Beer Cans -The solution.

 a) and b) Brian won and Eric's can was the slowest.

c) Why?
After having passed the tilting table the position energy has transformed into two kinds of movement energies: transitional (speed) and rotational. These energies sum up to the original energy given of the position (height) of the high end of the table.
If we neglect the mass of the can as compared with the mass of the beer (that doesn't rotate as the beer is easy flowing) all energy of the beer filled can is translatory. All of the other cans has some rotational energy and thus less speed. Thus Brian was right!

The empty can has all its mass (neglecting the ends of the can) in a thin layer around its border. This means a larger part rotational energy at a given speed and thus a smaller translatory speed. Eric who had an empty can had thus the slowest.

A more precise analysis follows:

i) For the beer filled can all energy at the end of the slope is lateral and given by
 
 
 

Mv2/2 = Mgh    ... (1)

where M is the mass (in this case of the beer), v is the speed and h is the height it has fallen. Thus
 
 
 

v2 = 2gh    ... (2)

ii) For the empty can the rotational energy is
 
 
 

Iw2/2 = MR2w2/2    ... (3)

since the inertia I is given by the mass (M) that is concentrated at the surface of the can.
R is the radius of the can and w is the rotational speed (radians/sek). Thus the transitional speed is

v = wR
and thus the lateral movement energy
Mv2/2   ... (4)

and the total movement energy is given by

MR2w2/2  + Mv2/2 = Mgh ... (5)

or


Mv2/2  + Mv2/2 = Mgh 

that finally gives us the speed in this case

v2 = gh ... (6)

Note that this is exactly half the speed square of the beer filled can!

iii) For the case with the sand filled can the rotational energy is

Iw2/2     ... (7)

where I is the inertia of the solid cylinder (sand filled can).
For a thin shell of a cylinder with height H , radius r and density r,  the inertia is
 
 
 

dI = d r2  = r 2prH dr r2 ....  (8)

or I =  r 2pH r4 /4   ....  (9)


But now the total mass is given by

dM  = r 2prH dr

or M =   r 2pr2H/2   ....  (10)

Thus I =  M R2/2   ....  (11)

and the rotational energy can be expressed as
 
 
 

Iw2/2 = MR2w2/4 = Mv2 /4   ... (12)

Again the lateral movement energy is is Mv2/2.

The total energy in this case can thus be given as
 
 
 

Iw2/2 + Mv2/2 =  Mv2 /4  + Mv2/2 = 3Mv2/4 = Mgh ... (13)

or

v2 = 4/3 gh    ... (14)

Summary: The squared speed of the three cans after having fallen any height h are thus 2gh for the beer filled can, gh for the empty can, and 4/3 gh  for the sand filled can.
 

Sense moral: Never drink a beer can and you will be a winner!


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Last updated: 2003-06-03 by Ingvar Jönsson