Discuss Reverse Physics and a Car here

[mith]

Puzzle link

[What if...]

For the first one,

if one is a semicircle w/nearly all of the mass at the ends so that the center of mass is outside the object (in the center), than moving the object closer to the center of mass from the middle of the object will increase the gravitational force, while increasing the smallest distance to the object.

For the second,

I was either very smart or very stupid in guessing that they were both wrong because, unless the scales are the same, (so that numerically v=t/2 and they are both right) the semicircle is not a true semicircle because the time may be 15 and the maximum velocity 79, and only using a scale, where one unit of distance/unit of time is the same as one unit of time, would either be true.

Edited because I wrote it when I was half-asleep.

[This message has been edited by What if... (edited 04-17-2003 05:31 PM).]

[Lucky Wizard]

2. What What If... said. The answer would actually be (pi/2)*v*(t/2), since it's actually a semiellipse regardless of how you distort it, and the formula for the area of an ellipse is pi*a*b, where a and b are the semimajor axis and the semiminor axis.

And can you clarify on your answer to #1? Not sure I understand it...

[Griffin]

For number one, I believe this is basically the same idea What If has...

If there were a hole through the center of the earth and you were to fall through it, you would be getting closer and closer to the center of mass, but gravity would be affecting you less and less (at the exact center no gravitational force would act on you at all).

[RedNifre]

Is it normal, that the posted solutions are censored? (I'm new here, so I don't know)

My solution for the first one:
The objects have negative masses.

[Applebyd]

Posts aren't censored

People are using an INV /INV ag to
hide the text so not to spoil the
puzzle for anyone who doesn't want to
know the answer.

DaveA

[Quailman]

And to see the invisible text, highlight it with your mouse or hit [ctrl] a.

[the.cynic]

So then, the actual answer to number one is -->YES<--

[This message has been edited by the.cynic (edited 04-16-2003 02:02 PM).]

[Rollercoaster]

2. Looking at the units of the equations will lead to the results already posted. (Distance cannot be proportional to the square of velocity, OR the square of time. It must be proportional to the product of velocity and time.)

Also, it should be pointed out that there is an infinite acceleration issue at the beginning and end of this trip. That's going to create a problem, unless this car is massless.

[Edit to invis]



[This message has been edited by Rollercoaster (edited 04-16-2003 02:58 PM).]

[What if...]

Sorry, I was in a hurry before, and somehow removed the very important word:semicircle. Here's my idea.

What I was picturing was a hollow semicircle/hemisphere (they are the same if you just rotate the semicircle), w/nearly all the mass on the ends. If an object is near (or touching) the midpoint of the semicircle (1), then the gravitational force could be smaller than if the object is slightly farther away overall, but nearer the (more massive) end (2). Of course, the shortest distance from the object to the curve can be smaller in the first case (very close to much less massive part) than the second (slightly larger distance to more massive end).
In retrospect, I could have used a much simpler example for this alone, but this example also exhibits the property mentioned by Griffin, for some reason was the only one I mentioned in my previous post, and which I do not think is the key to the puzzle since it does not mention the center of gravity. Since the center of mass of a semicircle is at the center of the corresponding circle(3), if you move the object from near an end(2) to the center(3), it will feel less of a force, despite being nearer to the center of mass. I think the first part (not using center of masses) is the answer because the second is common, and the puzzle just says closer, which does not necessarily mean the center of masses.

code:

---

                                                    


                           ________                 


                          /    1   \                


                         /          \               


                        /            \              


                       |              |             


  most mass is here -> |   2    3     | <- and here 


                                                  

---

Please just pretend the diagram looks good (and like a semicircle)!

Edited to fix some things that made little sense.

[This message has been edited by What if... (edited 04-17-2003 03:04 PM).]

[What if...]

I found another (simpler) solution to one, in which the point (1) w/lesser force is closer to the object and

[ChienFou]

No.2 pi x v x t / 4. Boring. compare the area of the rectangle with the area of the semi-circle inevitably the engineering solution.

[CrystyB]

(2) i don't get it: why do you disregard the info given?? it is said to be a semicircle after all, so v=t/2. Therefore both answers are correct. Scaling was already taken into account.

PS Your answers of pi*v*t/4 is also correct - like in the 0.FFFF...=0.9999...=1 debate

[ZutAlors!]

I think a good arguement for ignoring the "given" information is that the equation is to be determined symbolically. Although the graph implies that v = t/2 there are two problems here: 1) there's nothing preventing the two axes of the graph from being scaled differently, so that even though the distance corresponding to v equals the distance corresponding to t/2, v doesn't equal t/2. 2) Moreover, the equation(s) determined from the graph have no units attached. Obviously, if you alter the units in any way from whatever the graph represents, then you destroy the numerical equality.

[Nienscecco]

Reply to Griffin about #1 :

I think your example with the hole into the earth is wrong. When the corpse (Hoping the guy is dead when he reaches the center of the earth) is at the center of the earth (and that we consider this place the center of masses also). There will be many gravitational forces applied the corpse. The results of these forces will be equal to zero : meaning that the corpse will stay at the center only due to the gravity. But At this place the gravity will be maximum, infinite in fact... Considering the corpse at 0 distance from the center of gravity of the earth, the force would be infinite (divisionby zero).

So the best possible example is a massive object that is not symetric, with his center of gravity away from a less massive volume...Like What if suggested in the first place.

Please be indulgent with my english, I'm new here and I speak french by birth so it might not be very clear after all



[This message has been edited by Nienscecco (edited 04-17-2003 12:09 PM).]

[ZutAlors!]

You're saying a couple contrary things, Nienscecco. At the center of the Earth, the net effect of gravity will be zero, but the gravity won't be infinite. I suspect you're thinking of the gravitational law g = GMm/r². That treats the mass of the Earth as a point mass, and doesn't apply anymore one you're inside the Earth. I believe you could calculate the actual gravitational forces on the corpse by triple integrating a force calculation over the volume of the Earth, but this will come out to be zero. So "the Earth with a hole through it" should still be a valid answer.

[What if...]

Since people aren't points either, if your center of mass corresponded with the Earth's, parts of your body would be pulled outward (because the mass is around you). Your center of mass, of course, would not move, because (assuming the Earth is almost perfectly spherical) you would not be pulled more in any direction, so net force is zero. Otherwise, this embedded in a sphere solution is the same as having two equal massed objects (very distantly connected) on a line (A), and an object exactly in between them feeling no force:

code:

---

A.  _________    


   |         |   


   |         |   


  _|         |_  


 |_|    1    |_| 


                 


-----------------


B.               


  2              


  _       1      


 |_|-------      


                 

---

In the second picture is the simplest solution to the problem based on shortest distance rather than distance to center of mass. The line coming out of the mass (box) in B is a stick having very little mass. My second solution was a combination of A and B.

Edited to fix picture.



[This message has been edited by What if... (edited 04-30-2003 02:52 PM).]

[Nienscecco]

ZutAlors! Right, I agree. Thanks for leading me back on the right trail. I was indeed thinking of g = GMm/r². And it is ovious that it doesn't apply in the "hole inside the earth" situation. Sorry for my silly comment

Now let's discuss this scale matter for the #2 :

I really think that we have to take the given information from the graph as is. Anyway, the big bug according to everyone (including me at the beguining) seems to be that v cannot = t/2 because of the units : m/s cannot = s. My point is : whatever scale factor that we apply to the 2 axes of the graph, that won't solve the bug with the units. Scale is definitly not an issue here.

The v = t/2 only applies to the numeric values. And it is not true to say that we destroy this equation by trying to work with the units.

Lucky wizard was right with his comments on the Ellipse : Area = pi * a * b where a and b are the 2 axis of the ellipse.

Since time and speed are of different nature, the correct way to calculate the surface and fit the units in is to calculate the surface of the semicircle like if it was an semi-ellipse. So : 1/2 * Pi * 2v * t .

Conclusion :

We can either say that both scientists are right because in this case, v = t/2 numerically and they both calculate the surface of a semicircle in a correct manner : they both find the correct numerical value for the answer.

Or we can say that they are both wrong because each of them finds his answer from an equation that doesn't consider the physical units, and that each of their expression screw up with the units.

Anyhow, to calculate the surface of the semicircle as a semi-ellipse will result in finding the same numerical value that both scientists will also find.

BUT!!! and there is the very question : RollerCoaster is absolutly right about the fact that there is an infinite acceleration problem and that both scientists are discussing an impossible matter in the first place...


Sorry if my novel was too long



[This message has been edited by Nienscecco (edited 04-17-2003 04:02 PM).]

[The Cruciverbalist]

~decides, against his better judgement, to make another comment about a math-ish puzzle and more than likely make himself look like a fool~

I can't see why m/s can't equal s. When I simplify, I get m=s². It seems to me that any number and its square would fill the bill here.


------------------
Um... Seriously. - Homestar Runner

There needs to be a better word for weird. - Strong Sad

[What if...]

Cruciverbalist, meters can't equal seconds squared, because this would mean that when you square a time, you would suddenly get a distance! Fundamental units like meters and seconds do not have such a relationship, and what relationship they do have involves constants like the speed of light.

Nienscecco, I agree w/ everything, except the scale thing is an issue because if the intervals are 10 seconds on the x-axis, and the velocity has 1 m/s intervals, then what looks like a semi-circle would really be a semi-ellipse, and the two mentioned formulas would be wrong numerically, even. The v times t solution would survive these troubles, though.


[This message has been edited by What if... (edited 04-17-2003 05:32 PM).]

[pokerfaced]

I think that what is being said is that the fact that we have been told the graph is a semicircle is enough to assume that 2v and t are numerically equal. Symbolically:

v = x m/s
t = 2x s

Therefore we would have a semicircle, although the measurements aren't equal, for velocity can never be measured as time. However, the graph itself would represent a semicircle.

However, the area under the curve is meant to represent the total distance traveled. If we square time, the units are s^2. If we square velocity, the units are m^2/s^2. Neither result in a distance.

This means that although both are numerically right, neither is right in a practical sense.

[Nienscecco]

yes, we have to assume the semicircle pattern. It's part of the given info. Anyway, it's also part of what makes this problem a puzzle. Remove the notion of semicircle geometric representation and this will no longer be a puzzle, it will be a plain physics - cinematic problem including some algebra to express the solution.

The "essence" of solving a puzzle concerns "that tricky thing" that makes everyone wondering...or put in evidence a nonsense fact of reality contronted to a farely known kind of representation.

In this case, it is basically a conflict between the "geometric representation" of the problem and its application to fit the reality.

And reading everyone's posts, I think we've made the complete résumé of the problem and we already have the solution. It's all about understanding what the geometric representation of the problem lacks of...and then explain why both scientists can be right at the same time and also what is not very "scientifically correct" about their answer, considering the units into the equation...

Anyway, I'm hurried to see the official solution though.

[This message has been edited by Nienscecco (edited 04-18-2003 09:39 PM).]

[Nienscecco]

By the way, How many posts does it take to level up from Icarian member to another title? Just wondering...

[Lepton]

30 - it'll be gone in no time

[TomP]

Question: Can you find an example where the gravitational force decreases when two bodies are brought closer together?

Example: if a point "particle" were brought near an object shaped like a tire/donut along a line extending through a point that is in the middle of the tire/donut (equidistant to the edges), then the closer the point particle got to the center the less "force" would be "felt" because in the exact middle of the tire/donut, the apparent "force" would be null. This would be true of both the point particle and the tire/donut. Though one can quibble that this is only an "apparent" effect, well, so is gravity, which is not really a force.

[CrystyB]

#2

[What if...]

quote:

---

Though one can quibble that this is only an "apparent" effect, well, so is gravity, which is not really a force.

---

Since when? In relativity it is not treated as a conventional force, but this does not mean it is merely an "apparent" force, or somehow illusory. The same goes for the effect you mention, assuming the particle has zero diameter.
Whether or not the unit thing matters depends on how strict you want to be about your answer. I like the combination v and t because it gives correct answers and

[Jatan]

I don't think that the units issue is relevant. In Physics it is a common practice to operate with normalized unit-less quantities, acctually if you want to calculate the example properly and integrate you MUST first introduce normalized t and v (ie. : x = v/vmax, y=t/tmax) Quantities x and y are now without any units. Thus both scientists are right numericaly, yet they are way off if one wants to evaluate the calculation procedure as such.

[CrystyB]

i know that the antiparticles behave electrically reversed (so the e and a-p would reject each other, both being negative) but i thought that since an e and an a-e would collide in a little explosion/implosion/whatever, perhaps e and a-p wouldn't want to meet each other... It seems that i was wrong.

[cha]

For 1: There is no need to use two masses connected by a string. One would do. Imagine a large ball of twine with a diameter "d", with a string leading upward for some distance, say "10d". An object that begins at distance "d" to the right of the ball portion, then moves upwards toward the end of the string would experience decreasing gravitational force throughout almost all of the journey.

[Rollercoaster]

I guess my issue with #2 is that the two answers given were NOT calculated values of distance. They were equations. At least that's how I interpreted their responses. And whether or not those equations lead to the right answer when the calculations are carried out is irrelevant - the equations themselves are still wrong because of the mis-matched units.

What if I said, "What a coincidence - my height is exactly equal to my weight". Does that really mean anything at all? I don't think that statement is usable in any way without more information.

[What if...]

Cha's answer is what I meant by B. I like my two mass answer because getting closer to the center of mass can decrease the gravitational force, so it is an answer whether you define closer by distance to the object, or distance to center of mass.
i cant spel!

[This message has been edited by What if... (edited 04-21-2003 02:49 PM).]

[ckjgirl]

for the second one i got a little confused wouldn't it be t/2*pi

[Lucky Wizard]

Why would it be t/2*pi? I can't see how... ah... I think I see where you're confused.

You are supposed to be finding the area under the curve, not the length/circumference of the curve. The graph is not a picture of the path the car took; it is a graph of the car's velocity (speed) at any point in time. For instance, at the very beginning and the very end, the velocity was zero, since on the graph, a hypothetical marker at which the time was zero or t intersects with a hypothetical marker at which the velocity was zero. And from the graph, you can see that in the middle of the journey, the velocity was v. So if you had a graph of an object's velocity versus time and you wanted the distance the object had traveled in that time, how would you find that? You'd check the area under the curve. (The reason for this is complicated and requires being able to handle stuff like limits, the slope of the tangent line, Riemann sums, and the constant of indefinite integration. For now, simply accept that area has this property.)

So the area is what we're working on, which is why the puzzle specifically asks you to find the area. If the graph were indeed a (semi)circle of radius t/2, and the puzzle asked you to find the circumference, then yes, your formula would be the answer. But the puzzle is to determine the correct area.

[OcularGold]

1. I don't consider going closer to one point of an object yet getting away from the center of mass of the object as actually getting closer to it, since youre getting further away from most of it. but maybe im just being nitpicky about it.

my answer? a white hole with just about anything else would work.

2.

[What if...]

Would a white hole work? I have almost no knowledge regarding them (despite my interests), so I don't know...

[wolfinabag]

Of course, if we want to discuss solutions as inelegant as anti-matter or white holes here, we could also imagine, say, hypothetical objects who's masses decrease when other objects are brought nearer to them.

Prob 2
It seems many here quibble about the question of units. There is no fault in the quesiton in this regard, however, since units don't define distance. If 2 objects are 1 meter apart, they are also 0.001 km apart, but the distance is the same regardless. The point being, the two embarassingly dull self-proclaimed "scientsts" (lab technitions, I'd prefer to address them ass) would arrive at the same answer regardless of the method they used to calculate it. Just as you could find the area of a square by squaring the lengths of it, or taking the length of the diaginal and calculating the area of the two trangles, or some other infernally long winded method, you will arrive at the same answer in the end. It's that simple, really.

[OcularGold]

[mikegoo]

Disclaimer: Nothing of improtance will be found in this post.

[pk]

I agree that units are essential in the speed/distance puzzle. When you multiply numbers you are multiplying the numerical values AND the units together (this is why you can't add or subtract quantities that have different units). Often you leave out the units because it's obvious, but here it is critical. Bearing this in mind, if we look at what these two mathematicians have actually done:

the first one has (pi/2)*v^2 metres^2/seconds^2, which is not a distance!

and the second has (pi/2)*(t/2)^2 seconds^2, which is also not a distance!

So they are both wrong! They do, of course, both get the same numerical answer. If you treat it as a semicircle:

(pi/2)(v)(t/2) (metres/second)(seconds)=(pi/2)(v)(t/2) metres ---which is a distance.

By multiplying only the values, the mathematicians are in fact finding the area ON THE PIECE OF PAPER (or, okay, yes, on the screen!) under the semicircle (we know that it is a semicircle in the diagram beacuse we are told so). However, this is NOT the area in V/T space, which is what the diagram is representing, and it is the area IN V/T SPACE under the curve (whatever shape that curve may be [eg circular, elliptical] by virtue of the scales chosen in order to represent it diagrammaticlly) which is the distance required.


That's what I think anyway!

For the gravity one:

Imagine the Earth, Sun, and a third, much smaller body all in a straight line in space. There is a Lagrangian point between the Earth and the Sun, which is a point where the gravitational forces cancel out. Imagine that the third body is a "test mass" (ie. it can experience gravitational forces but doesn't have enough of its own mass to upset the rest of the system) and is on the surface of the Earth to start. Move it away from the Earth and as it moves towards the Sun, and whilst it is between the Earth and the Lagrangian point, it will experience less and less force, until it experiences no force at the Lagrangian point. Is this any good?




[This message has been edited by pk (edited 04-29-2003 12:21 PM).]