# Two balls on different paths

1. Aug 11, 2014

### Oculatus

I found this picture somewhere:

My question being, as in the picture, which ball gets to the end faster? I guess you could naively say that ball A reaches the end in less time, but then again, there is the force of gravity which would accelerate the ball going downwards (and deaccelerate it going upwards, respectively). In essence, the question really might be: does the gravitational pull accelerate the ball to a high enough velocity to compensate for the increased distance? I hope I'm going into the right direction with this, but I certainly lack the mathematical/physical knowledge to solve this (could the conservation of energy be used?).

I'd appreciate any input.

2. Aug 11, 2014

### phinds

Can you visualize roughly what the velocity vs time graph would look like for B as it goes down the trough and then comes back up?

3. Aug 11, 2014

### Staff: Mentor

I used to think it was obvious -- ball B has a higher average velocity, since its velocity is always greater than or equal to ball A. But on the other hand, ball B has to travel farther. Hmm.

So try picking a couple simple dip geometries to calculate the time it takes for the ball to make it across. Show us your calcs -- we are all interested in the correct answer...

4. Aug 12, 2014

### Staff: Mentor

I figured that the increase in distance traveled is a linear function of the depth of the dip, as is the increase in the ball's energy as it drops that distance. Speed goes as the square root of kinetic energy, so the distance traveled should increase more rapidly than the speed and A wins the race.

5. Aug 12, 2014

### willem2

The question is, can the horizontal component of the velocity vh ever become smaller than v0?
The only force that can reduce vh is the normal force. On a down-slope the horizontal component of the normal force points forward, so vh can only increase. If the hole was symmetrical, we would now be done, because vh on the up-slope would have to be the same as on the corresponding point on the down-slope, so vh >= v0 everywhere.

It seems possible to think of a real steep up-slope, that could reduce vh[/SUB to smaller than v0, but if that happens, the vertical component of the velocity would become so large that the ball would jump up above the starting level.

6. Aug 12, 2014

### bahamagreen

Is there anything changing the horizontal component of either of the balls' velocites?

Unless there is, it looks like a tie, to me.

7. Aug 12, 2014

### A.T.

You don't think that ball B will have a higher horizontal velocity than V0 at the bottom of the dent? See willem's post.

8. Aug 12, 2014

### Delta²

Yes there is, it is a component of the normal force (it is normal to the surface orientation and not normal to horizontal velocity) from the cavity surface that increases horizontal component $v_{h_B}$ to a velocity $v_{hmax}>v_0$. That is during the descent of ball B. During the ascend we have again the component of the normal force that decreases the horizontal velocity from $v_{hmax}$ back to $v_0$.

The total horizontal displacement is the same for both cases but because ball B has a time interval where its horizontal velocity is bigger than $v_0$, ball B wins the race.

Last edited: Aug 12, 2014
9. Aug 12, 2014

### Lok

I do not like the tie, as that would imply that a deeper hole will still end up in a tie, yet gravity is limited and a slightly deeper hole will clearly slow B regarding to A.

10. Aug 12, 2014

### willem2

There's a limit to how deep the hole can become without the ball B losing the contact with the ground. As long as the normal force points upwards it will also point forward on the down-slope, and the horizontal velocity can only increase. If the normal velocity points downwards, the horizontal velocity could decrease, but this can only happen if ball B can't leave the ground. (if it's in a tunnel for example). If the horizontal velocity of ball B can't decrease, ball B can never be slower than ball A.

11. Aug 12, 2014

### Lok

On a better thought this problem can have 2 solutions (if the never breaks contact statement holds) affected by the "ratio" of vo and g.

A very low speed with normal g will favor B.

A very high speed with normal g will favor A. As the hole will act as a bump (never breaks contact).

12. Aug 12, 2014

### A.T.

"Act as a bump" is not a explanation. You have to justify this using forces that act to reduce the horizontal veloctiy of B.

13. Aug 12, 2014

### Lok

As willem2 said above, at high speed (or very low g) ball B would lose contact to the ground, yet having a "does not break contact" constrain that does not necessarily imply a relationship between gravity and speed the hole would only act to create a bigger path.

While I know this is not the scope of the problem, it is there.

14. Aug 12, 2014

### jbriggs444

As I read the picture, the balls are rolling on the surface of the ground. Accordingly, the normal force can never be negative. Together with the "does not break contact" constraint, this limits the set of possible paths and starting velocities and limits them differently depending on gravity.

Unstated, however, is an important assumption that the balls roll without slipping.

Edit: My mistake on the rolling without slipping. "Negligible friction" covers this.

Last edited: Aug 12, 2014
15. Aug 12, 2014

### A.T.

That's how I understand it too. The assumption that B never breaks contact doesn’t imply an actual mechanical constraint that prevents that. It just limits the cases to be considered. And in these cases A can never be faster.

Last edited: Aug 12, 2014
16. Aug 12, 2014

### olivermsun

It's kind of tangential to the problem, but I think rolling without slipping is the opposite case from negligible friction. Otherwise, should the ball start rolling faster when it accelerates in translation? It could just continue rotating at whatever rate while also sliding along the slope.

17. Aug 12, 2014

### Delta²

Er are we talking for ways that the horizontal velocity of B can become smaller than $v_0$? The only way for this to happen is that the normal force is negative (i.e pointing towards the ground) which simply cannot happen, not in this case.

18. Aug 12, 2014

### bahamagreen

If I drive a car off a cliff, the horizontal component of motion remains constant.

But, if I roll the car down a long flat incline, it will increase... because a normal force is present.

I think I see that...

So, since the horizontal component of the distance is the same for both balls, but it is ball B that spends part of its travel in the segment of the dip, within which its horizontal component velocity is always greater than ball A (because ball B's entry speed into the dip equals its exit speed, but is above Vo throughout the dip)... so ball B advances over ball A as B enters the dip and maintains the advantage it holds when exiting the dip to the end, finishing before ball A.

19. Aug 12, 2014

### voko

Let's simplify this. Say $V_0$ is very close to zero. What happens then?

20. Aug 12, 2014

### haruspex

Yes, this is the critical observation: that during ascent the horizontal component cannot increase, yet on completing the ascent it must be back at Vo.

Here are some more interesting questions:
What shapes of dip (a) minimise, (b) maximise the time difference?

21. Aug 12, 2014

### voko

How is that so?

At a deepest point of the track, the entire velocity must be horizontal. If the deepest point is below the starting point, conservation of energy implies that the horizontal velocity is greater (greatest?) at the deepest point.

22. Aug 12, 2014

### A.T.

23. Aug 12, 2014

### voko

... and we can always go to a low $V_0$ by choosing a reference frame ...

24. Aug 12, 2014

### ZetaOfThree

Why does $V_0$ need to be low? As long as the ball stays in contact with the ground, $V_0$ shouldn't matter right?

25. Aug 12, 2014

### haruspex

That accords with my statement. Maybe you misread it.