salaw
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Posts posted by salaw
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Wikipedia says: " ... then each point on the surface will have an actual gravitational potential proportional to the value of Φ at that point. As a result, an object constrained to move on the surface will have roughly the same equation of motion as an object moving in the potential field Φ itself."
Rolando
I have one quick observation here. If the potential function for the marble constrained to remain on the surface of the funnel matches the potential function of an object, which we call the "planet", traveling in space near a gravitating body, which we call the "sun", then the marble's path projected into a plane will certainly not match the planet's path. Note particularly that since the path of the planet remains in a plane, and the path of the marble will not, the lengths of the two paths are not the same.
Just look at it as a problem in potential energy, and think about the case where the marble is "orbiting" in the funnel. The speed of the marble at each point on its orbit can be determined just from its distance from the center of the funnel. The speed of the planet can also be determined solely by looking at its distance from the center of the body it's orbiting. However, the length of the marble's "orbit" will be longer than the length of the planet's orbit, since the planet's orbit is a simple planar projection of the 3-d orbit of the marble. In particular, consider the marble's path as it approaches very close to the center of the funnel, and hence dips very far down the throat of the funnel -- the path length of its "orbit" becomes arbitrarily long, even if we hold its "aphelion" fixed. Conversely, as the planet's orbit is altered to let it approach the sun more and more closely while holding its aphelion fixed, the total path length of its orbit decreases. Since the speed of the marble and the planet are identical at identical distances from the respective foci of their orbits, the result must be that the marble will take longer than the planet to orbit the funnel, and in the limit it must take arbitrarily long.
Of course, it's also the case that a "rolling marble" is a bad choice of object to start with, as some of the energy it gains going down the funnel goes into its rotation. For a better comparison one should assume the object in the funnel slides frictionlessly, rather than rolling. In either case the above remarks still apply.
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Pete doesn't seem to have followed up with the diff eq's he was going to post, and since he also failed to mention something significant about the assignment I'll chime in here.
The thing he didn't mention is that the homework assignment here is "past tense" -- he handed it in last week, sans reduction to quadratures, and beyond saying more work should have been done, the prof neglected to provide a solution. So Pete is left scratching his head over the solution, as I am, since Pete talked to me about it, and I didn't see it either (uh, duh...). In any case this isn't "homework help" being requested here, at least not in the usual sense. At this point a complete solution would be much appreciated and would not be "cheating" in any sense of the word ... and so this is posted in general physics rather than the homework-help section.
Anyhow if we start with
[math]
{{\partial L}\over{\partial q_i}} =
{{d}\over{dt}}{{\partial L}\over{\partial \dot q_i}}
[/math]
where [math]L = T - V[/math], then, starting with the defs Pete gave (above, previous post) for T and V, and after taking derivatives and fiddling a little we get to
[math]
f'_i (q_i)\dot q_i^2 + 2 f_i(q_i) \ddot q_i + V'_i (q_i) = 0
[/math]
Of course the equations have separated, for sure, as each equation has just one coordinate in it. Next step is to "reduce it to quadratures" which apparently means rewriting it in terms of integrals of elementary functions.
Now, I am no differential equation whizz, to say the least, but this looks to me like a non-linear second order equation and I am pretty seriously unclear on how to get to *any* kind of solution from here, short of sticking the thing on a computer and solving it numerically (which we can't do, of course, since f is unspecified).
Goldstein says it can be reduced to quadratures, or more precisely he asks the student to show that it can always be reduced to quadratures (not exactly the same thing, eh?).
If anyone has a clue how to proceed, Pete would appreciate it, and since I got sucked into this, I'd appreciate it too.
Thanks in advance...
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Non-Uniform Motion in a circle
in Classical Physics
Posted · Edited by salaw
multiple post merged
This is a cool problem! I'm not sure I follow the OP's comments on it, but the problem itself is pretty nifty. Note the salient points:
Unfortunately I haven't got time to pursue this tonight, but if nobody else posts a solution I'll fiddle with it a bit this weekend. (Not that I'm at all sure I can solve it -- the first question I listed looks straightforward, but the second and third look hard!)
*********************************************
The foregoing text was written Friday night; the following additional analysis was appended the next morning:
I played around with this one some more this morning. Sure enough, finding the point where P1 breaks away from the cylinder is straightforward.
If we set the potential energy when [math]\theta[/math]=0 to 0, then we have
[math](1)\ V = mgR(\sin\theta - 2\theta)[/math]
where g is the acceleration due to gravity. The kinetic energy is
[math](2)\ T = {1 \over 2} m [R^2\dot \theta^2 + 2R^2 \dot \theta^2]
= {3 \over 2} mR^2\dot\theta^2
[/math]
and the Lagrangian is
[math](3)\ L = mR [{3 \over 2}R\dot\theta^2 + 2g\theta - g\sin\theta][/math]
Taking the usual partials,
[math]
(4)\ {{\partial L} \over {\partial \theta}} = gmR [2 - \cos\theta]
[/math]
[math]
(5)\ {{d} \over {dt}} {{\partial L} \over {\partial \dot\theta}} = 3mR^2\ddot\theta
[/math]
and setting [math]{{\partial L}\over{\partial\theta}}={d\over{dt}}{{\partial L}\over{\partial \dot\theta}}[/math], we find the equation of motion up to the moment when P1 breaks away from the cylinder:
[math]
(6)\ 3R\ddot\theta = g(2 - \cos\theta)
[/math]
If we can solve that, then we'll have the angle and velocity as a function of time. Unfortunately, I can't. But as it happens, to find the angle at which P1 breaks away, we don't need to solve (6) anyway; we just need the velocity as a function of the angle, and we can find that just from the total energy. The system is conservative, and we start with T=V=0, so we can equate T with -V, which gives us exactly what we want:
[math]
(7)\ {3 \over 2}mR^2\dot\theta^2 = g [2mR\theta - mR\sin\theta]
[/math]
or
[math]
(8)\ \dot\theta^2 = g [{4 \over {3R}} \theta - {2 \over {3R}} \sin\theta]
[/math]
And now we need the centrifugal "force". As usual I've forgotten the expression for that, so we'll take a moment to rederive it:
At breakaway, the central force must equal the (fictitious) centrifugal "force". Plugging in the mass of P1 times the expression for the centrifugal acceleration in polar coordinates, to get the centrifugal "force", and comparing it with (8) above, we find that it's equal to:
[math]
(9)\ mR\dot\theta^2 = mg [{4 \over 3}\theta - {2 \over 3}\sin\theta]
[/math]
The central force acting on P1 must be equal to that at the moment of breakaway. The central force is due to gravity, and is
[math]
(10)\ F_c = mg\sin\theta
[/math]
Equating (9) and (10) we find
[math]
(11)\ \sin\theta = {4 \over 3}\theta - {2 \over 3}\sin\theta
[/math]
or
[math]
(12)\ \sin\theta = {4 \over 5}\theta
[/math]
A little fiddling with a calculator indicates the breakaway point is at about 1.1311 radians, which is about 65 degrees. That's well before the top of the cylinder!
Note that the string from P1 is still wrapped partly around the cylinder at that point, so P1 is being pulled toward the cylinder by the force on the string. Consequently, it may very well hit the cylinder again before it finally falls free; we can't be sure from what I've written here. Unfortunately, answering that question, and finding the shapes of the paths of P1 and P2, are both well beyond my limited differential equation skills; to go from here I'd have to use a numerical simulation and the amusement value of such a thing wouldn't pay for the time it would take
.