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        riddles >> medium >> Stick in the bowl
        
(Message started by: Eigenray on Jul 9th, 2008, 9:53am)
Title: Stick in the bowl
Post by Eigenray on Jul 9th, 2008, 9:53am
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?
Title: Re: Stick in the bowl
Post by Qaster Qof Qeverything Q42 on Jul 9th, 2008, 10:00am
My guess says that the smallest it can be is the square of 2.
Title: Re: Stick in the bowl
Post by Eigenray on Jul 9th, 2008, 11:10am

on 07/09/08 at 10:00:13, Qaster Qof Qeverything Q42 wrote:
My guess says that the smallest it can be is 2 squared.

I don't understand.  Surely the length of the stick is not more than the diameter of the bowl, which is 2.
Title: Re: Stick in the bowl
Post by towr on Jul 9th, 2008, 11:56am
Maybe he means the square root.
Title: Re: Stick in the bowl
Post by Eigenray on Jul 9th, 2008, 11:57am
And hamburgers eat people!
Title: Re: Stick in the bowl
Post by towr on Jul 9th, 2008, 12:58pm

on 07/09/08 at 11:57:24, Eigenray wrote:
And hamburgers eat people!
Only in Soviet Russia.

[hide]In Soviet Russia, tired old jokes make you.[/hide] ;)
Title: Re: Stick in the bowl
Post by mikedagr8 on Jul 9th, 2008, 7:07pm
In (Communist) Cuba, cigars smoke you.
Title: Re: Stick in the bowl
Post by Grimbal on Jul 10th, 2008, 6:56am
My guess says that the smallest it can be is the square root of 3.
Title: Re: Stick in the bowl
Post by SMQ on Jul 10th, 2008, 7:17am
Neat!

[hide]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.[/hide]

[hide]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, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 - d2) / (1 + d), equals the slope of the tangent, d / http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 - d2).  Thus, d(1 + d) = 1 - d2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 2d2 + d - 1 = 0 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif d = (-1 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif3)/4 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif d = 1/2 (since d = -1 is not a useful solution).[/hide]

[hide]So the rod rests at a 30o angle, and has a length of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3; slick![/hide]

--SMQ
Title: Re: Stick in the bowl
Post by Eigenray on Jul 10th, 2008, 9:19am

on 07/10/08 at 07:17:26, SMQ wrote:
Since the horizontal forces need to balance, the angles are the same

But they don't have to have the same magnitude, do they?
Title: Re: Stick in the bowl
Post by SMQ on Jul 10th, 2008, 9:38am

on 07/10/08 at 09:19:10, Eigenray wrote:
But they don't have to have the same magnitude, do they?

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?

--SMQ
Title: Re: Stick in the bowl
Post by Eigenray on Jul 10th, 2008, 9:46am
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 07/10/08 at 07:17:26, SMQ wrote:
The force of gravity on the rod is resisted by equal upward forces at the two ends.

Why must they be equal?
Title: Re: Stick in the bowl
Post by Eigenray on Jul 10th, 2008, 9:50am

on 07/09/08 at 12:58:44, towr wrote:
Only in Soviet Russia.

And in [link=http://www.snpp.com/episodes/2F13.html]Rand McNally[/link].
Title: Re: Stick in the bowl
Post by SMQ on Jul 10th, 2008, 9:58am

on 07/10/08 at 09:46:09, Eigenray wrote:
Why must they be equal?

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.

--SMQ
Title: Re: Stick in the bowl
Post by Eigenray on Jul 10th, 2008, 10:10am

on 07/10/08 at 09:58:14, 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.

But the horizontal components also contribute to the torque.
Title: Re: Stick in the bowl
Post by SMQ on Jul 10th, 2008, 10:13am

on 07/10/08 at 10:10:35, Eigenray wrote:
But the horizontal components also contribute to the torque.

...

and in the same direction, too.  Phooey! :-[  But, but, but ... it's so elegant! :P

--SMQ
Title: Re: Stick in the bowl
Post by towr on Jul 10th, 2008, 10:25am
Aha, if the stick is not uniformly dense, but has all it's mass concentrated on the low end, then the minimum length is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2 !
Title: Re: Stick in the bowl
Post by Eigenray on Jul 10th, 2008, 12:13pm
The stick is dense uniformly.  (And its length is somewhere between http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2 and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3.)
Title: Re: Stick in the bowl
Post by SMQ on Jul 10th, 2008, 1:03pm
Ooh!  With proper scaling the force diagram has a marvelous geometric interpretation showing that [hide]the horizontal extent of the stick is 4/3[/hide] so the length of the stick is [hide](2/3)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif6[/hide].  That's almost as cool as my first attempt. 8)

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
Title: Re: Stick in the bowl
Post by Eigenray on Jul 10th, 2008, 3:03pm

on 07/10/08 at 13:03:55, SMQ wrote:
Ooh!  With proper scaling the force diagram has a marvelous geometric interpretation

Indeed!  I hadn't noticed that.  (You might want to double check that length though.)
Title: Re: Stick in the bowl
Post by SMQ on Jul 10th, 2008, 3:20pm

on 07/10/08 at 15:03:11, Eigenray wrote:
(You might want to double check that length though.)

Right, I can math.  Fixed. :-[

--SMQ


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