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Topic: Stick in the bowl (Read 1177 times) |
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Eigenray
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A stick is resting in a level, frictionless, hemispherical bowl as shown, so that it just barely sticks out over the lip (if it were any shorter, it would sink into the bowl). If the bowl has radius 1, how long is the stick?
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Random Lack of Squiggily Lines
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Re: Stick in the bowl
« Reply #1 on: Jul 9th, 2008, 10:00am » |
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My guess says that the smallest it can be is the square of 2.
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| « Last Edit: Jul 9th, 2008, 11:45am by Random Lack of Squiggily Lines » |
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Eigenray
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Re: Stick in the bowl
« Reply #2 on: Jul 9th, 2008, 11:10am » |
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on Jul 9th, 2008, 10:00am, Qaster Qof Qeverything Q42 wrote:| My guess says that the smallest it can be is 2 squared. |
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I don't understand. Surely the length of the stick is not more than the diameter of the bowl, which is 2.
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Eigenray
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Re: Stick in the bowl
« Reply #4 on: Jul 9th, 2008, 11:57am » |
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And hamburgers eat people!
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| « Last Edit: Jul 9th, 2008, 11:57am by Eigenray » |
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towr
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Re: Stick in the bowl
« Reply #5 on: Jul 9th, 2008, 12:58pm » |
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on Jul 9th, 2008, 11:57am, Eigenray wrote:| And hamburgers eat people! |
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Only in Soviet Russia.
In Soviet Russia, tired old jokes make you. 
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mikedagr8
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Re: Stick in the bowl
« Reply #6 on: Jul 9th, 2008, 7:07pm » |
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In (Communist) Cuba, cigars smoke you.
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Grimbal
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Re: Stick in the bowl
« Reply #7 on: Jul 10th, 2008, 6:56am » |
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My guess says that the smallest it can be is the square root of 3.
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SMQ
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Re: Stick in the bowl
« Reply #8 on: Jul 10th, 2008, 7:17am » |
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Neat!
It's a simple balance-of-forces problem with an elegant result. The force of gravity on the rod is resisted by equal upward forces at the two ends. At the upper end, this force is applied normal to the rod while at the bottom end it is applied normal to the bowl. Since the horizontal forces need to balance, the angles are the same; a tangent to the bowl at the point the rod touches has the same slope (in the opposite sign) as the rod itself.
Let d be the horizontal distance from the center of the bowl to the bottom end of the rod. We then have that the slope of the rod,  (1 - d2) / (1 + d), equals the slope of the tangent, d / (1 - d2). Thus, d(1 + d) = 1 - d2  2d2 + d - 1 = 0 d = (-1  3)/4 d = 1/2 (since d = -1 is not a useful solution).
So the rod rests at a 30o angle, and has a length of 3; slick!
--SMQ
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Eigenray
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Re: Stick in the bowl
« Reply #9 on: Jul 10th, 2008, 9:19am » |
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on Jul 10th, 2008, 7:17am, SMQ wrote:| Since the horizontal forces need to balance, the angles are the same |
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But they don't have to have the same magnitude, do they?
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SMQ
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Re: Stick in the bowl
« Reply #10 on: Jul 10th, 2008, 9:38am » |
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on Jul 10th, 2008, 9:19am, Eigenray wrote:
But they don't have to have the same magnitude, do they? |
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Eh? Since there are only the two forces with horizontal components, if they had different magnitudes the rod would be experiencing a horizontal acceleration, n'est-ce pas?
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Eigenray
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Re: Stick in the bowl
« Reply #11 on: Jul 10th, 2008, 9:46am » |
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Oh, I thought you meant that the normal forces must have the same magnitude. Looks like I read your post a little too fast:
on Jul 10th, 2008, 7:17am, SMQ wrote:| The force of gravity on the rod is resisted by equal upward forces at the two ends. |
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Why must they be equal?
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Eigenray
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Re: Stick in the bowl
« Reply #12 on: Jul 10th, 2008, 9:50am » |
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on Jul 9th, 2008, 12:58pm, towr wrote:
And in Rand McNally.
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SMQ
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Re: Stick in the bowl
« Reply #13 on: Jul 10th, 2008, 9:58am » |
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on Jul 10th, 2008, 9:46am, Eigenray wrote:
I guess I'm making the unstated assumption that the rod is of uniform density. If that's the case (or in any case where the density is symmetrical about the rod's center), then the vertical components of the two forces at the ends must be equal or the rod would be experiencing a torque about it's center of gravity.
So yes, for a rod of uniform density, I am indeed claiming that the magnitudes of the two forces acting on the ends will be the same (since considered separately the magnitudes of their horizontal and vertical components are the same) if the rod is at rest.
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Eigenray
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Re: Stick in the bowl
« Reply #14 on: Jul 10th, 2008, 10:10am » |
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on Jul 10th, 2008, 9:58am, SMQ wrote:| the vertical components of the two forces at the ends must be equal or the rod would be experiencing a torque about it's center of gravity. |
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But the horizontal components also contribute to the torque.
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SMQ
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Re: Stick in the bowl
« Reply #15 on: Jul 10th, 2008, 10:13am » |
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on Jul 10th, 2008, 10:10am, Eigenray wrote:
But the horizontal components also contribute to the torque. |
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...
and in the same direction, too. Phooey!  But, but, but ... it's so elegant! 
--SMQ
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towr
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Re: Stick in the bowl
« Reply #16 on: Jul 10th, 2008, 10:25am » |
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Aha, if the stick is not uniformly dense, but has all it's mass concentrated on the low end, then the minimum length is 2 !
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| « Last Edit: Jul 10th, 2008, 10:27am by towr » |
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Eigenray
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Re: Stick in the bowl
« Reply #17 on: Jul 10th, 2008, 12:13pm » |
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The stick is dense uniformly. (And its length is somewhere between 2 and 3.)
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SMQ
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Ooh! With proper scaling the force diagram has a marvelous geometric interpretation showing that the horizontal extent of the stick is 4/3 so the length of the stick is (2/3)6. That's almost as cool as my first attempt. 
Edit: set out to verify the picture with actual math and obtained the same solution. Sweet!
Edit 2: and made the same silly math error both times as noted below; fixed now.
--SMQ
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| « Last Edit: Jul 10th, 2008, 3:21pm by SMQ » |
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Eigenray
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Re: Stick in the bowl
« Reply #19 on: Jul 10th, 2008, 3:03pm » |
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on Jul 10th, 2008, 1:03pm, SMQ wrote:| Ooh! With proper scaling the force diagram has a marvelous geometric interpretation |
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Indeed! I hadn't noticed that. (You might want to double check that length though.)
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SMQ
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Re: Stick in the bowl
« Reply #20 on: Jul 10th, 2008, 3:20pm » |
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on Jul 10th, 2008, 3:03pm, Eigenray wrote:| (You might want to double check that length though.) |
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Right, I can math. Fixed.
--SMQ
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