A Tale of two cheeses - Adrian Oldknow a_oldknow@compuserve.com

Chapter 5: Limits to growth: the falling Cheddar
Chapter 1 used a model of acceleration which might be appropriate in a vacuum. However Cheddar is not very aerodynamic (drag coefficient = ??), and the nice warm summer's air would resist the cheese's motion, giving a deceleration which reduces the acceleration due to gravity. This resistance increases with the speed of the cheese and acts in the opposite direction to its velocity. At some critical speed the deceleration due to air resistance will cancel out the acceleration due to gravitational attraction entirely and the Cheddar will then have a constant velocity (called the terminal velocity). The refined model for the acceleration a of a cheese travelling at a velocity v in a resisted medium is: a = g - k. vp where k is some constant depending upon the shape and roughness of the cheese and upon the stickiness (viscosity) of the air, and p is some power depending upon the sort of speed at which the cheese is travelling. Experiments have shown that the terminal velocity of Cheddar cheeses dropping through hot summer's air is about 53 m/s and that the power p is about 2. With these values can you show why k is approximately 0.0035?
Assuming that this acceleration stays roughly constant in each small time interval you can now make up a modified table of velocity v m/s against time t seconds:
t 0 1 2 3 4 5
v 0 9.8 19.26

How long, approximately, will it take for v to reach 50 m/s? How much does this change if you compute the velocities every 0.1 s instead of every 1 s? (You may need a computer!) What does the graph of v against t look like now? Can you extend the table to include the distance fallen d? What should/does the graph of d against t look like? How far will the Cheddar have fallen when its speed reaches 50 m/s? A recent newspaper article stated that a Cheddar cheese will reach 99% of its terminal velocity of 53 m/s after about 14 seconds in which time it will have fallen about 570 m. See if you can find a value for p (and the corresponding value of k) for this data. If the balloon was at a height of 4000 m above the ground when the accidents occurred can you estimate the number of Listeria bacteria in the Camembert at the time the Cheddar hit the ground, and the speed at which the Cheddar was going?

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The differential equation is now [image] with v = 0 when t = 0. The difference equation is again non-linear (unless p = 1) and there is no analytic solution. The acceleration is zero when [image] and the corresponding value of v is called the terminal velocity.

We'll start with the numerical solution: [image] [image]
[image] [image]
[image] [image]

[image] [image]

[image] The analytic solution is given by:

[image] which we'll derive soon. [image]

[image]
Link to Wikipedia free-fall:

[image] [image]

[image] Again we just quote the analytic solution - read on!

[image]
[image]

Analytically we need to evaluate the definite integral:

[image]
With p = 2 this becomes, writing tv for terminal velocity: [image]

With p = 1 this is: [image]

And with p = 3 it is: [image]
Ouch!!