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Tipping Tube (Posted on 2005-06-23) Difficulty: 4 of 5
A man places a circular tube upright on a table. He then places a solid ball into the tube followed by a smaller solid ball. The tube stays upright. He removes the balls and places them again with the smaller ball being placed in first. The tube tips over. In both cases the man holds onto the tube until the balls come to rest and then lets go. The radii of the balls are 2.6 and 3.4 centimeters. The length of the tube is 18.0 centimeters and its thickness (external radius minus internal radius) is 0.1 centimeter. The balls and tube are made of the same material - so their weights are proportional to their volumes. Assume the points of contact between the balls, table, and tube are frictionless. What are the minimum and maximum values for the internal radius of the tube?

  Submitted by Bractals    
Rating: 3.7500 (4 votes)
Solution: (Hide)
The upper ball is trying to tip over the tube while the
weight of the tube and the lower ball are trying to keep
it from tipping over. Both balls exert an equal horizontal
force (say F) on the tube, but in opposite directions.
Therefore, the tube will tip over if

  F*[l + (u+l)*sin(theta)] > F*l + T*(r+t)             (1)

where u and l are the radii of the upper and lower balls
respectively and T, r, and t are the weight, internal radius,
and thickness of the tube respectively. Theta is the angle
between the table top and the line determined by the ball's
centers and is given by

  cos(theta) = (2*r-w)/w                             (2)

where w = u+l (also, one third of the length of the tube).
To determine the force F, we look at the horizontal and
vertical forces exerted on the upper ball

  F = B*cos(theta)
                                                       (3)
  U = B*sin(theta)

where B is the force exerted on the upper ball by the lower
ball along the line determined by their centers and U is the
weight of the upper ball. The equations in (3) combine to give 

  F*sin(theta) = U*cos(theta)                          (4)

Combining the inequality of (1) with equations (2) and (4) we get
that the tube will tip over if

  U*(2*r-w) - T*(r+t) > 0                              (5)

We have the following relation between the weights

                4 
               --- PI*u^3
   U            3                       4*u^3
  --- = ------------------------ = ---------------     (6)
   T     3*w*PI*[(r+t)^2 - r^2]     9*w*t*(2*r+t)

Combining equation (6) with inequality (5) we get that
the tube will tip over if

  9*w*t*(2*r+t)*(r+t) - 4*u^3*(2*r-w) < 0              (7)

This is just a quadratic inequality in r.

Plugging in u = 2.6 we get the tube staying upright if

  r < 4.90437...    or    r > 7.96488...

Plugging in u = 3.4 we get the tube tipping over if

  3.41938... < r < 25.54468...

Therefore, since the physical problem requires r < w = 6
(or both the balls will come to rest touching the table),

  r minimum = 3.41938...   and r maximum = 4.90437...

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Weebles wobble but...Jils2005-06-29 13:34:29
re: there's a catch i supposeBractals2005-06-27 13:12:56
there's a catch i supposearun2005-06-27 10:02:17
re(3): What I think I knowSachin2005-06-25 03:40:04
re(2): What I think I know (displacement)Jer2005-06-24 12:58:39
re: What I think I knowHugo2005-06-24 10:04:38
re(2): What I think I knowLeming2005-06-24 00:43:26
Some Thoughtsre: What I think I knowLeming2005-06-24 00:12:18
re: Is cylindar bottomless?ajosin2005-06-23 23:25:00
QuestionIs cylindar bottomless?Larry2005-06-23 19:34:41
What I think I knowJer2005-06-23 18:47:55
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