In the spirit of RaisingBrightKids
, perhaps we can collect here some questions that can start lines of enquiry and stimulate interest in science and mathematics.
Perhaps we should especially look for questions that can lead to investigations, experiments, predictions, tests, discussions, and generally to get people, and children (who are, after all, intelligent alien life-forms) to engage in the process of science.
An irresistible force.
In it's short form: The Moon is getting further away. What provides the force to pull it forward?
OK, now for the more complete version.
We know that the moon causes tides on Earth. Tides create friction, and hence dissipate energy. That energy is the rotational energy of the Earth, so the Earth's rotation is slowing.
And we account for that with the occasional leap-second.
But that means the angular momentum of the Earth is decreasing, but the angular momentum of the Earth-Moon system must be constant. Therefore the angular momentum of the Moon must be increasing. If you do your sums, the Moon must be getting further away.
And it is - about 3.8cm per year.
But if it's getting further away, something must be pulling on it. Obviously the Earth's gravity holds it in orbit, but if it's going into a higher orbit it must have more energy, so it must be being sped up.
How? Where is the force accelerating the Moon?
Gravity - from the oceans; see http://en.wikipedia.org/wiki/Tidal_acceleration (which gives a better description than the above, incidentally).
. This can be a really good one to talk about events having unequal probability. Too often, when confronted by two choices, people unthinkingly say it's "50/50".
Weighing the options
Consider a good, old-fashioned spring-balance. The more you hang from it, the larger the reading. Suppose it's been calibrated, and when you hang one kilogram from it the dial reads 1kg.
Now look at the arrangement we have here. We have two frictionless pulleys and a light, inextensible string such as is only ever found in a physics textbook. We put the spring balance in the middle, and we hang a kilogram from each end of the string.
What reading do we get?
The diagram doesn't make sense - the balance can't be supported only by horizontal strings. How much the strings would sag depends on the weight of the balance, which you don't give.
I don't think the spring-balance is meant to be supported by the strings. My interpretation is it's a fish scale or similar device, strung horizontally between the left and right strings. In that case, the string sag is irrelevant.
Is a fish scale exceptionally light? Any scale has some weight and so cannot be supported by horizontal strings.
I have a 1-22kg fish scale that I use for tensioning automotive belts. It weighs at most an ounce or two, and hasn't sufficient resolution to show its own weight. It is quite capable of indicating approximately 0.5kg and above, however. The diagrammed scenario would easily support it, especially if the horizontal string extensions are short. Anyway, the scenario is meant to be hypothetical, with all the usual assumptions of ideal strings, frictionless pulleys, weightless scales, perfectly spherical cows of infinite density, and so on.
In that case, the orientation is a red herring and the scale reads 2kg.
(The orientation is irrelevant. Are you sure about that reading? ;-)
The spring balance consists of a spring and a scale to measure its amount of extension. Consider simply putting a spring in that location, oriented horizontally so it is stretched horizontally, and ask how much it is stretched by. Is it stretched by as much as it would be if 2kg were hung from it vertically?
Is that sufficiently clear?
Hmm. On further consideration (and without use of the clarification, which I'd assumed anyway), I'll change my mind and say the reading is 1kg.
Also interesting is what the scale reads if the two weights are not equal (at least until the lighter weight is pulled off its pulley). Could this set-up be used to test the difference between gravitational mass and inertial mass?
Hmm. What are the precise initial conditions for that supplementary question? Also, what other info is given if you want to confirm the equality of gravitational mass and inertial mass?
If you look at the moon near one of its quarters, neither full nor new, but somewhere in between, you can clearly see the line separating the light side from the dark side.
Sometimes the moon is seen as a crescent, either recently departed, or now approaching the New Moon.
Sometimes the moon is seen as a Gibbous moon, close to the Full Moon, and either approaching or receeding.
And there is a point when the moon is exactly a half moon. Sometimes the exact moment comes when it's on the other side of the Earth and you don't see it, sometimes it's overhead. This is the moment when the terminator, the line separating light from dark is a straight line from your vantage point.
So the question is this:
- From which parts of the Earth's surface can you ever see the terminator as a vertical line?
- Under what conditions?
For extra credit:
Imagine standing on the North Pole at the moment the sun crosses the Earth’s Equator. Gravitational 'up' points in the same direction as ecliptic north.
Suppose the Moon is at its quarter, with the Earth-Moon-Sun angle being 90 degrees. Then the lunar terminator runs from ecliptic north to ecliptic south, and so the terminator will appear to run vertically.
If you now walk (briskly) towards the Earth's equator along the sunrise (or sunset) line, towards the moon, the moon will appear to rise in the sky until it is above your head. During all this time the terminator will continue to appear vertical.
If on the other hand you walk towards the equator at 90 degrees from the sunrise line, the moon will stay above the horizon, but will appear to twist. When you arrive at the equator the moon will have twisted by 90 degrees and the terminator will be horizontal. If you then walk along the equator towards the moon it will rise in the sky until it is above your head.
- From which parts of the Earth's surface can you ever see the terminator as a horizontal line?
- Under what conditions?
All correct, but still incomplete. See MoonTerminatorSpoiler
if you want a more complete answer. In essence, you are not describing the circumstances. Based on what you've said here you obviously could, but somehow you don't seem to understand what I'm asking for. I'm looking for answers like:
- Anywhere except the tropics
- At 15:37
Obviously, these are deliberately wrong ...
Moon and tides
Everyone knows that the tides are caused by the moon. But if that's true, why is it that the moon goes overhead once a day, and yet there are two tides per day?
How can one moon cause two tides?
Coffee Cup Physics
Take a coffee cup and tap it in different places around the rim. If you get it right, there is a clear difference in the "clinks" that result.
- Apparently, there is a higher-pitched vibration mode in which the handle moves less. I don't know enough to predict the amount of the difference (but it's about a tone for a large cup, and a semitone for a small cup).
- How would you test if your explanation is right?
Now use your explanation, your "theory", to make some predictions. Don't cheat - make the predictions first!
What are your predictions for the following questions:
- Will it work on every cup?
- No. A paper cup wouldn't necessarily clink at all.
- Will all cups give the same pitch difference?
- No. I have two cups that give a different pitch difference. (Okay, I admit that's not a prediction.)
- Will it matter whether the cup is full or empty?
- Filling the cup lowers the pitch of any clink.
- What happens when you tap opposite the handle?
- What if there is no handle?
- I'll have to try. . . I'll use this one labelled "Holy Grail - handle with care." Oops, I've broken it! Oh well, I don't s'pose it was worth much.
Now test your predictions - are you right? The act of making predictions and then testing them with experiments is the heart and soul of science. By testing your predictions you are testing your theory, which is the basis of your understanding. Without testing your predictions you may easily be fooled by coincidence.
Like a candle in the wind ...
A candle has its distinctive flame shape because of the convection currents set up as it burns. It heats the air which expands and rises, drawing the flame upward with it. Cooler air is drawn in from the bottom, feeding oxygen to the flame and shaping the lower part.
So what happens in zero gravity?
Be careful. The steady state solution is easy to find, but is it stable? Maybe the steady state solution is not what happens in practice. If not, what does happen, and how can you tell? How can you test your prediction?
I tried using Google to see if a candle has been lit in zero gravity. Apparently, it has. However, I read no description of such an experiment that confirmed a normal candle was used, except perhaps for some remarks about failures to light the candle and get a flame at all. For some experiments, it was specifically stated that a special candle was used. None of the experiments investigated what happens when a candle is lit in normal gravity and the gravity is then reduced to zero.
Here's a really annoying question I saw on an IQ test:
- A clock showing 3:45 is upside-down. Does the minute hand point to the left or right?
There are at least 5 answers. The first and correct one is: There is no answer. The problem doesn't mention HOW the clock became upside-down, which is crucial. If you turn (rotate) the clock upside down, it points to the right. If you flip over the clock (on an axis), it points to the left.
Also, the perspective of the viewer can preserve/invert all choices (if observer is behind the clock). Incidentally, I've never seen a single IQ test without at least one ambiguous or incorrect question.
another thought is that if the clock is "showing 3:45" it must be a digital clock, so it doesn't have a minute hand!
[If the clock is "showing" anything at all, the observer must be in front of it. If so, the minute hand points to the right. The question doesn't state the observer wasn't also upside-down, so which direction is "to the right"?
]. Also, was it the clock's or the observer's point of view?
Kudos to DougMerritt
and Udo Stenzel
A Thought Experiment
Dynamic equilibrium and Real Work: Gravitation to Rotation by Inertial Displacement.
Imagine an internally geared sprocket with the teeth off radial by 70 degrees.
Place a smaller sprocket inside, at the top of the larger where the larger's gear teeth are horizontal.
Now place a roller chain link on the teeth of both.
Both wheels are overbalanced, move, and the link falls.
Now imagine lenthening the roller chain symmetrically, then to the same horizon.
Then imagine the chain is <r(pi+2) long and endless.
This endless roller chain is hung on raked internal gear teeth on one side of a vertical wheel.
Where the outer wheel's teeth let the chain slide off, the chain runs up over a sprocket and down to the outer wheel's teeth again.
The axles are fixed, the sprocket and wheel free to rotate, the chain to run.
The vertical run of chain is around 2\3 the length on the outer wheel.
Will the approximately 1/5 portion of chain overcome summed resistances and overbalance both wheels and chain?
see also GravitationtoRotationbyInertialDisplacement