It was not always so. Books and published papers of 17th and 18th century science pioneers often recounted the creative process in detail, with all its false starts, dead ends, and failed hypotheses. Nowadays that is not considered proper style.
There are physics problems and physics puzzles. I like the puzzles. This collection encourages one to seek simple, clever and insightful methods to arrive at the answer. I especially like those that yield to only elementary mathematics. Some can also be solved by tedious formal mathematical analysis, but only as a last resort.
Only brief answers are given here. More discussion and elaboration is needed for some of these.
A few of these are quite old, with obscure sources. Some are of my own invention. I have tried not to include puzzles whose correct answers can easily be found on the internet.

 Going around in centripetal circles. Two identical masses (black) are connected by cords T_{1} and T_{2} of equal length, and swung around a "fixed" axis. If T_{1} is given, what is T_{2}? Explain your answer without using the word "centrifugal". First make a quick guess. Then solve it correctly.
Nitpickers may note that, due to gravity, this motion won't lie in a single plane, for the balls will orbit lower than the swinger's hand. Ignore this distraction.
Answer:
The outer ball has the inward cord tension force T_{1} acting on it. The other ball has T_{1} acting outward, and T_{2} acting inward. They both have the same angular velocity, ω. The radial force on a body moving in a circle is mω^{2}R. The balls have equal mass. In this case the radial force on the outer mass is twice that on the inner mass since it is twice as far from the axis. The other mass has net inward radial force T_{2} − T_{1} = T_{1}/2. So T_{2} = T_{1} + T_{1}/2 = (3/2)T_{1} = 1.5T_{1}.
The masochistic solution is to use a rotating (noninertial) frame rotating with the same angular velocity ω as the balls. In this frame everything is at rest, and if you didn't know it was rotating you'd say that the net force on each part of the system must be zero.
Let T_{1} be the tension in the left string and T_{2} be the tension in the right string.
The net force on the left ball is zero, so k (2 R) = T_{1} where k = m ω^{2}.
The net force on the right ball is zero, so T_{2} − T_{1} + k R = 0.
Then 2 (T_{2} − T_{1}) = T_{1} and T_{2} − T_{1} = T_{1}/2, and therefore T_{2} = 3T_{1}/2 = 1.5 T_{1}.
On examination, the algebra and result are the same as for the inertial frame (of course). But there's more opportunity for blunders and the centrifugal force language just obscures the logic.
Both methods show that the answer doesn't depend on angular velocity.
 The water bridge. In Europe there are a few unusual bridges. Normally bridges (with a street or railroad on top) cross a river. But there are exceptions. Sometimes a canal runs over the bridge that spans over a highway, or even over a river. These are called water bridges.

The Magdeburg Water Bridge. Photo from Sandra Sandrok Douglas. 

Bridge designers take into account the maximum weight of traffic a bridge can support. How does that apply to water bridges? Assume that the level of water on such a bridge remains nearly constant as ships cross it, and so does the mass of water above the bridge—it might be 10,000 tons. A ship that travels the canal has a weight of 1000 tons. What is the extra load on the bridge when the ship is in the middle of the bridge?
The Magdeburg Water Bridge is a navigable aqueduct in Germany that connects the ElbeHavel Canal to the Mittelland Canal, allowing ships to cross over the Elbe River. At 918 meters, it is the longest navigable aqueduct in the world.
Water bridges are rare in the USA. Stanton de Riel informs me of one.
The D&R Canal (Delaware and Raritan) has what you might consider a water bridge, over a tributary of the Millstone River, just north of Princeton, NJ. The canal is navigable, if currently only by canoe (commercial shipping having halted some time ago; the locks are not maintained any more).
Here's the exact Google Maps coordinates.
https://www.google.com/maps/@40.3429314,74.6306373,19.55z .
Puzzle submitted by HansPeter Gramatke.
Answer:
The water displaced by a boat is only a small part of the much larger body of water in the entire canal. Only if there were gates at either end of the bridge, preventing water loss, would the level on the bridge rise, all the weight of displaced water would still be supported by the bridge, and you'd also have the added weight of the boats to support.
But without gates there is no sudden increase in load as the boat moves onto the bridge. A boat displaces its own weight of water all the way along the canal, and that does not change when it crosses the bridge. The bridge was already supporting the extra weight of the boat before (and after) it crosses the bridge. The water level rise due to all the boats on the canal is still only a very small fraction of the water level in the canal without boats.
 The Mighty Muscus. A railroad train travels at 100 mph. A fly on collision course travels at 5 mph exactly opposing the train. The fly hits the window pane of the locomotive, gets stuck (R.I.P.), and continues to move with the train—of course at 100 mph, the speed of the train. Since the fly reversed its direction of movement, there must have been a point where it had a speed of zero, just for an instant. Then, just at that instant the train's speed must have also been zero. Explain how this strong fly could stop (even for a short instant) a train traveling at 100 mph.
Puzzle submitted by HansPeter Gramatke.
Answer: The fly problem was stated deceptively. The fly does not slow the entire train to zero, only a tiny part of it. The impact of the fly with the windshield does indeed slow the fly, and a very small portion of the windshield, to zero speed for an instant during the impact and compression, but that impulse does not immediately communicate (through an elastic compression pulse) to the entire train.
One could pose the same question with the collision of two billiard balls, one moving slowly and another moving rapidly. The slow ball's velocity reverses, and the fast ball's velocity is slowed, but continues in the same direction. So at some point the slow one must have had zero velocity. At that instant, what was the velocity of the fast moving one?
Velocity of what? We have so far been careless with language. We were correctly thinking that it is the conservation of momentum that relates the ball's velocities. Implicitly we were thinking of the velocities of the balls' centers of mass, and should have said so. The center of mass of the slow ball must be zero at some instant for its velocity goes from positive to negative and must pass through zero. But the velocity of the contact point between the balls is not zero at that instant, due to elastic compression going on there in both balls. And since the velocity of the fast ball does not change sign, its velocity vs. time curve never passes through zero. The momentum vs. time curve of the fast ball doesn't pass through zero either.
This reminds us that in nature there are no perfectly rigid bodies. Everything is compressible. And nothing propagates over distance instantaneously. These two facts ought to be stated and emphasized in physics textbooks, but they aren't.
 Drag race.
Dick and Jane are doing a lab experiment, measuring friction by timing wooden blocks sliding down an inclined plane at constant speeds. The blocks measure 3 x 4 x 5 inches. Dick suggests a race, and predicts that if a block is sliding on its 3x4 face it will have smaller contact area and smaller friction than if the block is sliding on its 4x5 inch face, so with smaller frictional drag it will win the race when both slide down the plane. Jane disagrees, arguing that they still weigh the same, so the race will end in a tie. Who is right, and why?
Answer: Jane is correct, the blocks will experience the same retarding force due to friction. We specified that the blocks slide at constant speed, so, as they aren't accelerating, the net force on each block as it slides is zero. The contact area must support the same weight in both cases, so the normal force is the same for both, and so are the forces due to friction.

Rolling down an incline. 

 Off to the races.
A standard physics problem (and demo) races cylinders rolling down an inclined plane. The cylinders are constructed to have the same mass and the same outer radius, but one is solid wood and the other is a metal hoop. The hoop, having the greater moment of inertia, accelerates less under the gravitational force, and loses the race.
But what if we handicap this race differently. Make two solid cylinders of the same length and radius, but of materials of very different density. They will have very unequal masses. Now which one will win, and why?
Now race two spherical balls of the same radius but different masses, say one of steel, one of wood. Which will win?
As usual in these puzzles, you may assume ideal materials, negligible friction and rolling without slipping (in spite of the absence of friction!). These puzzles may be resolved without explicit use of mathematics. Galileo could have solved them. Perhaps Archimedes could have.
Answer: The races will all end in a tie.
Make two cylinders of identical material, with the same outer radius, but one shorter than the other, say only half as long. Let's do a thought experiment. Two identical cylinders of equal length will obviously take the same time to roll a given distance down the track, for all conditions are equal. Now glue them together end to end. The composite cylinder, now twice as massive, will still descend in the same time. So mass doesn't matter here.
This is an example of the use of Galileo's superposition principle, which he used to explain the famous experiment of dropping light and heavy balls from a high tower (an experiment he probably never performed, and wouldn't have been the first to do anyway). The outcome of these experiments depends on the absence of frictional effects. If we race two blocks of different mass sliding down a frictionless plane, the result will also be a tie. But with friction, clearly the one with the least energy lost to friction will win.
Someone may nitpick this solution, noting that the length of the stack and the single disk are not the same. So just imagine taking a stack of many very thin disks on an axle of negligible weight. Now remove every other disk without changing the overall volume. We've reduced the density by 1/2, keeping the overall size the same. Or just imagine removing every other atom in the volume.
As we promised, this puzzle is solved without mathematics, but we used principles and methods that are at the foundation of mathematics. This trick would have been familiar to Archimedes. If you want the full mathematical treatment, see Hoop and Cylinder Motion at the Hyperphysics website.
Now apply your skills to the problem of two solid spherical balls of equal size but different mass. Perhaps one of steel and one of wood. Which would win the race? Again, we can observe that two identical spheres of the same mass would descend the ramp in the same time. Now imagine two such spheres made of many thin slices, but with space between the slices equal to the width of each slice. Now imagine the two spheres being consolidated, the slices of one fitting between the slices of the other. The single composite sphere now has twice the mass as each had before, and descends the ramp in the same time as before. If this seems crude, take the limit as the slices and the spaces between them become infinitisemially thin. Conclusion: this experiment is not mass dependent.
It follows that the period of a sphere, or a cylinder, rolling on a concave surface is also independent of the mass of the sphere.
It does not follow that two spheres of different size will have the same acceleration, even if their masses were the same.
Superposition principles apply only to systems where two or more influences are linearly additive. Many problems in classical mechanics and waves qualify. If two influences obey superposition, we have this condition: If influence A alone causes a result A', and influence B alone causes a result B', then the two acting together produce a result A' + B'. In mechanics, the additivity is usually vector addition.
Now for a stinker puzzle. Obtain a cylinder and a sphere of the same radius and mass, and roll them down an incline. Which will win the race? What if they had different masses?

The confounded tackle. 

 The confounded tackle..
Every so often ingenious tinkerers try to improve on Archimedes. Here's a clever variation on standard pulley systems. Calculate its mechanical advantage, assuming frictionless, massless pulleys and perfectly flexible rope of negligible mass.
Answer: This is an inverted "fool's tackle". If you try to assemble it, it will collapse. Perhaps this picture is a snapshot taken just before it collapsed.
 Soap Box Derby.
A kid is building an unpowered downhill racing car. He has the brilliant idea of using, instead of four wheels, only three wheels, to reduce the friction on the car. Will this modification increase the car's performance in a downhill race? Why?
Answer: Probably not much. Friction is given by F = −μN where N is the normal force on the bearings. A similar situation exists at the point of contact of a wheel with the ground. If this idealized law were actually applicable to real bearings and surfaces, then the total friction would be the same with three wheels as with four. But with lubricated bearings there are departures from the idealized law. Also, one must consider the fact that a rotating wheel is storing rotational energy that does not contribute to the downhill motion. (Bicycles have lightweight spokes to reduce their moment of inertia for this reason.) So three wheels might have a small advantage over four.
 A very fair race.
Some years ago someone had the bright idea to hold a hovercraft race. Hovercraft are supported above the ground by a large fan that forces air downward. The craft moves as if it were on a very low friction surface. These vehicles also have another fan for propulsion. They are rather large vehicles, so the race promoters designed a circular track that was banked so that vehicles at the inside of the track wouldn't have an advantage. Apparently the racetrack was never built, and no such races were ever run. Why?
Answer: To bank the track so there's no advantage for the inside track would require a bowl shaped track. Such a track would be curved such that the faster the hovercraft moved tangentially, the higher it would rise up the banked slope, and the greater its radius of motion around the track. So gaining speed would not get it around the track more quickly. Not only would all vehicles have the same advantage due to the track itself, none could gain any advantage by speeding up. However, equiping the vehicles with something like a sail on a sailboat could make this race interesting.

Pulling a yoyo. 

 Yoyo.
A yoyo toy rests on its edge on a level table.
 If the string comes out above the axle, what will happen when you pull the string parallel to the table? Will it roll left, or right?
 If the string comes out below the axle, what will happen when you pull the string parallel to the table?
 If the string is pulled straight up, which way will it roll?
 At what angle can you pull the string to cause it to slide along the table without rolling? The coefficient of sliding friction at the interface with the table is 0.5.
Support your answers with analysis of forces.
Answer: Consider torques. In the last case, the angle is one whose line of action passes through the contact point between the spool and the table. That's the same point through which the friction acts. So in this case the net torque is zero. Hence, no rotatation. This result is not dependent on the coefficient of friction.
 Impact. Two identical round pucks rest on a frictionless level table. One puck is propelled toward a stationary puck. They collide, and the collision is perfectly elastic. Neither puck rotates before or after the collision. Prove that after collision, the velocities of the two pucks are at right angles, no matter how the collision occurred.
Answer: Write the conservation of momentum equation. The factor m drops out. Write the conservation of kinetic energy equation. The factor m/2 drops out. The momentum equation can be shown as a vector triangle. The kinetic energy equation tells us that its side lengths satisfy the Pythagorean theorem for a right triangle.
The special case of a direct impact causes the moving puck to stop completely, so its velocity after collision is zero. Is this an exception? In vector algebra, two vectors are defined to be perpendicular if they satisfy A•B = 0. So any zero length vector (null vector) is perpendicular to any other vector. That implies that a null vector had indeterminate direction. It can be any direction. So the assertion of perpendicular final velocities is still valid. Isn't mathematics wonderful?
 Levitation.
Why doesn't this happen?

Morning levitation. 

Answer: The center of mass rule applies only to spherically symmetric bodies. The coffee cup is not. This effect is best seen after too many cups of Irish coffee.
 Burning the candle at both ends.

Harpo Marx burns the candle at both ends.
Horse Feathers (1932). 

A candle is trimmed at the bottom so that both ends of the wick are exposed. A nail or long needle is pushed through the middle of the candle, and is supported on the rims of two glasses. The candle is then lit at both ends. Typically the candle will oscillate around the axle. Is the resulting motion simple harmonic motion or just periodic? Is its period constant?
Answer: This could be a messy problem if you get fussy about details. Do the two flames burn at a constant rate even as the candle tilts back and forth? Not likely. As the candle burns, its length decreases, and its moment of inertia decreases, so one can expect its period of swing to change over time. Therefore its motion cannot be periodic (all periods equal in length).
You can see this experiment in this
video..
But is the swing nearly sinusoidal?
The candle is initially balanced at its center. Both ends are lit. The wicks are likely not equal length, so the flame heats wax at one end more. That end drips more wax, lightening that end and causing it to rise. At the higher end the flame is not as close to the wax, and the flame at the lower end is closer to the wax. Soon the lower end becomes lighter, so that end rises. The process repeats.
Another process enhances the motion. When one end dips to its lowest position, it dumps the molten wax accumulated in the previous cycle, so, with reduced mass (and length) it abruptly rises. Watch the video carefully, and you will see this dripping wax.
There's no reason to expect this motion to be sinusoidal, for after each swing the candle is briefly at rest, and suddenly loses mass at the lower end. Nor is the candle's motion periodic, for the period changes as the candle's mass decreases. Does the period decrease or increase? Why?
The period decreases, as the video shows. This would be expected, as the candle gets shorter, so its moment of inertia decreases.

Pressure Paradox.
An oldfashioned bottle of nonhomogenized milk is left undisturbed. The cream in the milk rises to occupy the narrow neck at the narrower top of the bottle. Is the pressure of the milk on the bottom of the bottle now the same, greater, or less than before?
You know this puzzle is old, for those milk bottles are seldom seen today. Nor is unhomogenized milk common. However, many food products come in similar narrownecked bottles. For a modern version imagine such a bottle of oilandvinegar salad dressing, shaken up. Then the oil slowly separates and rises to the narrow neck of the bottle.
Note: Many materials, when mixed, occupy a volume different than their total volume when separate. This is generally a small effect. In this problem, this volume difference will be ignored. It would, in fact, have negligible contribution to the changes of pressure being considered here.
Answer: The total force that the bottom of the bottle exerts on the table stays the same. After all, the total weight of bottle and its contents hasn't changed. However, the pressure (force/area) of the liquid acting on the bottom of the bottle decreases. It pays to read problems carefully.
Consider a vertical column of liquid at the centerline of the bottle, of diameter equal to that of the narrow bottle neck. It originally contained a mixture of milk and cream. Cream is less dense than milk. After separation the cream is on the top, but the central column now has a higher cream/milk ratio. This decreases the pressure at the bottom of the column. But by Pascal's law, the pressures in a liquid depend only on height, so the pressure is decreased all across the bottom of the bottle. This tells us that the net force on the bottom of the bottle has decreased.
Do the forces on the bottle add to zero? Yes, because that cream that rose into the neck of the bottle exerts forces on the bottle's neck. The shape of the bottle neck ensures that those forces have upward vertical components, and these are reduced just enough to match the decrease of force on the bottom of the bottle.
If the bottle were a perfect cylinder this would not be an interesting problem.
Another way to look at this:
The weight registered by the scales depends only on the mass of the bottle and its contents, not on the shape of the bottle. The puzzle (which was not my inventionit is a very old classic) tries to distract and confuse the reader by looking at the pressure exerted by the liquid. The pressure of liquid on the bottom of the bottle does depend on the shape, because it changes the height of liquid in the bottle. The height could be changed by making a cylindrical bottle of smaller cross section, but that would also decrease the area of liquid on the bottom of the bottle. Increasing the height decreases the area, so the pressure, F/A remains the same at the bottom.

Pascal's Vases © U. Minnesota. 

But by narrowing the neck we increase the liquid height (and the bottom pressure) without decreasing the bottom area, so it seems as if the force of the bottle on the scale pan might increase. The force of liquid on the bottom of the bottle DOES increase. But the force on the scale pan is unchanged. The hasty conclusion neglects the upward force of pressure on the neck of the bottle, which compensates exactly for the pressure increase on the bottom of the bottle, no matter what the specific shape of the narrowing at the neck. It is an example of the standard physics demonstration called "Pascal's Vases". .
But suppose the bottle had a shape consisting of two cylinders, a smaller diameter one above a larger one. And suppose that the bottle design was such that after separation, the cream exactly fills the upper cylinder and the milk the lower one. This demolishes our comment about the forces on the neck of the bottle, for they now have no vertical components. Get us out of this apparent dilemma.
 A pendulum has a bucket for the bob. It is halffilled with water. The water freezes. What happens to the period of the pendulum?
Answer: Its period decreases because the water expands upwards in the bucket as it freezes and the center of gravity of the bucket and its contents is higher than before. This decreases the effective length of the pendulum suspension and shortens the period of swing.
However, if the bucket were of a flexible material, would the water still expand upward? Yes, for the water expands while still liquid, reaching its maximum volume (and minimum density) at 3.98 °C (39.16 °F), just before it freezes.
The density of ice is 0.9167 g/cm3 at 0 °C, whereas water has a density of 0.9998 g/cm³ at the same temperature. Liquid water is densest, essentially 1.00 g/cm³, at 4 °C and becomes less dense as the water molecules begin to form the hexagonal crystals of ice as the freezing point is reached. The Wikipedia.
 Thermal pressure.
A solid cube rests on a level surface. The cube is heated by a large amount. Does the pressure the cube exerts on the surface,
 increase?
 remain the same?
 decrease?
Disregard relativistic effects and stick to classical physics.
Explain your reasoning.
Answer: The pressure decreases. The cube expands uniformly in all dimensions. The increase in height and width do not change its weight, so the net force the cube exerts on the table remains the same. But the bottom surface area increases, and pressure is force/area so the pressure decreases.
 A large ship moored at the dock has a rope ladder hanging over its side all the way into the water. Its steps are 30 cm apart, and 20 steps are above the water. The tide comes in at the rate of 15 cm/hour. After 6 hours how many steps are above water?
Answer: Twenty steps are still above water, for the ladder and boat both rise with the tide.
 Stubborn Ball.
A smooth ball rests at the junction of the floor and a tilted wall. When bodies are in contact, there's a force at the interface, directed along the normal to the contact surface. We show the force due to the tilted wall (green) at B and the force due to the floor (blue) at A. The blue vector has no horizontal component, so it doesn't cause rolling of the ball. But the green vector does have a horizontal component. Why doesn't that force cause the ball to roll away from the wall?
Answer: The force at the wall has zero size, and its components are zero. We always tell students to draw pictures of problems to help them organize their strategy. However, figures can mislead, as in this example. Just by drawing a force vector there we are tempted to assume it represents a nonzero force.
Is the zero force really perpendicular to the sloping wall? In vector algebra, the definition of perpendicularity of two vectors, A and B is whether they meet the condition A•B = 0. Any vector of size zero satisfies this rule, so we can say that the null (zero) vector is perpendicular to any other vector. Isn't math fascinating?
 A simple pendulum has a small mass (B) attached to a string of negligible mass suspended from a fixed support (F). The tension of the string is not constant during the swing. For a pendulum with string length of L = 30 cm swinging in an arc of θ = 10° on either side of the vertical, how much work is done by the force of the string tension acting on the mass during one period of the pendulum?
Answer: The string is always perpendicular to the arc of the swing. So its tension force is perpendicular to the bob velocity at any time, and therefore does no work.
If there were no air drag, would the pendulum ever slow and stop, or would it swing perpetually? Perhaps in a vacuum chamber? We are frequently reminded that there are no perfectly rigid bodies, a principle that ought to be in every textbook, in bold font and surrounded by a box.
We know that real pendulums don't swing forever. They gradually slow down and eventually stop. A pendulum must lose energy somehow, and this is usually due to the pendulum doing work on something else. How? Can it be due to stretching and contraction of the pendulum string? No, the string tension acts only perpendicular to the bob's velocity. Or does it? Consider a slightly nonrigid support. It can move just a bit under the action of the string tension. If there's any flexing of the string's support, the string is not always exactly perpendicujlar to the bob's velocity. So work can be done on the support. The pendulum bob itself must lose kinetic energy as it slows its motion. Another obvious process of kinetic energy loss is the fact that the bob does work on the surrounding air as it swings.
See the rigid bodies problem.
 Rigid bodies.
Newton's laws are said to be universal, that is they apply everywhere and at all times, at least for macroscopic (large scale) phenomena. Nearly every mechanics textbook has a chapter dealing with rigid bodies. Those are bodies that maintain their physical shape exactly during interactions. Show that perfectly rigid bodies cannot exist, for they would violate Newton's laws.
Answer: Consider two identical steel balls colliding. They must both decelerate for a short time while they are in contact, until they both have zero speed for an instant before separating. During the deceleration they are moving in opposite directions, and so they must be compressing, otherwise they would copenetrate, which is not allowed. If they did not compress elastically at collision, they would have to change velocity instantaneously. This would be an infinite acceleration, and therefore the impulsive force at that instant would be infinite, which is also not allowed. So the assumption of perfect rigidity is not physically realistic. There are no perfectly rigid bodies.
 Weighty matters.
Textbooks often define the weight of a body as the force that gravity exerts on a body at the earth's surface. But later on, they speak of situations where a body is fully or partially immersed in a liquid, and speak of "the loss of weight" of a body immersed in liquid. Then when discussing orbiting manned earth satellites they speak of "weightless astronauts". It is said that physics is a "precise" science, but it seems the language used in textbooks is far from precise. Resolve this dilemma.
Answer: Some textbooks define weight as "the gravitational force exerted on a body near the earth's surface", But the "weight" assumed in buoyancy problems is defined differently, as the weight registered on a balance scale supporting the immersed body in equilibrium. The astronauts orbiting at 200 miles above the earth's surface still experience a gravitational force of about 95% of what it would be on the surface of the earth. So clearly when we speak of astronauts experiencing weightlessness, it is not the same definition of weight given early on in the textbook. Some textbooks do not clearly warn readers about these three different definitions of weight.
The whole matter could be easily resolved if weight were defined as "the force required to support a body in equilibrium in its frame of reference". At the earth's surface this would be the force registered by a spring scale supporting the object at rest just above the earth's surface, and would be equal in size to the force gravity exerts on the object. The same definition would apply for buoyancy problems in their reference frame within a liquid. The astronauts and their spacecraft frame of reference are both in free fall, falling with the same acceleration due to the attractive force of the earth, and no additional force is required to keep them at rest in their spacecraft's reference frame.
 Leonardo's goof 1.
Leonardo da Vinci's notebooks have a number of errors.

Source: Leonardo da Vinci, Codex Arundel, folio 1030, drawing no. 68: London, British Museum 

This drawing shows his military tank, powered, as shown at the left, by gears and cranks.
Two people inside turned the cranks that drove the wheels. The gearing is the common "lantern gear" of the time. Ignoring the trivial observation that it would take two very strong men to power this, why wouldn't this work? There's no record that it was ever built and used.
Answer: The gearing would turn each pair of wheels in opposite directions. The tank would just sit there spinning its wheels. Some speculate that Leonardo deliberately included errors in some of his drawings, to discourage others from succesfully copying his ideas.

Leonardo's goof 2.
Lenardo da Vinci's notebooks have a number of errors. Here's one showing water streams from holes at various heights in a water tank.
What's wrong with this diagram? How should it look?
Answer: The stream near the bottom would have the greatest velocity, but its distance of fall to the table is smallest. The stream at half height would go farthest before hitting the table. This is also incorrectly illustrated in many textbooks.
The error here is to simplify a situation in which several variables influence the result. The speed of the stream from an orifice is given by Bernoulli's equation, and is s = (h_{1})^{1/2}. The horizontal distance of the stream from the reservoir depends not only on the horizontal component of velocity, but how far it falls h_{2} under the force due to gravity. The result is correctly given and illustrated in Richard Manlife Sutton's book Demonstration Experiments in Physics. American Association of Physics Teachers, McGrawHill, 1938. Here's the accompanying text:
A tall reservoir, equipped with overflow tube to give constant head, is supplied with water from the mains. Water may flow from any one of several horizontal orifices, equally spaced above the table on which the reservoir stands (Fig. 97). The (theoretical) distance d at which any jet strikes the table is 2(h_{1}h_{2})^{1/2} where h_{2} is the height of the orifice above the table and h_{1} is the pressure head. This distance is maximum when h_{2} = h_{1}, i.e., for a hole halfway up the reservoir.
For more details see Didakticogenic Physics Misconceptions, scroll down to "The Leaky Reservoir".

Leonardo's aerial screw. Codex Atlanticus. 

 Leonardo's goof 3.
Leonardo da Vinci proposed several ideas for manpowered flying machines. One, called the "aerial screw", had a rotating screwshaped airfoil, powered by two men on the platform below, turning cranks. Aside from the trivial observation that even two men wouldn't provide enough power, this idea has a serious flaw of physics that would prevent it from staying aloft. What is it? Obviously this idea didn't fly.
Answer: If the screw did spin fast enough to lift the contraption off the ground (which it wouldn't), the platform below would then spin in the opposite direction, conserving angular momentum. That would also make those guys turning the cranks quite dizzy. To prevent this you'd need two propellers counterrotating, or an outboard small propeller such as the one on a helicopter's tail.
 Textbooks often say that when an object is at the focal plane of a converging lens, the light from it, passing through the lens, forms a real image "at infinity". However it can equally well be said that it also forms a virtual image "at minus infinity", easily seen by looking through the lens toward the light source. So a single lens is producing two images. How can this be?
Are we playing fast and loose with the word "infinity" here? In some mathematics courses teachers used to say, "parallel lines meet at infinity". More careless language, it seems. Resolve this confusion.
This raises another question. Any lens can produce real images or virtual images, depending on the location of the object, and given by the lens equation 1/p + 1/q = 1/f. But is this all? Does a lens produce any other images?
Answer: Mathematicians cringe when they hear someone say, "parallel lines meet at infinity". They know that parallel lines never meet. Infinity is much abused in some textbooks. For one thing, infinity is not a number, and should not be used in algebraic expressions except in the symbolic form of a limit, such as "→∞".
The lens equation is often written in terms of image and object distances measured from the lens. They are related by the lens equation 1/p + 1/q = 1/f, where p and q are the object and image distances and f is the focal length of the lens. For a converging lens the graph of q, vs. p, is a hyperbola. The graph has an infinite discontinuity at f. Approached from one side of f the graph increases toward "infinity" and from the other side, it decreases toward "minus infinity".
This equation has a nonintuitive sign convention that gives students grief. For a converging lens forming a real image of a real object, p, q and f are all positive.
A better representation of the lens equation uses the concept of "convergence" where convergences of object, image and lens are O=1/p, I=1/q, and D=1/f, in the unit "diopters". A converging lens of focal length 1 meter has a convergence of 1 diopter. Parallel light rays have zero convergence. The lens equation is then O + D = I. A negative convergence is called a divergence. This has a very intuitive advantage. Light diverging from a source with divergence O is converged by an amount D, emerging with convergence I toward the image. If a real object lies within the focal length, it emerges less divergent; the divergent rays extended backward would meet at the image point—a virtual image.
If you graph I vs. O for a simple thin lens, the graph is a straight line with no discontinuities.
One textbook, years ago, used this better form of the lens equation. Unfortunately, physics teachers, with the old method fossilized in their bones, did not adapt and would not adopt, so later editions of the textbook returned to the old convention.
A single lens produces quite a few images. Reflections from lens surfaces produce images also. Obviously light can reflect from the first lens surface it encounters. Light passing into the first surface of the lens can reflect from the second surface and refract again from the first surface. These images involving at least one reflection are fainter and are sometimes called "ghost images". This is why camera lenses have antireflection coatings to minimize such ghost images. But antireflection coatings are never perfect. You have probably seen photos and motion pictures where the camera is aimed too close to a bright object like the sun, and many ghost images show up as circles of light. The shapes of these are sometimes polygons, in the shape of the iris diaphragm of the lens.
These "spurious images" are often seen by students doing experiments with an optical bench. They usually ignore them. I used to suggest that those images might also be worth studying. The suggestion was usually ignored. One student took my advice, and figured a way to use ghost images to find the index of refraction of the glass lens from the location of the images. How can this be done?
Moral: Don't be too hasty to discard data until you have investigated its cause and exploited it fully to learn something new.
 The image you see when you look in a mirror appears reversed left/right, but not up/down. If you are righthanded, your mirror image is lefthanded. If you touch your right ear, your image touches its left ear. But your image is not standing on its head. At first this seems paradoxical for the mirror is symmetric about its normal. You can rotate the mirror around its normal axis, and the image does not rotate. So why isn't the image also symmetrical about this normal? Resolve this confusion with a simple argument. You must be careful and precise in your use of language.
Answer: The reversal of the image is along the axis from you to your image, an axis normal (perpendicular to) the mirror. Right and left are not reversed, up and down are not reversed, but forward and back are reversed. If you touch your right ear, your image scratches the ear to your right. There is no left/right reversal if you stick to a fixed coordinate frame, anchored on your side of the mirror. But since the mirror image is reversed front to back we are tempted to think of it in a coordinate system rotated 180° about the vertical. So the paradox is partly semantic, a result of unconsciously shifting coordinate frames.
 Virtual image rotation. A Dove prism has the interesting property that when you look through it and rotate it, the image rotates through an angle twice as large as the prism was rotated. Explain.
If you don't have such a prism, use an equilateral prism, looking through it, as shown, so that the light has internal reflection at one side of the prism.
Answer: The light refracts from one face and then reflects from the base of the prism to the other face and then out of the prism.
 Up periscope. Submarines played an important role in WWII. You have seen those movies where the captain looks for enemy ships through a periscope, a long narrow tube extending upward to just above the water surface. Those were days before TV and fiber optics, so the periscope used only lenses and reflecting prisms. You know that looking through a long, narrow tube you cannot see more than a very narrow field of view, yet periscopes could see a much larger field. These periscopes could be 30 feet long and six inches in diameter. Looking through such a tube you'd see a field of only one degree. Yet periscopes typically had fields of 9° or more. How can this be done using only an optical system with glass lenses?
Answer: Since the question was about WWII submarine periscopes, I consulted a textbook of that period, Fundamentals of Optical Engineering by Donald H. Jacobs (McGrawHill, 1943).
Periscopes back then used a "relay lens system", consisting of converging lenses spaced two focal lengths apart. Each lens forms an image in the lens plane of the next lens, which converges the light toward the next lens. More lenses with shorter focal length could increase the field of view. This same system was used in medical endoscopes, long flexible tubes that could be snaked into bodily orifices to view your inner self. Nowadays tiny digital cameras and/or flexible fiber optics serve the same purposes.
 The physics of falling. Every introductory physics textbook tells you that in the absence of air drag, two bodies of different mass fall with the same acceleration, that is, they will fall equal distances in equal times. Galileo is usually mentioned in this context, though others did the experiment before him, and he probably never did the experiment with freely falling bodies (certainly not at the leaning tower of Pisa). But Galileo had a simple logical argument to conclude that the mass of the falling body does not matter. Remember that in Galileo's time algebra had not been invented, and calculus came along even later. The concept of gravity and Newton's law F = ma also came later. So how did Galileo conclude this important result, using only a simple logical argument?
Answer: Use Galileo's principle of superposition with two identical objects falling side by side. They fall in the same manner, remaining side by side as if they were tied together. Bring them close together so they touch. They still fall the same way. Then tie or glue them together, and the composite mass, larger than either one, still falls the same as each did before. One can split a mass in two pieces in any manner to establish the same conclusion. So whatever the mass, they will all fall equal distances in equal times.
This argument is similar to that used by Galileo in his book De motu (On motion), which was never published.

Weighing a moving system. 

 Weight reduction?
We are often told that if we keep moving we'll lose weight. But does a moving object's weight depend on its motion? A classic physics laboratory experiment is an Atwood machine: two unequal masses on the end of a string passing over a pulley. The system can be made to accelerate slowly enough to easily measure its acceleration, and with a little mathematics, determine the value of the acceleration due to gravity. The Atwood machine shown is suspended from a spring balance. The mass on one hanger is M, that on the other is (M+m). Suppose the heavier side (right side) hanger is fastened to the hook of the spring balance by an additional thread, preventing the masses from moving. The scale reads (2M+m).
The restraining thread is burned or cut and the system is set in motion, the left side rising and the heavier right side falling. While the masses are in motion the spring balance reads
 the same as before.
 more than before.
 less than before.
 zero.
Explain why.
Answer: Ernst Mach spent many pages discussing such problems in his book The Science of Mechanics (1893). Of course the mass of the system remains unchanged by its motion. We usually define weight as "the force required to support a system at rest near the surface of the earth." So one might argue that this moving system is not at rest, and does not satisfy this definition of weight.
This reminds us that spring balances, used to weigh unmoving systems, should never be marked in "grams", which is a unit of mass. They measure weight, in newtons or pounds.
Setting these minor quibbles aside. We didn't ask for the system's weight. We only asked what the reading on the spring balance would be. The spring balance will indicate a lower value with the system in motion, much to the surprise of many students who see it for the first time. As the weights move, the system's center of mass is moving downward.
We can easily set up such a system with laboratory hardware and a low friction pulley. Suspend it from a spring balance or place it on a balance scale and perform the experiment as described while observing the balance reading.
We can also demonstrate this on a bathroom scale. Stand at rest on the scale, then do a rapid knee bend to lower your center of mass briefly, then straighten up again, noting the scale reading as you do this. The scale reading responds to the acceleration of your center of mass. As your CM goes down, the reading decreases to less than your resting weight, but as it rises again, the reading increases to more than your resting weight.
The center of mass of the moving masses of the Atwood machine is accelerating downward, so the balance reading is smaller than when the system is at rest. The upward force due to the spring balance is less than the force due to gravity.
 When discussing kinetic theory, textbooks often model an ideal gas as a box with infinitely massive walls containing very tiny particles bouncing from the walls. Part of the argument considers one such particle bouncing from the wall. We are told that the collision is perfectly elastic and the particle rebounds from the wall with the same speed it had before hitting the wall. That tells us that the ball rebounds with unchanged kinetic energy, which students are all too willing to accept uncritically. We reasonably conclude that no energy was lost to the wall. But what about momentum? The particle had momentum mv before the collision, and momentum −mv after, since momentum is a vector. So there's a −2mv change in the particle's momentum, and from conservation of momentum, there must have been a change of +2mv change in the momentum of the wall.
So how can the wall gain momentum without gaining any energy? Are textbooks deceiving us again? Resolve this with an energy and momentum calculation.
Answer: Momentum is a vector quantity. Kinetic energy is a scalar. Momentum is proportional to speed. Kinetic energy is proportional to the square of the speed. These are the reasons for this apparently nonintuitive result.
Consider a collision between two billiard balls, assuming perfect elasticity. Write the equations for kinetic energy and momentum. Now take the limit of this result as one mass is made very large (or the other one made very small). As we approach the limit we have a negligible mass colliding with a huge mass. Take the limit all the way as the small mass goes to zero (or the large one goes to infinity) and you have the massive wall case. The small body rebounds from the massive wall with undiminished kinetic energy.
But what does the result mean? The wall gains momentum 2mv, but gains no kinetic energy? What, then, is its velocity? Its velocity is zero, and its mass is infinite—an indeterminate case. Limiting cases can sometimes be tricky. This also illustrates the dangers of treating infinity as if it were a number and using it as a number in equations.
 Elastic definitions.
Textbooks tell us that a perfectly elastic body is one which, when deformed, returns to its original shape without loss of energy. They also tell us that a perfectly elastic collision is one in which the participating bodies conserve both kinetic energy and momentum.
But consider a bell, made of brass with a brass clapper. Bells and their clappers are made of nearly elastic metals, and both preserve their shape after many collisions. A perfectly elastic collision is one that conserves mechanical energy without loss to dissipative processes. The collision of clapper and bell is not a perfectly elastic collision, for considerable energy is lost as sound, radiated away from the bell. Also the swinging bell and clapper soon come to rest, so you know their energy was dissipated somehow. So how can elastic bodies undergo inelastic collisions? Resolve this apparent contradiction.
Idle question: Would a bell and clapper made of perfectly elastic materials make any sound?
Answer: The contradiction is a semantic one, not a physical one. The two uses of the word "elastic" are defined differently. Also, perfectly elastic materials are not possible in nature. But we didn't say the bell was perfectly elastic, did we? The amount of energy lost as sound and thermal energy when a bell is struck is actually only a very small fraction of the kinetic energy of its parts. The bell is "nearly" elastic, not perfectly elastic.
If bell and clapper were perfectly elastic, they would not deform when struck. The collision duration would be zero, and the force at the contact point infinite. This is physically unrealistic. That would mean that no energy would be available to propagate through the body of the bell, and no subsequent vibrations of the bell, and therefore no ringing sound. Would there be an impulsive wave in the air as the bell and clapper instantaneously reversed direction? Would the bell swing forever? We have probably pursued this speculative analysis of an impossible event too far already.
 Equivalence?
Textbook treatments of relativity sometimes illustrate the "equivalence principle" with the example of a person in an elevator. The elevator cable breaks and the hapless occupant falls with the elevator, experiencing a "weightless" condition in which he floats freely in his elevator frame of reference as if there were no external forces acting. Textbooks often say that the person inside would be unable, by any experiment, to determine that there was a gravitational field in his elevator. This example is, of course, flawed, for with sensitive instruments a person in the elevator could detect the gravitational field. How?
Answer: If the gravitational field were constant in size and direction over the volume of the elevator, it would be indistinguishable from simple constant acceleration. But to engineer such a gravitational field is impossible without a gravitational source consisting of an infinite mass of infinite extent. An acceleration field in empty space is neither divergent or convergent, but gravitational fields are always divergent from finite massive objects. Sufficiently sensitive instruments could detect the divergence of the gravitational field inside the elevator.
 Ellipse or Parabola? Physics textbooks spend much space discussing trajectories of projectiles in the earth's gravitational field. These obey the law d = v_{o}t + ½gt^{2}, which is the equation of a parabola. But Newton tells us that the path of a cannonball (in the absence of air drag) is a portion of an ellipse with the center of the earth at one focus. The famous picture "Newton's mountain" illustrates this.
So if you were asked, "What is the path of a projectile, an ellipse or a parabola?", which answer would you give? Resolve this apparent contradiction.
Answer: The path of a projectile that falls back to earth is a portion of an ellipse, as Newton's picture showed. But if we make a graph of height above earth as a function of distance using Cartesian coordinates, we are distorting the actual shape by plotting distance on a straight axis (rather than on the curved earth), and plotting height, on a rectangular grid (rather than on a polar coordinate grid). This transformation of coordinate systems warps the ellipse into a parabola. So which is it, really? That's an unfair question.
See Secrets Your Physics Textbook Didn't Reveal for more details.
 Newton's third law says: If body A exerts a force on body B, then body B exerts and equal and oppositely directed force on A. Newton's other laws would be useless without this important law. Newton's laws are said to be universal, applying everywhere and at all times. But Newton's third law cannot be correct in all cases, even in classical physics. Show why, with a simple example.
Answer: Newton implicitly assumed that gravitational attraction propagates instantaneously over distance. But now we realize that that is not so, at least in classical largescale phenomena. So if two bodies are very far apart and one of them was to suddenly move toward or away from the other, it would take the other one a while to "know about it". So during this short time the forces they exert on each other would be unequal in size.
 Newton's law of gravitation, F = GMm/R^{2} is accepted without question by freshman. But a little thought reveals that it cannot be true in all cases. When R = 0 the force becomes infinite, an unphysical result. Give an argument why that is not a serious issue.
Answer:
The gravitational force of an extended object is measured from a point within the body. For a spherically symmetric body, the distance, R, is measured from its center. Two spheres therefore cannot reach an R=0 separation due to their size. But shrink them smaller, and their mass gets smaller. In the limit as R goes to zero, their mass goes to zero also and their attractive force goes to zero.
 Floating idea. A beaker of water sits on a scale used to measure its weight. A ball, less dense than water, would normally float on the water. But it is tied down, completely submerged, by a string fastened to the bottom of the beaker. The ball is surrounded by water and does not touch the beaker walls. The string obviously exerts and upward force on the bottom of the beaker. The string breaks, and the ball rises to the surface, floating there. The string no longer exerts that upward force on the beaker. Does the scale now read more, less or the same as before? Support your reasoning with a free body diagram.
Answer: As the ball rises in the water, the water level in the beaker decreases, decreasing the pressure on the bottom of the beaker by the same amount as the previous upward force due to the string. Of course, the beaker and its contents are the same in both cases, so you knew that the weight of the system did not change.
 Holey physics.
Physics problems are often framed with highly idealized situations. Here's a classic problem of that kind. If a straight hole were drilled all the way through the earth right through the earth's center, and a stone dropped down the hole, how long would it take to return?
To keep this simple, ignore the fact that the hole could not be drilled through the hot material in the earth, and if it were, it would fill immediately with magma. Then there's the pesky complication of the earth's rotation, so we must halt that, for the stone would collide with the wall of the hole. Which wall, by the way? Drilling the hole along the NS rotation axis of the earth would be one way to avoid this issue.
To complete the idealization, assume the earth's density is homogenous.
And to extend the problem, after you have found the previous answer, suppose that a straight tunnel were drilled from New York to San Francisco. Now install a railway track through the tunnel. How long would the trip take in an unpowered railroad car, without being given any push, neglecting friction, etc.?
As usual we seek the simplest solution, preferably not even requiring calculus.
Answer:
The round trip takes about 84.6 minutes. Remarkably, the hole need not be drilled through the center of the earth. Any straight hole joining any two points on earth, say London and New York City, could serve as an underground railway. If there were no friction at the tracks, and no air drag, a train at one end would "fall" to the other end in 42.3 minutes. This very old physics problem appeared in Lewis Carroll's 1889 children's book "Sylvie and Bruno". It was also mentioned by Arthur Conan Doyle in his short story The Stone of Boxman's Drift in Boy’s Own Paper, Christmas number, 1887. It also appeared in Unknown Stories: The Unknown Conan Doyle, Doubleday, NY, 1982. Even earlier, Leonard Euler describes such a problem in his "Letters to a German Princess." Clearly this idea must have been widely circulated. It is still often seen in physics textbooks as a homework problem.
You can find my noncalculus solution here Physics problems to challenge understanding, emphasizing concepts and insight.
The noncalculus solution of the hole through earth center uses Kepler's law, treating the "orbit" of the falling ball as an ellipse with nearzero minor axis. For obvious reasons, the previous result is the same as the orbital period of a low orbit earth satellite, skimming just above mountain tops.
This raises another question. Suppose we had such a straight tunnel between New York and San Francisco. Is this
42.3minute trip the best we can do without continuous power input? Is there another kind of tunnel connecting these two cities that would get us there faster?
Yes, there is. There are curved tunnels that are longer, but would achieve shorter transit times. Prove that this must be possible, and describe the geometry of the curve that would give the shortest time. For a mathematician's take on this and related problems see Andrew J. Simpson "Sliding along a Chord through a Rotating Earth" in The American Mathematical Monthly, Vol 113, No. 10 (Dec., 2006), pp. 922928.
 Forever is a long time.
On an infinite frictionless plane could a perfect cylinder, given an initial push, roll forever?
Answer: Well, you'd need quite a large budget to make that infinite frictionless plane. The word "rolling" assumes there's some gravitational force to preserve contact between the object and the plane. Making an infinite, uniform gravitational field would really break the research budget.
But, setting such trivial distractions aside, the question is really asking whether removing friction could prevent the rolling object from dissipating any of its energy.
Consider the practical issue of the rigidity of the cylinder and plane. In the real world, including a gravitational field, both cylinder and plane would deform at the region of contact. Even if there's no slippage there, the cylinder is in a depression in the plane, and is continually rolling up its forward slope. The normal forces of contact there are normal to the deformed surface on the forward slope of the depression, and do not pass through the center axis of the cylinder, but a bit in front of it. So they have torques that will decelerate the cylinder as they do work to slow its motion. In the engineering literature this process is called "rolling resistance" or sometimes "rolling friction", to distinguish it from the more familiar static and sliding (kinetic) friction.
But what if we specify perfectly rigid plane and cylinder? Of course this is not possible in the real world. However, material objects are made of atoms and atoms consist of charged particles. When charged particles accelerate they lose energy through electromagnetic radiation. Each part of a rolling object is accelerating, as does any body that isn't moving in a straight line. So our idealized rolling object still loses energy (very slowly) by radiation.
 Friction is a drag.
Students sometimes suppose that friction always opposes a body's motion, tending to reduce its speed. But there are many everyday examples showing that friction can be necessary to initiate and sustain motion. Give some examples. State the definition of friction so that it cannot be misinterpreted.
Answer:
Example 1: Walking. The force due to friction of the floor must act forward, in the direction you are walking. Without it, as on very slippery ice, you have great difficulty getting anywhere.
Example 2: The Automobile. If you really believe Newton's law, F = ma then the external force accelerating the automobile can only be due to the force of the roadway acting on its wheels. Without friction, there would be no horizontal force there. One could quibble that rolling resistance is a better description of the forces at the interface between tires and roadway.
In fact, without friction our industrialized world would be impossible, and our daily life would be unthinkable, if we even existed at all. See if you can devise a functioning world without friction and without other energydissipative processes.
 Racing photons.
Consider light passing through a converging lens from a point source to a point image. The light rays passing through the lens near its edge must travel a greater distance from source to image than do the rays passing through the center of the lens. Wouldn't this make the rays arrive at different times and possibly cause destructive interference at the image? Explain.
Answer: The rays through the center of the lens are slowed in passing through glass. Rays through the edge pass through less glass. In fact, in a welldesigned lens, all rays from a point object, passing through the lens take the same time to travel from object to image, and are in phase at the image point. That's the ideal case.
Actually both destructive and constructive interference occur near the image point, forming a microscopic diffraction pattern of concentric rings. But at its very center all the rays are in phase, and took the same transit time from object to image.
 Unweaving a spectrum.
Sir Isaac Newton (16421727) is famous for his experiments with light and prisms. He showed that the light passing through a prism separates (disperses) into a colored fan (spectrum). He also showed that if that colored light is then passed through another prism, properly arranged, it can be recombined into white light. Thus, he argued, the colors are actually in the white light, not created by the prism. Here's a gallery of examples from the web, supposed to illustrate this experiment.
Textbooks and web pages frequently illustrate this experiment with such pretty pictures—and get it terribly wrong! Google prism recombine white light and view the images. Most of the images will be wrong in one or more serious ways. This is a telling example of why the web is called "the misinformation highway", for it is dangerously compromised by potholes. If you tried to duplicate this experiment in the lab, following these examples, you would surely fail. Identify the errors in each of these. What is a correct way to decompose white light into colors and then recombine it into white light? There are several ways.
I once had a student who wanted a project for extra credit to raise his unimpressive average. I suggested he go into the lab and duplicate this experiment. He copied textbook illustrations and failed every time. He was frustrated. Finally I suggested he might find out where the college library was, then locate Newton's "Optiks". There he found out one way to do it successfully.
Answer: The divergent spectrum from the first prism must be reconverged to pass through the second prism, duplicating the same geometry as the first prism. Newton used a lens to do that (see last diagram above).
This can also be done using four prisms. Also with a cylindrical lens or mirror.
 The Soda Can.
Here's a puzzle from Martin Gardner's collection. It is an old problem, but the method is still instructive.
Assume that a full cylindrical can of soda has its center of gravity at its geometric center, half way up and right in the middle of the can. As soda is consumed, the center of gravity is initially lowered. When the can is empty, however, the center of gravity is back at the center of the can. There must therefore be a point at which the center of gravity is lowest.
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?
To devise a precise problem assume that the empty can weighs 1.5 ounces. It is a perfect cylinder and any asymmetry introduced by punching holes in the top is disregarded. The can holds 12 ounces (42 gram) of soda, therefore its total weight, when filled, is 13.5 ounces (382 gram).
Answer
The formula for the center of mass is:
m*H/2  h^2*M/H
x =  (1)
m + (h/H)*M
h is the height of liquid in the can. If you graph x vs. h when m = 1.5 oz, M = 12 oz, H = 10 cm, you get a curve that has a minimum at exactly h = 2.5.
This equation is the weighted average of the first moments of the empty can, and its liquid contents. For a cylindrical can, these two masses have their centers of gravity at their centroid, i.e., at H/2 for the empty can, and h/2 for the liquid with level h.
As the liquid level lowers, the center of mass of the system is at first within the volume of the liquid. But the lower it gets, the closer the center of mass moves to the surface of the liquid. At some point it is exactly at the surface of the liquid. As the liquid level lowers more, the center of mass rises and eventually reaches the center of mass of the can when the can is empty.
When the center of mass is exactly at the liquid surface, adding more liquid will raise the center of mass. But taking away a bit of liquid will also raise the center of mass. So this is the critical condition when the center of mass is lowest. Therefore the answer is that the center of mass is lowest when it is at the same height as the surface of the liquid.
This is, perhaps, the most profound and useful fact about this problem, for it also applies to containers of any shape. If the puzzle had been posed for an oddly shaped container, we probably would have gone through much mathematical agony to get an answer, and might not have discovered this simple generality.
But this is probably not the form of answer you wanted. We want to know where is that critical point, in relation to the height of the soda can?
Using the notation above, we can equate the moments of liquid and can.
(h/M)*M(xh/2) = m*(H/2x)
when x=h we get a quadratic equation:
(M/2*H)*x^2 + m*x m*H/2 = 0
The negative root is physically meaningless. The root we want is:
x = (m + sqrt(m^2 + m*M))/(m/H)
Substituting values, we get x = 2.5 for a can of height 10. That is 1/4 the height of the can. That simple fraction is a consequence of the ratio of the mass of the can to the mass of its contents when full. Most other mass ratios do not give simple fractions.
Now (2016) empty aluminum 12 oz soda cans weigh about 0.5 ounce (14.7 gram). One wonders whether the person who invented the problem chose 1.5ounce cans to make the arithmetic easier. Or maybe the problem dates from an earlier time when the cans weighed three times as much as they do now. Anyway, with today's lighter cans, the lowest center of gravity of the partially filled can is 5H/30.
A lengthy discussion of this can be found in Norbert Hermann, The Beauty of Everyday Mathematics (SpringerVerlag, 2012). A straightforward calculus solution is also included. It is too long to show here.
The calculus solution begins with a general expression (Eq. 1 above) for the system center of mass, x, as a function of the amount of liquid in the can (or the height of liquid in the can, h. Then set dx/dh = 0. Solve the quadratic equation to find the minimum value of x. This is a lengthy and messy solution. You will need to use L'Hospital's rule.
These web documents use intuitive approaches.
Top 10 Martin Gardner Physics Stumpers.
Problem #71 . Missouri State.
 Reverse Osmosis.
A correspondent from New Zealand sends us this ingenious idea that he saw in the Dec. 1971 Scientific American Amateur Scientist column. We'll let him describe it:
Osmosis is a process where water flows through a semipermeable membrane from a less concentrated to more concentrated solution. Reverse osmosis is where water flows through the membrane from a strong solution to a weak one. Of course you must have pressure behind the membrane to make it flow the "wrong" way. To get fresh water to flow from seawater through a membrane takes a pressure of about 20 atmospheres. This is the basis of desalinating devices used on large ships.
Anyway, what you do is get a very long pipe with a semipermeable plug in the end and fill it with fresh water and lower it over the side of a boat so that one end is a few metres above the surface and the other with the plug is at the bottom at the deepest ocean trench you can find—say 12 km down or however deep the ocean gets. Now, at this depth the head of salt water in the ocean around the end of the pipe is more than 20 atmospheres, say 21 atmospheres, so fresh water flows out of the ocean salt water into the fresh water pipe. The fresh water will rise about 10 m (21  20 = 1 atmosphere) above the surface somewhat like an artesian well. You may have to adjust the depths a bit depending on the density of the sea water but the principle seems plausible.
Not only will this device give an endless stream of fresh water but can be used to run a small generator.
The figure shows the tube in the ocean, its top end curved to direct water to the little water wheel, W. You've gotta love perpetual motion proposals that are so simple, with no moving parts, and hold promise of solving our world energy problems and our fresh water resource problems as well. That is, if only we can get enough of these machines running at once.
Pressure in the ocean varies linearly with depth, increasing by about 1 atmosphere for each 10 meters of depth. So the pressure in the ocean at a depth of about 200 meters (700 feet) is 20 atmospheres above atmospheric pressure. This fact may or may not be helpful.
This seems to be a great idea. But it won't work. Why not? An answer is given in the April. 1972 Scientific American, but it is a bit indirect. See also the June 1971 issue.
Answer:
The deception is partly in the wording of the problem, and the description that supposes that you lower the tube already filled with fresh water. If the tube were filled with water up to the ocean surface, the pressure difference across the plug would be much too small to initiate reverse osmosis.
To initiate reverse osmosis the pressure across the membrane or semipermeable plug must be approximately 20 atm. The pressure at a depth of 200 meters below sea level is about 20 atmospheres greater than atmospheric pressure. These approximate values will serve our present purposes.
So, to initiate reverse osmosis across the membrane, the fresh water in the tube must be at least 200 meters lower than the surface of water outside the tube, to have a 20 atmosphere pressure difference across the membrane. This is true no matter how long the tube is or how deep it goes into the ocean. End of story. The idea will not work. It's that simple.
What's amazing is how easily people become distracted by irrelevant considerations, water salinity, stratification of salinity density, engineering difficulties of making a long pipe, the necessity of cleaning the filter membrane, economic considerations, storms at sea, etc. etc. They are so obsessed with what they see as "practical" considerations that they are blinded to the simple logical flaw that really does make the idea unworkable.
But still, one must be careful of the assumptions used to reach a conclusion. In this case we assumed that the density of fresh and salt water were the same. That's actually not true. Does it matter? No.
 Which egg is boiled? This is a very old problem. Two eggs are on the table, one is fresh and one has been hard boiled. How can you determine which is boiled without breaking their shells?
Answers:
1. Spin one egg on the table top, then touch it to bring it to a stop. A fresh egg will resume spinning very briefly after the shell's motion is halted, because the fluid inside did not stop its rotation when the shell stopped rotating.
2. If both eggs are given the same spin, the boiled egg will spin longer. The fresh egg's white and yolk dissipate energy due to viscous forces.
 Which is hollow? Two spheres have the same diameter, weigh the same, and are painted the same color. One is solid, of lightweight material. The other is a hollow shell made of denser material. Without damaging them, how can you tell which is hollow?
Answer: The hollow one has a greater moment of inertia. Roll them down an inclined plane. The one that reaches bottom first is the solid one. (More must be said here.)
 An attractive puzzle.
This puzzle is often criticized for perceived ambiguity. Here's a version with most of the ambiguity removed.
You are given two iron bars, identical except for the fact that one bar has been magnetized, and the other is not magnetized. Using nothing other than the two bars and your hands, how can you determine which is the magnet? We will allow gravity to operate as usual on you and the bars."
Carelessly worded versions of this problem lead to answers such as these:
 Suspend one by a thread tied around its center and observe whether it tends to point north.
 Heat one of the bars very hot and let it cool. If the bars no longer attract as strongly, then the one you heated was the magnet.
 Drop one repeatedly on the floor. If the attraction between the bars is reduced, then the one you dropped was the magnet.
But we ruled these out by specifically requiring that you must use only the bars and your hands. No string or wire can be used, no other metal, and nothing to heat a bar. You can't even use the magnetic field of the earth. So what is the simplest way to identify the magnetized bar?
One answer, well known, is the "T" test. Place the bars touching in a T configuration, with the end of one at the center of the other. If they attract, then the one which is the upright of the T is the magnet, for the other has its poles at either end and no pole at its center.
But magnets of high permeability materials can be made with many poles, for example one with a [N SS N] arrangement. Such a magnet would not tend to point north when suspended and might fail the "T" test. What's the simplest way to identify the magnet, no matter how that magnet's poles are arranged?
Answer: Some people think that magnets "pull" unmagnetized iron or steel towards them. This overlooks Newton's third law, which tells is that two bars exert equal and oppositely directed forces on each other, whichever is magnetized. So when placed close enough together, both move, and this provides no clue to which is the magnet.
The original problem is so old that it was probably assumed that a magnetized soft iron bar would always have a N pole at one end and a S pole at the other. So the "T" test is the answer usually given.
A better answer: Let the bars attract end to end. Then reverse one bar, say bar A. If the other end of the reversed bar A still attracted the same place on the other one, B, then B must have been the magnet. This answer is still valid, no matter how the poles are arranged in the magnetized bar.
This puzzle gets stickier when you have a flat sheet refrigerator magnet. These have hundreds of poles arranged in parallel stripes of alternating polarity. Put one in a sealed envelope. In an identical envelope put a thin sheet of unmagnetized steel sheet of the same size and weight. (Cut a piece from a "tin" can lid, which is actually tin plated steel). Now how can you determine which envelope contains the magnet? Go ahead. Try it.
Solution. Refrigerator magnets have thir magnetized surface on only one side, and being so closely spaced, the pole stripes have short effective range. So if you have the magnet sheet and the steel sheet placed together, then flip one over, and if that increases the attraction or decreases it, that one is the magnet. The plastic sheet of the refrigerator magnet increases the separation of magnets and the metal sheet.
 Which is longer? Prepare two metal tubes. Mine are cut from 1 inch diameter aluminum tubing from the hardware store. One tube is 11 inches long. The other is 1/4 inch shorter. Try to ensure that the tubes have no scratches or imperfections that could distinguish one from the other.
Hold them up, one in each hand, and ask if anyone can visually see that one is shorter than the other. Of course no one can. Hold them side by side, touching, and the difference is obvious. Ask someone to take them, then turn around to hide them from your sight, choose one, and then hand it back to you. You pretend to judge its length between your hands, touching the tube at its ends, only with your fingertips. Set it aside and ask for the other, doing the same, then announce, "This one is (shorter/longer as the case may be)."
You could do this blindfolded, but that's probably overdoing it for a physics demo. What is your secret?
Answer: Secret. As you handle the tube, you slide your fingers around the rim. If the room is quiet you can hear the tube "sing" at its resonant frequency. Probably no one else will hear it. The tube with the lower pitch is the longest. Your ears can distinguish the length difference better than your hands. The pitch difference is about one semitone, easily noted by most people. However, not everyone has pitch memory, and cannot do this demonstration. As always, practice first.
In physics class, you should then demonstrate this for all to hear. Hold a tube between thumb and third finger about 2.5 inches from one end (a vibrational nodal point). Experiment to find the optimal position of the node. Then tap the other end with a fingernail, or strike it with a hard object.
 Rolling paradox.
Physics textbooks define the force due to friction as a force tangent to two surfaces at their point of contact. Consider a ball or cylinder rolling without slipping on a perfectly flat and level surface. We expect it to slow down. We naively assume that friction is the reason it slows down, eventually stopping. Certainly the friction is opposite to the ball's velocity, and would therefore decelerate the ball's motion by Newton's second law. But that force due to friction has a torque, and this vector torque around the center of mass of the ball is in the same direction as the ball's angular velocity vector. This would increase the ball's angular velocity, making it roll faster and faster. Resolve this apparent contradiction.
When inventors first proposed railway transportation, using steel wheels on steel rails, some skeptics said "The wheels will just spin in place, and the contraption won't go anywhere." Maybe this paradox was in their minds.
Answer: It isn't simple friction that slows the rolling ball or cylinder. It is "rolling resistance", due to the elasticity of materials. Consider a rolling cylinder. At the point of contact with the plane, the cylinder and plane both deform, slightly flattening the cylinder there and causing a depression in the plane. The wheel is therefore continually rolling "uphill" in this depression. The forces there are greater on the forward side, and have torques that oppose the forward rotation of the wheel.
The rolling resistance is due to the normal forces at the surface of contact. Friction is due to the tangential forces, given by f = μN where μ is the coefficient of friction and n is the normal force. The coefficient of friction for rigid materials like steel on steel is much less than one, so the normal forces are much larger than the tangential forces due to friction, and their effect on the cylinder's rotation is much larger.
In some engineering texts this is called "rolling friction".
 The bootstrap principle. The tall tales of Baron Munchausen include the story of his narrow escape from a sticky situation when he was mired in a bog. The resourceful Baron reached down and lifted himself up by pulling on his bootstraps. We know that is impossible, but can a person, using physics and a pulley system, lift himself using only his own strength?
Consider the system shown. A lightweight chair is used, with an overhead pulley. Can this work? Are there any limitations on this system? Show the vector analysis with free body diagrams.
Answer: Yes, it will work, if the chair isn't too heavy. How heavy? Construct a freebody diagram.
 At rest.
Rare is the physics book that doesn't say something like "The net force on a body at rest is zero" in the chapters on statics.
And it also says that if the net force is zero, the acceleration of the body is zero.
Then, in the dynamics chapters, we may see "A body thrown straight upward is momentarily at rest at the highest point of its trajectory". The student then logically concludes that at that point the net force on the body is zero (at least for an instant) and therefore its acceleration at that point is zero. This is the "at rest → zero net force → equilibrium → zero acceleration" fallacy. Can we blame students for taking textbooks at their word?
Can you resolve this apparent contradiction?
Answer: What do "at rest" and "momentarily at rest" mean, and are they different? Most people understand a body at rest to be unmoving for some finite period of time. Can a body even be "momentarily" at rest? One colleague suggested "at rest at a point in time." I think that compounds the problem, for the definition of velocity involves a limiting process of calculus, starting with a finite time interval.
Rather than resolve this sticky semantics issue, the cure is simple. When describing a situation where the body's velocity changes sign and therefore must pass through a value of v = 0, just say "the velocity is momentarily zero" at that point. This happens at the top of a trajectory, at the extremes of pendulum motion, and can happen at some point during the collision of two bodies. Avoiding use of the phrase "momentarily at rest" derails this logical fallacy before it begins, putting it to rest for good.
One reader suggests "instantaneously zero" as preferable to "momentarily zero". I don't think this helps much.
 Lost energy?

The capacitor paradox. 

This capacitor paradox has been discussed on the web and in published papers, yet people still argue about it.
Obtain two identical capacitors. Charge one of them. Then connect them together so that the charge is shared equally by both. A simple calculation shows that the energy of the two charged capacitors after this operation is only half that of the single initially charged capacitor. What happened to the lost energy?
Of course, one immediately suspects that energy is lost by heating the connecting wires. So we idealize the problem and use resistanceless connecting wires. Still, we must consider energy radiated away by the accelerating charges during the initial process of closing the switches and in the subsequent acceleration of electrons during the redistribution of charge. Yet published papers argue about the details of these processes.
So what's going on? Is circuit theory and classical electromagnetic theory wrong? Can you resolve this simply?
Answer: Here are some web links:
A Capacitor Paradox by
Kirk T. McDonald.
Definite solution of the two (many) capacitors paradox byVladan Pankovic.
The capacitor paradox.
A problem of missing energy when charging a second capacitor.
These represent the tip of an iceberg of published papers that agonize over details of this infuriating problem.
The problem is ill posed and deceptive at its outset. (1) It assumes idealized conditions not achievable in the real world. (2) The initially charged capacitor shares its charge with the other capacitor (a kinetic process) then we calculate the electrostatic energy of the pair in a static condition. But in fact, the charges oscillate back and forth between the two capacitors. If there are realworld energy dissipative processes (such as heating or radiation) the charges will settle to a static situation and those dissipative processes mentioned above will have been responsible for the energy loss. If we imagine all such losses are magically (unrealistically) absent, then the charges must oscillate back and forth forever, their kinetic plus potential energy equaling the potential energy of the initially charged capacitor.
A similar problem can be posed by considering two identical water tanks, one full, one empty, connected at the bottom by a pipe with a stopcock. The stopcock is opened and the water flows into the empty tank until both are half full. The potential energy of the two tanks is now half that of the initially full tank.
Viscosity of the flowing water is the culprit in this situation, accounting for the energy "loss", heating the water. If viscosity were absent, the water would oscillate back and forth, filling the second tank, then flowing back to fill the first tank and so on ad infinitum. Somehow this example seems less mysterious than the capacitor case. But the principle is the same.
But we should never stop thinking about a problem when we arrive at an answer. What if we have a valve in the connecting pipe? At the instant when the water in the tanks reaches equal levels we close the valve, preventing the oscillation of levels that we previously said accounted for most of the energy loss. And, even in this case, half the energy seems to have been lost. Now where did the energy go?
 Grasping straws.
1. We have all done this demonstration, using a drinking straw and a glass of water. Insert the straw in the water (A), close off the top of the straw with your finger, then raise the straw, keeping the top closed. This lifts a column of water inside the straw (B) in spite of the open end. What physics is being demonstrated?
2. We do not normally look at details of this simple demonstration, but what about the lower end of the straw? There's a surface of water there, exposed to the air. What is its shape?
 It bulges downward.
 It bulges upward.
 It is nearly flat.
Support your guess with a valid physical argument.
3. Now let's make it more interesting. Make a hole in the drinking straw at about two inches from the bottom. Make the hole as large as the end openings of the straw. Now immerse the straw in the water glass. The side hole must be below water level. Now close the upper end of the straw with your finger. Lift the straw until it is entirely out of the water (C). What do you predict will happen? Support your answer by an argument based on physical laws. Specifically discuss what's going on at the side hole. Now try it.
Answers:
1. The naive answer is "the water is held up by the reduced pressure above it". But that air pressure certainly can't exert an upward force on the water to balance the force due to gravity. That air is often said to be "sucking up" the water, but it cannot exert an upward force on the water.
Aside: We were, however, correct when we said, "This lifts a column of water inside the straw." The straw and its contents are both "lifted" upward. But it isn't the reduced air pressure inside the straw that is doing the lifting. What is?
As you lift the straw, the volume of air above the column of liquid expands, and its pressure lowers below atmospheric pressure. The pressure on the liquid surface in the glass is still above atmospheric pressure, and the pressure in the liquid increases with depth. So water pressure (exerting upward force) at the bottom of the straw exceeds the pressure above the liquid in the straw (exerting downward force), and this allows a column of liquid to be supported in the straw. Equilibrium is established when the net upward force on the liquid column due to the pressure difference is equal to its weight.
When the straw is lifted completely from the water, atmospheric pressure acting at the bottom of the straw supports the water column. So far this qualitative description tells us nothing about the shape of the water meniscus at the bottom of the straw.
As the straw is lifted completely out of the water, the pressure at the bottom is still as before—atmospheric pressure. No additional force is required to support the liquid inside and the upper water level in the straw remains unchanged. So the water surface at the bottom of the straw can be nearly flat. But try it, and you may be surprised to find it is not flat. Why?
Water clinging to the outside of the straw runs down the straw and this water may collect at the lower end of the straw and form a downward bulge. This additional drop of water is supported by water cohesion, and adhesion to the straw. If you touch your finger to the bottom edge of the straw, you can remove this additional water and then the lower end of the water column is nearly flat. In this process the upper surface of the water column doesn't change, as you would expect. Try it!
But if the straw's diameter is too large, this doesn't work. Nothing said so far explains why water does not simply run out of the straw, to be replaced by air. Surface tension must be playing some role to prevent this. If you plunged a soda straw into a glass full of fine sand you don't expect that sand would be retained in the straw when it is lifted out. Air could permeate the spaces between sand grains, but nothing like that happens with water because of cohesive forces between water molecules. As the tube radius is increased, the force of surface tension around the circumference of the straw increases in proportion to R^{2}, but the volume (and weight of water) in the tube increases in proportion to R^{3}.
But we should try the experiment with a drinking straw and fine sand. Surprise! The straw lifts the sand. The round sand particles wedge against each other and the sides of the straw. You can remove your finger from the upper end and the sand still stays in the straw.
The reduced pressure above the water is important. If you remove your finger, air rushes in and its downward pressure on the liquid forces it out of the straw. When you initially closed the upper end with your finger and lifted the straw, you prevented that happening.
A good discussion is found at the
Mad Scientists website.
2. In the second demonstration (C) the water will stay in the straw above the hole, much as before. It will not run out of the hole in the side of the straw, because the water pressure inside that hole is lower than atmospheric pressure. How do we know this? If you covered the side hole, the air pressure at the bottom of the straw is at atmospheric pressure. Water pressure in the water column increases with depth due to the weight of water. Therefore the pressure everywhere in the water column is less than atmospheric pressure, just as it was in the first demo. So air enters the side hole when it is opened and the water below that runs out the bottom of the straw. Try it!
Finally, let's look at the question of the process of lifting the straw from the water. This accelerates the straw and its contents briefly, while the straw is still in the water. During this brief acceleration, the water level in the straw does not rise immediately, and the trapped air above it expands, reducing the pressure there. So the water level doesn't rise immediately, but quickly does begin to rise, before the straw exits the water in the glass. The acceleration of both straw and water is likely zero again before the straw leaves the water.
 A slippery slope.
If you were descending a slippery slope in a car, would you retain better steering control if your front wheels or your rear wheels locked up?
Answer: Stanton de Riel sent this problem. He says:
"This is a nice problem because students tend to assume that the front wheels can steer you in all situations. Having passed (slowly) by many 4wheel drive trucks in the ditch on slippery, hilly Maine roads, while driving a frontwheeldrive Dodge Caravan, on one particularly nasty morning, I can exactly relate to this situation!"
 Powerful magnets?
One often hears strong magnets described as "powerful". But are they a source of power? I often hear people argue that magnets must be an inexhaustible source of power. They cite the lowly refrigerator magnet, saying, "It supports its own weight on the wall of the refrigerator forever, or at least for many years. So magnets must be a source of considerable energy." I hear this frequently from people who think they can devise a perpetual motion by arranging magnets in a rotating mechanism to extract their stored energy.
What is wrong with their argument?
Answer: Work is the product of force and the displacement that force causes. No displacement, no work. The refrigerator magnet is not moving, so no work is done on it or by it. The absurdity of the "powerful magnet" assumption is easily seen by considering a nail in the wall, supporting a heavy framed picture, year after year. Is the nail a source of energy? Of course not.
The refrigerator magnet certainly has some stored energy, as does any magnetized material. It was stored when the magnet was magnetized at the factory. But none of that stored energy is "used up" as it clings to the refrigerator.
Energy must be expended to manufacture a magnet, and the magnet does represent an example of stored energy in the alignment of its internal magnetic domains. How much? I posed this question on my website, and got an answer from Rick Hoadley (The Magnet Man) rhoadley@execpc.com
Rick considered an Alnico5 magnet, in the form of a bar magnet, 6” x 1” x ¼” in size, with a volume of about 2.5 × 105 cubic meters. The energy stored in an Alnico5 magnet bar of that size is 1.2 Joules. [If you used a stronger magnet, say an NdFeB magnet bar, it would store about 14.7 Joules.]
"1.2 Watts for 1 second is = 1.2 Joules = 1200 Watts for 1 ms. A typical hair dryer uses 1200 Watts while it is running. If there is an easy way to take the energy out of a permanent magnet with 100% efficiency, it could run the hair dryer for 1 ms. That’s all the energy in the magnet, 1.2 Joules.
If, however, you had a similarly sized NdFeB magnet, it could run the same hair dryer for almost 13 ms! Wow, one hair might get dry!"
 Gravity enhancement.
Henry Cavendish (1731–1810) measured the force of gravitational attraction between two lead balls in a laboratory setting. He used a sensitive torsion suspension to measure such a small force. Suppose we have a liquid in a Utube, in equilibrium, and then place a heavy lead ball (red) just under the left side of the tube. How will this affect the liquid levels in the tube?
Answer: The water level in the left side rises, lowering the level in the right side. This may seem counterintuitive, for the water in the left tube experiences a greater downward force from the heavy sphere. But don't forget the horizontal component of force, especially significant for the water in the horizontal section of the Utube.
Imagine the Utube tilted in the earth's gravitational field (B). The water shifts to the left side, as Archimedes says, "seeking its own level". The original tube with the heavy ball under one side experiences the combined force of earth and ball, a slightly "tilted" field, biased toward the left side.
A similar problem is posed in Göran Grimwall's excellent book Brainteaser Physics (Johns Hopkins, 2007), problem 2.8. It considers a mountain below the ocean surface, and asks whether the water surface above the mountain is affected. Would the water bulge downward, or upward there, or not at all?
The mountain, like our lead ball, exerts not only downward force components, but also horizontal force components on the water. The latter pull water into the region above the mountain, raising a slight upward bulge. This is too small to see with the naked eye, but can be detected by earth satellites, which use radar to map the ocean surface contours. Average sea levels are slightly higher over underwater mountains than over plains and trenches. Do a web search for "bathymetry". Another way to think of this is to note that, in equilibrium, the water surface must be perpendicular to the local gravitational force. The mountain causes slightly converging gravitational field lines toward the mountain, hence the upward bulge.
This problem also appears in the excellent collection 200 More Puzzling Physics Problems: With Hints and Solutions
by Péter Gnädig and Gyula Honyek. Cambridge, 2016. Here's their hint and answer.
H. 163. Our probable first idea – that the heavy ball tends “to pull down”
the liquid above it – is false. In fact, just the opposite occurs, and the liquid in
the left arm rises.
It is probably helpful to first analyse the extreme case in which the gravitational
field due to the heavy ball is much larger than that of the Earth.
S. 163. If the gravitational attraction of the ball were much larger than
that due to the Earth, then the two liquid surfaces would coincide with the
same equipotential surface of the ball’s gravitational field. These surfaces are
spheres centred on the ball; so the length of the liquid column in the left arm
would clearly be greater than the corresponding length in the right arm.
When the gravitational effects of both the Earth and the ball are significant
the equipotential surfaces are neither horizontal planes (as measured in the
Earth’s frame) nor spheres centred on the ball. Rather, the free water surfaces
coincide with an equipotential surface that is intermediate between these two
– in general terms an “incline” with an increasinglydownward inclination from
left to right. We can therefore conclude that the level of the liquid in the lefthand
arm will rise, whilst that in the right will sink.
It is possible to strengthen the qualitative reasoning given above with a
semiquantitative analysis, as follows:
The original situation (in which the liquid levels in the two arms are the
same) cannot be the new equilibrium state because we can find another arrangement,
only marginally different from the original, in which the energy of
the system is lower.
Let us transport – hypothetically – a small amount of the liquid, let us say
a layer of height ε, from the top of the right arm to the top of the left one. The
mass Δm of this amount of liquid is proportional to ε, and so the increase in
its gravitational potential energy in the essentiallyuniform field of the Earth
(using the wellknown formula mgh, with ‘h’ increasing by ε/2) is proportional
to ε^{2}.
However, the action also changes the (negative) gravitational energy of the
same amount of liquid in the field of the heavy ball. As the ball, of mass M,
is very close to the tube this field is not uniform and the change in potential
energy is given by −GMΔm(1/r_{2} − 1/r_{1})
where r_{1} and r_{2} are the distances of the
transferred liquid from the centre of the
ball before and after the transportation, respectively. As r_{2} < r_{1} this change
represents a decrease in the potential energy, and because Δm ∝ ε, this decrease
is proportional to ε.
Since, in this thought experiment, ε can be made arbitrarily small, the
quadratic gain can be made smaller than the linear loss, whatever the relative
sizes of the ball and Earth. Therefore the energy of the whole system decreases in
such a process. This shows the nonequilibrium nature of the original situation
and why the liquid level moves upwards in the left arm until it reaches a position
where any further mass realignment would not decrease the total energy.
 Negative reaction? Usually when we pull on something it moves toward us in the direction of the applied force (unless it is nailed down). Can you think of, or devise, a simple system that moves away from you when you try to pull it toward you?
Answer: At first this seems counterintuitive, but actually we are surrounded by examples of such "contrary" behavior of devices when we pull on them.
A block and tackle does that, lifting a heavy weight upward as you pull downward on the rope. Many of the classic "simple machines" of the ancients have this property. Even the simple lever does.
A yoyo toy rests on edge on a table, with its string emerging below the axle. Pull horizontally on the string and the yoyo rolls away from you.
I hope this didn't puzzle you for long. Sometimes the wording of a problem can mislead us—away from the solution.
 Foucault's pendulum.

Foucault pendulum at the Panthéon in Paris. 

Léon Foucault (1819–1868) installed a large pendulum in 1851 at the Panthéon in Paris, to demonstrate that the earth rotates. It was 220 feet long with a 62pound bob. When set swinging it slowly precessed because it maintained its initial plane of swing while the earth rotated underneath it. This was easily observed over the course of a day as its plane of swing changed with respect to the floor underneath it. Science museums around the world have such pendulums, and some university physics buildings do also.
But why does the pendulum maintain its motion in the original plane? After all, its suspension wire is attached at the top, and surely the rotation of the building will exert a twisting torque on the wire. Wouldn't this cause the pendulum's motion to follow that of the building it is in? Some explanation is needed.
Then there's the question of initial conditions. When the pendulum bob is pulled back in the morning and released, this process is done in an already rotating reference frame—the building itself. Shouldn't this initial motion bias the pendulum to retain that motion for the rest of the day, so its plane of motion wouldn't change at all with respect to the building? Therefore no apparent precession would be observed.
As a university student I was once given some good advice about physics. "Do you understand all you know about it?" These simple questions posed by the Foucault pendulum nagged me many years before I discovered the answers. Textbooks and professors avoid this by seldom raising such questions.
Answer: The torsion on the wire does rotate the bob, but doesn't affect its plane of swing. If a marker were painted on the side of the bob, it would continually face the same direction in the building.

Path of the pendulum as it would be traced on the floor below. 


The path of the pendulum bob doesn't lie in a plane. It retains the initial tangential velocity it had before it was released. After release, the pendulum is subject only to the tension in the wire and the gravitational force. The "restoring force" of the pendulum motion is the horizontal component of the suspension wire tension, directed toward a point just below the pendulum's suspension point. The pendulum's path relative to the floor is shown in the figure to the right. Each segment of the swing is a portion of an ellipse. The diagram is exaggerated; the path's deviation from the central point is so small that spectators don't even notice it. If the pendulum bob had a narrow pointer on its bottom, and there were a narrow peg on the floor at the central point, the pendulum pointer would never hit the peg.
However, any slight perturbations of the motion also cause a small ellipsoidal precession of the path. This is troublesome for short Foucault pendulums, which is why most are so large. Clever tricks can be used to eliminate this, but this causes some folks (who don't want to believe the earth rotates) to suspect that those improvements actually bias the pendulum's motion to produce its precession.
See:
A Short Foucault Pendulum Free of Ellipsoidal Precession and A Short, Driven, Fouocault pendulum.
Finally, on a philosophical note, why should a moving object retain its initial state of motion? How does it "know"? What reference frame should its motion be measured "with respect to"? The philosophy of science literature has many discussions of this and it is still debated. Do a web search for "Newton's bucket". One could argue that the question is illposed. Newton's second law tells us that the velocity of a body remains unchanged when no forces act upon it. That fact "defines" an initial state of a body, and any change in that state is a result of forces acting on it. The zeroforce situation defines a frame of reference, convenient for analysis, but that doesn't say that that frame has any "reality" outside of the problem at hand.
 Going around in circles. Mankind, sometimes called "a crawling disease on the face of the earth", affects the earth in many ways. But one effect of human activity is seldom mentioned. In most countries automobiles travel on the right side of the road. Traffic circles are traversed counterclockwise. Most automobiles and trucks return home after they take a trip, so their motion is net counterclockwise. In the USA carnival carousels (merry gorounds) also turn counterclockwise, and races, human, horse, dog and auto, are run counterclockwise. One exception is Great Britain (and a few other countries), where all these go clockwise, including auto traffic and roundabouts.
Does this rotational motion on earth's surface alter the rotation speed of the earth, if only just a smidgen? Might this speed up or slow down the earth's rotation? Should we be concerned? And what is the effect of all those earth satellites we have put into orbit, most of them launched toward the east?
Answer:
Every rotating object on the earth is set in motion by forces and their torques. Newton's third law tells us that every force is accompanied by an oppositely directed reaction force.
And every torque is accompanied by an equal and oppositely directed reaction torque.
The actionreaction pairs are internal forces within the system of the entire earth and its inhabitants, so the net force and net torque on the entire system is zero. But this wasn't the question. Internal forces can and do change the relative motions of internal parts of a system.
You may think this was a silly question, and you are right. Surface traffic's effect on the earth's motion is infinitesimal and certainly unmeasurable. But what direction might the effect be? Most vehicle traffic occurs in the northern hemisphere, due to the accident of there being more land there, and the greatest human population.
We can dismiss automobile traffic easily. Any round trip that traverses the same path going and coming has zero net angular momentum. Only paths with significant enclosed area have net angular momentum, and likely there are as many of these clockwise as counterclockwise.
The other motions, carousels, traffic circles and racetracks with clockwise motions, are, in the Northern hemisphere, opposite to the counterclockwise motion of the earth, and you might think this would slow the earth, and maybe tilt its axis by gyroscopic effects. But let's not jump to premature conclusions.
Every rotating body eventually comes to a stop, so its angular momentum is then returned to the earth, and the net effect on the body of the earth is zero.
What about orbiting earth satellites? One of these, when launched toward the east in the northern hemisphere, gains its angular momentum at the expense of the earth's angular momentum, slowing the earth's rotation just a bit. The net momentum of the system, including the earth satellite, is unchanged, but the angular momentum of the earth is decreased a bit while the satellite is in orbit. When the satellite eventually crashes to earth, or burns in reentry, that same amount of momentum is returned to the earth.
Space probes that never return to earth represent a loss of earth's momentum, slowing its rotation. Still, we needn't be concerned about it.
Irrelevant observation: Figure 8 auto race tracks were popular in the late 1940s in the USA. See figure 8 racing. Some auto race tracks are still in this form. This was intended to provide more excitement for spectators, as collisions at the intersection were inevitable. After all, that's what spectators pay to see—mayhem. In earlier decades, horse races were often run on a figure 8 track. Any suggestion that these were intended to minimize altering earth's rotation is absurd.

Illustrating centripetal force. 

 A circular argument. A ball is on the end of a string. Holding the other end of the string you swing the ball in a large circle. Textbooks often present this as a problem, asking you to relate the angular speed of the ball to the string tension, using the well known formula for centripetal force, F = mω^{2}R. But is the tension really equal to the centripetal force?
Due to air resistance the ball will slow down. To keep it going something else must supply energy in the form of work. But if the string is radial, and the ball's motion is tangential to its circular path, the force and displacement are perpendicular to each other. So how can the string do any work on the ball to sustain its motion?
Answer: To sustain the circular motion you must move your hand in a small circle. The string is always exerting a small component of force tangent to the circle, in the direction of its motion. This is the force that does work on the ball. The tension is slightly larger than its radial component, and that component is the centripetal force.
The string tension is not equal to the centripetal force for another reason. The string must supply a vertical component of force to balance the gravitational force on the ball. So if the ball moves in a horizontal plane, the string angles slightly below that plane.
 Pendulum perplexity. Every physics textbook tells us that the period of a simple pendulum does not depend on the mass of the bob. But these books rarely address the question "Why is the period independent of mass?" If you follow the derivation of the period formula you will see that the mass drops out of the calculation. But there's an easy and insightful way to prove this without even doing mathematics. Can you?
Answer: Consider two identical pendulums swinging side by side. They have the same period. Bring them close together and they swing in synchronism as if they were attached to each other. So let's attach them by gluing them together. They still have the same period as before, though they now swing as a single pendulum of twice the mass of each. Conclusion: the motion and the period do not depend on the mass of the pendulum bob.
 Leaning ball. A uniform sphere of mass m and radius r hangs from a string against a smooth, vertical wall, the line of the string passing through the ball's center. The string is attached at a height h = √(3r) above the point where the ball touches the wall. What is the tension T in the string, and the force F exerted by the ball on the wall? If the ball is rough, with coefficient of static friction μ_{s}, how are these forces increased or reduced?
Answer: This is problem 9 from the book Physics With Answers, 500 Problems and Solutions by A. R. King and O. Regev (Cambridge U. Press, 1997). It's a nice problem, but their answer to the last question is wrong. In fact, whether the wall is rough or smooth, there will be no force at the wall due to friction, for there's nothing to cause the ball to exert a torque at the wall.
The given answers to the first part are correct. T = 2mg/√3 and F=mg/√3. But the answer uses N in place of F suggesting to me that the problem and the answer were written by different persons, and the one who wrote the answer didn't "get" the trick question of the person who devised the problem.
The skeptical student might ask, "How do we know there's zero tangential force at the wall?" Most would think that to be obvious, but might be stumped how to prove it using physics. Consider torques about the center of the ball. The string tension is directed through the ball's center, so it has zero lever arm and zero torque. The normal force at the wall also has zero torque. The only other force that could exert a torque would be a tangential force at the wall. But we know the system is in static equilibrium, so the net torque about any torque axis must be zero. Therefore the tangential force component of the net force at the wall is zero.
 Action and reaction. Textbooks often tell us that Newton's law is something like "For every action there is an equal and opposite reaction." Of course this is carelessly worded. How can any two things be equal and opposite? One should say: "For every action there is an equal size and oppositely directed reaction."
But what is the definition of "action"? One might argue that "reaction" is a negative "action". If so, the original statement might be correct, but it is still confusing. Seriously, when you have an action/reaction pair how can you tell which is the action and which is the reaction?
Answer: This statement of Newton's third law is one of many misleading "slogans" that persist in physics textbooks, originating way back in history when we didn't understand things very well. They should be banned from all textbooks, in my opinion. Originally "action" meant the product of energy and time, as in "Planck's quantum of action." Nowadays the word seems to stand for a "force", at least in this version of Newton's third law.
Newton's third law is better stated "If body A exerts a force on body B, then body B exerts an equal and oppositely directed force on A." It is good to emphasize that these two forces always act on different bodies.
But which force is the action and which the reaction? Is one the cause and the other an effect? These are meaningless questions, for the two forces arise together and have equal status. Nothing is gained by labeling them with action/reaction language. Neither force can be thought of as the "cause" of the other. Neither one precedes the other in time. In fact the slogans "Every effect has a cause" and "Causes precede effects" are not useful principles in physics and certainly aren't profound. They just say that if two events are related by some physical law, and one happens earlier, then the first event may be at least part of the cause of the other. Or both could have a common cause. This language doesn't add anything to our understanding.
And if you still think that cause and effect language is meaningful, and that causes must precede related effects then what is the preceding cause of an action/reaction force pair? Or is it a case of two things "causing" each other?
The physics version of "Which came first, the chicken or the egg?" is "Which came first, the action force or the reaction force?".
 Putting the cart before the horse.
A horse is hitched to a cart. The horse exerts a forward force F on the cart and the cart exerts the same size force backward on the horse by Newton's third law. So the horse and cart will not go anywhere.
What is the flaw in this argument?
Answer:
This is an old problem designed to expose student misapplication of Newton's laws.
The two forces in Newton's third law act on different bodies, never on the same body. They are never added in any physical law. Newton's second law requires us to sum all forces acting on the same body (and only that body). The external forces that move horse and cart are those of friction at the ground.
 Lunar attraction. Standing on the earth, are you closer to the sun at high noon at the time of new moon, or at high noon a halfmonth later at the time of full moon? Why?
Answer:

Positions of earth and moon
at time of new moon.
C is the center of earth.
B is the earthmoon barycenter.
A is observer at midnight, D at noon.
Schematic. Not to scale. 

The center of mass of the earthmoon system (the barycenter) is within the body of the earth, at a distance of 4671 km from earth's center. The earth's radius is 6378 km. The barycenter makes a near circular orbit around the sun. The earth and moon both orbit the barycenter, this being a small perturbation on their orbits, so both earth and moon move in nearly circular orbits around the sun. At noon during full moon, the center of the earth is 2×4671=9342 km closer to the sun than it is during new moon. So you are also that much closer to the sun at noon at the time of full moon, when the moon is on the far side of earth from the sun.
 Going around in circles. The Ptolemaic model of the solar system was geocentric (earth centered) and based entirely on circles (which were considered the perfect figure). To agree with observations of planetary positions, it became extremely geometrically complex, with circles (cycles) and smaller circles (epicycles), deferents and equants and other gimmicks to make it agree with observation.

The Ptolemaic system, simplified. Not to scale.
Adapted from Van Allen, James A. 924 Elementary Problems and Answers in Solar System Astronomy. U. of Iowa Press, 1993. 
Copernicus attempted to simplify this, using a heliocentric (sun centered) model. But he still insisted on a geometry based on circles. His system still needed epicycles, but, he claimed, fewer of them. Less important than the number of epicycles is a property of the particular epicycles that his system eliminated. Six of the abandoned cycles and epicycles had, in Ptolemy's system, one important thing in common. What was it?
Answer: Three epicycles, associated with the orbits of the planets Mars, Jupiter and Saturn all had periods of 365.25 days, exactly the length of a year on earth. The planets Mercury and Venus also had cycles of 326.25 days "tied to" the orbit of the sun. And of course, the orbit of the sun, with period 365.25 days was eliminated, making a total of six fewer circular orbits. The radial lines positioning Mars, Jupiter and Saturn on their epicycles were all parallel at any time. These were also parallel to the earthsun line, like the minute hands on a collection of wall clocks. Notice that this fact is correctly rendered in the diagram. Unless you assume some mystical cosmic connection that links all of these, you might suspect that they are artifacts arising from some one simpler cause. Copernicus found that cause—the misplaced center of the system.
Textbook illustrations of the Ptolemaic system are never to uniform scale. Historical manuscript illustrations aren't either. This is one of the reasons this system was inadequate. It was only conceptual, with no attempt to correctly indicate planetary distances from earth, for the distances were not known back then. If you try to make an accurate drawing the Ptolemaic system with our modern knowledge of distances, you'll get a diagram with overlapping planetary paths, and you will be unable to avoid internal contradictions. For that matter, you seldom see a properly scaled picture of the modern solar system model, for if you correctly scale the orbits of the outer planets, the orbits of the inner planets are too small to see clearly.
 The persistent bug.
An infinitely stretchable elastic band connects a tree with the rear bumper of an automobile. As the auto moves away with constant speed the band stretches. A bug on the band crawls slowly toward the auto. Can the bug ever reach the auto, given enough time?
Answer: Yes. See Martin Gardner.
 The holey sphere.

The holey sphere. L = 6".
From mathworld.wolfram.com. 

Browsing Martin Gardner's books I stumbled on this diabolical puzzle. Gardner calls it "an incredible problem". He traces it back to Samuel I. Jones' Mathematical Nuts, 1932, p. 86.
It is seen on the web in various forms, often ambiguous in wording, along with endless discussions often leading nowhere. I have tried to restate it to remove ambiguity (which isn't easy).
A hole is drilled completely through a sphere, directly through, and centered on, the sphere's center. The hole in the sphere is a cylinder of length 6 inches. What is the volume of the remainder of the sphere (not including the material drilled out)?
You'd think there's not enough information given. But there is. The solution does not require calculus. Gardner gives an insightful solution that requires only two sentences, including just one equation.
Answer.
Assuming the problem is fair, then there must be a solution, and since the size of the sphere and the diameter of the hole were not given, the answer must be independent of these dimensions. So consider a hole of infinitesimal diameter and length 6 inches. In that case, the remaining portion of the sphere must be the same as the entire volume of a sphere of diameter 6 inches. That is 36π, which is the correct answer, as you could verify with more mathematical drudgery.
The deceptive feature of this puzzle is that one is tempted to assume that the sphere is of fixed size and remains that size as one drills different diameter holes in it. The requirement that the cylindrical hole be 6 inches in length, whatever its diameter, forces the sphere to be a different diameter for each hole size considered. One might worry that the limiting case of hole diameter zero might a problem. It isn't. There's no discontinuity in the function there.
 Oblate earth.
Due to its rotation, the earth isn't spherical. It is an oblate spheroid, bulging at the equator. Is its radius of curvature greater at the equator or at the poles?
Answer.
The radius of curvature is greatest at the poles. This is called the "polar flattening". Equatorial fattening and polar flattening go hand in hand. :)
The radius of curvature at the equator can be measured in two ways: in the equatorial plane, or in a meridion plane (passing through the poles). These don't give the same result. But whichever you choose, this radius of curvature is larger than at the poles.
 The resistor chain.

The resistor chain. 

Each resistor in this chain has resistance of 1 ohm. A power source is connected to the terminals A and B. The current in the rightmost two resistors is 1 ampere. What is the potential difference across the input terminals A and B of this chain? What is the resistance of the entire chain as measured at points A and B? What current does the power source supply to this circuit?
This problem is straightforward, though tedious, for the chain has only four "links". It isn't worthy of the label "puzzle". But what if the chain had 500 links? Extending the chain further is of no practical use, but it makes a nice puzzle to solve it for an infinite number of links, for a surprising pattern develops as you work it out.
Hint 1: Sometimes it helps to solve a puzzle if you approach it from the other end.
Hint 2: Sometimes it doesn't.
Hint 3. How might this relate to Fibonacci?
Starting at the far end, the last three resistors are a combination of two in series in parallel with one. So the current in the second from last vertically oriented resistor is 2 ampere, by Kirchoff's laws. The current in the second from last horizontally oriented resistor is therefore 3 ampere. Working backwards down the line we generate a series 1, 2, 3, 5, 8, 13, and 21 ampere. So the current input to the network is 21 ampere. This just happens to be the Fibonacci sequence. Nothing mystical here. The problem was invented to generate that sequence to mystify the person who perversely insists in tackling this problem from the far end. This is fine, until you try the same method on an infinite chain of resistors.
With an infinite number of links, since the series diverges, the required input current would be infinite to provide 1 ampere of current to the last resistor. Not surprising. That tells us that the equivalent resistance of such an infinite chain would also be infinite.
But that doesn't make sense. If the chain were infinitely long, our common sense tells us that the current in the "last" resistor would actually be zero. Working backwards from the far end, this would lead us to the conclusion that the current in each successive link is also zero, and the input current also zero. We get contradictory results depending on which end we work from. Has mathematics failed us?
Working from the left end we are on safer ground. We find that the resistance, for a finite input potential of V is a wellbehaved convergent series as we add resistor links. After about 50 links, additional links contribute negligbly to the total resistance.
This is another hazard of treating infinity as a number, or as something physically accessible to measurement. It reminds me of a little poem by Martin Gardner:
The Unending Mystery of π
π goes on and on and on,
And e is just as cursed.
I wonder, "How does π begin
When its digits are reversed?"
—Martin Gardner
 Skinning a catenary.
A power cable is strung between two utility poles. Of course, it sags, in the shape of a curve called a catenary. At each
end where it is attached to a pole, the cable makes an angle of 10° to the horizontal. The weight of this section of cable is W. What is the tension in the cable at its lowest point? What is its tension at each of the poles?
Answer:
There's nothing really tricky about this problem except the fact that so little information is given. We don't know the length of the cable, or the separation of the utility poles. We don't know the amount of cable sag. We'd have to look up the equation for a catenary. But we don't need all that information.
Mentally divide the cable in half at its midpoint, the lowest point of its curve. The tension at the lowest point is T_{1}. The tension at the utility pole is T_{2}. The weight of the halfsegment is W. Draw the free body diagram of this segment, and you get the equations T_{1} = T_{2} cos 10° and W = T_{2}sin 10°. Solve these and you get T_{1} = W/tan 10°.
The tension at each pole is T_{2} = W/2 cos 10°.
 Finding a center of mass.

Fig. 1. Find the center of mass. 

Fig. 1 shows an Lshaped flat sheet of metal of uniform thickness and composition. Can you find its center of mass, using only an unmarked straightedge?
We don't give you any dimensions because you won't need them.

Euclidean geometry. 

Answer.
In Euclidean geometry propositions are solved using only a compass and an unmarked straightedge. This problem doesn't even require a compass. Since the figure is made of a uniform sheet, the center of mass of any rectanglular piece is at its geometeric center, easily found by draing lines between its opposite corners and marking where they cross. The Lshaped figure can be divided into rectangles in two distinct ways, shown in the second and third diagram. The center of mass of two rectangles taken together is somewhere on the line joining their centers. Draw a line joining these centers (dotted). Then divide the figure into two different rectangles (third figure). Find their centers, and the line joining them. The center of mass must also lie on this line. The two lines cross somewhere (fourth figure) and that must be the true center of mass of the Lshaped figure.
Is it possible that with this method for some dimensions of the "L" the two lines would never cross? No, and you can prove it.
Reference: Apostol, Tom M. and Mamikon Mnatsakanian. Finding Centroids the Easy Way. Math Horizons, Sept 2000, p. 712,
 Falling Slinky. Suspend a Slinky ™ spring from one end. If you release that end, how will the spring fall?
 The entire spring falls, retaining its stretched length until the lower end hits the floor, then the rest of the spring falls, compressing as it goes.
 The entire spring falls, compressing as it goes.
 The lower end rises to meet the upper end, then the spring falls in compressed state.
 The lower end maintains its position until the rest of the spring compresses, then the spring falls in compressed state.
Followup question: What is the initial acceleration of the upper end of the spring as it falls?
 The acceleration due to gravity, g
 An acceleration greater than g.
 An acceleration less than g.
And another question: If a weight were attached to the bottom of the suspended slinky, how would that affect our previous answers?
Oh, just one more thing: If the spring constant or speed of compression pulse in the spring were different, could the lower end rise briefly just after the upper end of the spring is released?
As always, explain your answers.
Answer.
The bottom end is stationary as the rest of the spring compresses. This is so even if there's a weight attached at the bottom. The upward force of tension on the lower end is initially equal to the gravitational force. on that end. This does not change until the stretched portion of the string near the bottom unstretches. One can also observe that the bottom of the spring doesn't "know" what's happening above it until the spring coils near the bottom relax, i.e., until the compression pulse travels from the top to the bottom of the spring.
During this process, the acceleration of the top of the spring is greater than g. We know this must be so since the center of mass of the unsupported spring falls with acceleration g as the spring is shortening by compression, so the upper end must have greater acceleration than does the center of mass. So what is the initial acceleration of the top coil after release? Approximately 2g. Derive this result.
When the spring is suspended, the center of mass is
not at its midpoint, but is lower. This is because each part of the spring must support
the weight of everything below, so the separation of the coils is greater
in the upper part than in the lower part. Does this affect the outcomes? Why,
or why not?
The quantitative results above are not affected.
Since the force on the bottom coil of the suspended spring is zero, the upward force on it due to tension is equal to the gravitational force downward. When the spring is released, the spring tension decresases from top to bottom, never increasing anywhere along the spring. So the bottom coil cannot rise in any case.
These results are confirmed by high speed videos at Derek Muller's Veritasium website falling slinky. Watch all three videos.
 Snap!
We all know how to "snap" our fingers, which is easier to do than describe in words. Press the thumb and middle finger together forcefully, letting the finger slide suddenly off the thumb, and you will hear a snapping sound. Without doing it, explain exactly where the sound comes from.
Answer: The sound is that of the tip of your finger striking the base of your thumb. Now try it, and listen carefully.

Tug of war. 

 Tug of War.
Two equal weights, W, are arranged as shown, An oldfashioned spring balance is connected in the middle of the horizontal cord, and is supported so that it does not cause the cords to sag. (Perhaps, use a weightless spring balance.) What is the approximate reading of the spring balance?
 3W
 2W
 W
 Zero
 W/2
We hope you found this an absurdly simple question, but classroom studies have shown that many students, seeing this simple demonstration, predict the wrong answer. Other studies have shown that many physics students "lose it" when vectors are introduced. Similarly, many students in elementary school fall by the wayside in mathematics when fractions are intruduced, or later, when "x" is introduced as a symbol for an unknown.
The bodies are at rest in equilibrium. So the forces acting on each end of the spring balance are equal in size, but opposite in direction. The spring balance is designed to be used hanging vertically, and is also in equilibrium in that position, with a weight hanging from it. The spring extension is calibrated to measure the suspended weight, which is the force on its lower end. The force on its upper end is not the same size. It is larger, for it includes the weight of the balance scale itself.
Does it matter that in our puzzle the scale is in an unnatural position, lying horizontally? Yes, slightly. In its vertical position, the calibration of the scale includes the fact that the spring supports not only the load suspended from it, but also the weight of its own movable hook attached to the end of the spring. That's why we inserted the word "approximate" in the text of the problem. (Of course this wouldn't be an issue if you used a massless spring balance from the Ideal Scientific Equipment Company) For the same reason, it is not good practice to use a spring scale hanging upsidedown.
Answer:
The answer is (c). The system is in equilibrium, so the string tension is W. The spring scale measures that.
 15 puzzle magic squares.

Wooden 15 puzzle. 

The classic "15 puzzle" is still found in toy stores. It consists of square tiles numbered 1 to 15 in an enclosure 4×4 tiles in size. When filled there's one empty space, allowing tiles to be shuffled into different orders. Typically one tries to get the tiles in numeric order leftright by sliding them, never lifting them from the box.
Puzzle master Sam Loyd claimed he invented this toy in 1891, but he wasn't the first with the idea. Noyes Chapman applied for a patent on it in March 1880. Loyd did describe a prank one can play with it: just interchange two tiles so that it cannot be solved into numeric order leftright. He called it the 1415 puzzle because he interchanged those two tiles, but interchange of any two tiles would have the same result. The puzzle could still be solved by devious methods.
1 5 9 13
2 6 10 14
3 7 11 15
4 8 12
Or turn the puzzle sidewise:
4 8 12
3 7 11 15
2 6 10 14
1 5 9 13
But there are more opportunities for puzzles. Can you shuffle the tiles of the standard 15 puzzle to make a "magic square" in which the tiles in each row, column and diagonal sum to 30?
Answer:
15 1 2 12
4 10 9 7
8 6 5 11
3 13 14 The empty square is taken as "0".
Of course, you can assume the empty square is "16", and make a magic square with sum of 34.
2 3 13 The empty square is taken as "16".
5 11 10 8
9 7 6 12
4 14 15 1
This solution is obtained by adding "1" to each entry in the previous solution.
 Speedy eclipse. Seen from above the earth's north pole, the earth revolves around the sun counterclockwise. The moon orbits the earth counterclockwise. The earth spins on its axis counterclockwise. Then why does the region of totality of a solar eclipse move across the earth from west to east? For example, in the U.S. solar eclipse of 2017 the totality region arrived on the West Coast in Oregon about 1.5 hours later was seen on the East Coast in Charleston, S. C. Check your answer by calculating the time it took the region of totality to cross the U.S.A.
Answer:
You could get bogged down with data. The rotation speed of earth about its axis is 1000 mi/hr (1600 km/hr), it orgits the sun at 67,000 mi/hr (110 km/hr), and the moon moves around the earth at 2,288 mi/hr (3.683 km/hr). There's an easier way, if you consider the angular speed of moon and sun across the sky.

Slip or slide? 

 Slip or slide? Imagine that a new process can produce perfectly frictionless solid materials. A solid cylinder is placed at the top of an inclined plane, both made of this material. The cylinder is released, being careful not to give it any push or rotation. Will the cylinder roll down the plane without slipping, or will it slide down the plane without rotating? Or will it both slip and slide?
Answer:
No one can doubt that the cylinder will descend the plane because there's an unbalanced force component of its weight, acting down the plane, and there's an unbalanced torque due to its weight. The weight vector and the normal force of the plane both act through the center of mass of the cylinder, so these alone won't initiate rotation. However, there are no perfectly rigid materials in nature. Therefore there will be a slight deformation (depression) of the plane where it contacts the cylinder. This will allow a torque due to rolling resistance, and the cylinder will rotate as it moves down the plane? But will it also be slipping all the way down? Not an easy question.
This was an unrealistic problem in several ways. Perhaps one could solve the case where there is friction and then take the limit as the friction goes to zero. This still might yield no clear answer.
 A seasonal puzzle. In the Northern hemisphere, summer is warmer than winter. Why? The usual superficial answer is, "Because the earth's axis has a fixed direction in space, and in summer it tilts toward the sun, but in the winter it tilts away from the sun." That's true but isn't a complete answer. Two important processes aren't
mentioned. Can you explain why the tilt affects the seasonal temperatures?
Answer: Two processes are at work. (1) The tilt toward the sun causes the sun's radiation to be more effective at warming the earth. When the sun is low in the sky in the winter, its radiation is spread over a larger surface area. (2) The tilt causes the hours of daylight to be greater in the summer, increasing the daily warming.
© 2017 by Donald E. Simanek.
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