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2009-12-19, 20:42
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#1 |
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Random Account
Aug 2009
Not U. + S.A.
54528 Posts |
A Wheel
A wheel, simulated below, is attached to a bus traveling at 50 MPH. The center of the wheel, the hub, travels at the same speed as the bus. The 0° point on the wheel is not moving, relative to the roadway. The 180° point would move forward twice as fast as the bus, again relative to the roadway. There would be two points, one on each side, where the rotation speed matches the forward speed of the hub. Where are they?
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2009-12-19, 22:23
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#2 |
Jul 2003
So Cal
2,663 Posts |
What do you mean by "rotation speed?" If you mean the linear speed relative to the hub, then the entire outer rim has a linear speed relative to the hub that matches the hub's speed relative to the ground. If you mean the linear speed relative to the ground, then there's an entire circular arc (including the hub) pictured below on which the linear speed of the wheel on those points relative to the ground matches the hub's speed relative to the ground.
Edit: Just realized, you're probably looking for at what points on the outer rim does the linear speed of the rim relative to the ground match the hub's speed relative to the ground. That's just the intersection of the arc with the rim, at 60° and 300°. Last fiddled with by frmky on 2009-12-19 at 22:28 |
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2009-12-20, 01:15
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#3 |
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Random Account
Aug 2009
Not U. + S.A.
2·1,429 Posts |
Interesting. A friend attempted to demonstrate 90 and 270. I had a feeling this was not correct, but didn't labor the point.
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2009-12-20, 01:56
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#4 | |
Jun 2003
23×683 Posts |
Quote:
However, I suspect, that is not the question asked. |
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2009-12-21, 08:45
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#5 | |
May 2008
3·5·73 Posts |
Quote:
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2009-12-21, 11:41
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#6 | |
"Lucan"
Dec 2006
England
2·3·13·83 Posts |
Quote:
In the context of the question, "speed" means "magnitude of velocity relative to the roadway". "rotation" is a hint that the motion of the wheel relative to the roadway is rotation about a fixed point on the road located at P at this instant. By "motion" I mean velocity of each point on the wheel, NOT acceleration. Since the hub is the only point on the wheel which has constant velocity, it is a good idea to say that the velocity of a point on the wheel relative to the road is the velocity relative to the bus (hub) + velocity of bus. (Velocity is a vector of course). David Last fiddled with by davieddy on 2009-12-21 at 12:00 |
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2009-12-21, 22:14
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#7 |
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"Lucan"
Dec 2006
England
194A16 Posts |
Let your P be P and O the hub.
http://www.youtube.com/watch?v=c3j9i4EYTm8 i is the unit vector to the right. j is the unit vector up. Velocity of bus = vi Let C be a point on the circumference at an angle t degrees (as in your diaphram). Velocity of C relative to bus = -cos(t)vi+sin(t)vj I give up. YOU ARE NOT LISTENING ANYWAY |
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2010-06-25, 10:29
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#8 |
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Jun 2010
816 Posts |
very interesting but great.
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