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Re: [Phys-L] quickest route



OK, now do it for a real ball (rolls), and then check it in the lab.


________________________________________
From: Phys-l [phys-l-bounces@phys-l.org] on behalf of John Denker [jsd@av8n.com]
Sent: Saturday, July 27, 2013 12:18 PM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] quickest route

On 07/27/2013 06:42 AM, LaMontagne, Bob wrote:

Neat!

Yeah, I really like this one.

They take the same time!

That's not the answer I get.

Since the problem was ill posed, I guessed a pitch for the ramp of 38
degrees and then generalized.

Guessing i.e. hypothesizing is fine, but then we need to /check/
the hypothesis. When I do the check, I find that the times are
equal for a 3:4:5 triangle, i.e. θ = 36.87° ... and also equal
for θ = 90° ... but not otherwise! Generalizing from these
two examples doesn't work. For details, see below.

The problem is significantly ill-posed because I asked a yes/no
question ("Is the indirect route BCA faster than the direct
route BA?") but the right answer is neither yes nor no; it
depends on the shape of the triangle. This is typical of the
real-world questions that arrive at my desk: often there is
a yes/no question, to which the only correct answer begins
with the famous words, "It depends ....."

I assume no friction, including no air friction and no sliding
friction. Therefore the spherical cow does not roll; it merely
slides along the path without rolling. I did not use the word
"roll" in the original problem statement. As David Bowman points
out, analyzing a realistic rolling motion would be messier.

In the absence of friction, you can solve the problem in your
head as follows: There is a characteristic speed "V". This is
the speed the particle has when it reaches point A at the end
of the route. If we hold fixed the height of the triangle,
this is invariant with respect to route and invariant with
respect to shape, by conservation of energy.

The /average/ speed over the leg BA is the same as the average
speed over the leg BC ... namely V/2. So the time for leg BA
is longer than the time for leg BC by a simple scaling factor:
the time-ratio equals the length-ratio.

Meanwhile, the speed over the horizontal leg CA is not V/2
but rather the full maximum speed V. So a unit of length
on this leg counts only half as much as a unit of length on
the other legs.

For a 3:4:5 triangle with altitude 3, we have time 3 + 4/2 for
route BCA, versus time 5 for route BA, so it's a wash.

For anything shallower than a 3:4:5 triangle, the indirect
route wins. When the slope is very small, the indirect
route wins by a huge margin, a factor of 2, as you can
figure out in your head by considering the almost-degenerate
ε:1:1 triangle.

We now consider triangles steeper than 3:4:5.........

For an equilateral right triangle, we have 1 + 1/2 for route
BCA, versus 1.414 for route BA, so the direct route wins
by about four percent.

For a 3:4:5 triangle the steep way, with altitude 4, we have
time 4 + 3/2 for route BCA, versus 5 for route BA, so the
direct route wins by about ten percent.

The steeper the the slope, the more the direct route wins,
up to some point when the margin starts decreasing again.
At θ = 90°, the two routes are obviously identical.

The maximum occurs at θ = atan(2). You can show this via
simple calculus. In the algebra-based introductory physics
course, you can find the maximum by plotting the function
sin(θ) + cos(θ)/2
and eyeballing the maximum, or using the spreadsheet
program's root-finding tool. I didn't evaluate this
arctangent in my head, but my computer tells me it is
63.43°, at which point the direct route wins by about
12 percent.

==============================

Suggestion: Check the work! This is the cornerstone of
critical thinking.

At this point the question arises, how many checks are
needed? If we thought it was a linear relationship,
checking the hypothesis at two points would be a valid
proof, but in this case the relationship is nonlinear.

The hypothesis that the two routes take the same time
works for angles near 37° and it works for angles near
90° ... but it fails by a wide margin for angles near
zero. The third check tells the tale.
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