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A Floating Ball (Posted on 2003-04-04) Difficulty: 4 of 5
A hollow plastic ball floats in a tub of jello (the flavor is irrelevant, but Iíll tell you: itís key lime). The jello sets, and when the ball is removed, the saucer-shaped depression in the jello measures 6cm across and 1cm deep. If the density of the plastic is twice that of jello, what is the average thickness of the plastic shell?

  Submitted by Bryan    
Rating: 2.8000 (5 votes)
Solution: (Hide)
Draw a circle with radius R, and add a line segment representing the jello line (length of 6, 1 unit above bottom of circle). Now draw two radii, one straight down, and another to one end of the line segment. This gives us a right triangle where

(R-1)^2 + 3^2 = R^2

Solving, R = 5 (the good old 3 4 5 triangle), and an equation for the circle is
x^2 + y^2 = 25.

Just as the volume of the sphere can be found by integrating an infinite series of flat discs:

Vsphere = integral(pi*x^2 dy) from -5 to 5

similarly, the volume of the depression can be found from

Vdep = integral(pi*x^2 dy) from -5 to -4 (or, by symmetry, from 4 to 5 to keep the minus signs at bay)

Vdep = integral[pi*(25 - y^2)]dy from 4 to 5
Vdep = pi*[25y - 1/3*y^3] from 4 to 5
Vdep = pi*[(125 - 1/3*125) - (100 - 1/3*64)]
Vdep = 14pi/3

Archimede's principle tells us that the weight of the floating ball = the weight of the displaced jello. Since the plastic is twice as dense as the jello, it follows that

Vplastic = 1/2*Vdep = 7pi/3

If the inside of the plastic shell has radius a,

Vplastic = 4pi/3*(5^3-a^3) = 7pi/3


a^3 = 125-7/4
a = 4.9766

Therefore, the average thickness is

THICKNESSave = 5 - 4.9766 = 0.0234cm

Comments: ( You must be logged in to post comments.)
  Subject Author Date
CharlieP C2004-02-04 20:51:39
re(3): solutionfriedlinguini2003-04-04 11:49:02
re(2): solutionCharlie2003-04-04 11:24:18
Solutionre: solutionCharlie2003-04-04 11:00:19
SolutionsolutionCharlie2003-04-04 10:52:10
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