Tuesday, November 13, 2012


Wonka Physics

            The movie, Willy Wonka, is full of physics defying stunts that tend to impact all of the golden ticket winners. For my physics news, I decided to look specifically at the fizzy lifting drink scene. Both Charlie and Grandpa Joe become “lighter” than air, allowing them to float.
            Due to the fact that the idea derives from soda, we can attribute the “floating” to the carbonation. In order for both individuals to float, their buoyancy force needs to be greater than their gravitational force.

Using the p= m/v equation, we can rework it so that all knowns are on one side of the equation.

Volume of gas required for Grandpa Joe

V=Mjoe/Dair  72.6kg/1.293kg/m^3= 56.1m^3

Volume of gas required for Charlie

V= Mcharlie/Dair 45.4kg/1.293kg/m^3= 35.1 m^3

            Similar to a ballon, both Charlie and Grandpa Joe would need to have a similar shape. Using the Calculated volumes, we can determine their sizes.
X=radius

Grandpa Joe
4/3 (pi) (x)^3= 56.1m^3
x=2.37                                                            diameter= 4.74m

Charlie
4/3 (pi) (x)^3= 35.1m^3
x=2.03                                                            diameter= 4.06m

Grandpa Joe would be 4.74m wide and Charlie would be 4.06m wide, much larger than Violet to put it on Wonka scale.

            Using volumes calculated, we can relate forces and determine the buoyancy force of each individual.

Grandpa Joe
Fbuoyancy= PVg

(1.293kg/m^3)(56.1m^3)(9.8m/s^2)= 710.9N

Charlie
(1.293kg/m^3)(35.1m^3)(9.8m/s^2)= 445N

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