Welcome Back!
Welcome back to all the Augie students. Classes started yesterday here, but my first classes meet today: Quantum II directed study and Modern Physics.
This is the first time in a long time that I haven't taught 201. I keep thinking of ideas for examples and demos, but I have nowhere to use them. Maybe I can turn them into blog contest material.
Speaking of which, our most recent blog contest is still open. I am now setting a deadline of September 22 for you to answer the last burning question of that contest and win Whitey's:
If a tire in a vacuum, which is already rolling, rolls onto a perfectly horizontal, perfectly smooth, perfectly frictionless surface, what will happen? The tire is not perfect, so it is deformable. Will the deformation of the tire rob it of kinetic energy and thus slow it down? Slowing down on a horizontal surface requires a horizontal force; without friction or air resistance, what could that force be?
Welcome back to physics club, too. A few people have been asking about our first meeting. It will be Thursday of week 2 (Sep 13). There will be pizza and cookies, so be there!
And be sure to welcome our new physics faculty: Jim van Howe (he's teaching E&M and advanced lab this term) and Joe Owczarzak (he's teaching 201 this term).
4 comments:
I just randomly stumbled on this blog while looking for another one, but here's my answer
In this problem the rotational kinetic energy and the translational kinetic energy are different things and should be addressed separately.
In the frictional case, the no slip condition of the tire couples them together, so the deformation of the tire leads to an overall slowing.
In the non-frictional case, the tire can be spinning much faster than it is moving forward.
This means that the tire will eventually stop spinning (due to the deformation of it), but it will never stop moving forward (there is no lateral force to slow it down)
Ryan
Good job, Ryan.
The changing deformation of the wheel does rob kinetic energy, but only rotational kinetic energy. The rotation slows down and stops, while the translational motion continues at a constant speed. Eventually the tire will just be skidding along (frictionlessly).
If you're ever in the Quad Cities, contact me (ceciliavogel@augustana.edu) for your free Whitey's ice cream cone. If you're not, I guess the gift certificate wouldn't do you any good. Oh, well, at least there's the glory!
I sort of regret answering the question given that I have no way of collecting it, and have robbed an _actual_ Augustana student from getting it.
Just roll my prize over into the next contest (make it worth two cones, I guess)
Thanks, Ryan. That'll let us have an extra contest this year.
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