Tricky word prob. from trig class: A train rolls down a trac

anna_sims

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I don't even know how to start this one, and my teacher isn't exactly forthcoming with hints.

A train rolls down a track at 60mph. It has wheels with radius = 15 in. The wheels contain a flange designed to hold the train on the track that extends 1 in from the radius of the riding surface.

Since Point A (on the flange) is farther from the center of the wheel than Point B (on the edge of the wheel), it is moving faster than B.

Due to this phenomenon, there is portion of the rotation that Point A is moving backward with respect to the ground.

Find the duration of this time period in seconds (Accurate to 10 decimals).

Hint: The answer is very near one hundredth of a second.


Any procedure needed to get the answer would be appreciated.
 
I think I get it. A very interesting question.

60 mph = 1056 in/sec -- Linear Speed of Train

(1056 in/sec)/(2*pi*15 in/Rev) = 11.20451 Rev/Sec -- Angular Speed of Wheel

(11.20451 Rev/Sec)*(2*pi*16 in/Rev) = 1126.40020 in/sec -- Linear speed of Flange (Point A)

This makes: (2*pi*16 in/Rev)/(1126.40020 in/sec) = 0.08925 sec/Rev

Now, the question is this: In general, the forward speed of the train is greater than any negative horizontal component of any point on the wheel. Even though points on the wheel may be moving backwards with respect to the train, the forward speed of the train causes them to move forward with respect to the ground, or at least stand still momentarily. However, since the flange is faster than the wheel where the wheel contacts the track, there is some time where the flange is going faster backwards than the train is speeding forward. Thus, for a moment, an observer on the ground would observe tha flange moving backwards.

Clear as mud?

The flange is moving WITH the train at the top of its rotation, so we need to consider only the bottom 1/2 the rotation -- When the flange is moving backwards with respect to the train.

Well, we just need the horizontal component, so: acos(1056/1126.40020) = 0.35542

The horizontal component of the flange speed is greater than the forward speed of the train for 2*0.35542 radians = 0.71084 radians. That is 0.35542 radians descending from the front and, by symmetry, another 0.35542 radians ascending toward the back.

(0.71084 radians / (2*pi*radians))*(0.08925 sec) = 0.01010 sec

Look's like we're in the neighborhood.

Note: What a delightful problem. I never had considered such a thing, before. What a great thing it is to have mathematics and thinking working together to solve a new problem type!

Note: If this is some challenge problem for prizes or for glory, I'm going to be very upset that it was submitted to a public forum. That sort of thing would just be wrong.
 
Ya; that's sure interesting. Check me, TK; is this approach sensible:

Make the wheel = 16"
Make point B inside wheel, 1" from edge; point A is on edge.
Draw circle radius 16", and inside it circle radius 15".
At bottom, draw horizontal line through B, then through A:
"black out" the space between the 2 lines:
in other words, the eye sees point B doing its full rotation, but does not
see point A when it travels the arc hidden behind the blacked out area.
So:
calculate length of that arc: in this case, theta being ~40.72827 degrees,
arc length = ~11.37348 inches.

At A's speed of 1126.4 inches per second, time to travel the arc is:
11.37348 / 1126.4 = ~.01009719 : really same as your .0101

Steps:
1: calculate A's speed (from the given B's speed)
2: calculate theta (radius = wheel's radius + flange length)
3: calculate arc length
4: time = arc_length / A's_speed


Found this:
Illusion of the rolling wheel
==================
Many people see the light going in a circle because they mentally follow the moving bicycle, and see the wheel just rotating, but the actual path involves both the circular motion of the wheel plus the linear displacement of the wheel as it rolls along the ground. The result of the combination of the two motions is a cycloid.
The axle of the wheel moves in a straight line.
And points on the wheel between the axle and the rim move in paths that are also cycloids.
The classic question is, Is there a point on a train that is moving backward? The answer is yes, a point on the flange of the wheel beneath the top of the rail moves backward.
If you slide a bead down a wire shaped like a cycloid the bead will slide from point A to point B in the shortest time. This problem was solved by John Bernoulli. Galileo thought the least time shape was the arc of a circle, he was wrong. This problem of least time is one of the most important problems of physics. It is called the brachistochrone problem.

Copernicus modeled the planets as moving in circular orbits about the sun. This was heresy at the time. As more accurate measurements of planetary motion became available Kepler added other circles, epicycles to the circles in order to match the model the the motion of the planets. An epicycle is drawn by a circle rotating on another circle just similar to what you are drawing here. Eventually, he had to add so many circles that he abandoned epicycles and shifted to conic sections, ellipses, for planetary orbits. This is one of the greatest moments in all of science.
 
...and your 40.72827º looks an awful lot like my 0.71084 radians.

Thanks for the background. I almost feel elevated, having blindly wandered some of the same paths of some early Greats.
 
Ahhh....I see...we're doing same thing...
but my way is better, of course :roll:
 
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