## Rolling to the top of a hill

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ihope127
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### Rolling to the top of a hill

Suppose I have a ball on a hill. The ball is rolling directly up toward the top of the hill, such that its total energy is equal to what it would be if it were at rest at the top of the hill. So, it has enough energy to get to the top, but does it actually do so in a finite amount of time? Does it then roll back down? Which way?
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scarecrovv
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### Re: Rolling to the top of a hill

Do you mean a ball and hill that could physically exist, made out of actual atoms? Or do you mean an idealized, no friction ball and hill made of wonderflonium, existing in a perfect vaccum that behaves like the things you learn about in an high school physics class? There is a difference. If you have a real ball on a real hill, then the final result depends on exactly what they're made of, and the exact geometry we're discussing. If the top of the hill is flat, and the hill (or the ball) deforms ever so incredibly slightly under the weight of the ball, and there's a little friction, the ball may come to rest at the top if it starts with precisely the right velocity and you have a little luck. If the top of the hill is pointy, I can't imagine it stopping under any circumstances, and the direction it falls towards is anybody's guess.

In the wonderflonium scenario, yes, the ball will come to rest at the top of the hill. Whether it does so in finite time depends on the exact geometry.

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### Re: Rolling to the top of a hill

in the real world there will be to much chaos to predict i.e. you may or may not get a gust of wind that makes the difference.
in an ideal world, if it only changes its kinetic energy to gravitational potential energy it will reach the top, and in finite time (i think you may be trying to reinvent zeno's paradoxes ), if it changes any kinetic energy to any other form e.g. sound or heat, it won't reach the top
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antonfire
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### Re: Rolling to the top of a hill

What makes you so sure it will reach the top in finite time?

Spoiler:
It needn't. In fact, it won't.

One way to convince yourself of this is to note that the less kick you give an object sitting at the top of the hill, the longer it will take to roll off, and there is no upper bound on how long it can take. Then reverse time.
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### Re: Rolling to the top of a hill

antonfire wrote:What makes you so sure it will reach the top in finite time?

Spoiler:
It needn't. In fact, it won't.

One way to convince yourself of this is to note that the less kick you give an object sitting at the top of the hill, the longer it will take to roll off, and there is no upper bound on how long it can take. Then reverse time.

Spoiler:
You've argued that you can make it take an arbitrarily long amount of finite time. That's different from infinite time.

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### Re: Rolling to the top of a hill

hmm if you simplify to a point mass on the top of an inclined plane then clearly any amount of initial movement results in the point falling in a fixed time as it is just falling with some fraction of g. what's not clear to me is if you even need a push, i.e. can the point mass rest on the infinitely sharp point of the top of the slope, i think that is probably not definable. I think the same hold if you go for a ball instead of a point on a inclined plane.

if there is a finite distance along the flat to the edge from the balls initial starting place, then i think you are right, an infinitely small push would take an infinite amount of time to even reach the edge, so the exact nature of the slope and the starting position may make a difference here
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### Re: Rolling to the top of a hill

crazyjimbo wrote:
Spoiler:
You've argued that you can make it take an arbitrarily long amount of finite time. That's different from infinite time.

Spoiler:
Suppose the potential energy at the top is E. antonfire argued that given any time T, there is a small amount of energy dE such that a ball with kinetic energy E+dE at the bottom will take longer than T to reach the top.

So if the ball with kinetic energy E at the bottom takes a finite time T to reach the top, then by adding more kinetic energy we can make it take longer than T, which is a contradiction.
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ThomasS
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### Re: Rolling to the top of a hill

If the integral curve of a Lipschitz vector field ends at a point where the vector field is zero, then the integral curve does not reach the point in finite time. (Actually, this might be true when the field is only continuous.. trying to remember now.) Perhaps the easiest way to see this is by time symmetry. If it would never start moving from a point, it will never reach the point either. All bets are off if you allow the field to be discontinuous, or simply non-Lipschitz.

So yeah, it takes infinite time to stop, unless the hill is a bit odder than one normally imagines. Actually, the roundness of the ball probably makes the gradient of the energy function Lipschitz essentially no matter what you do with the hill itself.

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### Re: Rolling to the top of a hill

In an absolutely perfect universe in which the only objects that exist are this ball and this curve, where each object is perfect and smooth even when infinitely magnified, where values are known to an infinite number of sig figs, in which there are no quantum effects... A perfect universe, the ball will never actually reach the top of the hill. It will just continue to make an ever smaller amount progress towards the top.

So if it never gets there it will also never fall.

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### Re: Rolling to the top of a hill

I have to agree that the ball will never stop at the top in this ideal universe.

Basically, the reason the ball slows down/speeds up is because the normal force is not pointing straight up, but at an angle. As the ball approaches the top, the slope is shallower, and normal force is closer to being parallel with gravity. At the top of the hill, the slope is 0, and there is no force pushing the ball left or right, so the ball has to /enter/ the infinitesimally small point with a velocity of zero, and would have had leave the previous point with a velocity of zero. the problem, of course, is that in order for this to be true, the ball cannot move at all, and so has to /start/ at the top at rest to /end/ at the top at rest.

If, at the point next to the top, the ball's speed ended at zero, then it would roll back where it came from. If it ended that point at a velocity greater than zero, it would be on the top with a non-zero velocity, and would roll past the top.

With friction, however, the ball can stop at the top at rest.

RogerMurdock
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### Re: Rolling to the top of a hill

Isn't this essentially the same question as what happens when you fire off an object at EXACTLY escape velocity? It gets going infinitely and infinitely slower, but never actually stops. Except in this case it's a ball and a hill. The difference is that the object fired off can NEVER stop, otherwise it would eventually be pulled back through gravity. With the ball and the hill, the ball has a place to stop that makes physical sense...but I'm not sure if I actually ever gets there. (Assume perfect universe and both bodies are ideal and whatnot and blah blah).

Am I thinking correctly?

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### Re: Rolling to the top of a hill

RogerMurdock wrote:Isn't this essentially the same question as what happens when you fire off an object at EXACTLY escape velocity? It gets going infinitely and infinitely slower, but never actually stops. Except in this case it's a ball and a hill. The difference is that the object fired off can NEVER stop, otherwise it would eventually be pulled back through gravity. With the ball and the hill, the ball has a place to stop that makes physical sense...but I'm not sure if I actually ever gets there. (Assume perfect universe and both bodies are ideal and whatnot and blah blah).

Am I thinking correctly?

Sure, kind of. Except the ball is doing with position and velocity (approaching and approaching but never reaching) what the rocket is doing with just velocity.
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antonfire
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### Re: Rolling to the top of a hill

crazyjimbo wrote:
antonfire wrote:What makes you so sure it will reach the top in finite time?

Spoiler:
It needn't. In fact, it won't.

One way to convince yourself of this is to note that the less kick you give an object sitting at the top of the hill, the longer it will take to roll off, and there is no upper bound on how long it can take. Then reverse time.
Spoiler:
You've argued that you can make it take an arbitrarily long amount of finite time. That's different from infinite time.

Hence, "one way to convince yourself of this", not "one way to prove mathematically that this is the case".
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### Re: Rolling to the top of a hill

evilbeanfiend wrote:hmm if you simplify to a point mass on the top of an inclined plane then clearly any amount of initial movement results in the point falling in a fixed time as it is just falling with some fraction of g. what's not clear to me is if you even need a push, i.e. can the point mass rest on the infinitely sharp point of the top of the slope, i think that is probably not definable. I think the same hold if you go for a ball instead of a point on a inclined plane.

if there is a finite distance along the flat to the edge from the balls initial starting place, then i think you are right, an infinitely small push would take an infinite amount of time to even reach the edge, so the exact nature of the slope and the starting position may make a difference here

I think you're correct. You will end up with the center of gravity of the ball exactly above point at the top of the hill.

If the infinite time were the case, though, the ball would have an infinitely small amount of additional energy over and above the energy from rolling down the slope. Any amount of movement along the perfectly flat surface at the top requires addition energy that isn't in the initial question. It specifically mentions the total energy of the system is equivalent to the energy of the ball at rest on the top of that hill.

I think, from an analysis point of view, we can treat the hill as two straight inclines coming to a sharp point, and the result is that the ball ends up balanced on the point. In the real world, this is of course a messy solution, but there's nothing more wrong about it then ignoring friction.
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### Re: Rolling to the top of a hill

that's funny cos I now think i was wrong. my solution presupposes motion in the time reversed scenario which is at odds with the 'at rest' description in the question, balancing on a point isn't very physical but then we already know we are ignoring lots of other physics
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### Re: Rolling to the top of a hill

If there's no friction, absolutely. Time symmetry, motherfuckers!. You really just reverse it. If you take a shape of hill for which the motion is known analytically, you can find the reverse motion analytically just as easily. For instance, inclined plane. At the top you'd have E=mgh. In order to reach the top, you start at the bottom with that amount of energy, all kinetic, pointing up the ramp. Works for any other shape.

With friction, not even close. Friction breaks time symmetry.

What is absolutely impossible is asymptotically approaching the hill top. You will either reach it in finite time, or turn around and fall again after a finite time.
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### Re: Rolling to the top of a hill

doogly wrote:What is absolutely impossible is asymptotically approaching the hill top. You will either reach it in finite time, or turn around and fall again after a finite time.
No, what is impossible[1] (with the energy specified in the OP) is reaching the hill top in finite time. If you do, you will stay there. Reverse time, and you have a ball sitting on a hill for a while, and then at some point it suddenly starts rolling one way or the other. Oops.

Of course, the two straight inclines coming to a sharp point case has multiple solutions to the relevant differential equations, so that argument doesn't apply. A point mass sitting on the tip can suddenly start accelerating in either direction for no apparent reason whatsoever and still be consistent with the laws of of physics. So in this situation it can and does reach the top in finite time, and the laws simply don't tell you what it does afterwards.

On the other hand, if the thing is actually a rolling ball instead of a sliding point mass, the potential gets "smoothed out" near the tip, so you have unique solutions again, which means it can't reach the top in finite time. (When it's near the tip, it is essentially an upside-down pendulum.)

[1]: In a situation with time symmetry and unique solutions to the relevant differential equations, which is true if the shape of the hill near the top is "smooth enough".
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### Re: Rolling to the top of a hill

if the slope is frictionless then the ball won't roll (dont think this changes much though)
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### Re: Rolling to the top of a hill

antonfire wrote:
doogly wrote:What is absolutely impossible is asymptotically approaching the hill top. You will either reach it in finite time, or turn around and fall again after a finite time.
No, what is impossible (with the energy specified in the OP) is reaching the hill top in finite time. If you do, you will stay there. Reverse time, and you have a ball sitting on a hill for a while, and then at some point it suddenly starts rolling one way or the other. Oops.

Unstable maxima aren't oopses, they're just unstable.
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### Re: Rolling to the top of a hill

Assuming the hill is smooth, I think it does take infinite time.

Example:

Let's say our "hill" is a path shaped like a unit semicircle, and our "ball" is a nice sliding point of mass m. Our coordinates shall be A, for angle ( ) and t for time. It starts at A=0 with E=m*g, and as A changes, E=mg(1-sin(A)). Going through kinetic energy, we get dA/dt = sqrt(2*g*(1-sin(A))).

Despite the idea that this problem would be nice to solve, we end up with a nonlinear differential equation. Blech.

Okay, how about a parabola? That should give a better result, right? Particle starts at y=-1 with E=-mgy on path y=-x^2. sqrt(dy/dt^2 + dx/dt^2) = v, where E=1/2mv^2. Since dx/dt=dy/dt*dx/dy = dy/dt*1/(2*sqrt(y)), v^2=dy/dt^2*(1+1/(4*y)).

So sqrt(2*g*y/(1+1/(4y)))=dy/dt. Dammit, this is also a bad equation to solve. *le sigh*
Last edited by Charlie! on Tue Feb 02, 2010 6:48 pm UTC, edited 1 time in total.
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### Re: Rolling to the top of a hill

The problem is that the constraint that you stay on the hill fails to reverse if you consider rolling up. If I consider rolling up and to the right as the reverse of down and to the left, I can take my initial velocity at the bottom as the reverse of the final velocity from when I fell down, but the problem is, you won't stay on the hill!

So there are lots of cases where you can't just roll back up to the top. For a simple inclined plane, you can.

What can never happen is the infinite time. You will always have some finite amount of deceleration, and it will win. You can't just asympotically approach a peak if there is gravity around - you either make it somewhere you can rest at (and if it's unstable you still fall) or you turn around in finite time. If you think about it should be abundantly clear. You can't stalemate gravity.
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### Re: Rolling to the top of a hill

doogly wrote:The problem is that the constraint that you stay on the hill fails to reverse if you consider rolling up. If I consider rolling up and to the right as the reverse of down and to the left, I can take my initial velocity at the bottom as the reverse of the final velocity from when I fell down, but the problem is, you won't stay on the hill!

So there are lots of cases where you can't just roll back up to the top. For a simple inclined plane, you can.

What can never happen is the infinite time. You will always have some finite amount of deceleration, and it will win. You can't just asympotically approach a peak if there is gravity around - you either make it somewhere you can rest at (and if it's unstable you still fall) or you turn around in finite time. If you think about it should be abundantly clear. You can't stalemate gravity.

Of course you can stalemate gravity. That's what escape velocity is, for instance. How long does something that was traveling at escape velocity take to come to rest?

Of course, you might say something like "gravity becomes infinitely small for the rocket ship, but gravity is still finite for the ball rolling up the hill." The trick is that the component of gravity perpendicular to the surface DOES become infinitely small as you roll up the hill - the normal component is just supported by the hill and doesn't matter.

Okay, anyhow, I updated my last post with disappointing results, but I now see how to do it, so I may as well say it here. The trick is to revisit the circular path, and approximate sin(A) about A=pi/2 with its second-order approximation, 1-A^2! When you do that, you get the nice result that dA/dt = sqrt(g)*A, which is solved by an exponential approach, meaning that our object does indeed take infinite time to reach the top! The approximation only works in a small neighborhood around the top, but it takes an infinite amount of time to pass through that neighborhood.
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### Re: Rolling to the top of a hill

doogly wrote:The problem is that the constraint that you stay on the hill fails to reverse if you consider rolling up.

Charlie, you're still failing to consider this part. If, at any point along the circle, the mass is moving *up* with enough energy to reach the height of the top of the circle, it won't stay on the circle.

And if it's on a frictionless parabola, then enough energy to get it to the top will send it right over, since a parabola is the path it would take even if there were no hill at all.

Nope. I'm afraid doogly's right: It reaches the top in finite time, or it starts falling back down in finite time, or it leaves the hill entirely. Thinking it does take an infinite amount of time is really just falling for a more sophisticated version of Zeno's paradox. And any argument claiming it takes infinite time could also be used to "prove" that a ball you throw up in the air will never come back down.
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### Re: Rolling to the top of a hill

gmalivuk wrote:And if it's on a frictionless parabola, then enough energy to get it to the top will send it right over, since a parabola is the path it would take even if there were no hill at all.

That is not a relevant example. We are trying to get it to come to rest (at least momentarily) at the top. In your example the ball still has lots of kinetic energy when passes over the top (due to the horizontal velocity), so you gave it too much kinetic energy to start with. If you had started the ball with less, the parabola is no longer the path it would have taken without the hill, and it then will almost come to a standstill at the top. It will either go across and roll down the other side, or not quite reach the top and roll back to where it started. The better you judge the correct starting speed/energy, the slower it will go when it crosses the top, or the closer to the top it will reach before moving back again. If you were ridiculously accurate at not rolling it too fast, you could get it exceptionally close to the top before it reversed. Its return journey is then no different that just placing the ball at that point and letting it go. The closer you get it to the top of the hill, the longer it will take to get up to speed and roll down again.

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### Re: Rolling to the top of a hill

gmalivuk wrote:
doogly wrote:The problem is that the constraint that you stay on the hill fails to reverse if you consider rolling up.

1Charlie, you're still failing to consider this part. If, at any point along the circle, the mass is moving *up* with enough energy to reach the height of the top of the circle, it won't stay on the circle.

2And if it's on a frictionless parabola, then enough energy to get it to the top will send it right over, since a parabola is the path it would take even if there were no hill at all.

Nope. I'm afraid doogly's right: It reaches the top in finite time, or it starts falling back down in finite time, or it leaves the hill entirely. Thinking it does take an infinite amount of time is really just falling for a more sophisticated version of Zeno's paradox.3 And any argument claiming it takes infinite time could also be used to "prove" that a ball you throw up in the air will never come back down.
(numbers mine)

Um... Did you notice the differential equation? I thought the differential equation was kind of neat. Did you notice how it took an infinite amount of time? That was nice.

Anyhow, on to the numbers.
1: Not true above a certain angle. Remember that "sliding off a sphere" problem that's real popular in mechanics classes? It's like that, but in reverse. Besides, I said in advance that I was changing "hill" for "path."

2If you wanted something to follow a parabola oh height 1 (potential energy m*g at the top), you'd give it m*g of kinetic energy in the y direction AND some extra kinetic energy in the x direction. By forcing a lower-energy object (total energy m*g) to move along that same parabola you'd change what happened.

3That's fairly insulting. Perhaps the differential equation should have been mentioned here, instead of baseless conjecture (I just now spent 10 minutes doing the same energy method on a ball thrown in the air - it gives a differential equation for parabolic motion, big surprise. So that claim at least is incorrect).
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### Re: Rolling to the top of a hill

Consider the parabola [imath]y = h - (a-x)^2[/imath] for positive constants [imath]a[/imath] and [imath]h[/imath], in constant downward gravity [imath]g[/imath], with a perfectly frictionless, sliding object of mass [imath]m[/imath] beginning from the origin with initial velocity [imath]v_0[/imath] along the direction of the parabola.

Identify [imath]v_e[/imath], the initial velocity above which the mass will leave the track and below which it will stay on the track.

Identify [imath]v_c[/imath], the initial velocity above which the mass will reach the top of the hill and below which it will not.

Determine which of [imath]v_e[/imath] and [imath]v_c[/imath] is greater, and comment on the physical meaning of this inequality.

Calculate [imath]t_1[/imath], the time at which the object reaches its maximum altitude, as a function of [imath]v_0[/imath].

Examine the limit of [imath]t_1[/imath] as [imath]v_0\rightarrow v_c[/imath] from above and from below.

Prove or disprove that, for any [imath]T > 0[/imath], there exists a [imath]\delta > 0[/imath] such that if [imath]|v_0 - v_c| < \delta[/imath] then [imath]t_1 \geq T[/imath].

Comment on what this means physically.
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### Re: Rolling to the top of a hill

doogly wrote:You will always have some finite amount of deceleration, and it will win.
Not if the amount of deceleration is decreasing fast enough.

gmalivuk wrote:And any argument claiming it takes infinite time could also be used to "prove" that a ball you throw up in the air will never come back down.
The situations are different because a ball thrown in the air has a constant downwards acceleration, whereas a ball rolling up a hill has a downward acceleration which can go to zero.
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### Re: Rolling to the top of a hill

gmalivuk wrote:
doogly wrote:The problem is that the constraint that you stay on the hill fails to reverse if you consider rolling up.

Charlie, you're still failing to consider this part. If, at any point along the circle, the mass is moving *up* with enough energy to reach the height of the top of the circle, it won't stay on the circle.
The constraint forces necessary to keep the ball on the hill when you reverse time are identical. If it doesn't leave the hill going one way, it won't leave it the other way.

gmalivuk wrote:And any argument claiming it takes infinite time could also be used to "prove" that a ball you throw up in the air will never come back down.
Wow you really didn't read any of the arguments did you.
doogly wrote:Unstable maxima aren't oopses, they're just unstable.
Hm, neither did you. Oops.

The fact that it's unstable is irrelevant. The fact that if it reaches the hill in finite time you will have multiple solutions to the differential equation (with the same initial conditions) is the point. For any "smooth" hill, this won't happen.

Just solve the freaking differential equation if you're still unconvinced.

Hell, I'll do it for you. Mass, gravity, etc. are all 1 for convenience. Consider a point mass sliding on a hill whose shape is a cycloid, so that its height below the top of the hill is proportional to half the square of the distance to it (measured along the hill).

If s is that distance along the hill, V(s) = -s^2/2, so the Lagrangian of the system is L(s,s') = s'2/2 + s2/2. The differential equation is s'' = s. One solution to this differential equation is s = e-t.

To preemptively put down any complaints about the constraint forces, as you get closer to the origin, the necessary constraint force approaches 1, pointing straight upwards, so certainly in a neighborhood of the tip, the hill is pushing the thing out, not in, so it won't fly off. (Which is intuitively obvious anyway since it's going so slow.)

Seriously you guys. It's not that hard.
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### Re: Rolling to the top of a hill

Yeah, sorry. I guess the problem with trying to intuit this without actually running the numbers is that my intuition is going behind my back and secretly including real-world things like friction and non-point masses and such, so that what I picture in my head as happening isn't actually what would happen in the idealized situation we're setting up here.
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### Re: Rolling to the top of a hill

It's worth noting that the ball only takes an infinite amount of time to arrive if we assume that the top of the hill has a convex slope. (I.e. a slope defined by a curve with a negative second derivative.) If the net force on the ball from gravity isn't approaching zero as the ball reaches the top, then the ball gets up just fine. It's reasonable to assume this, since most hills in the real world look like this, but the OP doesn't provide any information about the slope of the hill, and we're not in the real world. If we assume, for example, that the slope of the hill is a straight line, then Gmal's comment about this being the same as throwing the ball up in the air is correct.

Also, is this at all affected by the rotational momentum of the ball? My intuition is that in our perfect Newtonian world the transfer of energy between the rotation and the motion of the ball is one for one no matter how screwy the slope is or whatever else you do to it, but I don't have the brainpower right now to actually do the math.

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### Re: Rolling to the top of a hill

firechicago wrote:It's worth noting that the ball only takes an infinite amount of time to arrive if we assume that the top of the hill has a convex slope. (I.e. a slope defined by a curve with a negative second derivative.) If the net force on the ball from gravity isn't approaching zero as the ball reaches the top, then the ball gets up just fine. It's reasonable to assume this, since most hills in the real world look like this, but the OP doesn't provide any information about the slope of the hill, and we're not in the real world. If we assume, for example, that the slope of the hill is a straight line, then Gmal's comment about this being the same as throwing the ball up in the air is correct.

No, it is specified as a ball. A ball could not even roll to the top of a perfect needle in finite time and stay there. Since the ball is an infinitely differentiable shape the force has to approach zero when going towards an unstable equilibrium.

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