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Physics problem

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Jun 03, 2007 smittens link
No it's not homework. I made it up while spinning my ship around during a very boring afternoon. So it is quite possible that there is no solution, in which case please let me know and I'll try to fix it. It uses standard cartesian coordinates.

A 1kg particle starts at the origin and accelerates in the +x direction at 3 m/s/s (meters per second squared). From the moment it starts moving, a force pushes perpendicular to the particle's path from below (so at the very beginning, the force pushes in the +y direction, and then starts bending counter-clockwise as the particle moves counter-clockwise). If every time the particle crosses the x-axis, it is 1m further to the right on the x axis, what is the force?

Edit: 2 mil (possibly more) to the first who figures this out. If it even can be figured out.

Edit 2: Okay, in that problem it just goes in a circle. So instead, the acceleration of the original force is increasing at 3m/s/s/s (3 meters per second cubed)

Edit 3: I think as long as the perpendicular force is less than 3N it WOULD spiral, so I'm changing it back to a constant acceleration.
Jun 04, 2007 toshiro link
Acceleration is meters per second squared.

1 kg, 3 m/s^2 thus Nx = 3N
Ny = obviously f(Nx). Is Nx constant?

So, the particle moves in a spiral? Your problem seems a bit oddly worded.

"If every time the particle crosses the x-axis, it is 1m further to the right on the x axis, what is the force?"

Do you mean, it rotates around the origin, and moves radially, as well?

Also, do you mean 'crosses the x-axis' only when seen from above, so it does not really touch the x-axis?

This seems pretty easy, but I'd need more info. One thing I can say: stop using cartesian coordinates and change to spherical ones. It will make everything *much* easier. Plus, you can still transpose them back at the end.
Jun 04, 2007 smittens link
Hmmm I don't really know spherical coords...

I guess I do mean acceleration. Nx (the original force) is constant at 3n
Yeah it moves in a spiral, going counter-clockwise around the origin
Jun 04, 2007 Midknight Warrior link
You would probably have more luck in an XYZ plane.

then again, this is just a 10th grader speaking :P
Jun 04, 2007 toshiro link
Behold the power of the wiki.

Basically, spherical coordinates remain a triplet (as should anything in a 3D space), but with radius, planar angle and elevation angle. It facilitates vector calculations and especially cinematics and/or dynamics a great deal.

If by tomorrow evening no-one has supplied a solution, I'll dig out my old stuff and look it up.
Jun 04, 2007 Surbius link
I think spherical coordinates would complicate it for some people if not then most people and it is completely not needed since the particle only has a initial +x force with a constant perpendicular force based upon the line of force and no +/-z force is presented to change the particles path to require a 3rd coordinate number.

From the data presented the object would either be constantly circling in one path in a circle or while time progresses the speed of the particle increases with the given acceleration and the movement would progress along a spiral.
Jun 04, 2007 smittens link
Yes! It spirals! That's why it gets further and further away from the origin!
Jun 04, 2007 Surbius link
Now try to replicate it.

For all of this is just theory.
Jun 05, 2007 toshiro link
Er, yes. Spherical coordinates are not needed, I mistook the axes (I feel like a damn newb). I guess the 'from below' thing put me off-guard, causing me to disregard the complete absence of z-values.

Instead, polar coordinates would be best.

Anyway, a stab in the dark: Ny = 1 N?
Jun 05, 2007 Surbius link
Not exactly sure what you're throwing out there tosh. Is it a answer you are unsure of or is it a problem you came up with?

Anyway, physics is fun! Now just add some chemistry and a little more math and you get this. http://www.youtube.com/watch?v=IHeI4TRoKvo&mode=related&search=
Jun 05, 2007 toshiro link
I was doing some calculations & diagrams by hand and head, and thought that perhaps 1 N might be the force applied, but upon reviewing the problem, that is impossible.

I'll dig up my books on cinematics and solve it after work in the evening. That is, if I'm not too tired.
Jun 05, 2007 Surbius link
Now from what I understand with physics and force is that acceleration never reaches a max speed, it just keeps on going faster and faster. Now whether Enstein's theory that matter becomes heavier when it accelerates is true then this spiraling particle will eventually have to reach a 0m/s^2 acceleration and thus start looping in a circle.

But if matter doesn't become heavier when it accelerates then this constant force of 3N is driving this 1kg paricle in a constant spiral to no end.

I'm sorry if I am completely wrong or just confusing you with babble.

I will draw up some grids and directions of motion of what the particle will do in if it always travles 3m/s or every second it travles it gains 3m/s.

EDIT: This was edited after tosh posted after this post.

The grid shown here is a 100m grid which means to me that each quadrant is 50m by 50m making a span from -50x to +50x 100m.

This set is showing what I calculated to be the particle traveling at a constant 3m/s (3N) with a 1m/s (1N) push that is perpendicular to the forward movement of the particle.

http://i2.photobucket.com/albums/y25/Surbius/PathofMotion.png
http://i2.photobucket.com/albums/y25/Surbius/PathofMotion2.png
http://i2.photobucket.com/albums/y25/Surbius/PathofMotion3.png

Let me explain the above three links to the pictures I took.
The first picture is showing the progression of the particle using +x 3m movement with a +y 1m movement making the paricle change position from 0,0 to 3,1. As the particle progresses the path loops back. I color coded the paths, RED= 1st loop, Yellow= 2nd loop, and Magenta= 3rd loop. As shown, particle is moving along the path of a perfect circle which is outlined in both the second and third picture.

This next set is showing what I calculted to be the particle traveling at a constant acceleration of 3m/s^2 which means (not sure) that it would be gaining 3m/s every second of movement, this also includes the perpendicular push is still a constant 1m/s.

http://i2.photobucket.com/albums/y25/Surbius/PathofMotionA.png
http://i2.photobucket.com/albums/y25/Surbius/PathofMotionA2.png

This set of pictures show what the particle might do if it accelerated an extra 3m/s for ever progressing second. For example, the intial starting speed is 0m/s, one second later it is 3m/s then 6m/s, 9m/s, 12m/s, and so on and so forth. As you can see the particle is climbing quite drastically and with the faster speed increase for every second it doesn't seem to be sprialing at all. Even if I were to progress far beyond the 500m grid I has set up previously (100m grid shown) the angle of climb would lessen every second of acceleration. The change would progress to be so little that it would appear to be (to the human eye from a far viewing distance) climbing at 90 degress and not going beyond that threshold.

I will admit that I am probably wrong but from what I calculated and projected to you, it makes sense in my mind. If you know I am wrong then please by all means prove to me and possibly others reading this what should happen in this scenario.
Jun 05, 2007 toshiro link
Hm, yes. I just assumed that we were talking newtonian physics. Too tired to solve anything, though.
Jun 05, 2007 smittens link
So what you're saying is that it can't spiral?
Jun 05, 2007 Surbius link
Pretty much, yeah.

EDIT: Crap. I hate it when I second guess myself.
EDIT: Actually I countered myself with that lengthy post from my first hypothesis. Now I'm not sure whether my second hypothesis was right.

/me cries
Jun 06, 2007 Roda Slane link
I don't know the exact mathematics, or how to prove anything, but I can tell you several things about the formula from just looking at it.

A) You specified the the +x force was... well.. +x, but the the +y force would always be perpendicular to the objects vector. If you assume the the +x is always parallel to x of origin, then the object will never cross the x axis. For a force of +x > +y, you will end up with some diagonal line of +x/+y to infinity. For a force +x < +y, the object will rotate until +y is exactly opposite of +x, and the the object will travel to infinity -x ( in reference to the origin) at some relatively fixed point of +y. If however, you meant the +x was alway parallel to the objects vector, then we may continue.

B) You specified the +x force in m/s/s, but didn't specify the intensity of the +y force (which would also amount to a m/s/s force). The ratio would not really affect the final geometry, but it would affect the size of that geometry.

C) At this point, I can not predict the exact geometry of the path, but I predict that it will always cross the point of origin, and will always retrace it's original path. (assuming that the object was stationary when both forces simultaneously came into play.)
Jun 06, 2007 toshiro link
Okay, let's re-word smittens' initial post.

We know that there is a constant force with vector u(1,0,0) and its multiplier (correct term?) being 3 kg*m*s^-2 in cartesian coordinates; let this force be Fu. We know that we have a Force perpendicular to the object's trajectory (I shall henceforth refer to the object as P, assuming it to be an ideal point mass, which will become important later on, of mass m = 3 kg). If we assume the force to be constant (which, I think we cannot, eventually), it would look something like this:

Let p be the momentary movement vector of P. Then the vector of the second force mentioned above (let's call it Fv, with its vector being v) must satisfy the equation p*v = 0 (note: this is the dot product, not the vector product) at all times t.

We have a further condition, namely that the trajectory periodically crosses the x-axis (I am assuming only the positive part... am I correct, smittens?), and at every time, the x-value of P's location is increased by 1 m.

So much for the rewording.

I agree with Roda generally on point A, except that it failed to mention or include momentum, point B is partially wrong in that force is measured in Newton (1 N = 1 kg*m*s^-2, not in m*s^-2, which is acceleration. And yes, this is very important), but otherwise correct as I see it. Especially important is the part about whether Fu is only initially collinear with the x-axis.

Subsequently, we can say that if p equals u, i.e. they are identical save for scalar factors, we can reduce it to discussing said scalar factor, comparing it with Fv's. I will call them a for Fu and b for Fv.

If they are identical (a = b), I think P will in fact spiral, but the increments on the x-axis would not be identical, but steadily increasing (since P's velocity will be increasing).

Not enough processor time available for other computations atm.
Jun 07, 2007 Roda Slane link
I believe I may have made a mistake of my analysis. Specifically, that if +x is alway parallel to the origin, and +x < +y, you may in fact achieve a spiral.

toshiro, Any force of N, can be converted to m/s^2. Would I be correct in deducing that the correct formula would be: N/Mass = meters/s^2
Jun 07, 2007 toshiro link
Yes, you are correct about the formula. But a force is measured in kg*m*s^-2, not in m*s^-2. The latter is called acceleration and not the same as Force. They're like distance and speed (m and m*s^-1). Closely linked, in fact; one is derived from the other, but not at all the same. Or like energy and power (joule and watts).

So, if you divide a force exerted upon an object by the mass of that object, you get the translational acceleration that the object 'sees' (provided the force is applied at the object's center of gravity. If it isn't, we get another thing entirely, but I'm not going to hold a lecture here, partly because I couldn't).

Edit: So you see, acceleration is the consequence of an applied force. Just like altered distance from a point of reference is the consequence of velocity. Omitting the trivial cases of the respective values being equal to zero.
Jun 08, 2007 Fnugget link
if i get bored i might try and work it out for real. I need a refresher on this stuff anyway. but in the meantime, i don't think you can can get such a pattern of crossing the axis at constant intervals. that seems to be the consensus around here.