# Hills and valleys

| Comments (7) | Misc Hovav Shacham ran a small physics problem by me the other day. You have a body rolling along a flat, frictionless, track of length L, as shown in "A" above. Assume it's been given a single initial impulse so it's moving at speed S. We all know that it will take time L/S to traverse the track completely.

Now, consider the diagrams labelled B, C. These are two tracks with the same horizontal displacement as A, but we've added either a hill or a valley to the middle of the track. The diagrams aren't to scale and don't assume that shape of the hill/valley is an arc, even though it's shown that way, but you can assume that:

• There's only one hill/valley. It's not rolling.
• B and C are mirror image of each other. I.e., they're equally deep and high with identical slopes.
• All tracks are symmetrical around the dashed line "Y"
• The ball is moving fast enough to complete each track (in particular, to clear the top of the hill in B).

So, two questions:

1. What's the relationship between the time the ball takes to traverse B and C? You don't need a numerical answer, just equal to, less than, or greater than.
2. The same question as above, but for A and C.

This doesn't require calculus or detailed mechanical calculations, just simple qualitative physics and some intuition.

Answers after the break.

Question 1: This is relatively straightforward, though a lot of people's intuition leads them wrong, since intuitively it feels like B and C are symmetrical so the result should be the same. I'll work through this in enough detail that you can see that that's wrong.

Obviously, B and C are the same length, which we can call L' (L' > L). Because of conservation of energy, the ball must be rolling at speed S at both X and Z, and between the start and X and Z and the end. In other words, while the ball speeds up going down the hill it needs to slow down by the same amount going up. So, we can ignore the flat sections of the track because they're the same and focus on the non-flat sections.

Looking at track B, we can see that while the ball starts with speed S, it slows down as it climbs the hill till it reaches Y. Then it speeds up to the point Z, where it is again at speed S. So, if we write its speed as a function of horizontal displacement, d, we can say Sb(d) ≤ S. By contrast, if we look at track C, we can see the opposite. The ball starts with speed S, but slows down as it descends the hill till it reaches Y. Then it speeds up to the point Z, where it is again at speed S. So, if we write it's speed as a function of horizontal displacement, d, we can say Sc(d) ≥ S.

Since Sc(d) ≥ S ≥ Sb(d), and we know that at least at point Y, the ball in C really was moving faster (actually, it's moving faster everywhere but on the flat spots), and the lengths are the same, we can conclude the the ball traverses C faster than B.

Question 2: I originally told Hovav I would need a pencil and paper and probably calculus to work this problem, but you don't. The short answer is that B is always slower than A, but C can be slower, faster, or the same speed. To see that, consider the following diagram, D, which is has a clearly non-circular hill. Again, using only the conservation of energy, we can see:

• The ball travels with speed S from the start to X.
• From X to L it accelerates to speed S(L) > S.
• From L to M it moves at a constant speed S(L).
• From M to Z it decelerates to speed S.
• From Z to the end it travels with speed S (using the fact that the track is symmetrical around Y).

As before, we can ignore the regions start-X and Z-end, since they are the same in both diagrams. What we need to concern ourselves with is the horizontal component of speed. Now, consider the section between L and M, which is flat on both tracks. Clearly, the ball is moving faster on the lower track: i.e., it's moving at S(L) > S. So, the time the ball takes to travel between L and M in the lower diagram is less than the time it takes in the upper diagram. It's greater by (M-L)/((S(L) - S)). Since S(L) - S is a constant, we can make the difference arbitrarily small or large by making the L-M distance longer. Clearly, then, even if the time it takes to get from X to L and M to Z in D is greater than the equivalent time in A, we can compensate for this just by making the L-M distance larger. Clearly, then, the track with the dip in the center can be faster than the flat track.

But I said it depends, so let's look at how it can be slower. Finally, consider track E, shown below. In this diagram, the ball gets minimal additional horizontal velocity. It gets to go very fast, but it's almost all happening in the vertical direction. By making the well in the center arbitrarily deep, we can make the travel time to the bottom of the well arbitrarily high. The bottom line, then, is that depending on the shape of the dip, the travel time can be either greater than or less than the flat track. It it requires actual math to determine which one it is in the general case, though.

UPDATE: People are complaining, I think rightly, that there's something wrong with case (E). The problem is at least in part, as Bob McGrew observes, that the horizontal component of the ball never goes down. So, I've been thinking of the ball as following the track, but in this case it would clearly fly off and go bouncing into the far wall, at which point things get fishy. We could talk about a different system where it's attached to the track. In retrospect, I worry that that's a problem with this whole question unless (even if?) the shape of the curves is very carefully designed.

This smells wrong to me. Shouldn't the horizontal component of velocity be unaffected by the fall, given the assumptions? (Specifically, the ball has to continue rolling smoothly and frictionlessly - I'm imagining a very small ball riding down your well so it doesn't bounce back.) The horizontal component of the velocity should increase at least somewhat before decreasing. In which case, C should always be faster than A.

What am I missing?

I really don't think that is right. The potential energy is the same. Sapping it, obviously, leaves somewhat less potential, while providing it gravitational pull gives it back. Friction is a problem. Which is why point E is silly - you've made friction too hard to overcome at point Q. I could come up with a number of reasons why this is bad science, but really, I don't think you want to defend this.

If the surfaces are frictionless, why is the ball rolling and not sliding? It always annoyed me in mechanics problems in high school that they'd say the surfaces were frictionless, and then talk about balls rolling on them.

I didn't read Bob's post before sending my frictionless comment, but that's the key here. Your intution about balls speeding up is that rolling down a hill will make it spin faster. That's not true in a frictionless setup, so as Bob points out, the only change in velocity is in the y-axis; gravity will not change the x-axis velocity if there is no friction.

Clearly, you have to nail down a particular model of how the ball travels. If we assume the ball is rolling on a track with zero friction loss and zero ability to bounce or otherwise leave the track, then speed can be computed strictly by converting potential energy (height) to kinetic energy, i.e., mgh=0.5*mv^2. Thus, the speedup should be proportional to the square root of the change in height. You could then integrate that across the valley / hill.

Case E points out where the above is an approximation. The deeper you make the well, the longer the path traveled by the ball, so naturally it's going to take longer to go along the track. Conversely, if you assume that the height transition is insignificant relative to the length of the track, then the conversion of potential to kinetic energy happens immediately and the total distance traveled is (approximately) constant.

It's a fairly complex force problem that you'd have to use integration to solve, but in simple terms you just think about the x-axis acceleration:

The only force in the x-axis is provided by the normal from the slope (assuming frictionless sliding). This means that the ball speeds up due to the normal force being off-vertical on the downward part of the slope and similarly slows due to the normal on the upward part of the slope.

Where this breaks down is if the slope is forcing the ball down when the shape of the slope forces the ball to move downwards faster than gravity pulls it (ie pulling "negative" g's). In those cases the normal force actually slows the x-axis velocity as it is pulling backwards.

I disagree with the conclusion that E will be slower than A. Application of conservation of energy is not very intuitive here, it is easier to understand this using conservation of momentum in the horizontal direction.

If the movement is frictionless and the body remains in contact with the surface at all times, there are only two forces acting on it. The force of gravity is vertical and therefore never affects the horizontal component of the velocity. The force of surface reaction is always normal to the surface (there is no friction), so it accelerates the body horizontally on downward slopes and decelerates on upward slopes. Thus, it is obvious that in cases C, D and E the horizontal speed at the same horizontal distance from the start will be always greater or equal than horizontal speed in case A, so the total travel time will be less. The reverse is obviously true for B.