c. o. g. is right at the spatial location of the c. o. g. itself, which can only happen in this situation when the surface of the liquid is right at the c. o. g. .
In other words, the c. o. g. is at its lowest point as it passes through the falling surface of the liquid, and the rest is an algebraic calculation involving the density of the liquid and the mass of the can and volumes and heights and stuff.
posted by jamjam at 11:41 AM on November 27, 2014 [2 favorites]
We did the torus one in high school. To prove that the hole expands we had a metal ball that wouldn't fit through the hole when it was cool. What we learned was the teacher gets really pissed off if you let the torus contract around the ball as it cools, making it nearly impossible to remove.
posted by Miss Otis' Egrets at 11:52 AM on November 27, 2014 [2 favorites]
posted by Miss Otis' Egrets at 11:52 AM on November 27, 2014 [2 favorites]
If it was a flatbed, that would be different; for one thing the pigeons would escape.
If they were turkeys, we would be heading dangerously close to WKRP territory....
posted by GenjiandProust at 12:17 PM on November 27, 2014
If they were turkeys, we would be heading dangerously close to WKRP territory....
posted by GenjiandProust at 12:17 PM on November 27, 2014
I guess I could make that a little more qualitative: if you have an arrangement of masses and a plane through the c. o. g., removing mass from one side of the space divided by the plane will cause the c. o. g. to move out of that plane into the space on the other side.
In the situation here, as we remove liquid from the top surface, redrawing the plane through the c. o. g. at each step, the c. o. g. will progress downward until the only mass left to remove is on the other side of the c. o. g. plane, at which point the c. o. g. will start back up.
Therefore, the lowest point of the c. o. g. is where it passes through the surface of the liquid.
posted by jamjam at 12:22 PM on November 27, 2014
In the situation here, as we remove liquid from the top surface, redrawing the plane through the c. o. g. at each step, the c. o. g. will progress downward until the only mass left to remove is on the other side of the c. o. g. plane, at which point the c. o. g. will start back up.
Therefore, the lowest point of the c. o. g. is where it passes through the surface of the liquid.
posted by jamjam at 12:22 PM on November 27, 2014
Since the force of gravity is dependent on distance, an object inside the hollow moon near the shell would fall toward the nearest part of the shell, would it not?
The mass of the part of the shell that is closer is smaller than the mass of the part of the shell that is farther away.
posted by straight at 12:24 PM on November 27, 2014 [1 favorite]
The mass of the part of the shell that is closer is smaller than the mass of the part of the shell that is farther away.
posted by straight at 12:24 PM on November 27, 2014 [1 favorite]
Re mirrors: It strikes me that a significant and often-ignored part of it is that left/right and up/down are different sorts of direction. Left and right are relative to the way the observer is facing, up and down are not. Heck, just replace the mirror with another person standing facing you: that person has the same up/down as you, but different left/right.
posted by baf at 12:28 PM on November 27, 2014 [1 favorite]
posted by baf at 12:28 PM on November 27, 2014 [1 favorite]
P.M. has a critical typo:
a mirror flips handedness because if you hold up your left hand, your reflection holds up their hand
should say:
a mirror flips handedness because if you hold up your left hand, your reflection holds up their right hand
To be really pedantic we ought to specify a flat mirror. A curved mirror is a little different. For holiday fun take the spoon at the dinner table and look at your reflection in the convex side then flip the spoon around to the concave side and see your reflection go upside down and tell your drunken relatives about the amazing discovery you just made!
posted by bukvich at 2:26 PM on November 27, 2014
a mirror flips handedness because if you hold up your left hand, your reflection holds up their hand
should say:
a mirror flips handedness because if you hold up your left hand, your reflection holds up their right hand
To be really pedantic we ought to specify a flat mirror. A curved mirror is a little different. For holiday fun take the spoon at the dinner table and look at your reflection in the convex side then flip the spoon around to the concave side and see your reflection go upside down and tell your drunken relatives about the amazing discovery you just made!
posted by bukvich at 2:26 PM on November 27, 2014
Re mirrors: It strikes me that a significant and often-ignored part of it is that left/right and up/down are different sorts of direction. Left and right are relative to the way the observer is facing, up and down are not.
This is because the usual way of changing the way you face is by turning around a vertical axis, i.e., walking around to face the opposite direction. You can blame gravity for this. But you can also change the way you face by performing a handstand. In this case you have reversed top and bottom, but not left and right. If you print the letter "R" on a piece of paper and hold it up to a mirror, it is reversed left and right because you have rotated it that way to face the mirror without even thinking about it. If you flip it vertically (around a horizontal axis) instead, it appears reversed top to bottom but not left to right--again, because you performed that rotation. When we see our image in the mirror, we perform a mental rotation around the vertical axis, i.e., we imagine ourselves walking around the mirror to the other side. It's not quite right, though, because our wedding ring is now on the wrong hand, and why would a mirror arbitrarily perform that trick? A better way to picture how a mirror reflects an image is to imagine a rubber version of yourself running full speed into a mirror so fast that you reverse yourself front to back. The first parts of your body to hit the mirror, like your nose and toes, stick to the mirror, and your back parts continue on through until you are reversed.
posted by weapons-grade pandemonium at 8:22 PM on November 27, 2014 [2 favorites]
This is because the usual way of changing the way you face is by turning around a vertical axis, i.e., walking around to face the opposite direction. You can blame gravity for this. But you can also change the way you face by performing a handstand. In this case you have reversed top and bottom, but not left and right. If you print the letter "R" on a piece of paper and hold it up to a mirror, it is reversed left and right because you have rotated it that way to face the mirror without even thinking about it. If you flip it vertically (around a horizontal axis) instead, it appears reversed top to bottom but not left to right--again, because you performed that rotation. When we see our image in the mirror, we perform a mental rotation around the vertical axis, i.e., we imagine ourselves walking around the mirror to the other side. It's not quite right, though, because our wedding ring is now on the wrong hand, and why would a mirror arbitrarily perform that trick? A better way to picture how a mirror reflects an image is to imagine a rubber version of yourself running full speed into a mirror so fast that you reverse yourself front to back. The first parts of your body to hit the mirror, like your nose and toes, stick to the mirror, and your back parts continue on through until you are reversed.
posted by weapons-grade pandemonium at 8:22 PM on November 27, 2014 [2 favorites]
My thoughts on 10:
When filled halfway (where the can's center of gravity is when it's empty) there are six ounces of pop present.
When six ounces of pop (half the amount of pop in a full can) are present, the center of gravity is lower than the empty can's center of gravity. If we half the six ounces again, giving us three ounces, the center of gravity is lower still. If we half the 3 ounces again, we have 1.5 ounces, and at this point the weight of the liquid is equal to the 1.5oz weight of the can, giving us the lowest possible center of gravity before the lack of counterweight means the center of gravity begins returning toward the center of the can.
I'm currently failing business calculus (tackling it after 10 years since my last high school math class which was algebra), so it brings me great joy to have possibly figured out the answer without using any. One thing I have noticed in the midst of one of my worst academic performances ever is that problems asking to maximize or minimize things like area almost always involve multiplying or dividing by two.
posted by Perko at 1:53 AM on November 28, 2014
When filled halfway (where the can's center of gravity is when it's empty) there are six ounces of pop present.
When six ounces of pop (half the amount of pop in a full can) are present, the center of gravity is lower than the empty can's center of gravity. If we half the six ounces again, giving us three ounces, the center of gravity is lower still. If we half the 3 ounces again, we have 1.5 ounces, and at this point the weight of the liquid is equal to the 1.5oz weight of the can, giving us the lowest possible center of gravity before the lack of counterweight means the center of gravity begins returning toward the center of the can.
I'm currently failing business calculus (tackling it after 10 years since my last high school math class which was algebra), so it brings me great joy to have possibly figured out the answer without using any. One thing I have noticed in the midst of one of my worst academic performances ever is that problems asking to maximize or minimize things like area almost always involve multiplying or dividing by two.
posted by Perko at 1:53 AM on November 28, 2014
I'd never heard of Martin Gardner before. A cursory search suggests that's a problem I should remedy.
With the caveat that these are great fun and I appreciate the post, a few of the puzzles are kind of wonky.
For the Hollow Moon puzzle, I'd guess that there are very few people who both have the requisite skills to solve the problem and also don't already know the answer. As far as I can tell, there's no way to solve the problem using intuition. At least, none that doesn't lead to the wrong answer in slightly different scenarios, e.g., a cylinder, or a magic 1/r^3 force. (I'd love to be proven wrong!) A physics puzzle to which the solution is "take a mechanics or multivariate calculus class, and this will be one of the worked examples" isn't really a puzzle.
The Colliding Missiles scenario is a straight-up algebra homework problem. For a physics puzzle, it's mighty lean on both physics and puzzle. The Can of Soda puzzle has the flaw that the interesting part is in the question, rather than the solution.
On the other hand, the Mirror, the Iron Torus, and the Balloon in a Car are perfect: not obvious at first glance, and definitely soluble without having to know anything. Magnet Testing has the feature that the question itself neatly answers many of the objections about the detailed magnetization pattern that annoying people like me are likely to bring up.
posted by eotvos at 9:24 AM on November 28, 2014
With the caveat that these are great fun and I appreciate the post, a few of the puzzles are kind of wonky.
For the Hollow Moon puzzle, I'd guess that there are very few people who both have the requisite skills to solve the problem and also don't already know the answer. As far as I can tell, there's no way to solve the problem using intuition. At least, none that doesn't lead to the wrong answer in slightly different scenarios, e.g., a cylinder, or a magic 1/r^3 force. (I'd love to be proven wrong!) A physics puzzle to which the solution is "take a mechanics or multivariate calculus class, and this will be one of the worked examples" isn't really a puzzle.
The Colliding Missiles scenario is a straight-up algebra homework problem. For a physics puzzle, it's mighty lean on both physics and puzzle. The Can of Soda puzzle has the flaw that the interesting part is in the question, rather than the solution.
On the other hand, the Mirror, the Iron Torus, and the Balloon in a Car are perfect: not obvious at first glance, and definitely soluble without having to know anything. Magnet Testing has the feature that the question itself neatly answers many of the objections about the detailed magnetization pattern that annoying people like me are likely to bring up.
posted by eotvos at 9:24 AM on November 28, 2014
> it doesn't really reflect left and right, you just expect to face people with their right on your left, and their left on your right. It just reflects everything back directly.
That isn't really a good explanation - it's what one of my teachers would have called "fast talk". If you see everything "back directly" then why is text reversed left right, but not flipped upside down?
I read this list back in the day, and I remember it word-for-word (and the answers) - except, strangely, the last one, which I just worked through to great amusement. It's a good question because when you go through the math, you realize the solution doesn't need the math.
SPOILER.
The key idea is realizing that there are two cases - one where the center of gravity is below the surface of the water, and one where it is above the water. In the first case, lowering the water level lowers the center of gravity, because you've added air above the center of gravity. In the second case, lowering the water level raises the center of gravity, because you've added air below the center of gravity.
So the center of gravity decreases (linearly, it turns out, but you don't need to know that) until it reaches the surface of the water, and then increases again - so the minimum value happens when the center of gravity is at the surface of the water.
At that point, the weight of all the water plus the weight of the can below the water level equals the weight of all the can above the water level. If c is the weight of the can and w the weight of the water, then a little algebra gives you the equation c / (2c + w).
In the specific problem, c = 1.5oz and w = 12oz - and 1.5oz / (2 * 1.5oz + 12oz) = 0.1, so the can will be 10% full and the center of gravity 10% from the bottom (surprise!)
We can check our numbers - at that point there is 1.2oz of water and 0.15oz of can below the surface of the water, which is 1.35oz; above the can there is 0.9 x 1.5oz, which is also 1.35oz.
BentFranklin fell into the trap of believing that these two cases are symmetrical - but they aren't. The center of gravity slowly goes down until you reach the top of the water, and then shoots up - in this case, four times as fast as it went down...
posted by lupus_yonderboy at 9:38 AM on November 28, 2014
That isn't really a good explanation - it's what one of my teachers would have called "fast talk". If you see everything "back directly" then why is text reversed left right, but not flipped upside down?
I read this list back in the day, and I remember it word-for-word (and the answers) - except, strangely, the last one, which I just worked through to great amusement. It's a good question because when you go through the math, you realize the solution doesn't need the math.
SPOILER.
The key idea is realizing that there are two cases - one where the center of gravity is below the surface of the water, and one where it is above the water. In the first case, lowering the water level lowers the center of gravity, because you've added air above the center of gravity. In the second case, lowering the water level raises the center of gravity, because you've added air below the center of gravity.
So the center of gravity decreases (linearly, it turns out, but you don't need to know that) until it reaches the surface of the water, and then increases again - so the minimum value happens when the center of gravity is at the surface of the water.
At that point, the weight of all the water plus the weight of the can below the water level equals the weight of all the can above the water level. If c is the weight of the can and w the weight of the water, then a little algebra gives you the equation c / (2c + w).
In the specific problem, c = 1.5oz and w = 12oz - and 1.5oz / (2 * 1.5oz + 12oz) = 0.1, so the can will be 10% full and the center of gravity 10% from the bottom (surprise!)
We can check our numbers - at that point there is 1.2oz of water and 0.15oz of can below the surface of the water, which is 1.35oz; above the can there is 0.9 x 1.5oz, which is also 1.35oz.
BentFranklin fell into the trap of believing that these two cases are symmetrical - but they aren't. The center of gravity slowly goes down until you reach the top of the water, and then shoots up - in this case, four times as fast as it went down...
posted by lupus_yonderboy at 9:38 AM on November 28, 2014
> As far as I can tell, there's no way to solve the problem using intuition.
All the problems in this problem set would have had an intuitive solution. These all appeared in the Mathematical Games column, and there was a hard rule - "no advanced mathematics". Calculus was specifically ruled out.
I remember bits and pieces about the solution - which involved pairing off bits of mass on one side of the sphere with the other side - but not the details. It's a pity, it was no doubt really slick and yet elementary.
posted by lupus_yonderboy at 9:43 AM on November 28, 2014
All the problems in this problem set would have had an intuitive solution. These all appeared in the Mathematical Games column, and there was a hard rule - "no advanced mathematics". Calculus was specifically ruled out.
I remember bits and pieces about the solution - which involved pairing off bits of mass on one side of the sphere with the other side - but not the details. It's a pity, it was no doubt really slick and yet elementary.
posted by lupus_yonderboy at 9:43 AM on November 28, 2014
(Sorry, BentFranklin - yours was not the incorrect solution to the can problem...)
posted by lupus_yonderboy at 9:46 AM on November 28, 2014
posted by lupus_yonderboy at 9:46 AM on November 28, 2014
It's kind of an implicit infinitesimal or approximation argument, but for the gravitational shell question, you can do the following:
No matter where you are inside, look in any direction and imagine a small solid angle of s steradians centered on that direction. This will intercept a small patch of the shell distance r1 away from you.
And imagine a solid angle in the opposite direction, of the same angular measure, which will intercept a patch at distance r2.
The patch in the first direction is roughly of mass:
r12 · s · shell_thickness · density
The inverse square law for gravity will cancel the r12 and give a force independent of r1. The calculation for the patch in the opposite direction has the r2 drop out, so it gives a force of the same magnitude but in the opposite direction. So net zero force.
A better argument would track the approximations better, and probably include a quick sketch, but that's the general idea I remember seeing.
posted by benito.strauss at 10:06 AM on November 28, 2014
No matter where you are inside, look in any direction and imagine a small solid angle of s steradians centered on that direction. This will intercept a small patch of the shell distance r1 away from you.
And imagine a solid angle in the opposite direction, of the same angular measure, which will intercept a patch at distance r2.
The patch in the first direction is roughly of mass:
r12 · s · shell_thickness · density
The inverse square law for gravity will cancel the r12 and give a force independent of r1. The calculation for the patch in the opposite direction has the r2 drop out, so it gives a force of the same magnitude but in the opposite direction. So net zero force.
A better argument would track the approximations better, and probably include a quick sketch, but that's the general idea I remember seeing.
posted by benito.strauss at 10:06 AM on November 28, 2014
SPOILERS ALERT
Thanks, Lupus. Both you and jamjam have convinced me the soda can problem actually is more interesting than it first appeared. Apologies to Gartner for slighting it.
Poking around a bit, I still haven't found a really compelling intuitive gravitational force from a sphere explanation. (On preview, benito.strauss' is pretty good, though it doesn't require much less detailed knowledge than more formal approaches. But perhaps I'm hoping for too much.)
The text in Gartner's The Colossal Book of Short Puzzles and Problems (link with unknown legality here) has this as the full solution:
posted by eotvos at 10:14 AM on November 28, 2014
Thanks, Lupus. Both you and jamjam have convinced me the soda can problem actually is more interesting than it first appeared. Apologies to Gartner for slighting it.
Poking around a bit, I still haven't found a really compelling intuitive gravitational force from a sphere explanation. (On preview, benito.strauss' is pretty good, though it doesn't require much less detailed knowledge than more formal approaches. But perhaps I'm hoping for too much.)
The text in Gartner's The Colossal Book of Short Puzzles and Problems (link with unknown legality here) has this as the full solution:
Zero gravity prevails at all points inside the hollow asteroid. For an explanation of how this follows from gravity's law of inverse squares see Hermann Bondi's Anchor paperback The Universe at Large, page ANSWER 102.The Bondi book doesn't seem to exist online. However, there's a copy in my library. I'll take a look on Monday.
H. G. Wells, in his First Men in the Moon, failed to realize this. At
two points in the novel he has his travelers floating near the center of
their spherical spaceship because of the gravitational force exerted by
the ship itself; this aside from the fact that the gravity field produced
by the ship would be too weak to influence the travelers anyway.
posted by eotvos at 10:14 AM on November 28, 2014
eotovos, I usually err on the side of too many details. Don't know if this is any more intuitive, but boiling it down a bit:
No matter what direction you look, there is a piece of the shell pulling on you. Its size (and thus mass) grow as the square of the distance away from you, but the gravitational force it exerts diminishes as the square of the distance away. So there's the same force pulling in every direction, leading to a net zero.
(The fact that the size of the patch pulling on you grows as the square of the distance away is the part that seems the least intuitive to me.)
posted by benito.strauss at 10:30 AM on November 28, 2014 [1 favorite]
No matter what direction you look, there is a piece of the shell pulling on you. Its size (and thus mass) grow as the square of the distance away from you, but the gravitational force it exerts diminishes as the square of the distance away. So there's the same force pulling in every direction, leading to a net zero.
(The fact that the size of the patch pulling on you grows as the square of the distance away is the part that seems the least intuitive to me.)
posted by benito.strauss at 10:30 AM on November 28, 2014 [1 favorite]
Oh, and I agree that the soda can problem is the most interesting. So interesting that I'm not going to read any of the explanations here until I've had more of a chance to play with it myself.
posted by benito.strauss at 10:32 AM on November 28, 2014
posted by benito.strauss at 10:32 AM on November 28, 2014
I want to know if the pigeons are spherical.Don't be silly. They're pigeons, not cows.
posted by Flunkie at 2:03 PM on November 28, 2014 [2 favorites]
the minimum value happens when the center of gravity is at the surface of the water. At that point, the weight of all the water plus the weight of the can below the water level equals the weight of all the can above the water level.
I'm afraid that's not correct. The center of mass is not the point at which the mass on both sides is the same. In other words, it's not a median, it's a weighed mean.
I turned the calculus crank and found a surprisingly simple result: The minimum occurs when the total mass of the can + soda is the geometric mean of the mass of the full soda can and the mass of the empty can.
Of course, doing it this way is no fun, but once you know the form of the answer it can be easier to find a natural argument that leads to it. I mean, look at it. Surely there is a geometric interpretation hiding in there somewhere. I've been trying to connect it to jamjam's elegant characterization of the minimum, but no luck so far.
(I'm also starting to have doubts about whether this really is a Gardner problem. It feels very different from the rest of them.)
posted by narain at 3:12 PM on November 28, 2014
I'm afraid that's not correct. The center of mass is not the point at which the mass on both sides is the same. In other words, it's not a median, it's a weighed mean.
I turned the calculus crank and found a surprisingly simple result: The minimum occurs when the total mass of the can + soda is the geometric mean of the mass of the full soda can and the mass of the empty can.
Of course, doing it this way is no fun, but once you know the form of the answer it can be easier to find a natural argument that leads to it. I mean, look at it. Surely there is a geometric interpretation hiding in there somewhere. I've been trying to connect it to jamjam's elegant characterization of the minimum, but no luck so far.
(I'm also starting to have doubts about whether this really is a Gardner problem. It feels very different from the rest of them.)
posted by narain at 3:12 PM on November 28, 2014
However, it's a moon, which means there's a planet nearby. Thus, you'll drift to the point inside the shell closest to the planet. This force can be surprisingly strong -- the moon's gravity, after all, is strong enough to cause the tides on Earth.
Hmm, the actual tidal forces on the Earth are only about 10–7 g, about a hundred times smaller than the gravity on the comet that Rosetta is visiting. (The height of the ocean tides is due to resonance.) I think that works out to an acceleration of about a centimeter per second per hour, so with that force I think you'd drift in a very slow, tiny circle as the hollow moon turns on its axis. But I guess that counts as drifting!
posted by mubba at 10:15 AM on November 29, 2014
Hmm, the actual tidal forces on the Earth are only about 10–7 g, about a hundred times smaller than the gravity on the comet that Rosetta is visiting. (The height of the ocean tides is due to resonance.) I think that works out to an acceleration of about a centimeter per second per hour, so with that force I think you'd drift in a very slow, tiny circle as the hollow moon turns on its axis. But I guess that counts as drifting!
posted by mubba at 10:15 AM on November 29, 2014
> The center of mass is not the point at which the mass on both sides is the same.
DOH! Of course, you're right, dammit! Right, it's when (roughly) the integral of the torque on one side equals the integral of the torque on the other.
No time to verify your solution but:
> The minimum occurs when the total mass of the can + soda is the geometric mean of the mass of the full soda can and the mass of the empty can.
In the specific case at hand, that means a total mass of 4.5oz, which means 3oz of water, which is certainly plausible. I don't believe the reasoning in the previous answer suggesting 25% is right, even if it got the right answer, because if true the result would be independent of the relative masses of the water and can, which it in fact is not according to your (plausible) result.
> (I'm also starting to have doubts about whether this really is a Gardner problem. It feels very different from the rest of them.)
This would explain why I remember the other nine word-for-word and don't remember number 10 at all.
posted by lupus_yonderboy at 11:57 AM on November 29, 2014
DOH! Of course, you're right, dammit! Right, it's when (roughly) the integral of the torque on one side equals the integral of the torque on the other.
No time to verify your solution but:
> The minimum occurs when the total mass of the can + soda is the geometric mean of the mass of the full soda can and the mass of the empty can.
In the specific case at hand, that means a total mass of 4.5oz, which means 3oz of water, which is certainly plausible. I don't believe the reasoning in the previous answer suggesting 25% is right, even if it got the right answer, because if true the result would be independent of the relative masses of the water and can, which it in fact is not according to your (plausible) result.
> (I'm also starting to have doubts about whether this really is a Gardner problem. It feels very different from the rest of them.)
This would explain why I remember the other nine word-for-word and don't remember number 10 at all.
posted by lupus_yonderboy at 11:57 AM on November 29, 2014
(I'm also starting to have doubts about whether this really is a Gardner problem. It feels very different from the rest of them.)
It appears in the collection Wheels, Life, and Other Mathematical Amusements (the tenth collection of his Mathematical Games columns from Scientific American), albeit with beer instead of soda. He credits the problem to a Walter van B. Roberts of Princeton, but the specific numbers given in the "precise" version are Gardner's. His instructions are specifically to solve the problem without using calculus.
posted by Wolfdog at 2:46 PM on November 29, 2014
It appears in the collection Wheels, Life, and Other Mathematical Amusements (the tenth collection of his Mathematical Games columns from Scientific American), albeit with beer instead of soda. He credits the problem to a Walter van B. Roberts of Princeton, but the specific numbers given in the "precise" version are Gardner's. His instructions are specifically to solve the problem without using calculus.
posted by Wolfdog at 2:46 PM on November 29, 2014
(I think it's fair to say it's the sort of problem that must have occurred to many people independently at various times.)
posted by Wolfdog at 2:51 PM on November 29, 2014
posted by Wolfdog at 2:51 PM on November 29, 2014
Well, one can certainly start with jamjam's formulation and turn the algebra crank, and get the solution without any calculus.
What I really want to see is a solution that uses only high-level principles and qualitative reasoning, not working out a bunch of algebra or calculus on paper. That's what makes this feel different from the other Gardner problems.
posted by narain at 2:58 AM on November 30, 2014
What I really want to see is a solution that uses only high-level principles and qualitative reasoning, not working out a bunch of algebra or calculus on paper. That's what makes this feel different from the other Gardner problems.
posted by narain at 2:58 AM on November 30, 2014
What I really want to see is a solution that uses only high-level principles and qualitative reasoning, not working out a bunch of algebra or calculus on paper.Well, then, let's look closer at the question:
Knowing the weight of an empty can and its weight when filled, how can one determine what level of soda in an upright can will move the center of gravity to its lowest possible point?"Via measurement."
posted by Flunkie at 5:15 AM on November 30, 2014
Actually, that's another interesting question! How would you measure the center of gravity of a partially filled soda can? The usual plumb line trick won't work because the soda will move around when you tilt the can. Freeze the soda first? Ah, but then it will expand...
(I'll stop threadsitting now.)
posted by narain at 7:49 AM on November 30, 2014
(I'll stop threadsitting now.)
posted by narain at 7:49 AM on November 30, 2014
Suspend the can vertically in a tall vacuum chamber. Release it. Immediately upon release, push it horizontally. Observe whether or not the can hits the ground verticallly. Keep repeating until it does, varying the vertical offset within the can's height that is pushed. That is the height of the center of gravity; the locations of the center of gravity within the other dimensions are obvious.
posted by Flunkie at 8:50 AM on November 30, 2014
posted by Flunkie at 8:50 AM on November 30, 2014
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