28 Aug 2006 (First Class)

Introductions. Josh, Eli, and Peter, too. Susie Hauger? Go through the class list.

Hand out and go through the course outline and syllabus and ask for questions.

Chapter 1 “Measurement”: Encourage them to read the chapter

• Emphasize SI units, but mention CGS and discourage British

• Precision and Significant Figures

• Dimensional analysis(!)

Example: What is the “radius” of a black hole with mass M? Have G with [G]=L3/T2M and can also use c, with [c]=L/T. Then need L=MxGycz=Mx-yL3y+zT-2y-z so x=y, 3y+z=1, and z=-2y. This gives y=1=x so z=-2, and “radius”=GM/c2. (Actually missing factor of 2.)

Laboratory Activity

Using the cart with the fan. Get used to using the motion detector, measure x(t) and show that it looks parabolic. Derive the velocity graph and value for the acceleration.

Extension: “How do you know the acceleration is correct?” See if they will get the idea of tilting the cart until the cart and fan are stationary. Measure the angle, and see if a=g*sin(angle). Good questions about the uncertainty in measurement should result.

31 Aug 2006

Do (1) vectors, (2) 1D kinematics, and (3) F=ma during “class” time. Then break, and do (4) discussion of analysis (including uncertainties) for first lab. Time to take more data.

VectorsRefer students to Secs.2-1 and 2-3. “You should be familiar with this.” For now, think of vectors as a shorthand for keeping track of orthogonal directions (“components”) which behave independently of each other, i.e. “bookkeeping”. Allude to a “greater meaning.”

One Dimensional KinematicsFirst stress x and then x(t). Draw a plot and show the slope v(t)=dx/dt. Introduce “x-dot.” Next do a=dv/dt=d2x/dt2. Introduce “v-dot” and “x-dot-dot.”

Emphasize “motion under constant acceleration” (Section 2-5). Do some integrals:

a =dvdt

Zdv =

Zadt v = at + v0

x =12

at2 + v0t + x0

Give them another way to integrate the kinematics. (Call it the “energy method?”)

v = a vv = va = xa

ddt

!12

v2"

=ddt

xa v2 = v2

0 +2da

Typical application: Freely falling bodies (Section 2-6)

Newton’s Second LawSome philosophy: Kinematics vs dynamics and “Equation of Motion”

F=ma=mx-dot-dot. “Differential equation”

“Free fall” F=mg; Weight vs Mass; Tell them about the equivalence principle

Free body diagrams: Show them a mass hanging from one or two strings

Discussion of analysis for “acceleration” labDo a free body diagram for mass on track, flat and also inclined. Indicate how this is used to check their answer.

Error analysis: (1) Reading the graph for acceleration and (2) error in vertical height of triangle for tilting to check answer. For (2) what about the error in the length? Get into a discussion of dominant uncertainties.

7 Sept 2006

Hand in homework! Check syllabus updated for next week’s assignment.

Read up on laboratory activity for Monday; some notes will follow.

Newton’s Laws in Three DimensionsMake use of vectors for bookkeeping. Projectile motion is a good example. It is “free fall” in y-direction, but motion with no force (in absence of drag!) in x-direction.

Drag ForcesFriction with a viscous medium. Force depends on velocity and acts in the opposite di-rection to motion. For many objects and media, good approximation is D=bv for drag force, where the constant b will depend on the object’s size and shape.

Integrate equations for motion with a drag force. It’s a little tricky: Start with F=ma

!"bx = mv !b(x! x0) = m(v! v0) mv0!b(x! x0) = mx This estab-lishes that x=x0 when v=v0. At this point, we might as well set x0=0. Now integrate:

dt =dx

v0!bx/m and so t =!m

blog

!v0!

bm

x"+C

where C is an integration constant. Want to have v=v0 when t=0. So choose C=(m/b)log(v0). Then

t =!mb

log!

1! bmv0

x"

. Check initial conditions. Also find x(t) and v(t):

x(t) =mv0

b

!1! e!bt/m

"

and v(t) = x(t) = v0e!bt/m. Discuss the kind of

motion this equation describes. Plot x(t) and v(t). Want to make a big deal out of the fact that this only makes sense if v0 is positive and nonzero.

Falling bodies with a drag force. This leads to “terminal velocity” which is simple to de-rive, namely mg=bv. This is a way to determine b. See table 4-1 in textbook. This is also a way to “check your answer” for deriving a formula for a falling body y(t). F=ma gives

mg!by = mvy , where “positive” is “down.” The “constant” term makes the integration

even trickier, so we will take a short cut to the velocity: dt =

dvy

g!bvy/m (Eq.4-20 in the textbook). Starting with zero initial velocity at time t=0, we have the same integration

as for x(t) above. That is vy(t) =

mgb

!1! e!bt/m

"

, Eq.4-22 in the textbook. Notice that as time increases to infinity, we get the correct terminal velocity!

Maybe take a break here? Want to finish up with enough time so that students can make it to the picnic, and I can get over to the meeting with Sam Heffner!

Uniform Circular MotionEmphasize velocity as a vector, which can change by changing its direction and not its magnitude. Change in velocity is acceleration, so there has to be some force involved. Figure 4-16 in the textbook shows it all. Draw two points on circle, each ϑ left or right away from vertical, then tangent velocity vectors at that point. Difference in velocity vec-tors is 2vsin ϑ. Note that the direction of the difference is radially inward. Time difference is path length s=2r ϑ divided by v. So, magnitude of average acceleration is

v2

rsin!

! . Then take limit as ϑ goes to zero. Spend some time on Taylor series? Lingo: Let them know that this is called “centripetal acceleration.”

The Conical PendulumNice example of “tension” as a force, vectors, and circular motion. Figure 5-18 in the textbook has all the details. The question to ask: “What is the period of this pendulum?”

Will get some demos from Scott to show them. One quarter length leads to half the pe-riod, doesn’t depend on mass, that sort of thing.

Draw the free body diagram (Fig.5-18b). Clearly Tcosϑ =mg. Also Tsinϑ =mv2/R. Would rather work with the length L of the pendulum bob, so take sinϑ=R/L. Cancel T, and also m, to get gsinϑ/cosϑ=v2/R. For period t this gives

t2 =!

2!Rv

"2

= (2!)2Lsin" cos"gsin"

= (2!)2 Lcos"g or

t = 2!

!Lcos"

g. This is

Eq.5-15 in the textbook.

Examine the limits. It is interesting for ϑ=0. (It is not zero, but is the same as for the or-dinary pendulum, which we’ll study later.) For ϑ=90o it gives t=0. What does that mean?

11 Sept 2006

This class is dedicated to acquiring data for an experimental investigation motion involv-ing frictional forces and two “free bodies.” It is modeled after Activity 04 for the regular Physics I classes, and you should review that write-up and download the logger pro file for that activity. You should also review the write-up of Activity 03 on motion with friction.

You will take data on the motion of a cart along a horizontal track, the same way you did with the cart/fan activity with which you worked before. In this case, however, the cart will move because it is attached to a string, which stretches over a pulley to a hanging weight, which falls. The hanging weight and the cart are couple together by the string, which exerts a tension force on both the cart and the weight. The hanging weight is also acted on by its gravitational force, and the cart is acted on by a frictional force.

You can change the mass M of the cart by placing one or more of the large rectangular bars into the tray on top. You may also choose one or more different hanging weights (mass m), which you attach to the string by hand. The acceleration a of the cart will de-pend separately on both M and m. Note that the frictional force on the cart (which op-poses the motion induced by the tension T in the string) depends on the coefficient of kinetic friction μ, which you should be able to determine from your measurements.

Work out the equations which give you the acceleration a of the cart, in terms of M, m, and μ. The tension T will disappear from the equation when you consider also the ex-pression for the acceleration of the hanging mass, which you can assume is the same as a. (What assumptions are you making when you use the same T and a for the cart and for the hanging mass?)

You should use your logger pro data to determine a two ways. The first way is the same as what you did for our first experiment. That is, use the program to follow the cart while it is accelerating, and determine a from the second derivative of the displacement as a function of time. The second way is to make use of our integrated expression which eliminates time and combines velocity v, acceleration a, and total displacement d:

v2 = v20 +2ad

This is possible because the hanging weight will hit the floor before the cart moves the length of the track. As soon as this happens, the cart stops accelerating, so you use logger pro to follow it afterwards to get the velocity v. The distance the cart travels dur-ing the time that it accelerates is the total displacement d. You will need to do some work to combine your various uncertainties to check that the two values you get for a are consistent with each other.

Once you decide which method give you the more precise value for a you can use your data for different masses M and m to get the coefficient of kinetic friction μ. Note that you can check this value using hints from the write-up of Activity 03.

14 Sept 2006

Homework: Hand it in!

Two topics for today: (1) Momentum, and (2) Center-of-Mass. Break in between. Will not cover “Systems with variable mass” (i.e. we will skip Sec.7-6)

(Linear) MomentumVector!! p=mv; Newton said F=dp/dt; Generally take m constant, so F=mdv/dt=ma

!J !Z f

i!Fdt =

Z f

id!p = !p f "!pi = !!p

Collisions: Draw F12(t) vs t; Then prove “Conservation of Momentum” in collisions, i.e.

!F12 =!!F21" !!p2 =!!!p1" !!p1 +!!p2 = !(!p1 +!p2) = 0

“Linear Momentum and Translation Symmetry”: Remind them that !r = xi+ y j + zk

!p = m!v = md!rdt

and!r!!r" =!r +!r0 # !p! !p" = !p “Symmetry!”

“Linear Momentum and Boosts” !v!!v" =!v#!v0 !p! !p" = m!v#m!v0

You can boost to a “frame” where the total momentum !p! = 0 Good example can be had in the collision of two bodies. For reasons to be seen (2nd period) this is called the “center of mass” frame, but better here it should be called “center of momentum”:

!p! = !p1 +!p2" (m1 +m2)!v0 = 0 so !v0 =

m1!v1 +m2!v2

m1 +m2

“Collisions”: Explain Fig.6-17, redraw, emphasize this is shown in CM frame, and also that different conditions all represent conserved momentum. (Energy? Later!!)

Center of Mass (aka Center of Momentum)Consider a system of N particles. The “center of mass” is defined as the position

!rcm !m1!r1 +m2!r2 + · · ·+mN!rN

m1 +m2 + · · ·+mN=

1M

N

!n=1

mn!rn

Note that !vcm =!v0 , i.e. the velocity needed to boost to the “center of mass” frame.

Now consider a solid object. Replace m with dm, and sum with integral sign. That is

!rcm =1M

Z!rdm =

1M

Z!r!dV

Example 1: Linear rod with length L and constant density. Cross sectional area A.

dV=Adx M=ρAL xcm =

1M

Z L

0x!Adx =

1M

!AL2

2=

L2 Right!!

Example 2: Linear rod with length L and density increasing from zero by x2. This is like having a (homogenous) cone, with the area increasing from zero to πR2 at the rod end.

First calculate the mass: M =

Z L

0dm = !

Z L

0"(xR/L)2dx = !"R2 L

3 Notice that the dimensionality of the result is correct!! Always keep this in mind!

Now calculate the center of mass. (Introduce the term “first moment”??)

xcm =1M

Z L

0xdm =

1M

!"R2

L2L4

4=

34

L

Note that this “halfway point” in terms of mass is 2L/3 (prove it!), which is farther to the left than the center of mass. That’s okay. If you imagine the the CM as the place from which you hang it and it doesn’t twist (we’ll talk about torque and rotations later), then you expect the CM to be to the right of the halfway mass point, because the lever arm can be shorter for the bigger mass.

18 Sept 2006There’s a lot of stuff to do today! Want to get through it all, then gravity on Thursday, with some time left over for a bit of review.

Work and Kinetic EnergyW=Fs=Fxcosϑ (for constant force); Figures; Work done “by” and “on”. Can be negative!

“Dot product”: See pg 233/234; Emphasize “projection” and use unit vectors to derive

!A ·!B = AxBx +AyBy +AzBz so we write W = !F ·!s

Variable force: dW = !F · d!sW =

Z 2

1dW =

Z x2

x1

F(x)dx for 1D case.

Example: Constant force (obvious)

Example: Spring force F=-kx; Talk about Hooke’s Law! W =!1

2kx2

Proof of the “Work Energy Theorem” in 1D. (Argue generality with parallel and perps.)

W =Z 2

1F(x)dx =

Z 2

1m

dvdt

dx =Z 2

1m

dvdt

dxdt

dt =Z 2

1

ddt

!12

mv2"

dt =12

mv22!

12

mv21

and we write K=mv2/2 so W=K2-K1=ΔK. This is called the “Work Energy Theorem.”

WE Theorem is independent of reference frame! Consider a boost to v′=v-v0:

!K! = !K" v0(mv2"mv1) and

W ! =Z

Fdx! =Z

F(dx" v0dt) = W " v0

ZFdt = W " v0!p

In other words, if then W=ΔK then W′=ΔK′. Aka “Gallilean Invariance.”

Recall “elastic collisions”: Were defined as initial and final speeds same in CM frame. Now define as kinetic energy before equals after. Independent of the reference frame!

Break Time

Potential Energy and Energy ConservationImagine there exists a function U(x) for which F(x)=-dU/dx. Then W=∫ Fdx=-ΔU. But does this make sense? Consider moving from 1 to 2 along two different paths, a and b. (Like Figure 12-4, but I changed notation to be consistent.) For W=-(U2-U1) to make sense, we must have W(Path a)=W(Path b) regardless of the paths. In other words, it only makes sense if the force F leads to “path independent” values of the work.

Let path b be traced backwards, going from 2 to 1. Then the value of W changes sign but the magnitude is the same. So, coming back to 1 results in zero net work. We write

that “the integral of the work around a close path is zero”, i.e.

I!F · d!s = 0

.

Examples: Gravity has U(y)=mgy and a spring has U(x)=½kx2.

Note: It doesn’t work for friction! Work around a closed path is obviously not zero!

“Total Mechanical Energy” E=K+U is a “conserved quantity”:

dEdt

=ddt

!12

mv2 +U"

= mvv+dUdx

dxdt

= v [mv!F ] = 0

Discuss motion in a general potential well:

Conservation of Energy (in general): Postulate some form of “internal” energy, different from mechanical. Two ways to realize this. One is “heat energy”, actually mechanical motion of the atoms and electrons in matter, aka “thermal energy.” Another is concept of “binding energy”, in fact a conversion of matter to energy through E=mc2. We then talk about the work done by all external forces giving rise to W=ΔK+ΔU+ΔEint. It is also pos-sible to transfer energy using “heat” instead of “work.” We will study this when we move on to the phenomenology and “laws” of thermodynamics.

21 Sept 2006

As usual, homework to Josh right away!

Note: HW Due Monday! (Two exercise on “gravitation”.) Exam #1 in one week.

Today: Newton’s Law of Gravity. Spend some time talking about the dual role of mass, equivalence principle, Einstein, and so forth. Say it in words first, i.e. “The force be-tween two bodies is proportional to the product of their masses, inversely proportional to to the square of their separation, and directed along the the line between them.” Then

give them !F = G

M1M2

r2 r where

r =!rr with a diagram.

Can emphasize in terms of “12” or “21” subscripts on F and unit vector for r.

Mention G as a “proportionality constant.” Also talk bout the unit vector in the r direction.

Mention that “inverse square” laws (like gravity, Coulomb) are now understood in terms of quantum field theory and three spatial directions. “It had to work out that way.”

Gravitation near the Earth’s surface. Assume a mass m. Then F=GMEm/r2 where r is really just the Earth’s radius RE. Therefore g=GME/RE2. Mention Cavendish experiment.

Variation with altitude? Sure. Remind them that (1+x)n=1+nx when x is very small. We will use this in a little while (today) after we have the concept of gravitational potential.

Shell Theorems. Won’t prove these, but they are very useful. Not hard to buy the idea, based on symmetry and vector forces and so on.

Shell Theorem #1: A uniformly dense spherical shell attracts an external particle as if all the mass of the shell were concentrated at its center. See Section 14-5!

Shell Theorem #2: A uniformly dense spherical shell exerts no gravitational force on a particle located anywhere inside it. See Section 14-5!

Gravitational Potential Energy. Spend some time talking about the path independence of gravity. See Figure 14-11. Pick a path which goes straight between points 1 and 2, so

W1!2 =Z 2

1!F!r ="

Z 2

1Fdr ="

Z r2

r1

GMmr2 dr ="GMm

!"1

r

"####r2

r1

= +GMm!

1r2" 1

r1

"

Now use W=ΔK and ΔK=-ΔU so ΔU=-W. Agree that U=0 at r2=∞. Then we write

U(r) =!GMmr

Now imagine the change in potential energy near the Earth’s surface. Imagine that you go from the surface up to a height h. Then

!U = U(RE +h)!U(RE) =!Gmm!

1RE +h

! 1RE

"=

GMEmRE

!1! 1

1+h/RE

"

(Note that the book uses y instead of my h.) Now h is much smaller than RE, so we can use 1/(1+x)=1-x to good approximation. So

!U =GMEm

RE

!1!1+

hRE

"= mgh

which we know is the right answer!

Interlude: Should we spend some time on Taylor expansions, and where these expres-sions for small things come from? Will do this on Monday for sure when we talk about the Euler formula for exponentials, but not sure now.

Escape Velocity. Launch a rocket from the Earth’s surface, straight upward. Does it fall back down or does it escape? Answer this with conservation of energy. At launch it has energy ½mv2 and potential energy -GMEm/RE. For “escape velocity” it will go fast enough so that when it is at “infinity” it is no longer moving with any velocity. Therefore it has no energy. In other words, the total energy is zero. We have

12

mv2escape!

GMEmRE

= 0 so

vescape =!

2GME

RE .

Obviously you can do this for planets and other things.

Circular Orbits. Just use F=ma but here a=v2/r is the centripetal acceleration, and F is

the force due to gravity. So

GMmr2 = m

v2

r and v =

!GM

r . This becomes Kepler’s laws, especially after doing it for elliptical orbits.

Aside: Dark matter and the behavior of galactic rotation curves. (See next page.)

25 Sept 2006

Collect homework! First exam is Thursday! Open Book, Open Notes, ...

Today we begin a study of “oscillations.” Expect some syllabus changes.

The Spring Force: Energy ConsiderationsRecall “Hooke’s Law”: F=-kx where x is the “extension of a spring.” Generalizable!! We

already have the potential energy:!U = U(2)!U(1) =!

Z x2

x1

Fdx =12

kx2x!

12

kx21

Measure from x1=”no extension”=0, i.e U(1)=0 and x2=x so U(x)=½kx2.

Now consider a mass m attached to the spring. E=½mv2+½kx2=constant. Emphasize that both x and v are squared. So v is max when x=0 and vice versa. “Oscillation!”

Draw the harmonic oscillator potential well and illustrate it that way.

The Equation of MotionMore information in the exact form of x(t). Want to solve equation of motion, i.e. F=ma,

or !kx = mx or x =!!2x where !2 ! k/m . The solution is the function x(t).

Try to solve it in your head. Think of some function f(x) (different x!!) so that when you take the derivative twice, you get the same function back with a minus sign. Then the solution will be x(t)=f(ωt). Get them to realize there are two such functions, sin(x) and cos(x). Of course! Second order equation, so need two “constants of integration.”

So x(t)=a cos(ωt) + b sin(ωt) Determine a and b from “initial conditions.” Start at t=0 with x(t=0)=x0 and v(t=0)=dx/dt=v0. So a=xo and bω=v0 and x(t)=x0cos(ωt)+(v0/ω)sin(ωt).

Different way of writing this same solution. Instead of a and b use A and φ where a=Acos(φ) and b=Asin(φ). (Easy to write equations that relate a and b to A and φ.) Now use the trig identity cos(X-Y)= cos(X)cos(Y)+ sin(X)sin(Y). This means we write x(t)=Acos(ωt-φ).

Discuss motion in these terms. Call A the “amplitude” and φ the “phase.” Easy to see the meaning of amplitude. Phase is a little harder. Draw two sine curves, one with zero phase and one with non-zero and positive phase. Point out that they will learn more about this in the “oscillation” laboratory work.

Break at this point. Will learn a fancier way (using Euler’s formula after talking about Taylor expansions in general) after the lab next Monday, so starting following Thursday.

Use remaining class time for questions, prior to the midterm exam, if they want.

5 Oct 2006

Note updated syllabus with homework assignment due next week.

Quick Review of Exam 1Don’t spend more than ten or fifteen minutes on this. Learn the stuff, and keep in mind the places you tripped up. It will appear again on the final exam.

1. “Free Body Diagram” plus “Centripetal Acceleration” plus “Conservation of Energy” plus “ΣF=ma.” Note that tension T does no work because it is always perpendicular to the direction of motion. Answer is T=3mg. Learn how to get this answer!!

2. We did part (a) in class. Again need “ΣF=ma.” Then dE/dt=-bv2. Energy is lost!

3. Know this one, example straight from class and homework. “Mass per unit length!”

4. Simple use of stuff from class. Emphasize that (b) and (c) are the same question!

Math: Taylor Expansions and “Applications”We will return to “oscillations” after break. For now, a math “review.” (Hopefully!)

Taylor expansion of y=f(x) about x=x0. Put y0=f(x0). Slope at (x0,y0) is (y-y0)/(x-x0)=f’(x0). Then y=f(x)=y0 +f’(x0)(x-x0)=f(x0)+f’(x0)(x-x0). “Linear Approximation for f(x).” Can extend this procedure to higher powers. (Won’t do it, but check a math textbook if you want.)

f (x) = f (x0)+ f !(x0)(x" x0)+12!

f !!(x0)(x" x0)2 +13!

f !!!(x0)(x" x0)3 + · · ·

Example: Expand f(x)=(1+x)a about x=0. Get f(x)=1+ax. Useful linear approximation!! Make it clear that this doesn’t depend on the sign or magnitude of a.

More Examples: Do sin(x) and cos(x) and ex and draw the similarities.

Now introduce i2=-1.Then get to eix=cos(x)+i sin(x).

Time for a break

Back to the Physics of Oscillations

Recall F=ma, or !kx = mx or x =!!2x where !2 ! k/m . Recall solutions written as x(t)=x0 cos(ωt)+(v0/ω)sin(ωt) or x(t)=Acos(ωt-φ).

Now we discuss a different way to write the solution. Emphasize that the derivative of ex is ex, so derivative of eax is aeax. Then second derivative of eiax is -a2eiax.

So, it is natural to write x(t)=a1eiωt+a2e-iωt. Work through initial conditions and show that it turns into x(t)=x0 cos(ωt)+(v0/ω)sin(ωt) so it is all the same as before. This will lead to the relations 2cos(ωt)=eiωt+e-iωt and so forth.

Probably good to mention “real part” is always assumed when you write the solution this way.

Let them know that this will be a very useful technique, and it will be used a lot in future course in physics and engineering.

Example: Damped harmonic motion from mx =!kx!bv =!kx!bx Rearrange

with some definitions !20 ! k/m and !! b/2m so x+2!x+"2

0x = 0 . Now try an “ansatz” x(t)=Aeiωt. (Is ω=ω0? Keep going forward and we’ll find out.)

A(!!2 +2i"!+!20)e

i!t = 0 so ! = i"±

!!2

0!"2. Continue the

discussion, including a few words about “real part”, “amplitude”, and “initial conditions”. Point students to Eq.(17-39).

Maybe we’ll do some “forced oscillations” and “resonance” next Thursday. Let’s see.

Homework Assignment due Thursday 12 Oct 2006

1. Perform a Taylor expansion of the function f(x)=x2 about the point x0=1, writing down all nonzero terms. Show that the result is is just a fancy way of writing f(x)=x2.

2. The “hyperbolic cosine” function is defined to be cosh(x)=(e+x+e-x)/2 and the “hyper-bolic sine” function is sinh(x)=(e+x-e-x)/2. Show cosh(ix)=cos(x) and sinh(ix)= isin(x).

3. Chapter 17, Problem 12. (Recall the lab we did on Monday 2 October.)

4. Chapter 17, Exercise 44. (Damped harmonic motion, with numbers.)

Laboratory 10 October 2006

This paper introduces the laboratory exercise on “oscillations” which will form the largest basis on which your lab books will be graded. We’ll start with a review of the laboratory exercises which you’ve already seen, with some idea of what should be in your lab books for these.

You are welcome, in fact encouraged, to do more work on the past labs some time prior to handing in your lab books for a preliminary look (Oct.26th) and final grade (Nov.30th).

Remember that the materials I handed out then, and the summaries below, are just meant to highlight guidelines. I am anxious to see what other things you discover that you can investigate in these laboratories.

Cart/Fan Acceleration Measurements (28 Aug)You first measured the acceleration of the cart on a track, with a fan attached to the cart which provided a force. The acceleration with the fan, call it af, was determined from a plot of position x(t) as a function of time taken with the sonic ranging device and your computer. The program doubly differentiates the position to get a constant (?) value for the acceleration.

You then also were asked to “check to see if you got the right answer” by raising the end of the track just enough so that the force due to gravity cancels out the force from the fan, and the cart stands still. The acceleration ag,=gsin(θ) should be the same as af,.

An important aspect is to use “uncertainties” when comparing af, and ag. The former has uncertainties from the way you read the data from your computer plot, or otherwise get it from the sonic ranging data. The latter has uncertainty from the precision with which you can measure the tilt angle θ. The two should agree “to within the uncertainties.”

Cart/Hanging Mass Acceleration Measurements (11 Sept)This exercise had you measuring the acceleration differently, by determining the velocity after the cart accelerates a little while, and the distance over which it travels, before the hanging mass hits the floor and acceleration ceases. With a zero initial velocity, there is a simple formula that relates the final velocity to the acceleration and distance traveled. I think I called this the “energy method” for reasons that should since have become clear to you. (If not, ask about it!)

Again, you determined the acceleration two ways. Once using the double differentiation of the x(t) data, and again using the “energy method.” Compare them, once more doing what you can to estimate the uncertainties.

Mass Hanging from a Spring (2 Oct)This was a “warm-up” exercise for what you’ll start today. Using the standard Physics I setup with a mass hanging from a spring, you set the system into oscillation (“simple harmonic motion”) and measured its motion with the ranger and computer. From this you could determine the (angular) frequency ω of the oscillation.

You were also able to predict the correct answer for the frequency by measuring the spring constant k. This is from observing how much the spring stretches for a certain mass. (Some of you learned that it’s a good idea to check to see if the spring stretches the same amount when you add 200g to the 50g hanger, and then another 200g to the 250g already there. Why?)

Do your measured and predicted values of ω agree to within uncertainties?

Oscillations of a Cart/Mass/Spring System (Today)Now you will start the most ambitious of the laboratory exercises. You can work with it today, do some analysis and think about it over the next week, and take more data with the same setup. Again, here are some guidelines that you should consider while taking and analyzing your data.

Your cart now has two springs attached, one on each side. A metal strip is magnetically attached to the cart, extending to one side. This is what will intercept the beacon of the sonic ranger. Some of you have spring with different spring constants, and there are more of everything in the bins at the front of the classroom. You also have the long, black metal weights which you can add to the cart.

This setup is fraught with sources of systematic error. Pay attention to your data and think about what you should be checking up on.

Your goal, ultimately, is to determine both the mass of the cart and the spring constant, just from a measurement of the motion of the oscillator. You can do this by observing the frequency as a function of the number of black metal weights added to the cart. Try to come up with a relationship between the frequency and the “added” mass, and use this to figure out how to plot your data so that you can determine the mass of the cart and also the spring constant.

Today is for exploration, for asking questions, and for starting to think about how to get to your ultimate goal. Don’t forget about sources of uncertainty, but it is okay if you’d like to see how things compare first. You’ll have a scale at your disposal next week to get the mass of the cart and of the black weights, but for now you can do everything in terms of “units of black weight” mass. That probably sounds strange to you, but think about and ask questions.

12 Oct 2006Homework in. Note new assignment posted, due next Thursday.

Today: Kinematics and Dynamics of “Rigid Bodies”Kinematics: Draw a figure of an axis of some body with an axis. Emphasize “one axis” for now. Now “position” variables x, v, and a become “angular” coordinates φ, ω, and α.

“Constant angular acceleration” just change the names! Note the new units, though.

Same ω as we used for harmonic motion? Do x=Rcos(φ)=Rcos(ωt).

Vector quantities?! Do the “rotate the book” demo to show that rotations don’t commute, so we can’t make φ a vector. However, small rotations look better. Insinuate that infini-tesimal rotations do commute. Hence we could make a vector out of dφ, so make ω and α vectors, with direction “along axis” via Right Hand Rule (sense is arbitrary, but axis is the only non-biased direction you can use.) Use pictures.

Relationship between linear s=rφ and angular φ variables. Emphasize tangential speed vT=rω and that “radial” speed is zero. Not for acceleration,i.e. aT=rα and aR=vT2/r=ω2r.

Dynamics: Start with F=ma on a point constrained to rotate about an axis, Fig.9-2. Want to study α so focus on aT. No effect from constraining force. Tangential acceleration is driven by FT=Fsin(ϑ)=maT=mrα. Define “torque” τ=rFsin(ϑ) so that τ=Iα where I=mr2 is called the “rotational inertia” or (to my generation) “moment of inertia.”

Torque as a cross product. Define, show magnitude correct, and direction consistent.

Break Here

Rotational inertia of a collection of particles is I=Σmr2. Present and prove the “parallel axis theorem” I=ICM+Mh2. Proof:

I = !i

mir2i = !

imi[x2

i + y2i ] = !

imi[(x!

i + xcm)2 +(y!i + xcm)2] = ICM +Mh2

after reminding students of the definitions of xCM and yCM.

Rotational inertia for solid objects: I=Σ(δm)r2 becomes I=∫ (dm)r2=∫ r2dm. Note: This is the “second moment” of the mass, whereas CM is the “first moment.” If time, do this for a rod with axis through center, through end, and show that parallel axis theorem works.

Important: Emphasize Fig.9-15 and that you will likely need these to do problems! I suggest that you spend some time deriving one or two or more of these.

“Center of Gravity is the same as Center of Mass:” See Section 9-4. (Time to derive?)

The Pendulum: Figure 17-10 and text, including “small angle approximation.” Point out correction term (Eq.17-25) and interesting exercise in laboratory.

Then do “Physical Pendulum”, Fig.17-11 by solving differential equation in angular coor-dinates. Result is T=2π(I/MGd)1/2. Point out that measuring period can give the rota-tional inertia about any axis.

Laboratory 16 Oct 2006This is a continuation of the work you started in the laboratory on 10 October last week.

Based on the data you took last week, and the analysis of that data, you should have the following results so far:

• For one pair of springs, measurements of the frequency for different “added” masses

• A plot of 1/ω2 as a function of “added mass” showing a straight line dependence

• Values for the “spring constant” k and the mass of the cart, derived from the slope and intercept of the straight line above. Try to come up with values for the uncertainties on these quantities, from your estimate of the uncertainty on the slope and intercept.

• An independent measurement of k from the extension of the spring from some hang-ing mass(es). Estimate the uncertainty in k from the extension measurements.

• An independent measurement of the cart mass, using one of the scales in the class-room. (Maybe you used both scales. The difference is a measure of the uncertainty.)

If you didn’t get all this done, use some of the time during this class to get this far.

After this is done, you should have discovered two things. First, the spring constants you measured above should differ from each other by a factor of two. Why? You should explain your answer by using the equation of motion for the cart, and not just say “it’s because there are two springs.” Correcting for the factor of two, are the results the same within your uncertainties? Did you measure the spring constant of each of the two springs, or did you make the assumption that they were identical? If the latter, maybe you want to check and see how good is that assumption.

Your second discovery should be that the mass of the cart you derive from the straight line is more than that you got from the scale(s). Maybe 10% larger, maybe more, but in any case, well outside your experimental uncertainties. Can you think of a reason for that? Can you think of a way (or ways) to check your reasoning, at least to see if it is a plausible explanation?

19 Oct 2006Note: One problem assigned for next Monday! Prelim lab books due on Thursday.

This class: Basics of angular momentum, but then time to work on the lab again.

Angular MomentumVery important concept in all of physics. Just scratching the surface here.

First, one particle. Need to have an “axis.” Use origin: !" =!r!!p Magnitude is

l=rpsinϑ=r⊥p=rp⊥. Refer to Fig.10-1 and draw it. Next connect it to torque:

d!"dt

=!v!!p+!r! d!pdt

=!r!!!F =!"but emphasize order of derivatives and why the

first term is zero.

Nice emphasis that it is an “alternate description:” Do the freely falling particle using an impact parameter b for the origin and show that it all works. (Sample Problem 10-1.)

Now system of particles. Use “total angular momentum” by adding them all up. So

!L =N

!n=1

!"n

and

d!Ldt

=N

!n=1

!"n = !!"ext

since all the “internal” torques cancel. This is analogous to Newton’s second law, an is indeed just an “alternate description” but it gives counter-intuitive results because of “hidden” internal forces. Bike Wheel Demo!

Now rigid bodies. Focus on one piece of a rigid body (Fig.10-7) and the z-component of angular momentum. Look “from the top.” Then lz=rp=rmv=rm(rω)=mr2ω=Iω. Adding up all the pieces of the rigid body, ω is the same for all pieces and I becomes the moment of inertia of the entire body. That is, L=Iω.

Conservation of Angular Momentum: No external torques, then angular momentum is conserved. Analogous to linear momentum conservation. Read Sec.10-4. Do the demo with the turntable and weights in the hands of some willing volunteer.

On to the laboratory...

23 October 2006Turn in your homework! Preliminary lab books due on Thursday!!

Today: Waves, from a more or less formal point of view. (Basically following Chap.18.)

What is a Wave?Spend some time getting them to think of waves with which they are familiar. Water waves, sound waves, light waves in that order. Go from “bobbing up and down” to the idea of “pressure as a wave” (including seismic waves, which could also be transverse!) to the (rather bizarre, when you think about it) phenomenon of light as waving (but inter-connected) electric and magnetic fields. We will only discuss “mechanical” waves now.

What do they have in common? Energy is transported from one place to another, but not the “stuff” that is waving. (Read DaVinci quote on page 401.) Maybe they’ll have some fun figuring out what the “stuff” is for light waves, but leave them hanging on it.

General Properties of Waves1. Direction of particle motion, i.e. transverse or longitudinal.2. Dimensionality. String/spring (1 trans/long); water (2); light/sound (3 trans/long)3. Periodicity, i.e. is it periodic or not? A “pulse” is an extreme example of “not”4. Shape of wavefront. Do 3D, i.e. planar or spherical. Introduce idea of “ray”

Good figures (18-1,2,3) in textbook.

Mathematics of a Traveling WaveUse the string paradigm for this discussion. Shape of string is y(x) at any given time t, but it will change with time. So, better to talk about shape as y(x,t). At this point, restrict the discussion to an “ideal string” with no losses due to friction.

Suppose shape at t=0 is f(x)=y(x,0). Wave moves to the right with speed v. Then it will have the same shape, but measured from x=vt instead of x=0. That is, the shape will be f(x’) where x’=x-vt. Draw a picture to explain, i.e. textbook Fig.18-4.

So, y(x,t)=f(x-vt) is a rightward “traveling wave.” Obviously, y(x,t)=f(x+vt) is a leftward traveling wave. Discuss y(0,t) a little, just like Fig.18-5, with an asymmetric shape.

Sinusoidal WavesGeneral language is based on this, but “Fourier Analysis” gets us back to general. Give amplitude and wavelength with y(x,0)=f(x)=ymsin[2πx/λ], so y(x,t)=ymsin[2π(x-vt)/λ]. Now have time “period” T for any fixed x point to make a full oscillation, i.e. T= λ/v. Generally instead speak of “frequency” f (or ν) =1/T. That is, v=fλ.

Final “wave speak” words: “Wave number” k= 2π/λ and “(angular) frequency” ω= 2πf. Then we write a sinusoidal wave as y(x,t)=ymsin[kx-ωt]. You will see this forever more. If there is time, talk about the “phase constant” ϕ, and relative phases of waves.

Break

Speed of Waves on a Stretched StringNow going for more “physics.” What is the “dynamics” of the mechanical wave set up on a stretched spring? (Take some time with this.)

First, how do you describe the string? It is infinitely long, right? Consider a small length of the string δl. It has a mass δm=μδl where we call μ the “linear mass density.” Also, this small length of string is being “pulled” by a tension F to the left and to the right.

Do some dimensional analysis! [μ]=M/L and [F]=ML/T2. So, v=(const)×[F/μ]1/2

Now let the string deform with a “circular bulge” at the top. (Figure 18-8.) Imagine that you are following the bulge, so the string is moving past you. The force in the y direction is Fy=2Fsinϑ=2Fϑ=Fδl/R, and the string is moving through the top at the wave speed v. It has a centripetal acceleration downward v2/R caused by this force. So, use Newton’s Second law and get (Fδl/R)=(μδl)(v2/R) and so v=[F/μ]1/2. (The constant is unity.)

Good suggestion: Check out Sec.18-5 on “The Wave Equation.” Do you know how to use partial derivatives? Good idea to get used to the idea. Important concept!

This is as far as we got. Will do more “waves” on Thursday.

Energy in Wave MotionWork with sinusoidal waves. Consider a “piece” of the string. The kinetic energy is

dK =12

dmu2y =

12(µdx) [!ym!cos(kx!!t)]2

How much does the kinetic en-ergy change in the time dt it takes the wave to move a horizontal distance dx=vdt? The

answer is

dKdt

=12

µ!2y2mvcos2(kx!!t)

. For the potential energy, need to figure out how much the string stretches due to the tension. This will give you the work done by F.

dU = F!"

dx2 +dy2!dx#

= Fdx!"

1+(!y/!x)2!1#

Now we make a critical

assumption, namely that the “slope” is always small. Hence dU =

12

Fdx(!y/!x)2

. With F= μv2 we find that dU/dt=dK/dt, so dE/dt=2dK/dt. From this get the average power transmitted is Pav=(1/2)μω2ym2v.

This may be too much to get through in the second half.

Save the topics of Superposition, Interference, and Standing Waves for another day, perhaps next Thursday and push “Fluids” and “Sound Waves” off for a while.

26 October 2006Hand in lab books. Second exam next Thursday. Material through today will be on the exam. One HW problem due on Monday (on waves).

Review from Monday1. General form of a wave y(x,t)=f(x±vt)2. Sinusoidal waves y(x,t)=ymsin[kx±ωt] with k= 2π/λ and ω= 2πf3. Dynamics example: The stretched string, giving speed v=[F/μ]1/2

(My apologies for blowing you away with the “wave equation.” Save it for another day.)

Energy in Wave MotionWork with sinusoidal waves. Consider a “piece” of the string. The kinetic energy is

dK =12

dmu2y =

12(µdx) [!ym!cos(kx!!t)]2

How much does the kinetic en-ergy change in the time dt it takes the wave to move a horizontal distance dx=vdt? The

answer is

dKdt

=12

µ!2y2mvcos2(kx!!t)

. For the potential energy, need to figure out how much the string stretches due to the tension. This will give you the work done by F.

dU = F!"

dx2 +dy2!dx#

= Fdx!"

1+(!y/!x)2!1#

Now we make a critical

assumption, namely that the “slope” is always small. Hence dU =

12

Fdx(!y/!x)2

. With F= μv2 we find that dU/dt=dK/dt, so dE/dt=2dK/dt. From this get the average power transmitted is Pav=(1/2)μω2ym2v.

Superposition and InterferenceGive them some words about “linear” differential equations. Could show them using the wave equation that if y1(x,t) and y2(x,t) are both solutions, then so is y(x,t)=y1(x,t)+y2(x,t).

Show them Figure 18-12. Mention “Fourier Analysis”, and that this is the reason why we can get away with talking about waves as sinusoidal only.

Now interference: Point out the usefulness of discussing the “phase” of a wave. That is, for two waves with same k and same ω but Δϕ=π, you get zero. Remind that power goes like square of the amplitude, enhancing interference. This is a common theme!

Break

Standing WavesWatch: y(x,t)=y1(x,t)+y2(x,t) with y1(x,t) =ymsin[kx±ωt] and y1(x,t) =ymsin[kx±ωt]. You end up with y(x,t)=[2ymsin(kx)]cos(ωt). This does not travel! It is a “standing” wave. Note that the amplitude varies as a function of position x. In fact, the amplitude is zero whenever kx=nπ, n=0,1,2,..., or x=nλ/2.

Energy considerations. Show figure 18-8. Draw the analogy with “oscillations.”

Think about this “the other way around.” That is, take a string and fix one end at x=0, and the other end at x=L. This will “support” standing waves with λ=2L/n. Frequency is related to wavelength by fλ=v, where v is the speed of the wave. Therefore, frequency of the standing wave is f=nv/2L.

Demo (Mechanical vibrations of a string): Start with some discussion of resonance. Can show that “higher mode” frequencies go up with n. You can also increase the weights on the end. Factor of four increase in tension should increase v (and so f) by factor of two.

30 October 2006Turn in homework assignment. Exam #2 is on Thursday.

Today a lightning discussion of “fluids” before the break, and “sound waves” after.

Fluids: Statics and DynamicsActually our first discussion of “continuum mechanics”, aka “classical field theory.”

Solids can support tensile and shear forces. Fluids cannot. Liquids can support com-pressional forces, but gases not even that so well. “Fluids do not have any shape.”

Express all this with math. Density is ρ=Δm/ΔV or =m/V for “uniform density.” Material response to pressure is “bulk modulus” B=-Δp/(ΔV/V). Bulk moduli for liquids and solids are some 10,000 times as for gases. (See book, Sec.15-2.) “Incompressible.”

Fluid at rest. Thin horizontal slide, area A and thickness dy. Fig.15-2. Not moving, so sum of vertical forces is zero. Find dp/dy=-ρg. This is how pressure changes vertically. Nice example in atmosphere, assuming ideal gas and uniform temperature. Follow book page 335 to find p=p0e-h/a where a=p0/gρ0=8.55 km. “Scale height” for the atmosphere.

Pascal’s Principle: p=pext+ρgh so Δp=Δpext for incompressible fluid, like water or oil. Demonstrate using example of the hydraulic lever, Figure 15-8.

Archimede’s Principle: Figure 15-10: Weight of blob of water counteracted by pressure of surrounding water. Replace blob with something else, so get “buoyant force.”

Now dynamics: Discuss “fluid flow.” Mass δm=ρδV crossing distance vδt is ρAvδt. Mass conservation implies that ρAv is a constant. Called “mass flux.” Illustrate with a simple example of faster moving fluid in a constricted pipe.

Bernoulli’s Equation: Conservation of energy for a fluid flowing in a pipe with vertical motion against gravity. See Fig.16-6 and Eq.16-5. External work done on the fluid is Wext = p1A1!x1 +(!p2A2!x2)[!!mg(y2! y1)]. Set that equal to the change in kinetic en-ergy K=½δmv2. Find p+½ρv2+ρgy=constant.

Good reading in book: Examples, and also “fluid flow” as a prototype for learning about vector fields. (Maybe Peter will do this next semester?)

Break

Sound WavesTechnically, these are mechanical waves in any three dimensional medium. Solids can support transverse and longitudinal waves since they can support shear forces, as well as compressional forces. Fluids can only have longitudinal waves since no shear forces.

Only present results here and not do the quantitative analysis. Some discussion in Sec.19-2 and 19-3, but it i a bit difficult to follow. Better left for more advanced courses in continuum mechanics.

Sound is a “pressure wave” or, alternatively, a “density wave.” These are related one way or another based on the fluid properties. (Example is “ideal gas law.”) For a typical sound wave, pressure and density waves are in phase with each other. (Fig.19-2.)

For “plane wave” of sound in pressure, write p=p0+Δp where Δp(x,t)=Asin(kx-ωt). The speed of the wave is v=ω/k=[B/ρ0]½. See Table 19-1 for the speed of sound in different materials. Faster in liquids and solids than gases since B is larger even though ρ is, too.

Sound and hearing. See Figure 19-5 and Table 19-2. Normal conversation is 10-6W/m2. Equation 19-19 says this is (Δpm)2/2ρv. For air have density ρ=1.2kg/m3 (Table 15-2) and speed v=340 m/s (Table 19-1). This gives (Δpm)2=816 (in whatever are the right SI units, which happen to be Pa2=(N/m2)2.) One atmosphere is p0=105N/m2 (page 332). Therefore the fluctuation in pressure that one hears in normal conversation is just a few parts in 10,000 over normal atmospheric pressure. Your ears are very sensitive!

Standing sound waves, obvious application are pipe organs. One important difference with mechanical waves on a string. A closed end of the pipe is an “antinode”, whereas an open end of the pipe is a node. See Section 19-6.

Assuming time is left, open the floor to questions prior the exam. That’s it for now!

6 November 2006Today we do “special relativity.” The in-class activity counts as homework. Note that there is a homework assignment (fluids; special relativity) due on Thursday.

Objective for today: Gain an appreciation for what the subject is, and do the graphical exercise to get a physical feeling. Lots more we won’t cover, like addition of velocities, relativistic momentum and energy, and so forth, also relativistic four-vector notation and the mathematics of boosts, but you will see that in later physics courses.

Problems with Classical PhysicsNewton inconsistent with Maxwell. Something had to give, and it was Newton. Einstein came up with the simplest way to state it, namely (1) physics is the same in all frames of reference (“relativity”) and (2) the speed of light is the same to all observers (!).

Neat consequence: Bouncing light clock (Figures 20-4 and 20-5): !t20 = (2L0/c)2 but

!t2 = (2L/c)2 = (2L0/c)2 +(u!t/c)2 or !t = "!t0 where ! = 1/!

1!u2/c2. This is known

as “time dilation.” Also “length contraction.” See your textbook (and the exercise.)

The Lorentz TransformationHow does an observer in one reference frame see “events” in another reference frame? The answer is the Lorentz Transformation. Not hard to derive, but we won’t bother to do it here. Just quote the answer. (See Textbook, Eq.20-14 and 20-17.)

x! = !(x"ut) and t ! = !(t "ux/c2) and the other way is x = !(x! +ut !) and t = !(t ! +ux!/c2)

Obviously, to go from one to the other, just change u to -u.

Natural UnitsWhy bother to carry around the c all the time? Think of it as a conversion factor between space and time. Measure time in seconds, but measure distance in light-seconds. This actually gives a suggestive form for the Lorentz Transformation:

x = !x! + !ut ! and t = !t ! + !ux! but !2! !2u2 = 1so put ! = cosh" and !u = sinh"

In other words x = (cosh!)x! +(sinh!)t ! and t = (sinh!)x! +(cosh!)t ! which is sort of like a “rotation” but using hyperbolic cosine and sine instead of circular. Hmm....

OK, so now move on to the worksheet, perhaps after a break.

Lecture 9 Nov 2006

Collect homework, including leftover “relativity” worksheets from Monday.

An apology: I never tried the “relativity” worksheet before, and there were obviously some problems. For whatever it’s worth, David Statman got in touch with me yesterday and told me that he has found it always takes longer and is more confusing than he thought it would! Oh, well, next year.

Coupled OscillationsFollow posted notes from today. Note homework assignment due next Thursday is at the end of the notes. This will all be reinforced by the laboratory exercise we do next Monday.

Laboratory 13 Nov 2006

This lab is on coupled oscillations, using the same setup that we had for demonstration in class last Thursday. You will have your own two-mass and three-spring setups, and you are encouraged to work in pairs in the same way as for previous labs. You will be able to take data using the motion sensor, following just one of the two masses.

Refer to last Thursday’s class notes for the mathematics of coupled oscillations.

In this, your last laboratory for the term, I encourage you to be creative. Here are some suggestions for data to take and results you can determine from your analysis.

1. Measure the normal mode frequencies and compare their ratio to expectations. It can be tricky to get the masses into “exact” eigenmodes, but do the best you can. You can take several trials, getting the period from the “zero” crossings, and use the average and standard deviation (or just the range) for the value and uncertainty. For identical masses and springs, you know what you should get. How do you calculate the ratio if the coupling spring has a different value?

2. Predict absolute values of the eigenmode frequencies. This will be similar to earlier labs. You’ll need the mass of the carts, and values for the spring constants.

3. Eigenmodes for different masses. How do you expect things to change if the two masses have different values? Try this, and see if you can set the system in motion in one or both of the two eigenmodes. Does this agree with what you expect from the equations of motion?

4. General motion of the system. After you determine the eigenfrequencies, try setting one of the masses in motion from rest but away from its equilibrium position. See if its subsequent motion agrees with the predicted motion from the notes. It’s probably easiest to try this first with equal masses and three identical springs, but you can of course see what happens in a more general case.

Keep all this in your lab book. A good job on this lab will help your final lab report grade.

Class Notes 27 Nov 2006Thanks to class: Prof Persans had a good experience! (Thanks to him, too!)

Today: (1) Review of Persans’ material on thermodynamics, (2) Some new stuff on thermodynamics, (3) Hand out the IDEA forms

Announcements: See the new syllabus, posted online. Third exam is this Thursday! No more homework is due. Final lab books are due one week from Thursday, which will be the last class of the term. We’ll do a “course review” that day. Next Monday, we’ll see a “glimpse of the future” by studying “The Principle of Least Action.”

Persans ReviewTemperature, Heat, and the First Law of Thermodynamics. Kinetic model of gases and the “ideal gas law.” Thermal expansion, heat capacity, and heat flow through slabs.

Temperature scales (F, C, and K) and units of heat (kCal and BTU, but also joule).

Ideal Gas Law: PV=NkBT=nRT

Kinetic theory: Temperature as molecular motion, i.e. 12

m!v2" =32

kBT

Specific Heat Capacity: ΔQ=CΔT=mcΔT See the tables for c in your textbook!

Latent heat of phase changes, i.e. transformation.

Heat flow thru cross sectional area A: dQ/dt=kA(dT/dx); k= specific thermal conductivity

Heat energy and work: Heat a gas and it can expand and do work against a piston.

Have “work done on the system” W =

Z V2

V1pdV

and !Eint = Q+W where Q is the heat added to the “working material.” This is the “First Law of Thermodynamics”.

Specific examples: Work done on an ideal gas under constant pressure Wp = p(Vf !Vi)

or constant temperature WT = NkBT ln(Vf /Vi) .

Some discussion (?) of cycles and engines.

Entropy and the Second Law: New StuffReversible, irreversible, and the “arrow of time.” This is profound stuff.

Entropy change is !S!

Z f

i

dQT “along a reversible path.” See the book for a “proof” that

S is a “state variable” for the case of an ideal gas. This is true in the general case.

Consider “free expansion”, Fig.23-22. Have dQ=-dW since no change in internal energy. In order to calculate the change in entropy, one needs to calculate the integral over some reversible path.

So, connect i and f with isothermal expansion so dW=-pdV=-NkBT(dV/V). (Note that the sign is negative because W is the work done on the gas by the piston, so dW must be

negative if dV is positive.) Then !S = NkB

Z f

i

dVV

= NkB lnVf

Vi . This is a positive quantity since the final volume is necessarily larger than the initial volume for free expansion.

This is an “isolated” system. There was no change in entropy outside the walls because there was no heat flowing through the walls. If we actually did carry out an isothermal expansion, though, then the heat flow would match what went in, and there would be an equal but opposite change in entropy in the space outside the walls.

We have just seen an example of the “Second Law of Thermodynamics”, namely that “When changes occur within a closed system, its entropy either increases (for irreversi-ble processes) or remains the constant (for reversible processes). It never decreases.”

Time for questions prior to Thursday’s exam.The test will not cover today’s material, but entropy may show up on the final exam.

Break and/or Hand out IDEA formsNeed a volunteer to pick them up and take them to the Physics office.

Class Notes 4 Dec 2006

Final class will be this Thursday, 7 Dec. Course review. Come with questions!

Final lab books are due on Thursday. Remember, this is 20% of your grade.

Final exam: Wednesday 13 Dec 6:30-9:30pm J-ROWL 2C30 (i.e. here). If you have some sort of conflict let me know right away!

Physics and the “Action Principle”John Cummings will introduce the subject. Want to get to the concept that “action” is the integral over time of some quantity (i.e. “the Lagrangian”) which depends on the “path.”

A Specific Example: Motion under Constant AccelerationDemonstrate that “the right path is the one that minimizes the action.”

Call x(t) a “path” through (one dimensional) space and time. A path has specific end-points, i.e. (x1,t1) and (x2,t2) where x1=x(t1) and x2=x(t2).

Consider a particle moving under constant acceleration between (x1,t1)= (0,0) and (x2,t2)= (1,1). We “know” that the right path is x=t2. (We measure position in meters, time in seconds, and the acceleration is 2 in these units.)

So S[(x(t)] =

Z t2

t1L(x, x)

. The “physics” is in figuring out what L(x, x) is supposed to be.

For mechanical motion, we now know that L(x, x) = K!U =12

mx2!U(x) . For U(x) take -dU/dx=F=ma=2m so U(x)=-2mx.

So S[x(t)] =

Z 1

0

!12

mx2 +2mx"

dt. Calculate for x(t)=t2:

S[t2] = 2mZ 1

0

!t2 + t2"dt =

43

m

Challenge to class: Calculate S[x(t)] for some other path x(t). The only restriction is that x(0)=0 and x(1)=1. Does anybody get a smaller value than 10m/3? Try to get them to work in pairs, and use different functions.

For me: S[tn] = m

Z 1

0

!n2

2t2n!2 +2tn

"dt = m

#n2

2(2n!1)+

2n+1

$

Another function would be x(t)=sin(πt/2). Use half-integral formula to integrate KE term. Answer works out to (π/4)2+(4/π)=1.89>(4/3).

Class Notes 7 Dec 2006Course review. Expect questions from students, but here is a list of topics that will be covered on the final exam.

Kinematics and Dynamics of Particles• Position, velocity, and acceleration as derivatives with respect to time• Newton’s second law as a differential equation• Uniform circular motion and centripetal acceleration• Free body diagrams for solution to statics and dynamics problems• Specific force laws for springs, friction, and drag

Momentum and Energy• Momentum as a vector quantity; Translational symmetry• Center of Momentum (aka Center of Mass)• Work, kinetic energy, and the work-energy theorem; Galilean Invariance• Potential energy and the conservation (or not) of total mechanical energy

The Law of Universal Gravitation• Newton’s force law for gravitation; The dual role of mass• Shell theorems• Gravitational potential energy• Circular orbits; Escape velocity

Mechanical Oscillations• Simple harmonic motion as derived from a mass and spring system• Oscillations as sines and cosines, or as exponentials of imaginary numbers• Meaning of initial conditions for position and velocity, in terms of amplitude and phase• Damped harmonic motion• Coupled oscillations (See separate notes); Normal modes as “eigen” modes

Waves• Waves as oscillations in time and space• Wavelength and wave number; Period and (angular) frequency• Traveling waves and standing waves; Normal modes of a standing wave• The Wave Equation; Superposition and interference• Energy in wave motion

Kinematics and Dynamics of Rigid Bodies• Angular coordinates, velocity, and acceleration• Tangential acceleration versus radial acceleration• Rotational inertia (aka Moment of inertia); Determining for solid bodies• Torque and Newton’s Second Law in angular variables; Center of gravity• The Parallel axis theorem; The Pendulum and the Physical Pendulum

Fluid Statics and Dynamics• Fluids vs Solids; Liquids (“incompressible”) vs Gases (“compressible”)• Density, pressure, and “bulk modulus”• Fluids at rest: Atmospheric “scale height”, Pascal’s Principle, Archimedes’ Principle• Fluids in motion: Mass flux (also, a little about Bernoulli’s Equation)• Sound waves: Longitudinal in fluids, but longitudinal and transverse in solids

Special Relativity and the Lorentz Transformation• The inconsistency between Newton and Einstein: Make c the same for everyone• Consequences: Time Dilation and Length Contraction• The Lorentz Transformation: Equations and graphically; Natural units (c=1)

Thermodynamics• Temperature and Fahrenheit, Centigrade (Celsius), and Kelvin scales• The Ideal Gas Law: PV=NkBT• Kinetic theory of ideal gases: Average molecular kinetic energy=(3/2)kBT• Specific heat capacity; Heats of transformation; Heat flow; Thermal expansion• Work done on an ideal gas• Heat and work: The First Law of Thermodynamics• Entropy and the Second Law of Thermodynamics

Principle of Least Action• Unifying principle for all of (classical) physics• The Action S[x(t)] as a “functional” of the path x(t)• Action as the integral between (time) “endpoints” of the Lagrangian L=K-U• Calculating the action for a given path

Topics in Mathematics• Basic differentiation and integration formulas• Integration as the addition of many small things; Integrating over solid bodies• Taylor series and approximation formulas• Euler’s formula for exponentials of imaginary numbers• Hyperbolic sines and cosines

Laboratory Techniques and Data Analysis• The role of redundancy in data taking• Making use of diagrams and graphs• Determining uncertainties