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Problem Set #1

1.
A rocket ship blasts off at t-0 and accelerates upward at a rate of $5 {\rm m/sec^2}$. A mile away, a news cameraman films the event using a tripod that tilts so that he can keep the rocket in the center of the field of view. What is the rate of change of the camera's angle with the horizon ( $\frac{d \theta}{dt}$)? What are the units of your solution? Justify your answer.


The rocket ship problem is a classic one, but there is seldom much effort placed on the units in the solution. Using elementary calculus, one quickly finds that the height of the rocket is

\begin{displaymath}y(t) = \frac{5}{2} t^2 {\rm\ (meters)}.
\end{displaymath}

The horizontal distance remains the same, and if we do a little checking, we find that a mile is about 1609 meters. From this information, we can find camera angle

\begin{displaymath}\theta(t) = \arctan\left(\frac{5}{2} \cdot \frac{t^2}{1609}\right).
\end{displaymath}

What are the units on $\theta$? Most would say radians, but why? Does it really matter? When you are computing the arctangent, it does not really matter what units are used. Most students are used to setting the mode on their calculator output whatever units are desired. However, when you change modes from say, radians to degrees, you are changing functions! Essentially, your calculator has several different versions of arctangent from which you can choose by selecting the right mode. Another way to think about this is that the relation

\begin{displaymath}\frac{d}{d\theta}\tan (\theta) = \sec^2\theta
\end{displaymath}

is only valid for radians. Mathematically, changing units is changing $\tan(\theta)$ to $\tan(c\theta)$ where c is some conversion factor for the new units. If you differentiate the latter function, you will not have exactly the same relation as shown above.

Differentiating, we find that

\begin{displaymath}\frac{d\theta}{dt} = \frac{5t}{1609+\frac{25}{4}\frac{t^4}{1609}}.
\end{displaymath}

The units are $\rm\frac{rad}{sec}$, of course.

2.
Suppose one wishes to characterize the terminal velocity of a water droplet falling in the atmosphere. The relevant variables would be the characteristic size of the droplet, the acceleration due to gravity, the viscosity of the atmosphere (with units M L-1T-1) and the density of the liquid. How many dimensionless combinations of the relevant quanitities are there? Find all dimensionless products of these variables.


To compute all the dimensionless quantities associated with a falling raindrop, one simply follows the procedure discussed in class.

Feature variable units
Characteristic length l L
Gravity g L T-2
Viscosity $\mu$ M L-1 T-1
Density $\rho$ M L-3
Velocity v L T-1

Next, one checks to see for which combinations of exponents

\begin{displaymath}l^a g^b \mu^c \rho^d v^e = 1.
\end{displaymath}

If you do it right, you will find that

\begin{displaymath}\left(\frac{l^3 g \rho^2}{\mu^2}\right)^{d/2}
\left(\frac{v^2}{l g}\right)^{e/2} = 1.
\end{displaymath}

Thus, there are two dimensionless constants. Specifically, they are

\begin{displaymath}\frac{l^3 g \rho^2}{\mu^2} \ \ \ \ {\rm and}\ \ \ \
\frac{v^2}{l g}
\end{displaymath}

Notice that the second one is the reciprocal of the Froude number which you read about in the dinosaur article! Thus, the ratio of kinetic to potential energy is important in this problem.

3.
A rule of thumb for cooking turkeys is that one should allow 20 minutes of cooking time per pound of turkey in a 400o oven. Using dimensional analysis, determine whether or not this is a good idea. If you think it is a bad idea, propose a better one.


To understand the turkey problem, we follow the same nondimensionalizing procedure as discussed in class. We assume that the heat capacity and the density turkeys does not change with size. There is more than one way to do this problem. For instance, you could use the turkey's total volume or weight rather than the turkey's characteristic size. Assuming constant density, both alternatives would have units L3. The important thing to understand is that turkeys are self-similar. That is, if a turkey is twice as tall as normal, it is probably twice as wide as well.

Feature variable units
Raw temperature $\Delta T_a$ M T-1 L-2
Cooked temperature $\Delta T_b$ M T-1 L-2
Thermal conductivity k L T-2
Turkey characteristic size l L
Cooking time $\tau$ T

After doing the algebra, you should find that the only nondimensional combinations in this problem are

\begin{displaymath}\frac{\Delta T_a}{\Delta T_b} \ \ \ \ {\rm and} \ \ \ \
\frac{k \tau}{l^2}.
\end{displaymath}

I am no cook, but I would say that the rule of thumb is no good because weight is proportional to l3, not l2. Thus, if $w = \rho
l^3$ where $\rho$ is the density of a turkey, see that

\begin{displaymath}\frac{k \tau \rho^{2/3}}{w^{2/3}}
\end{displaymath}

is a dimensionless constant, and the cooking time should not grow linearly with weight. Rather, the cooking time should grow sublinearly with w to the 2/3 power.

4.
Read the Scientific American article ``How Dinosaurs Ran'' by R. M. Alexander. Express the principle results of the first half of the article (summarized in the graph on page 132) in terms of Buckingham's $\Pi$ Theorem. Suppose you did not have access to the data in this graph, but you hypothesize that the only relevant variables for quadrupedal animals are speed v, stride length s, hip height l and gravity g. Use whatever resources you can to determine the functional relationship between the dimensionless parameters. (In other words, find animals representing data points with different dimensionless values. Otherwise, your graph will collapse to a point!)


Buckingham's $\Pi$ Theorem rounds up all the dimensionless combinations pretty quickly to reproduce the Froude number and the ratio of stride length to hip height as the author suggests. Furthermore, Buckingham's $\Pi$ Theorem also asserts that this is an exhaustive list. I was a little disappointed (and incredulous) that no one was able to acquire much data for this problem. I am still curious whether or not one can reproduce the functional relationship demonstrated in the problem.



 
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Louis F Rossi
2001-09-07