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How Good Is Your Intuition, 2? (Another Mind-Teaser)
I am told this question was on a recent SAT test.
Two coins lay flat on a table top . The small one has radius 1, the large one has radius 3 . The large coin is fixed, no motion (translation or rotation). The small coin is placed so that it's edge touches the edge of the large one. The small coin is then rolled clockwise around the edge of the large coin. The coins' edges do not slip where they touch (think of gears). How many times will the small coin rotate around its own center, before the center of the small coin returns to its original location? For all you weasels: - the coins are round and they cannot be deformed, Schumi - there is no effing thermal expansion, Randy - rotation and translation are measured relative to the tabletop, so no good telling me about how the earth rotates or the milky way rotates - Last edited by jyl; 01-18-2011 at 06:29 PM.. |
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Band.
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3 times.
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what he said.
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Steve '08 Boxster RS60 Spyder #0099/1960 - never named a car before, but this is Charlotte. '88 targa ![]() |
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It is not quite that simple . . .
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Band.
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Doh! 1.5 times maybe.
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ok ~3 times
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damn
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Cogito Ergo Sum
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Schfifty-five....
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< disregard >What am I doing wrong then?
C = 2πr or πd C of big coin = 6π C of little coin = 2π One rotation takes the point on the edge of the small coin so it's 2π or 1/3 of the way around the large coin's circumference, the second takes you around to 2/3, and the third rotation takes you around so that the starting point on the outside of the small coin has traveled to the same point on the outside of the large coin. If the two outside points have again met2, the center should be in the same location as it started. < /disregard, crap, it's all wrong >
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Steve '08 Boxster RS60 Spyder #0099/1960 - never named a car before, but this is Charlotte. '88 targa ![]() |
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C = 2 pi r, where r= 3
= 6 pi total distance around |
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4 times
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Steve '08 Boxster RS60 Spyder #0099/1960 - never named a car before, but this is Charlotte. '88 targa ![]() |
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Yeah, that's what I got, but it's not really relevant in this case. The question is how many rotations (360° or 2π) will the small coin need to get back to the point where it started. Well, the circumference of the small coin will wrap around the large coin 3 times, but that's not 3 rotations, thats 4 full rotations.
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Steve '08 Boxster RS60 Spyder #0099/1960 - never named a car before, but this is Charlotte. '88 targa ![]() Last edited by masraum; 01-18-2011 at 07:52 PM.. |
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Suppose the edge of the large coin were cut off, and straightened into a line segment. It has length 2 * pi * 3. The circumference of the small coin is 2 * pi * 1. So the small coin needs to roll 3 times to get from one end of the line segment to the other. Exactly as you said.
But - the edge of the large coin is not a line segment. It is a circular arc that makes a full 360 degree loop. Thus the small coin makes 1 full rotation just to follow this arc. Even if the small coin were slid around the large coin's edge (the same point on the small coin's edge is always touching the large coin), it would make 1 full rotation by the time it returned to its original location. 3 + 1 = 4. The small coin makes 4 full rotations around its own center. Test this. Take 2 quarters. Place them so that the 6 o'clock of the rotating quarter touches the 12 o'clock of the fixed (bottom) quarter. Orient the rotating quarter so that the top of Washington's head points at 12 o'clock. Now roll the rotating quarter around the fixed quarter until it is back at its starting point. How many times does the top of Washington's head point at 12 o'clock? Once, or twice? Now, imagine the small coin were rotated around the "inside" of the large coin's edge. What's the answer then? I'm told the professor who wrote this SAT question got the wrong answer himself. I think the thing here is, our intuition is familiar with round things rolling on straight things (wheels on a road). We are not accustomed to thinking of round things rolling on other round things. Unless you're an engineer designing planetary gears.
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Quote:
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I don't even want to know how you did the earth-wire problem . . .
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When I imagine it, on the first rotation (relative to the fixed) the small coin has rotated only 2/3... so 2.
edit: make that 4/3. so 4 total ...the small would have gone thru 4/3 to get to that first 2/3 position. should have read Steve's answer/ thinking, so I wouldn't have been so quick to answer. ...but what fun would that be?
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Everyone you meet knows something you don't. - - - and a whole bunch of crap that is wrong. Disclaimer: the above was 2¢ worth. More information is available as my professional opinion, which is provided for an exorbitant fee. ![]() Last edited by island911; 01-18-2011 at 10:39 PM.. |
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Quote:
Les
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Best Les My train of thought has been replaced by a bumper car. |
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Yes, 3 - 1 so 2.
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Isn't the center of the coin always in the same position, if it is rotating around the center? That's like saying how many revolutions must the Earth make before the North Pole is back in its original location - the North Pole is always in the same location.
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