Difference between revisions of "OSH/Example/1"
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< OSH
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| + | ====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===== | =====Comments===== | ||
Revision as of 21:50, 1 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) $
