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During the last family get together, my nephew was bragging about his newly acquired bowling skills. Apparently, he has this amazing talent to bowl a ball at precisely his desired speed. I was feeling a bit annoyed by all the fuss, so when it was time for ice cream for everyone, I came up with a challenge to settle it once and for all. He could have his ice cream, but only if he could bowl the shiny red ball up the slide in our backyard so that it came to a complete stop at the top of the slide. He learned about conservation of energy at school, so he made some calculations, found the desired speed, and bowled. Did he ever get his ice cream?

my nephew bowling the shiny red ball up the slide

Clarifications:

  • Assume all high-school physics idealisations. For instance, there’s no friction anywhere (not even air), no rolling, etc.
  • The floor is smoothly connected to the bottom of the slide so there are no bumps anywhere. But you shouldn’t need the details of the shape of the slide.

Source: I’ve heard the underlying idea behind this puzzle in a physics class, and I’ve since come to know that it’s pretty well-known, at least within physics community. I don’t think it was asked before on this site, but if I’m wrong, I’m happy to close the question as duplicate.

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    $\begingroup$ I guess if he secretly left his chewing gum at the top of the slide, he'd have a chance... $\endgroup$ Commented 7 hours ago

2 Answers 2

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This answer uses the smoothness of the system so I'm not sure it's the most general.

The high school physics setup describes a system that is

Time reversible. If q(t) is a trajectory, then q(-t) is another solution. In this case a ball that starts from rest and slides down the hill.

But staying at rest at the top forever is another valid solution to the initial conditions (at top, v=0)

These are two different solutions for the same initial conditions, violating uniqueness. Now I invoke the smoothness of the slide to say the solution must be unique (this is a standard theorem depending on Lipschitz continuity, which we have). Since permanent rest at the top is a fine solution, the other trajectory must be something else (one that takes infinite time). This is in contrast with Norton's dome which has a cusp at the top.

Poor nephew!

If the top of the slide is parabolic we get an example where traversing

a finite distance takes infinite time.

Let's suppose we trace the parabola y = x^2. When we start at some (x0, y0) we must have kinetic energy = gravitational potential energy: 1/2*m*v0^2 = m*g*(-y0). Notice that the velocity is proportional to the square root of the vertical distance remaining.

What does it look like when we're halfway there? Now we're at position (x0/2, y0/4). We have used up 3/4 of our kinetic energy. Now our velocity (v1) is given by 1/2*m*v1^2 = mg(-y0/4). After substitution v_1=v_0/2. Ah.

The key is to notice that the remaining parabolic trajectory, starting from (x0/2, y0/4) is exactly a scaled copy of our original situation. We have climbed, but we haven't made any progress.

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  • $\begingroup$ I wasn’t aware of these subtleties before. But that’s essentially what I had in mind. $\endgroup$ Commented 2 days ago
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    $\begingroup$ I think there's another argument reveal spoiler showing vertical distance decays exponentially if the top smoothly curves to horizontal that I didn't want to work out. If someone presents that they'll get +1 from me! $\endgroup$ Commented 2 days ago
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    $\begingroup$ @TimSeifert. A quick googling told me that smoothness implies locally Lipschitz and restricting to a compact interval makes that Lipschitz. Since the slide is smooth and compact, I think the forces you get by going along it are Lipschitz. $\endgroup$ Commented yesterday
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    $\begingroup$ @TimSeifert. Smoothness implies all derivatives are smooth, but that’s not the case with Norton’s dome. We need the forces to be Lipschitz and forces are some derivatives of the function tracing the slide. $\endgroup$ Commented yesterday
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    $\begingroup$ Note that even if the slope follows that of a Norton's Dome, you have to consider that the center of gravity of the ball is at a fixed distance of the slide. This smoothes out the shape of the dome. Around the top, the path of the center of gravity approximates to a parabola with no funny derivatives. $\endgroup$ Commented 29 mins ago
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This is the same answer explained in another way.

Start by assuming that the nephew succeeds and the ball stops at the top. Given that:

- If the ball is on an angled surface of the slide, it must still have some speed. If it was out of speed, then it would start to roll back down. It must still be rolling up the slide.
- Once the ball gets to the flat part at the top, it can now lose speed and stop rolling.
- There is no friction or air resistance so there are no forces that can slow the ball down.
- The ball has to have some speed at every point that is not flat but it has to have zero speed exactly when it reaches the flat spot.
- If the transition from the vertical to horizontal could be very, very stretched out so the ball slows down more and more, but it could never be long enough that the ball would actually have zero speed when it was finally flat.
- Whether you solve this by saying it doesn't stop at the top and rolls off the back of the slide or you say that the transition area is infinitely long and it takes infinitely long to stop, the result is the same. The nephew never gets the ice cream.

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    $\begingroup$ I don't think the reveal spoiler transition area needs to be infinitely long to get infinite stopping time. For instance we might launch a sled up a parabola and it spends an eternity decelerating near the top. $\endgroup$ Commented yesterday
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    $\begingroup$ Thank you. I read the other answer first, and while I'm sure it's correct and insightful, I didn't understand it at all. This answer made it clear to me. $\endgroup$ Commented yesterday
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    $\begingroup$ Agree with @n1000 - the ball can roll up the slide and slow down enough that it never reaches the top of the finite-sized slide, yet never rolls back down either. It's perfectly possible to take an infinite amount of time to traverse a finite distance if your speed decays fast enough. The "infinite transition" can help conceptualize it, but the final bullet seems to suggest that either 1) the ball rolls off the back, or 2) the ball rolls back down the front, or 3) the slide is infinitely long and the ball doesn't come back down. But the ball can also 4) spend forever ascending a finite slide. $\endgroup$ Commented yesterday
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    $\begingroup$ @Pranay Indeed, I thought the specific use of ice cream was clever - even if the nephew suggests you just wait "a bit longer", eventually his ice cream reveal spoilerwill melt! $\endgroup$ Commented yesterday
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    $\begingroup$ This sounds to me like you're arguing that there can be no smooth real function that has a nonzero derivative at some point but is still eventually constant. But this is definitely false (a standard example being exp(-1/x^2) for negative x and 0 else). But how would you rule out such a behaviour? $\endgroup$ Commented yesterday

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