OSH/Example/1

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?

Note that you are not expected to know the physics involved in this example. The point of this example is to demonstrate good mathematical writing.

The following two solutions demonstrate the most common problems with students answers: saying too little and saying too much. The third solution demonstrates a good balance in the middle.

Solution 1

$F_1= - \mu V$

$\mu=\pi r^2 k_1$

$F_1=\pi r^2 k_1 V$

$F_2= \frac{4}{3} \pi r^3 k_2 d g$

$\pi r^2 k_1 v= \frac{4}{3} \pi r^3 k_2 d g$

$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 and gestures while writing. Imagine how it would be for a student if math instructors just put the formulas on the board without saying a word. An additional minor point is that there is no punctuation. An equation like "F_1=-\mu V" is a sentence that could just as easily have been written "The drag force is equal to the negative of the drag coefficient of the sphere $\mu$ multiplied by its volume." Written out in words, it deserves a period at the end so in symbols it should as well. Although the punctuation is a minor point, the fact that mathematical expressions should read like grammatical sentences is an important point.

Solution 2

The drag coefficient is a dimensionless quantity that is used to quantify the drag or resistance of an object in a fluid environment such as air or water. It is used in the drag equation, where a lower drag coefficient indicates the object will have less aerodynamic or hydrodynamic drag. The drag coefficient is always associated with a particular surface area. The drag force on a sphere is given by the product of the sphere's velocity and its drag coefficient. Translating this relationship into mathematical form, we get the equation $F_d= - \mu V$ where $\mu$, the drag coefficient, for a sphere is proportional to the cross-sectional area. $\mu$ is given by $\mu=\pi r^2 k$ where $r$ is the radius of the sphere and $k$ is a factor that depends on the physical properties of the sphere and the fluid through which it moves. We can take the expression for $F_d$ above and rewrite it by replacing the constant $\mu$ with the expression $\pi r^2 k$ as discussed above. When we do this, we get a new expression for the drag force which is $F_d=\pi r^2 k V$.

The gravitational force is given by the product of the mass and the gravitational constant so that $F_g= m g$ where $m$ is the mass of the sphere and $g$ is the gravitational constant on the surface of the earth. The mass depends on volume and density, $d$, with $m=\frac{4}{3} \pi r^3 d$ where we have assumed that the sphere is of constant density.

Terminal velocity is the velocity at which all forces acting on the object are balanced. In this case, the sphere reaches terminal velocity when the drag and gravitational forces are equal. Setting these two forces equal to each other we get $F_d=\pi r^2 k v= \frac{4}{3} \pi r^3 d g = F_g$. This equation can be solved for $V$ to get the terminal velocity expressed in terms of the radius of the sphere, as requested $V=4 r d g / (3k) $.

Comments

The first problem with this answer is that it is too wordy. The second problem is that parts of it are directly copied from Wikipedia. This amounts to plagiarism and can land you in a heap of trouble. The third problem is that mathematical formulas are hard to find amongst all the text. This means that the marker will have a hard time quickly determining if you have all the necessary components in your answer. Making the reader's job easier means that your writing is more likely to actually be read. In the case of submitted homework, the reader is giving you marks so it's definitely in your best interest to make the reader's job easier. This is precisely a sign of good communication.

In a number of places, this solution repeats information including sentences and formulas that say the same thing (not necessary) and in some cases multiple sentences that say the same thing (even worse).

Solution 3

The drag force on an object is given by

$F_d= - \mu V$

where $\mu$, the drag coefficient, for a sphere is proportional to the cross-sectional area and is given by

$\mu=\pi r^2 k$.

$r$ is the radius of the sphere and $k$ is a factor that depends on the physical properties of the sphere and the fluid through which it moves. Rewriting $F_d$ using this expression for $\mu$, we get

$F_d=\pi r^2 k V$.

The gravitational force is given by

$F_g= m g$

where $m$ is the mass of the sphere and $g$ is the gravitational constant on the surface of the earth. The mass depends on volume and density, $d$, with $m=\frac{4}{3} \pi r^3 d$. The sphere reaches terminal velocity when the drag and gravitational forces are equal:

$F_d=\pi r^2 k V= \frac{4}{3} \pi r^3 d g = F_g$.

By dividing through by $\pi r^2/k$ we get

$V=4 r d g / (3k) $.

Comments

This is an ideal solution. The formulas are set off clearly from the text so that they can be found easily. In case the logic connecting them is unclear to the reader, there are explanatory comments that act as guides. However, the text is concise - only the necessary information is included.