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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