PHY 121/123 LECTURE NOTES - Week 1

Required materials:

1) Text: Eugene Hecht - Physics: Algebra/Trig, 2nd Edition

2) Course Outline

3) Lab manual

4) Standard Lab Notebook

Recitation instructors are responsible for insuring that your grade represents your accomplishment in the course.

1 m 11:35am Wijers

2 m 3:20pm Metcalf

3 m 4:25pm Metcalf

4 tu 8:20am Kuo

5 tu 3:20pm Nathans

6 w 11:35am Nathans

7 w 3:20pm Goldman

8 w 4:25pm Zahed

9 th 8:20am Goldhaber

10 th 3:20pm Nathans

There are many important things discussed in the course outline, which therefore deserves some time and attention soon. These items won't be duplicated here.

Physics has a lot to do with time and space - so does this course! There is not enough space for all the students who want to attend the morning lecture. I hope that some who come then will find a way to attend in the afternoon, where there is plenty of room, and therefore a more intimate atmosphere. On top of that, some of the wrinkles in the morning lecture will be ironed out by afternoon, so that lecture will be better. Another issue about time is how to conduct the lecture. Let's start on time, and begin winding down at the 50 minute mark. That does NOT mean this is the end of the lecture and the time to begin talking, just that questions one might have hesitated to ask earlier definitely are welcome then, as I try to wrap things up and indicate where we shall be starting the next time.

The text for this course is a `noncalculus' book, but we shall be explicitly using a bit of calculus in the lectures and homework. Before you panic, let me note a couple of things. First, virtually all of you do take a year of calculus because of other requirements of your programs. Therefore we are just using something you have learned or are learning. Secondly, there is no such thing as noncalculus physics. Calculus was invented precisely to deal with the questions we shall be facing. A teacher or a text can avoid using the words `calculus', `derivative', and `integral', but nevertheless these concepts must be used. All we are doing is being open about that fact.

Let's be more explicit. As mentioned in the Course Outline, physics is concerned with describing and predicting motion. The description part automatically includes description of the rates at which things are changing. Such a rate is precisely a derivative in the sense of calculus. The prediction part involves deducing from a situation at one time, including values and time derivatives of the important quantities, the situation at a later time. As we shall see, this process of deduction is nothing more than an integration in the sense of calculus. That is why the concepts of calculus are extremely valuable, not only for your physics class, but also for most other aspects of your lives. We would be doing a disservice to you if we did not help you see those connections.

A few words about the textbook: This volume is much bigger and heavier, but supposedly a dollar or two cheaper, than some of the competing books. There are several features which are quite attractive. There is a five-step approach to problem solving outlined on pp. 18-19 and supplemented later in the book, an approach which is useful not only for this course. It really does help, and as I do problems in class I shall try to point out where these steps come in. In addition, there should be a CD packaged with the book, including `walk-through' problem solutions, as well as supplementary material for those who are interested. The CD may be viewed on Sinc Site computers. The staff have found a couple of glitches:

1) Error message: Bad Image. The application is not a valid Windows NT image. Please check this against your installation diskette.

Students can continue and it seems to work OK, but we did not test the individual modules so we do not know the effect of this problem.

2) Additionally, even though this seems inconsequential, as the program exits, clicking rapidly on the closing screen causes the application to crash and Dr. Watson creates a program dump - so please exit slowly!

If you read material in the book in advance, you will be in a better position to follow the lecture, and to ask questions.

Martin Luther King's first sermon on record was quite apropos to this course, in several ways. He began by noting that everyone accepts the reality and inevitability of physical law. For example, only a crazy or suicidal person would jump out of a plane without a parachute. This applies even for people who do not know the mathematical details of exactly how gravity works. However, those who do can make use of the physical law much more effectively than just refraining from jumping out of a plane! Rev. King's second point was that moral law is just as real as physical law. I agree with him, but as he already noted, it is a bit harder to know exactly what moral law is. Nevertheless, he was ready to make two assertions, both of which seem to me very wise.

First, it is and always will be against the moral law to hate. Of course hatred is destructive to the person hated, but it also poisons and sickens the person who hates. When you find a way to dissipate the anger you feel against someone else, no matter how justified that anger was, you release a great weight from yourself, promoting your own well-being. Secondly, judging the rightness of an action only by whether you can get away with it - as King said, being ``slick'' rather than open and honest - also is against the moral law. Even if no one else seems to know about dishonest actions, the one who did them knows, and this secret is a burden which can weigh one down as much as hatred can. Just ask President Clinton! In that connection I commend to you the section on academic honesty in the ten-page packet of course material.

Problem: The Tortoise and the Hare

A hare starts at a speed of 5 m/s along a path for 20 minutes, then lies down for a nap. A tortoise begins at the same starting time to go at a rate of 0.5 m/s. Five hours after the start, the hare wakes up and starts dashing towards the goal again at 5 m/s. The goal is 9.3 km from the starting point. Which one gets there first?

Solution: This is a problem in which there are several time periods, but in each period the speed of the hare is a constant and the speed of the tortoise is a constant. To begin, 1 m/s is the same as

$${\rm 60 \ (s/min)\times 60 \ (min/hr) \times 1 / m/s =}$$

$${\rm 3600 \ m/hr =3.6 km/hr.}$$

In the first 20 minutes, or 1/3 hr, the hare goes $5 \times 3.6 {\rm \ km/hr \times 1/3 \ hr} =6$ km. Meanwhile, in five hours, the tortoise goes 9 km. Now the final part of the race begins in earnest! How long does it take the tortoise to get to the goal? The relation between distance d on the one hand and (constant) speed s and time t on the other is quite simple and intuitive:

d=st.

As we want to know the arrival time of the tortoise, we rewrite this equation to put the unknown time on the left, and everything else on the right: t=d/s. Of course, we are starting our clock at the moment the hare wakes up, which is also the moment the tortoise passes the 9 km mark. Therefore, we have $t = {\rm 300 \ m/(0.5 \ m/s) = 600 \ s}$ for the tortoise, and $t = {\rm 3300 \ m/ (5 \ m/s) = 660 \ s}$for the hare, which therefore crosses the line one minute later than the tortoise.

It is easy to see that if the goal had been $9.4 \ {\rm km}$ from the start then the hare would have won. This raises another question. At what point does the hare pass the tortoise? One could approach this problem by trial and error, but it is easy to get a direct and exact answer. Let the distance from original start to passing point P be $9+x \ {\rm km}$. Then for the the tortoise the time in seconds is $t = 1000x \ {\rm m}/(0.5 \ {\rm m/s)} = 2000x \ {\rm s}$, and for the hare, $1000(3+x) \ {\rm m}/(5 \ {\rm m/s}) = (600+200x) \ {\rm s}$. Setting these two expressions equal gives

$$(2000-200)x=1800 x =600 \ {\rm km},$$

or $x =0.33 \ {\rm km}$, which indeed is between 0.3 km and 0.4 km.

There is another, more philosophical, aspect of this story, going back to Greek thinkers more than 2000 years ago. This is what a philosopher named Zeno had to say about it. Imagine the hare running faster than the tortoise, and trying to catch up. It's impossible, said Zeno, because if initially the hare were, say 2 m behind, then after a while the distance would be cut in half, to only 1 m, and then again in half, to 0.5 m, et cetera, but this is an infinite sequence, so obviously it must take forever to go through all the steps and actually catch the tortoise. Perhaps the deepest thinker about physics in those days was Aristotle, who actually wrote a whole book with the title ``Physics". [It's a lot shorter than Eugene Hecht's book, but unfortunately really out of date!] While there was a lot he did not know how to figure out, Zeno's paradox just amused him. All Zeno had shown, said Aristotle, was that if the hare is behind the tortoise then the hare is behind the tortoise.

If we look at it from the viewpoint of our basic definition and equation d=st, then the matter becomes clearer. Let's suppose that the speed with which the hare closes the distance to the tortoise s=1 m/s. Then in the first second the distance is reduced to 1 m, in the next half second to 0.5 m, et cetera. What then is the total time for the hare to catch up? Answer:

$$t= 1+(1/2)+(1/2)^2 +(1/2)^3+.....=2 \ {\rm s}.$$

That answer of course is obvious:

$$t = d/s = 2 \ {\rm m/( 1 \ m/s) = 2 \ s},$$

but what a nifty way to write it! Indeed there is an uncountable infinity of points between the hare and the tortoise when they are any finite distance apart, but because the motion of the hare passes infinite numbers of points in finite time, this gives no problem in catching up. There is no contradiction of principle, because just as the interval in space contains infinite numbers of points, so does the interval in time, so each point in space has its corresponding point in time, even though both intervals are finite in duration.

I emphasize this old story for a simple reason. People who already have learned how to do elementary physics problems like this one can be quite amazed when someone new to the subject does not immmediately know the answers, or at least how to find them. On the one hand, the fact that smart people thousands of years ago had even greater trouble shows that the correct procedure is not obvious. On the other hand, the fact that a simple and correct procedure does exist means that it is possible for you to master it if you try. In other words, there is solid substance for you to learn in this course. It will take work, but it can be done. It requires you to develop new perspectives, and if you do they will enrich and strengthen your ability to deal with all kinds of issues.

Relative Observables, Vectors, and Motion in 2 Dimensions

In this last discussion we have sneaked in an important new idea - relative velocity. Let's develop that. Before, we were talking about distance, time, and speed.. In our example of the tortoise and hare we measured distance from a start, but took for granted that the motion is always in one direction. That means we really were talking about displacement. Displacement x is just distance with direction specified, so distance is just absolute value of displacement,

d=|x|.

By the same token, we talked about speed, but the motion again was always in one direction. That means we really were talking about a positive velocity v, where again the relation is between a directed quantity and its magnitude,

s=|v|.

These two notions lead directly to a further one, relative displacement $\Delta x =x_1-x_2$ and velocity $\Delta v = v_1-v_2$, where, for example, 1 is for hare and 2 is for tortoise. Let's redo the problem of the hare passing the tortoise this way. For the final phase of the race, initially the relative displacement is $\Delta x = -3 \ {\rm km}$, where the minus sign reflects the fact that the hare is behind the tortoise, and the relative velocity is $\Delta v = 4.5 \ {\rm m/s}$. The time it takes for passing is given by a familiar relation, now expressed in terms of displacement and velocity rather than distance and speed:

$$\Delta x(t_P) = t_P \Delta v + \Delta x = 0$$

yields a relative displacement of zero, meaning the hare and the tortoise are in the same place. The location of the tortoise (and therefore also the hare) at t_P is then

$$x_P 9 \ {\rm km} + v_2 t_P ,$$

with

$$v_2 t_P = v_2 \vert\Delta x / \Delta v\vert = 3\ {\rm km (0.5 \ m/s / 4.5 \ m/s) = 0.33 \ km} ,$$

just the answer we got before. If you go back through and compare both methods, you will find that the subtraction of velocities used to define the relative velocity is done at a later stage of the first method, so the logic is the same, just with a different order of steps. The second method surely is smoother, and so we should be prepared to find it useful in the future.

If we want to make our descriptions more realistic, we need to look at motion not just along a line, as in the tortoise-hare story, but in more dimensions. In many cases two dimensions are enough to get what we want. How do we describe displacement and velocity in two or more dimensions? We need to specify not only the magnitude but also the direction. In one space dimension the direction is just a sign - positive or negative, left or right. Obviously in more dimensions we need to say more. The tool for this is the vector. For example, displacement is given by

$${\bf \vec{x}}=(x_1,x_2) \ {\rm or} \ (x,y) ,$$

where the first coordinate gives displacement along the horizontal or x direction, and the second along the vertical or y direction.

An alternate way to express this is in terms of distance and angle,

$${\bf \vec{x}}= (r,\theta) ,$$

where r is the distance from one end of the displacement vector to the other,

$$r=\sqrt{x^2+y^2} ,$$

and $\theta$ is the angle of the vector from the positive x axis, with the angle increasing if a vector tilts from there towards the positive y axis:

$${\rm tan} \theta = y/x.$$

It's easy to check that these definitions are consistent, using Pythagoras' theorem, or equivalently the statement

$${\rm cos^2\theta + sin^2 \theta = 1 },$$

and the fact that the tangent of an angle is simply the ratio of the opposite to the adjacent sides of a right triangle.

We are ready to use vectors in a problem, and thus see how they work, but first let's talk a bit about calculus. What is constant velocity in terms of calculus? First, velocity is simply the time derivative of displacement:

v=dx/dt.

Recall that the derivative of a function f(t), df(t)/dt, is simply the slope of a straight line tangent to the graph of f (along the vertical axis) versus t (along the horizontal axis). If displacement is proportional to time,

x = x₀ + v₀ t,

then standard rules for derivatives say

v = dx/dt = v₀.

Indeed, the velocity is a constant, v₀ and that justifies the above description of motion with constant velocity. Now we want to go from description to prediction, saying that an object has constant velocity v₀, starting at the point x₀, and now we want to predict where it will be later. To do this we need the relation

$$\int_a^b dt (df/dt) = f(b) -f(a) ,$$

or in words, integration is just the reverse of differentiation. We know that if df/dt = v₀, that means

f(b) - f(a) = v₀(b-a) ,

but that is equivalent to

x = x₀ + v₀ t,

which we already knew. Therefore indeed integration, at least for the case of constant velocity, just states the answer or prediction for motion.

Finally, let's apply the tools developed to a practical, indeed topical, problem, air traffic control. At t=0 one airplane is 10 km north of an airport, flying west at 600 km/hr. A second airplane is 20 km west of the airport, heading north. The controller needs to tell the second pilot what speed s to fly, so that the two planes (assumed to be at the same altitude) will not collide. We'll approach this by computing what would be the bad speed that would give a collision, so anything substantially different from that would be OK. First, do it the commonsense way, like the first approach to the tortoise-hare problem. The collision, if there were one, would be at the point x=20 km, y= 10 km. Plane one takes a time $t= 20 \ {\rm km / (600 \ km/hr) = 2 \ min}$ to get to the collision point. The time for plane two is $t= 10 \ {\rm km /}s$, where s is the unknown speed which would produce a collision. This means

$$s={\rm 600 \ km/hr \times (10 \ km/20 \ km) =300 \ km/hr}.$$

Now let's do this over using relative velocity and relative displacement in two dimensions. Again, the condition for collision is

$$\Delta {\bf \vec{x_0}} + t \Delta {\bf \vec{v_0}} = 0.$$

This can only happen at positive t if the two vectors $\Delta {\bf \vec{x_0}}$ and $\Delta {\bf \vec{v_0}}$point along the same line in opposite directions: With $\Delta {\bf \vec{x_0}} = {\rm (-20,10) \ km}$ and $\Delta {\bf \vec{v_0}} = (600, - s) \ {\rm km/hr}$, we get again $s = (1/2) \times 600 \ {\rm km/hr = 300 km/hr} .$

If the directions of relative displacement and velocity are opposite, then a collision is inevitable, regardless of the magnitude of the velocity - that just determines the time for the collision. In more dimensions, the result is exactly the same. If there are more planes, one just has to look at every pair of planes and make sure they are not on a collision course in the same way we have done for one pair.

Some closing comments for the week:

1. In class I discussed the tortoise-hare problem in terms of Hecht's five steps for solving a physics problem. You can get practice in doing this from the CD-ROM that goes with the book. In the future I may sometimes include that analysis in my notes.

2. This coming week's lab involves studying the motion of a simple pendulum, especially the relation between the period for small oscillation and the length of the pendulum from the support point to the mass at the end. The theory for that only will come up a bit later in the course, but for future labs the lab and lecture should be in synch with each other. Therefore, for this first one, your focus should be both on careful experimental technique and on how to estimate the reliability of your measurements. For this it is very helpful to complete the writeup in the lab. That way, if something doesn't seem to be working right, you still have a chance to check the apparatus and figure out where the problem lies.

3. I did not discuss acceleration at all, but for purposes of CAPA this week all you need is that acceleration a=dv/dt, so positive slope of velocity v=dx/dt versus time gives positive acceleration, etc.