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Floating Copper Ball


Date: 02/26/2002 at 10:27:04
From: Amber
Subject: Floating Copper Ball

Find the wall thickness of a hollow copper ball (sphere) with an 
outside radius of 50.0 cm that "just floats" in water.


Date: 02/26/2002 at 10:47:13
From: Doctor Ian
Subject: Re: Floating Copper Ball

Hi Amber,

By 'just floats' I assume you mean that the top of the sphere is even 
with the top of the water.  Is that correct?

The key principle here is that if you submerge something in water, the 
water will provide a buoyant force equal to the weight of the water 
displaced.  

(This is known as Archimedes' principle, because according to legend, 
he came up with as a solution to the problem of figuring out whether a 
crown was made from pure gold or an alloy.)

Since you know the outside radius of the sphere, you can compute the 
volume of the sphere. That will tell the volume of the water 
displaced. If you look up the density of water, you can use that to 
compute the weight of the water displaced. And that is the buoyant 
force.  

Now, if the sphere 'just floats', it means that the buoyant force 
(i.e., the water pushing up) exactly cancels out the gravitational 
force (i.e., gravity pulling down). So what you need to do is figure 
out how thick the sphere has to be so that the weight of the copper is 
equal to the weight of the water displaced. 

If the outside radius of the sphere is R, and the inside radius is r, 
then the volume of copper used will be 

  V = volume of outside sphere - volume of inside sphere

    = (4/3) pi R^3             - (4/3) pi r^3

To find the weight of the copper, you'll need to look up the density 
of copper. 

Can you take it from here?

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Physics/Chemistry

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