Difference between revisions of "OSH/Example/1"

Revision as of 08:33, 2 September 2013

Example problem statement

The drag force on a sphere is given by the product of the sphere's velocity and its drag coefficient. The drag coefficient is proportional to the sphere's cross-sectional area. The gravitational force on a sphere is proportional to its volume. The velocity at which these two forces are equal is called the terminal velocity. How does the terminal velocity of a sphere depend on its radius?

Solution 1

$F_1= - \mu V$

$\mu=\pi r^2 k_1$

$F_1=\pi r^2 k_1 V$

$F_2= 4/3 \pi r^3 k_2 d g$

$\pi r^2 k_1 v= 4/3 \pi r^3 k_2 d g$

$V=4 r k_2 d g / (3k_1) $

Comments

Terms are not properly defined. Using $F_d$ and $F_g$ for drag and gravitational force would make it a bit more clear but, even then, a few words to say what they mean are essential to remove ambiguity. Although the steps manipulating the formulas are all there, the logic of how they connect is absent making it difficult to understand what is going on. The blackboard in a math class might often look like this but the difference is that the instructor usually speaks while writing. Imagine how it would be for a student if math instructors just put the formulas on the board without saying a word.

Solution 2

Comments

Solution 3

Comments