Which Path is Quickest? Investigating Ball Bearings on an Inclined Tube

[Jackie Ma]

Homework Statement

I don't know what's a homework statement.

Relevant Equations

Equations of motion.

I've found an interesting problem in an undergrad physics book which I've bought, and my high school teacher to whom I've showed it said it was intriguing, and didn't know the answer.

A tube in the shape of a rectangle with rounded corners is placed in a vertical plane, and inclined towards the left. You introduce two ball bearings at the upper-right corner. One travels by one path; the shortest side of the rectangle, and then the longest, and the other by the other path; the longest side of the rectangle, and then the shortest. Which will arrive first at the lower left-hand corner?

Since it's a question, there is no answer in the book.
I think the ball that goes first by the shortest path (on the right) will arrive first. I've also done literal calculations for the time on both paths, with x and y as the paths, and theta and beta as the angles, but I got 2 expressions, with which it is difficult to estimate which one actually gives the shortest time, but it seemed to me that it was what I had deduced previously. I can't be sure, though.

But I think that the ball going to the right will arrive first.
Is it the good answer?

*I typed the problem because I had thought that that was what was asked on this forum, but here's the problem.

[Mentor Note -- See post #5 for an improved version of this image]

 

[DaveC426913]

I don't follow the setup. Can you clarify?

placed in a vertical plane, and inclined towards the left

What does this look like?I assume this refers to the tube's cross-section:

A tube in the shape of a rectangle with rounded corners

[UPDATE] Nope. My assumption was wrong.

 

[berkeman]

I don't follow the setup. Can you clarify?

Agreed, I'm not following the setup either.

@Jackie Ma -- Can you upload a (good quality) diagram of the problem? You can use the "Attach files" link below the Edit window to upload a PDF or JPEG image of the problem. Thanks.

 

[kuruman]

There is now a sideways attachment. Shortest path is not the issue here because both paths are equal in length. Here is a hint: How does the speed of the two ball bearings compare at the exit corner? This will give you a clue as to which will exit in the shortest time if you also think about acceleration during each leg of the trip.

BTW, the homework statement is the full question that your are supposed to anwer including figures and diagrams.

A tube in the shape of a rectangle with rounded corners is placed in a vertical plane, and inclined towards the left. You introduce two ball bearings at the upper-right corner. One travels by one path; the shortest side of the rectangle, and then the longest, and the other by the other path; the longest side of the rectangle, and then the shortest. Which will arrive first at the lower left-hand corner?

 

[berkeman]

There is now a sideways attachment.

Blagg. I downloaded, rotated and enhanced the image:

 

[Orodruin]

It depends on the angle of inclination. However, taking the angle in the image, yes, the right path (CD) will be the faster.

 

[DaveE]

To see this intuitively, consider the cases where the rectangle is rotated very close to 0^o or 90^o. Are the velocities the same at the turn? How long did it take to get there and then how will the velocity change for the last side?

 

[DaveC426913]

To see this initiatively, consider the cases where the rectangle is rotated very close to 0^o

Yeah, that does it for me.

 

[kuruman]

To see this initiatively, consider the cases where the rectangle is rotated very close to 0^o or 90^o. Are the velocities the same at the turn? How long did it take to get there and then how will the velocity change for the last side?

Also, at 45° inclination the two times are equal regardless of the ratio of lengths. That's because the accelerations are the same all around so this is a constant acceleration situation over equal lengths.

 

[haruspex]

To see this initiatively, consider the cases where the rectangle is rotated very close to 0^o or 90^o. Are the velocities the same at the turn? How long did it take to get there and then how will the velocity change for the last side?

and in general, early acceleration is more useful than late acceleration.

 

[DaveE]

This is a great example of the sort of problem that you can quickly get a feel for by examining the extremes and symmetry. For me at least, the answer, although imprecise, is obvious if you take that arbitrary angle and look at the "interesting" limits. In this case 0^o, 45^o, and 90^o. Then you can write the equations for an arbitrary angle and solve it exactly if you need to.

 

[Jackie Ma]

This is a great example of the sort of problem that you can quickly get a feel for by examining the extremes and symmetry. For me at least, the answer, although imprecise, is obvious if you take that arbitrary angle and look at the "interesting" limits. In this case 0^o, 45^o, and 90^o. Then you can write the equations for an arbitrary angle and solve it exactly if you need to.

``Then you can write the equations for an arbitrary angle and solve it exactly if you need to.``; yes, I did that to verify what I thought was the answer.

``This is a great example of the sort of problem that you can quickly get a feel for by examining the extremes and symmetry.`` I know what symmetry is in physics, but what does it mean in this context, though?

 

[haruspex]

``Then you can write the equations for an arbitrary angle and solve it exactly if you need to.``; yes, I did that to verify what I thought was the answer.

``This is a great example of the sort of problem that you can quickly get a feel for by examining the extremes and symmetry.`` I know what symmetry is in physics, but what does it mean in this context, though?

In this case, setting the angle to 45°. As noted, that makes it equivalent to a single slope of that angle, no matter whether the short or long side is traversed first.

There is a slight flaw in the question though. It omits to say the tube is smooth, so you should expect the balls to roll. When they hit the corner, the roll is the wrong way, so some KE must be lost. At 45°, less will be lost if the shorter side is taken first.

 

[Orodruin]

Also, somewhat relevant:
https://en.m.wikipedia.org/wiki/Brachistochrone_curve

The functional to be minimized is
$$
\int \frac{\sqrt{dx^2 + dy^2}}{\sqrt{y_0-y}}
$$
which is relatively simple for the case $y = kx$.

 

[Lnewqban]

Forward vid to the 17:00 mark: