Expert: Josh - 2/24/2004

Question
will a 12 pound bowling ball with a circumference of 27 inches float in water?

Archimedes Principle: An object will float if it is lighter than a quantity of water with the same volume.

Density of pure water:  1.00g/ml

Answer
Hi Cheryl,

There are various ways of doing this. Why not just calculate the density of the bowling ball, which is, by definition, mass over volume.

Exercise: Convert these measurements into standard SI units, ie., from pound to kg, inches to m.
Consult: http://www.dgsgardening.btinternet.co.uk/convert.htm#mass
http://www.dgsgardening.btinternet.co.uk/convert.htm#length
[1 lb = 0.4536 kg]
[1 in = 0.0254 m]

Mass, m= 12*0.4536 kg
Circumference = 2*pi*r, where r=ball radius.

So, r=(27*0.0254)/(2*pi) meter ...USE CALCULATOR

Now, volume for sphere is v=(4/3)*pi*(r^3) ...USE CALCULATOR

The density, d=m/v is measured in [kg/m^3] ...USE CALCULATOR

For water, 1.00g/ml is equivalent to 1.00g/cm^3,
(1*1000 g)/(1000 cm^3) or 1kg/1000cm^3
But,    1 cm^3 = 0.01[meter]*0.01[meter]*0.01[meter]
              = 1*10^(-6) m^3.
Hence, 1000 cm^3 = 0.001 m^3
So, water has a density of 1000kg/m^3

Answers: You should get these approximate values.
r~=0.10914845 meter
m~=5.4432 kilogram
v~=0.0054468002 cubic meter
d~=999.339 kilogram/cubic meter

Which is less than 1000kg/m^3. So, what is the conclusion?

Cheers.