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Physics I Honors Monday 25 August Fall 2008

First class! Name and course on board, read roll, introduce TA’s.

Distribute outline, lecture, and lab schedules. Discuss them.

New material for today: Start “thinking like a physicist.” Two new ideas, namely(1) Solving problems with dimensional analysis, and(2) F = ma is a “differential equation”

Today’s Topic #1: Dimensional Analysis

Given a problem, ask “What is the important physics?”. Then list the quantities that havesomething to do with that physics. In many cases, you can combine these things to get theanswer, at least to within a factor of 2 or π or something like that.

Convenient notation:[distance] = L[time] = T[mass] = M[temperature] = K

Example: What is the radius of a black hole of mass M? This is a combination of gravityand relativity, so use M , G, and c. Newton’s constant G has units m3/kg-sec2, so [G] =L3M−1T−2. (Easy to see this from F = GmM/r2 = ma.)

Let the radius be R = MxGycz where we need to find x, y, and z. This gives

[R] = MxL3yM−yT−2yLzT−z = Mx−yL3y+zT−2y−z = L1 (1)

so x = y, 3y + z = 1, and 2y + z = 0. The second and third of these say that y = 1, and sox = 1 and z = −2. Therefore we get

R = GM/c2 (2)

The “correct answer” is 12GM/c2, called the Schwarzchild Radius.

(Hint: Useful website for units and constants ishttp://pdg.lbl.gov/2008/reviews/consrpp.pdf)

1

Practice Exercise

You’ve heard of “String Theory.” The length of strings is supposedly the “Planck Length,”since gravity is unified with quantum mechanics and relativity at this distance. Gravity isgoverned by Newton’s gravitational constant G, quantum mechanics by Planck’s constant h,and relativity by the speed of light c. Combine these to find the Planck Length. (Note thatthe dimensions of h are distance×momentum, or L2M/T .) Look up values of G, h, and cand compare the Planck Length to the size of a proton (10−15m).

2

Physics I Honors Monday 25 August Fall 2008

Today’s Topic #2: Newton’s Second Law as an “Equation of Motion”

The most important of Newton’s Laws is the second. For motion in one dimension, we writeF = ma. (We’ll get to ~F = m~a later this week.) Physicists refer to this as the “equation ofmotion” for a particle moving through time in one dimension of space.

Let x(t) be the position x of a particle at time t. Then the velocity is v(t) = dx/dt ≡ x.(Have you all seen this notation before??) Acceleration is a(t) = dv/dt = v = x. SoNewton’s Second law can be written as

x =1

mF (x, t) (3)

where we make it clear that the force F may itself depend on position and/or time.

Equation 3 is called a “differential equation.” Unlike an algebraic equation, whose solutionis a value for a quantity like, say, x, the solution of a differential equation is a function, sayx(t). Since x(t) describes the motion of a particle with mass m acted upon by a force F (x, t),we call Eq. 3 the “equation of motion.”

Simple example. Let F (x, t) = F0, a constant independent of both x and t. Then

x = v =dv

dt=

F0

m(4)

v(t) =dx

dt=

F0

mt+ v0 (5)

x(t) =1

2

F0

mt2 + v0t+ x0 (6)

where the last line is the solution to this equation of motion. This might look more familiarto you from your high school physics course, if we write a = F0/m. You probably calledthis “motion under constant acceleration”, but of course a constant acceleration implies aconstant force, so it’s the same thing.

Each step in the integration above involves a “constant of integration.” I called these con-stants v0 for the first step, and x0 for the second step. These are standard notations, andin fact good ones. The constant v0 is the velocity at time t = 0, also called the “initialvelocity.” Similarly, x0 is the “initial position.” In order to solve any equation of motion, wewill need to specify these “initial conditions.”

3

Practice Exercise

Let y(t) describe the one dimensional motion of a particle with mass m that moves vertically.Assume that the particle is acted on by a (constant) force F = −mg. (Isn’t that peculiar, aforce that is exactly proportional to the quantity m that appears in the equation of motion?)Assume that the particle starts out at y = 0 with an initial (upward) velocity +V . Find themaximum height to which the particle climbs, in terms of m, g, and V . Make a sketch ofy(t) as a function of t. What is the shape of the curve? (Mathematicians have a name forit, and I’m sure you’ve heard it before.)

4

Physics I Honors Thursday 26 August Fall 2008

This class will concentrate mainly on mathematics. First we will review vectors, and andapply them to three dimensional motion. Then we’ll talk about Taylor series, how they canbe used in approximations, and then apply them to complex numbers.

Today’s Topic #1: Vectors and Motion in Three Dimensions

Introduction: The vector equation F = ma is a shorthand for three equations, namely

Fx = max and Fy = may and Fz = maz

Notation: A in book, ~A on board. Unit vectors are n in book, n on board.The length of a vector is |A| = A, so |n| = 1.

If you add two vectors, the result is still a vector: C = A + B

A number times a vector is still a vector: C = bA This means that A = AA.

The “dot product” between two vectors is a scalar: A · B = AB cos θ where θ is the anglebetween the two vectors. This is the projection of A onto B or vice versa.

The “cross product” between two vectors is a vector: A × B = C (Watch out! Some dayyou’ll learn about polar and axial vectors!) The magnitude of C is AB sin θ. The direction ofC is given, by convention, by the “right hand rule”. This means that A×B = −B×A.

Vectors can be written in terms of coordinates in any given coordinate system or referenceframe. For an (x, y, z) coordinate system, define base vectors i, j, and k, all orthogonal (sotheir mutual dot products are zero) and with i× j = k, etc. Then

A = Axi + Ay j + Azk (7)

A ·B = AxBx + AyBy + AzBz (8)

A×B = (AyBz − AzBy )i + (AzBx − AxBz )j + (AxBy − AyBx)k (9)

(1) General Motion in Three Dimensions. A vector r(t) = x(t)i + y(t)j + z(t)k givesthe motion of a particle in three dimensions. The velocity vector is v(t) = r(t) and theacceleration vector is a(t) = r(t). The (vector) equation of motion is r(t) = F(x, t)/m.

(2) Examples. Work W = F·d is a dot product. Torque τ = r×F is a cross product.

(3) Uniform circular motion. The motion r(t) traces out a circle at a constant rate.Let ω be the number of radians per second traced out,i.e. ω/2π revolutions per second. (Review radians?)

r(t) = r(i cosωt+ j sinωt) (10a)

v(t) = ωr(−i sinωt+ j cosωt) (10b)

Note : v · r = 0

a(t) = −ω2r(i cosωt+ j sinωt) = −ω2r(t) (10c)

Note : a = ω2r = v2/r (10d)

The force that is responsible for this acceleration mustpoint back to the center of the circle!

5

Practice Exercise Find the period P of the “conical pendulum” pictured below, in termsof the hanging angle θ, the string length L, and the acceleration g due to gravity.

The mass m executes uniform circularorbits with radius R. If the speed is v,then the period is just the time it takesto execute one orbit, namely P = 2πR/v.So, you need to figure out v in terms of R,or perhaps in terms of the string length Land the hanging angle θ, since R = L sin θ.

Since we know that F = ma, we just needto figure out what the force F is. Clearly,this force is provided by the tension T , butonly by part of it.

6

Physics I Honors Thursday 26 August Fall 2008

Topic #2: Taylor Series, Approximation Techniques, and Complex Numbers

It makes sense that we can approximate any function by some high order polynomial:

0 1 2 3 40

2

4

6

8

10

Zeroth orderFirst orderSecond orderFull function

The function is expanded about the pointx0 = 1. Shown are the “zeroth order” (i.e.a constant) approximation, the first order,second order, and the full function. Thezeroth order is just the value of the func-tion at x0 = 1, and the first order is thetangent line at that point. The second or-der approximation comes close to the fullfunction in the neighborhood of x = x0.

f(x) = a0 + a1(x− x0) + a2(x− x0)2 + a3(x− x0)3 + · · · (11)

f(x0) = a0 f ′(x0) = a1 f ′′(x0) = 2 · a2 f ′′′(x0) = 3 · 2 · a3 (12)

f(x) = f(x0) + f ′(x0)(x− x0) +1

2!f ′′(x0)(x− x0)2 + · · · (13)

Examples : (1 + x)α = 1 + αx+1

2!α(α− 1)x2 + · · · (14)

ex = 1 + x+1

2!x2 +

1

3!x3 +

1

4!x4 +

1

5!x5 + · · · (15)

sin(x) = x− 1

3!x3 +

1

5!x5 + · · · (16)

cos(x) = 1− 1

2!x2 +

1

4!x4 + · · · (17)

Obviously, if |x| 1, i.e. “x is small”, then only a few terms (maybe just one) is needed inorder to get an accurate approximation.

Complex numbers. There is a number i such that i2 = −1. Numbers formed by a “real”number times i, like 2i, −i

√3, or xi where x is a “real” number, are called “imaginary”

numbers. Don’t let the words fool you, though. All of these are perfectly valid numbers. Theimaginary numbers are a group outside integers, rational, and irrational (real) numbers.

Numbers like 3 + 2i and z = x+ iy are called “complex” numbers. They have a “real part”and an “imaginary part”. We write, for z = x + iy where x and y are real, x = Re(z) andy = Im(z). The “modulus” of z is |z| =

√x2 + y2.

Now, here’s a neat thing:

eix = 1 + ix+1

2!(ix)2 +

1

3!(ix)3 +

1

4!(ix)4 +

1

5!(ix)5 + · · ·

=

[1− 1

2!x2 +

1

4!x4 + · · ·

]+ i

[x− 1

3!x3 +

1

5!x5 + · · ·

]= cos(x) + i sin(x)

This is called Euler’s Formula. It is very useful in physics and engineering!

7

Practice Exercise

First, a “math rule.” The derivative of an exponential is an exponential. That is

f(x) = eax =⇒ f ′(x) = aeax

Use this to find the solution to the equation of motion for a particle of mass m subject toa force F (x) = −kx where k is a positive constant. Write down the equation of motion asx(t) = F/m. Then show that x(t) = Ceiωt is a solution to the equation of motion, for anyvalue of C, so long as ω has one of two possible values. What are those values?

Next, write the “general solution” x(t) as the sum of two functions Ceiωt, one for each ofthe two possible values of ω and with two different values of C. (Call them C1 and C2.) Is itobvious to you that this sum of solutions is also a solution to the equation of motion?

Finally, determine the two constants C1 and C2 from the initial conditions x(0) = A andx(0) = 0. Simplify the result as much as possible. (Euler’s formula will be handy.) Whatkind of motion does this describe.

8

Physics I Honors Thursday 4 September Fall 2008

Let’s try to stick with the “20-20-20” plan, that is new material for 20 minutes, 20 minutesto work on a practice exercise, and then 20 minutes to talk about the solution.

Today’s Topic #1: Applications of Newton’s Laws

Discuss “Free Body Diagrams.” Statics first. Use example of one block A sitting atop anotherblock B (page 68 in K&K) and take care to note the weight of each block, the normal forceon B from the lower surface, and the force F1 (F2) which block B (A) exerts on block A(B). Hence F1 = MAg and N = F2 + MBg. Third law says F1 = F2 so N = (MA + MB)g,which is just what you expect, of course.

Pretty much the same thing for two masses hanging, connected by a wire under tension. Thetop wire hangs both masses! Tell the story of the Hyatt Regency disaster, July 17, 1981 inKansas City, Missouri: http://en.wikipedia.org/wiki/Hyatt Regency walkway collapse

Dynamics is very similar, just don’t set a = 0. See practice problem and also homework.

Let’s do the conical pendulum. (See Example2.8 in your textbook.) Find the period P interms of the angle θ, the string length L, andthe acceleration g due to gravity. The speedof m is v = Rω, so P = 2πR/v. To find v, useF = ma with a = ω2R = v2/R. (Recall Equa-tions 10.) Find F = T sin θ with T cos θ = mg.Putting it all together, we have

P 2 = (2π)2R2

v2= (2π)2R

a

= (2π)2 Rm

T sin θ= (2π)2Rm cos θ

mg sin θ

P = 2π

√R cos θ

g sin θ= 2π

√L cos θ

g

Practice Exercise

!"

!#

Find the acceleration a of the coupled masses on the left, interms of m1, m2, and g. The mass m1 slides on a horizontal,frictionless surface, m2 hangs freely, and the string thatconnects them is massless. The pulley is also massless, andgives no frictional resistance to motion.

Start by drawing free body diagrams of m1 and m2. Identifyall forces on each of the two masses, and then work with thecomponents that are in the direction of the acceleration.

9

Topic #2: The Everyday Forces of Physics

Start by making a list. Use vector notation!

• Tension, the pull of a “massless” string: T

• Weight, i.e. gravity near the Earth’s surface: W = −mgk

• Springs, i.e. “Hooke’s Law”: F = −kx

• Frictional forces

– Kinetic friction: fk = µkN “opposing the direction of motion”

– Static friction: fs ≤ µsN “as big as it needs to be to keep the object from moving”

Note that µs > µk for the same two surfaces. Demo with book on desk!

• Viscous forces: Fv = −Cv

• Inverse square “central” forces (Draw some pictures here.)

– Newtonian gravity: Fa→b = −(GMaMb/r2)ra→b

– Electrostatic attraction: Fa→b = +(kQaQb/r2)ra→b

Note: General central force is just Fa→b = f(r)ra→b

Let’s study the motion of an object under the influence of a drag force. (See Example 2.16.)Assume a mass m is moving through a viscous medium. The equation of motion is

−Cv = ma = mdv

dtor

dv

dt= −C

mv (18)

Now integrate. The “math trick” you need is d(lnx)/dx = 1/x. We have∫ v

v0

dv

v= −C

m

∫ t

0

dt or ln

(v

v0

)= −C

mt or v(t) = v0e

−Ct/m (19)

Note use of initial condition notation. The behavior is clear. The particle moves withvelocity v0 at t = 0, but then slows down exponentially.

Note: Study the “shell theorem” in Note 2.1. Basically, a spherical object attracts an externalmass gravitationally, where the equivalent object mass is all that is contained within a sphereor radius intersecting the external mass.

Practice Exercise

Express the gravitational acceleration near the Earth’s surface, g, in terms of G, and themass ME and radius RE of the Earth. Start by putting an “external mass” m on the surfaceof the Earth. Then calculate the gravitational attraction of the Earth on this mass, usingthe formula for Newtonian gravity.

Cavendish devised an sensitive experiment that measured G in 1798. People say he “weighedthe Earth.” What do they mean? (Note that Eratosthenes measured the Earth’s radiussomewhere around 200 B.C.)

10

Physics I Honors Monday 8 September Fall 2008

Today’s Topic #1: Momentum and Systems of Many Particles

Newton actually wrote the Second Law as∑

F = dp/dt where momentum p = mv. If massm doesn’t change with time then

∑F = mdv/dt = ma, so we recover the old version.

Now imagine an object colliding with a brick wall. The collision will change the momentumof the object. The force of the collision varies over the collision time. (Draw a figure; SeeExample 3.9 in K&K.) We can use Newton’s second law to write an expression involving theintegral of the force and the change in momentum:

∆p =

∫ t2

t1

F(t)dt ≡ J

We call J the “impulse.” The “average force” Favg is J divided by ∆t = t2 − t1.

Now imagine two objects colliding with each other. Each exerts an “equal and opposite”momentum on each other. That is J1 = −J2, or ∆p1 + ∆p2 = ∆(p1 + p2) = 0. In otherwords, the total momentum P ≡ p1 + p2 is conserved in collisions. Note that this doesn’twork if there is some “external” force acting on the masses.

Now look at two particles again, but not just collisions. Allow some external force Fext toact as well as internal forces. Then

dP

dt=dp1

dt+dp2

dt= m1a1 +m2a2 =

∑Fext

The last equality follows because the internal forces cancel in the sum. It is convenient towrite P = Mvcm where M = m1 +m2 and vcm = drcm/dt with

rcm =m1r1 +m2r2

m1 +m2

=1

M(m1r1 +m2r2) (20)

in which case Newton’s Second Law reads∑

Fext = M rcm = Macm. We call rcm the centerof mass. (Is “ center of momentum” better?) For more than two particles

rcm =m1r1 +m2r2 + · · ·+mNrN

m1 +m2 + · · ·+mN

=1

M

N∑n=1

mnrn (21)

Don’t forget that this is really two (or three) equations for xcm, ycm, and (if three) zcm. Thefigures below show how this all works out:

11

Practice Exercise

An “elastic” collision is one in which kinetic energy is conserved. (The kinetic energy of anobject with mass m and speed v is K = 1

2mv2. The total kinetic energy is the sum of the

individual object’s kinetic energies. We will learn more about this later.) Let two objects,one with mass m and the other with mass 2m collide elastically in one dimension. They areinitially heading towards each other with speed v. Find their speeds and directions after thecollision, in terms of m and v.

Topic #2: Calculating the Center of Mass of Solid Objects

If we are working on a solid object, we could sum over all the atoms that make it up, butthat would be nuts. Instead, we use calculus, and assume that the object is continuous.Then the sum becomes an integral. That is

M =

∫dm and rcm =

1

M

∫dm r

We usually write the infinitesimal mass dm as a density (usually called ρ or σ) times avolume element dV (or some other measure), i.e. dm = ρ(r)dV . This allows the density tovary as a function of position inside the solid object. The rest is in the details.

Here we can slice up an eggplant, making the point that∫dV is the sum of small pieces.

Here is a simple example. What is the volume of a right circular cone of height h and baseradius R? Let z measure the height, and slice the cone up into thin disks, each of thicknessdz. The radius of each disk is R(h − z)/h = R(1 − z/h), so the volume of each disk isπR2(1− z/h)2dz. Therefore, the volume of the cone is∫

dV =

∫ h

0

πR2(1− z/h)2dz = πR2

∫ h

0

(1− 2z

h+z2

h2

)= πR2h

3

or one-third of the base area times the height. Of course this is volume, not mass, but that’sbasically the same thing for constant density. If the density is not a constant, you’d need toinclude the functional form of ρ(r) and that would be a harder problem

Practice Exercise

Find the position of the center of mass of a long thin rod of mass M and length L. Thedensity of the material making up the rod changes linearly from zero at one end. Start byrealizing that the mass of a short segment of length dx is dm = σ(x)dx where σ(x) is thelinear mass density. Write a formula for σ(x), and determine any parameters by fixing themass of the rod to be M . Then, find the position of the center of mass.

12


Today’s Topic #1: Work and Kinetic Energy in One Dimension

Let’s use some calculus to manipulate F = mdv/dt, in one dimension:∫ b

a

F (x)dx = m

∫ b

a

dv

dtdx = m

∫ b

a

dv

dt

dx

dtdt = m

∫ b

a

vdv =1

2mv2

b −1

2mv2

a (22)

We write F = F (x) but it could be a constant, for example F = −mg:

−mg∫ yb

ya

dy = −mg(−h) =1

2mv2 − 1

2m02 =⇒ v =

√2gh (23)

for an object falling through a distance h = ya− yb starting from rest. We already knew thisfrom solving the equation of motion (Eq. 6) but here we’ve done this a different way.

A harder, but still easy, example is a spring force F = −kx. Equation 22 becomes

−1

2kx2

b +1

2kx2

a =1

2mv2

b −1

2mv2

a (24)

You will play around with this on the practice exercise.

These are both examples of something called the Work-Energy Theorem. Define thework (in one dimension) done by a force F (x) on a mass m through a path xa → xb as

Wba =∫ baF (x)dx. Define the kinetic energy of that object as K = 1

2mv2, for velocity v.

Then Eq. 22 says thatWba = Kb −Ka ≡ ∆K (25)

That is, the work done on the object equals the change in kinetic energy. Of course, thework done for a constant force is just the product of the force times the displacement.

Let’s use this to calculate “escape velocity.” A rocketstarts at ra = Re with (escape) velocity v. It ends upat very large distance, rb → ∞, with v = 0. (If theinitial velocity were larger, it would have a finite speed atinfinite distance. If smaller, it would fall back to Earth.)

∫ ∞Re

(−GMem

r2

)dr =

GMem

r

∣∣∣∣∞Re

= −GMem

Re

= 0− 1

2mv2 =⇒ v =

√2GMe

Re

(26)

We will consider all these ideas from a purely “energy” point of view on Monday.

Practice Exercise An object with mass m is acted on by a spring force F = −kx. It haszero velocity when x = A. Use Eq. 24 to find the velocity v = V when x = 0, in terms of k,m, and A. Explain why this is consistent with x(t) = A cosωt, which solves the equation ofmotion so long as ω2 = k/m.

13

Topic #2: Work and Kinetic Energy in Several Dimensions

How do we generalize Eq. 22 to recognize that force is a vector, i.e. F = mdv/dt? Assumingwe want to end up at the work energy theorem, we can try working backwards:

d

dt(v2) =

d

dt(v · v) = v · dv

dt+dv

dt· v = 2v · dv

dt= 2

dr

dt· dvdt

(27)

∆K =

∫ b

a

d

(1

2mv2

)=

∫ b

a

1

2md

dt(v2)dt =

∫ b

a

mdr

dt· dvdtdt =

∫ b

a

(mdv

dt

)· dr (28)

or ∆K =

∫ b

a

F(r) · dr (29)

That’s all well and good formally, but what does the last integral mean? The motion r(t)traces out some path in space, and the integral is “along the path.” (See the figures below.)Indeed, we call it a “path integral” or “line integral.” (The latter terminology is morecommon. The former has a larger connotation when studying quantum mechanics.)

A couple of simple things to notice:

• For a constant force F acting along a straight line path with displacement d = ∆r, thework done by the force is W = F · d

• For an object moving in a circular path, the “centripetal” force keeping it moving thisway does no work on the object, because F(r) · dr = 0 everywhere along the path.

• If the path is “closed”, like a circle or some other closed loop, then the line integraldone all the way around is written as

∮F · dr

Practice Exercise

Find the work W =∮

F · dr done around theclosed rectangular loop shown, for two forces (a)F(x, y) = yi, and then for (b) F(x, y) = xi. Treatthe closed loop as four, straight line paths I, II, III,and IV, and calculate the work done on each path.The total work done is just the sum of the workdone over each of the four straight line segments.The force in (b) is called “conservative” and theforce in (a) is called “non-conservative.” We willbe discussing all this a lot more.

14


Today’s Topic #1: Potential Energy; Conservation of Energy

Consider a mass m on a spring, i.e. F (x) = −kx, and take the time derivative of

E =1

2mv2 +

1

2kx2 (30)

dE

dt= mvv + kxx = v(mv + kx) = v(F + kx) = 0 (31)

We say that E is a “constant of the motion.” It is “conserved quantity.” Note thatd(kx2/2)/dx = kx. In general, let F (x) = −dU/dx for some function U(x). So we have

E ≡ 1

2mv2 + U(x) =⇒ dE

dt= mvv +

dU

dt= mvv +

dU

dx

dx

dt= v

(F +

dU

dx

)= 0 (32)

We call U(x) the “potential energy” and E the “total energy.” Relating this to work,

Wba =

∫ b

a

F (x)dx = −∫ b

a

dU

dxdx = −

∫ b

a

dU = − [Ub − Ua] (33)

That is, the work done from a to b is the negative of the change in potential energy.

Now generalize to three dimensions. This is simple, in principle. We just write

Ub − Ua = U(rb)− U(ra) = −Wba = −∫ b

a

F(r) · dr (34)

but the line integral can (in principle) depends on the path! If it does, then Eq. 34 makesno sense. We can only define a potential energy for forces with path independent work, alsoknown as “conservative” forces. These let us write down an energy that is conserved.

Non-conservative forces lead to a loss of (mechanical) energy. For F = Fc+Fnc, we have

Kb −Ka = W totalba =

∫ b

a

F · dr = − [Ub − Ua] +

∫ b

a

Fnc · dr =⇒ Eb − Ea = W ncba (35)

The energy lost in going from a to b is just the work done by the non-conservative forces.

Finally, let’s define power. Power is just the rate at which work is done. That is

P ≡ dW

dt=

F · drdt

= F · drdt

= F · v (36)

Practice Exercise

An object with mass m falls at a constant (that is, “terminal”) velocity v through theatmosphere. If the drag force is f = −bv, find b in terms of m, g, and v. Then, calculate (a)the change in total mechanical energy ∆E when the object falls through a distance h, and(b) the work W nc done by the non-conservative forces. Finally, show that ∆E = W nc.

15

Topic #2: Energy Diagrams

We gain insight about motion by looking more literally at energy conservation, i.e.

E =1

2mx2 + U(x) so x = v = ±

√2

m[E − U(x)] (37)

Velocity can be to the left or right, but “speed”(≡ |v|) is determined. The square root leadsto turning points to the motion, which limitthe position based on the energy. See the fig-ure, and discuss motions for different values ofE. Relate force plot to the energy plot. Dis-cuss behaviors of particles with various Ei, andmention oscillations. Draw the analogy with“up and down a hill.” Note that, in princi-ple, one can get the motion x(t) by integratingEq. 37. We’ll do this in a practice problem.

Example: The Pendulum. A simple pendulum is a mass m attached to a massless stringof length ` which swings in a plane through an angle φ. (See page 255 in your textbook.)Analyzing this in terms of “F = ma” is a little tricky, but we can do it. It is two dimensional,but the two “axes” change with time. One is radial (in which there is no motion) and theother is “angular” along the path, a variable we’ll call s = `φ. We have

Fs = −mg sinφ = ms so φ = −g`

sinφ (38)

Nobody knows how to solve this problem analytically. However, if the pendulum never swingsthrough large angles, then sinφ ≈ φ and we recover the “harmonic oscillator” equation ofmotion, with ω2 = g/`.

We can also look at this in terms of energy. The potential energy in terms of φ is

U(φ) = mgh = mg(`− ` cosφ) = mg`(1− cosφ) (39)

Plot this as a function of φ and oscillate about the minimum at φ = 0. (Also point out theequilibrium point at φ = π.) This leads into a discussion of “small oscillations”, i.e.

U(φ) = mg`

(1−

[1− φ2

2!+φ4

4!+ · · ·

])≈ mg`

φ2

2= m

g

`

s2

2(40)

which is a “harmonic oscillator” potential with k replaced by mg/`, i.e. k/m→ g/`.

Practice Exercise

Use Eq. 37 with U(x) = mgx and E = mgh to find x(t). Assume that x = h when t = 0.(This describes the motion of something falling from rest at a height h, right?) Split dx/dtso that you get

∫dx on the left and

∫dt on the right. It should be obvious after integrating

that the constant of integration is zero for this initial condition. You’ll need the followingobscure calculus rule: ∫

dx√a+ bx

=2√a+ bx

b(41)

where a and b are constants.

16


Today’s Topic: Conservation Laws and Particle Collisions

Textbook (Sec.14.4): “No new physical principles are involved; the discussion is intended toillustrate ideas we have already discussed.”

Pinitial = m1v1 +m2v2 Pfinal = m1v′1 +m2v

′2 (42)

Pinitial = Pfinal (43)

m1v1 +m2v2 = m1v′1 +m2v

′2 (44)

Elastic and inelastic collisions. If the (internal) forces that cause the collision are conser-vative, then mechanical energy wil be conserved. Far away from the colliding point, there areno forces, so no (internal) potential energies, and conservation of energy means conservationof kinetic energy. If non-conservative forces act, then energy can be lost. We write

Ki = Kf +Q i.e. Q = Ki −Kf ≥ 0 (45)

If Q = 0 then the collision was elastic. Using the same notation as above

1

2m1v

21 +

1

2m2v

22 =

1

2m1v

′21 +

1

2m2v

′22 +Q (46)

Example 4.18: Elastic Collision of Two Balls. (Recall Sept.8 practice exercise.) Putm1 = m and m2 = 3m, v1 = v and v2 = −v. Conserving momentum and energy,

mv − 3mv = mv′1 + 3mv′2 (47)1

2mv2 +

1

23mv2 =

1

2mv′21 +

1

23mv′22 (48)

Equation 47 givesv′1 = −2v − 3v′2 (49)

which we can then plug into Eq. 48 to find

2v2 =1

2(2v + 3v′2)2 +

1

23mv′22 (50)

= 2v2 + 6vv′2 + 6v′22 (51)

or 0 = vv′2 + v′22 (52)

17

This equation has two solutions for v′2, each of which gives v′1 by Eq. 49, namely

v′2 = −v =⇒ v′1 = +v (53)

or v′2 = 0 =⇒ v′1 = −2v (54)

The first of these is just the initial conditions. If nothing happens when the two balls passnear each other, then nothing happens. If they do in fact collide, however, then the secondball stops, and the first one fires backward at twice the velocity that it started with.

Collisions and the Center of Mass. The velocity of the center of mass is

V =m1v1 +m2v2

m1 +m2

(55)

Let’s call this the “laboratory frame.” We can change our frame of reference to the “centerof mass frame” by imagining that we are observing the collision by moving along with thecenter of mass. In this frame, the velocities of the two particles are observed to be

v1C = v1 −V =m2

m1 +m2

(v1 − v2) (56)

v2C = v2 −V =m1

m1 +m2

(v2 − v1) (57)

so that the momenta in the center of mass frame are given by

p1C = m1v1C =m1m2

m1 +m2

(v1 − v2) = µv (58)

p2C = m2v2C =m1m2

m1 +m2

(v2 − v1) = −µv (59)

where we have made the definitions

v ≡ v1 − v2 and µ ≡ m1m2

m1 +m2

(60)

The total momentum p1C + p2C = 0 in the center of mass frame, as we expect. It is easyto show that for elastic collisions in the center of mass, v′1C = v1C and v′2C = v2C . Yourtextbook has the details.

We call v the “relative velocity” and µ the “reduced mass.” You will come back to thesequantities over and over again as you study the physics of collisions, both classically andquantum mechanically.

Practice Exercise

Suppose that the two balls in Example 4.18 stick together after the collision, i.e. v′1 = v′2.Calculate the Q value for the collision.

18


First exam is this Thursday, in class, at 10am. Today, we will do some math.

A Review of Some Calculus Rules

f(x) f ′(x)∫f(x)dx f(x) f ′(x)

∫f(x)dx

xa axa−1 1a+1

xa+1 1x

— ln x

ex ex ex lnx 1x

—cosx − sinx sinx sinx cosx − cosx

Chain rule :d

dxf(u) =

df

du

du

dxProduct rule :

d

dxuv = v(x)

du

dx+ u(x)

dv

dx(61)

Partial Derivatives, Gradient, and Force & Potential Energy in 3D

The derivative (with respect to x) of a function f(x) is just the rate at which f(x) is changing,that is ∆f/∆x when the changes are very small. We write f ′(x) = df/dx, where the rightside of the equation is a much more descriptive notation.

Now suppose that f is a function of two (or more) variables x and y, that is f(x, y). Wecan ask how f changes with respect to x if y is held constant, or vice versa. We write thesederivatives as ∂f/∂x and ∂f/∂y, respectively, and call them “partial derivatives.” There areother notations, but they are all bad. In many cases df/dx and ∂f/∂x can be quite different,but we will not get into this kind of confusing situation in this course!

Suppose x changes by an amount dx, and y changes by an amount dy. How much doesf(x, y) change? The answer is pretty obvious, namely

df =∂f

∂xdx+

∂f

∂ydy (62)

So now, consider the work dW done by a force F(r) over a distance dr. The change inpotential energy U(r) = U(x, y, x) (assuming that one can be defined) is

dU = −dW = −F(r) · dr (63)

Write the left side using Eq. 62 and write out the right side in terms of components, so

∂U

∂xdx+

∂U

∂ydy +

∂U

∂zdz = −Fxdx− Fydy − Fzdz (64)

Clearly the x component of F(r) is just −∂U/∂x and so forth. It is nice to write this in asimple notation using vectors, by defining the gradient operator ∇ as follows:

∇ ≡ i∂

∂x+ j

∂

∂y+ k

∂

∂z(65)

In this case, Eq. 64 is really just a statement that

F(r) = −∇U(r) (66)

We frequently refer to F(r) as a “vector field” and U(r) as a “scalar field.”

19

A Useful Theorem about the Curl of a Vector Field

!"#$%

&"

&$

Imagine a tiny, closed rectangular path in the (x, y) plane, with sidelengths dx and dy, and with the lower left corner at the point (x, y).It is very simple to calculate the work done by a force field F(x, y)while going around this path, since the force is approximately con-stant along each of the straight legs. We have∮

F · dr = F (x, y)dx+ F (x+ dx, y)dy

− F (x+ dx, y + dy)dx− F (x, y + dy)dy (67)

Collecting some terms, and ignoring a change in y, say, when moving in the x direction,∮F · dr = [F (x+ dx, y)− F (x, y + dy)] dy − [F (x+ dx, y + dy)− F (x, y)] dx

=

[F (x+ dx, y)− F (x, y + dy)

dx− F (x+ dx, y + dy)− F (x, y)

dy

]dxdy

=

[∂F

∂x− ∂F

∂y

]dxdy = [∇× F]z dxdy = [∇× F] · dA (68)

We call ∇× F the “curl” of the field F, and have dA ≡ dxdy and the vector dA pointedperpendicular to the plane bounded by the tiny loop.

Now imagine an arbitrary surface A bounded by some curve C. You can “tile” A into abunch of tiny rectangles, and the loop integrals around any two adjacent rectangles cancelson the border. In other words, the tiny loop integrals (68) end up giving the loop integralaround C. Mathematically, this says that∮

C

F(r) · dr =

∫A

[∇× F] · dA (69)

This is called Stokes’ Theorem. Obviously, a force field F(r) is conservative if ∇× F = 0.This is automatically satisfied if F(r) = −∇U(r).

A Different Useful Theorem about the Divergence of a Vector Field

Now consider a vector field E(r) and a tiny box in three dimensional space, with sidesdx, dy, and dz. The “flux” out of the box near the point (x + dx, y, z) is pretty clearlyEx(x+ dx, y, z)dydz. Play a game similar to what we did with Stoke’s theorem, building anarbitrary volume V out of a bunch of tiny boxes. For two adjacent boxes, the flux out of onecancels the flux into the other, so you end up measuring the flux through the entire (closed)surface A that contains the volume. It isn’t hard to go from here to show that∮

A

E(r) · dA =

∫V

[∇ · E] dV (70)

Physicists call this Gauss’ Theorem; mathematicians call it the Divergence Theorem. (Thequantity ∇ · E is called the “divergence” of the field E.) We will find some important usesof Gauss’ theorem this semester when we study fluid behavior. Next semester, you will useGauss’ theorem in your study of electromagnetism.

20


Hand back Exam #1. Average was 74.1. Questions? Tell me. Grades ≤ 60 see me.

Angular Momentum of a Particle

Consider an object moving in two dimen-sions under the action of a central forceF = Fxi + Fy j. The figure shows that

Fy/Fx = y/x or Fy − yFx = 0

Now consider a quantity ` defined as

` ≡ m (xy − yx) (71)

This quantity is a constant of the motion:

d`

dt= m (xy + xy − yx− yx)

= m (xy − yx) = xFy − yFx = 0

If we define L ≡ r × (mv) = r × p, then we see that ` = Lz. We call L the angularmomentum of a particle, and it is conserved for motion in a central force field.

It is convenient to think of L in terms of parallel and perpen-dicular components of r. If we write r = r⊥ + r‖ then

L = r× p =(r⊥ + r‖

)× p

= r⊥× p + r‖× p = r⊥× p (72)

so that ` = Lz = |r⊥| |p| (= |r| |p⊥|) (73)

We call |r⊥| the “moment arm” of the angular momentum.

In 1609 Kepler told the world that planets “sweep out equalareas in equal times.” Newtonian gravity is a central force,and Kepler’s statement is a consequence of angular momentumconservation. Following the figure at the right

dA1

2(r + dr)(rdθ) so

dA

dt=

1

2r2dθ

dt=

1

2rds

dt(74)

But ds/dt = v⊥ = p⊥/m so dA/dt = rp⊥/2m = Lz/2m.

Practice Exercise

Angular momentum depends on the choice of coordinate system. Show this explicitly forsomething falling straight down, starting from rest at y = 0. What is L for something fallingalong the y-axis? Do it again for a line parallel to the y-axis, along the line x = A?

21

Torque

If angular momentum is constant for an object acted on by a central force, then how do weget the angular momentum to change? By brute force, we see that

dL

dt=

d

dt(r× p) =

dr

dt× p + r× dp

dt= r× F (75)

where we used F = dp/dt and v× p = 0 with v = dr/dt and p = v. We write τ ≡ r× F,where τ is the torque. It is convenient to think of torque as force times a moment arm.

Imagine a bunch of masses hanging from a string. The torque from gravity is (y is “up”)∑n

rn × Fn =∑n

rn × (−mngj) =

(∑n

mnrn

)×(−gj

)= rcm ×

(−gj

)(76)

If we hang from the center of mass, then all moment arms are measured from here andrcm = 0. So, there is no torque and the object is balanced. Sometimes we call the center ofmass the “center of gravity.”

Practice Exercise

Calculate the torque for each of the two cases in the previous practice exercise, due to amass m falling under the influence of gravity, and show that in each case, τ = d`/dt.

Angular Momentum and Fixed Axis Rotation

So what is the motion resulting from a torque? Consider one single particle, inside the solidobject, located at r and acted on by F. It can only move in a circle of radius r and the forcein that direction is F sin θ. If the path length it travels is called s, then Newton’s SecondLaw says F sin θ = ms = mrφ. Multiply both sides by r, and we have τ = (mr2)α, whereα = φ is called the angular acceleration. The quantity mr2 has replaced mass when wedescribe the motion in angular variables.

Since α is the same for all parts of the object, and since τ is either from one force at onepoint, or due to a sum of such things, Newton’s Second Law for a solid object becomes∑

τ = Iα where I =∑n

mnr2n (77)

is called the rotational inertia or moment of inertia. Notice that I is a property of the solidobject and the specified axis. (Do you get the feeling that I is not really a “scalar” but nota vector either? If so, you are right. It is actually something called a “tensor.”)

You can prove something called the Parallel Axis Theorem which states that the rotationalinertia of any body about an arbitrary axis equals the rotational inertia about a parallel axisthrough the center of mass plus the total mass times the squared distance between the twoaxes. Mathematically, we write I = Icm +Mh2. See Example 6.9 in your textbook.

Rotational Inertia of Solid Objects

Of course, for solid objects, we don’t do the sum over all the masses in Eq. 77 to get therotational inertia. Instead, we do the integral I =

∫r2dm. This is just like calculating the

center of mass, that is∫r dm, but with another factor of r. See Example 6.8 in your textbook.

We will be doing more of these sorts of things, include homework problem 6.8.

22

Physics I Honors Thursday 2 October Fall 2008

Solid objects both rotate and translate. There is a lot of physics associated with these mo-tions. See Chap.6. Rather than derive and review it all, here are the rules (Table 6.1):

Pure rotation (about a fixed axis) Rotation plus translation (“0” means “CM”)L = Iω Lz = I0ω + (R×MV)zτ = Iα τz = τ0 + (R× F)z with τ0 = I0αK = 1

2Iω2 K = 1

2I0ω

2 + 12MV 2

Good examples to review include the Atwood’s Machine (Example 6.10), Kater’s Pendulum(6.12), and the “Disk on Ice” (6.15). We’ll do two applications in class today.

Objects Rolling Down a Plane. (Refer to Example 6.16)

Consider some object with radius b, mass M , and I0 = βMb2, rolling down a ramp. Howdoes the shape (i.e. β) affect the time it takes for the object to get to the bottom of theramp?

First analyze it using just∑F = ma. This says that

W sin θ − f = Mg sin θ − f = Ma

Friction makes it roll, with torque bf , so

bf = I0α = (βMb2)(a/b) = βMab

Solving, a = g sin θ/(1 + β) (78)

A different approach uses torque and angular momentum.

τz = τ0 + (R× F)z = −bf + b(f −W sin θ) = −bMg sin θ

(The normal force cancels W cos θ so no net torque.)

Lz = I0ω+(R×MV)z = −βMb2ω−bMV = −(1+β)Mb2ω

Then put τ = Lz = −(1+β)Mb2α = −(1+β)Mba and get the same answer as before.

So, a larger of β gives a smaller acceleration and takes longer to get down the ramp.

A third way to look at the problem is in terms of energy. The potential energy at the startis Ui = Wh = Mgh and there is no kinetic energy. At the bottom of the ramp, the CM hasspeed V so Kf = 1

2I0ω

2 + 12MV 2 = 1

2(βMb2)(V/b)2 + 1

2MV 2 = 1

2(1 + β)MV 2. Then,

Ui = Kf =⇒ Mgh =1

2(1 + β)MV 2 =⇒ V =

√2gh

1 + β(79)

and, again, large β travels more slowly at the bottom. (Friction is irrelevant here. Why?)

Practice Exercise

Use one dimensional kinematic relations, for something moving through a distance d withacceleration a, to show that Eq. 79 is the same as Eq. 78.

23

Kinematics of an Ultraelastic Rough Ball

See Richard L. Garwin, American Journal of Physics 37(1969)88. It is posted on our website.(Discuss how to look up papers in “e-Journals” using http://library.rpi.edu/)

Play with the ball a little, as in Figure One and then as in Figure Three.

The analysis is based on two important assumptions about the ball and its collisions:(a) The collision is elastic, conserving the sum rotational+translational kinetic energy(b) There is no slip in contact between the ball and a surface, i.e. the ball is “rough”

The figure shows the coordinate system and nomenclatureused in the paper. We use “b” for “before” and “a” for“after” the collision. The abbreviation C = ωR is used forbrevity, where R is the radius of the ball. The moment ofinertia (about the CM) of the ball is I0 = βMR2 with β =2/5. (Garwin uses α instead of β, but I switched so that thereis no confusion with our notation for angular acceleration.)

The kinetic energy of the ball is just

K =1

2I0ω

2 +1

2MV 2 =

1

2βMR2ω2 +

1

2MV 2 =

1

2M(V 2 + βC2) (80)

so Kb = Ka. The angular momentum about the contact point (see the figure) is just

L = I0ω −R(MV ) = βMR2ω −MVR = MR(βC − V ) (81)

and since all of the forces in the collision pass through the contact point, Lb = La. This im-mediately gives a simple result, since Eqs. 80 and 81 imply, with conservation applied,

β(C2b − C2

a) = V 2a − V 2

b and β(Cb − Ca) = −(Va − Vb) (82)

These equations are divided to yield

Cb + Ca = −(Va + Vb) or Ca + Va = −(Cb + Vb) (83)

Garwin calls V + C the “parallel velocity,” the velocity tangent to the ball’s surface at thecontact point. We have shown that this velocity is conserved in magnitude, but reverses indirection after a collision with the wall or the floor. Subracting Eq. 83 from 82 yields

Ca =β − 1

β + 1Cb −

2

β + 1Vb and then Va = − 2β

β + 1Cb −

β − 1

β + 1Vb (84)

Let’s apply this to a super ball bouncing off the floor. Pause for a moment to realize thatthe velocity V⊥ perpendicular to the floor is conserved in the collision. (Why?) If a ball fallsvertically downward, then Vb = 0 and Va = −2β/(β + 1)Cb = −(4/7)Cb for β = 2/5, i.e. asolid sphere. That is, it bounces to the side with a transverse velocity that depends on howfast it is spinning.

If on the other hand, a ball that is not spinning (Cb = 0) bounces at a 45 angle so thatVb = V⊥, then Va = (1− β)/(1 + β)Vb = (3/7)V⊥. That is, it bounces up at an angle (withrespect to the vertical) θa = tan−1(Va/V⊥) = tan−1(3/7) = 23 < 45. Try it and see!

See Garwin’s paper for other calculations, including the trajectory for bouncing off theunderside of a table.

24

Physics I Honors Monday 6 October Fall 2008

Today we recognize angular velocity ω, angular momentum L, and torque τ as full-fledgedvector objects. The physical implications are interesting.

What exactly is a vector? Not angular position!

I have told you that vectors are “just bookkeeping” but that’s not really true. To be usefulin physics, they need to have some obvious (?) mathematical properties. In fact, if I try to

specify the angular position of an object with the construct θ?= θxi + θy j + θzk where θx,

θy, and θz are angles with respect to fixed axes, then I will get into trouble quickly.

Show the trick with the textbook. Rotations do not commute but vector addition must!

However, infinitesimal rotations do commute. (See Note 7.1 in K&K.) The trick is that thepart that that does not commute enters in second order in the angles. So I can write

ω =dθxdt

i +dθydt

j +dθzdt

k = ωxi + ωy j + ωzk (85)

for angular velocity. For its direction, see the following figures of a rotating rigid body:

Pick some particle in the object. Its posi-tion is r and its velocity is v = dr/dt. Themagnitude of v is (r sinφ)dθ/dt. In otherwords, we write

dr

dt= v = n× r

dθ

dt= ω × r (86)

where we define ω ≡ (dθ/dt)n. In particu-lar, ω points in the direciton of the rotationaxis, specified by the unit vector n.

Example: Angular Momentum of a Rotating Skew Rod

Find the angular momentum L of the two masses around therod midpoint. By definition, L =

∑(ri × pi). For each mass

r ⊥ p and r × p is perpendicular to the rod! Since r ⊥ p wehave |r× p| = lp = lmv = lmωl cosα so L = 2mωl2 cosα.

For this axis I =∑mr2 = 2m(l cosα)2 so L 6= Iω! The

equality is violated in both magnitude and direction, althoughLz = L cosα = Iω. This is because the “object” is asymmetricand our non-tensor formulation of rotational inertia is naive.

Practice Exercise

Suppose a particle rotates counter clockwise in the xy plane. Let θ be the polar angle,relative to the x-axis. Then, its angular velocity would be ω = ωk where ω = θ. Show thatequation 86 gives the correct vector velocity v = ω(xj− yi) = ωrθ. Draw a picture with theparticle at a couple of points, and sketch r and v.

25

Example: Torque on the Rotating Skew Rod

The (direction of the) angular momentum of the skew rod changes with time. What torquecauses this? To see this, let’s find dL/dt and interpret the result. We can do

L(t) = L sinα cosωti + L sinα sinωtj + L cosαk (87)

τ =dL

dt= ωL sinα

(−i sinωt+ j cosωt

)so τ = ωL sinα (88)

Alternatively, we can use a geometric approach. Looking downthe axis from the top, on the horizontal (“h”) plane, we see that|∆Lh| = Lh∆θ. This is the only component of L that changes.Therefore

τ =dL

dt=dLhdt

= Lhdθ

dt= (L sinα)(ω) (89)

so we get the same result as before. The torque is zero if α = 0, and also if α = 90 (L = 0).The torque is from forces which hold the vertical bearing onto the rotation axis. Can you seewhy the torque zero in both cases? (Imagine you’re holding the bearing in your hand.)

The Gyroscope

A simple gyroscope is a spinning wheel on a horizontal axis, supported on one end only:

Contrary to our intuition, it does not fall down. Instead, gravity exerts a torque which causesit to precess about a vertical axis which passes through the support point.

Formally, this is easy to see. The angular momentum vector L is along the spinning axis. Thetorque about the support point is only the weight W, since the normal force N has no momentarm about this point. The torque τ has magnitude τ = lW and direction perpendicular toboth W and L. So L changes in direction, not magnitude, tracing the circle in the horizontalplane, shown on the left. Calculating the precession angular frequency Ω

τ = lW =dL

dt= LΩ so Ω =

lW

L=

lW

I0ωs(90)

A simple model makes it easier to see, physically, why the gyroscopeprecesses. Consider two masses on an arm, rotating in the horizontalplane. Instantaneously apply a torque with opposing forces F up anddown on the two masses, so there is no net force and the center of massdoesn’t move. If the force acts for time ∆t, then ∆p = m∆v = F∆tfor each mass. One mass moves up and the other moves down. Thus

∆φ ≈ ∆v

v=F∆t

mv=

2lF∆t

2lmv=τ

L∆t so

dφ

dt=τ

L(91)

which is the same result that we got in Equation 90.

26


Different format today. We will spend most of the period doing “worksheets” which you willturn in. These will be counted together as one homework assignment. The point is to helpyou get up to speed on the material we’ve been covering, especially the use of vectors todescribe motion, both translation and rotation, in two and three spatial dimensions.

Worksheet #1: Angular velocity for motion in a plane.

Collect the worksheets, and work the problem out on the board.

Rigid Bodies and Conservation of Angular Momentum

For a symmetric rigid body, we now know that the angular momen-tum vector is parallel to the angular velocity vector. We write

L = Iω where I =∑

mir2i

is the rotational moment of inertia about the rotation axis. We saythat L is conserved if there are no external torques but nobody canprove this is so, starting from Newton’s laws! See the diagrams onthe right. The torques τ ij = ri×fij cancel only if fij works along theline ri−rj. That is, fij = −fji (Newton’s Third Law) is not enough.There is some higher reason for angular momentum conservation.

Worksheet #2: Conservation of Angular Momentum

Collect the worksheets, and work the problem out on the board.

General Angular Momentum of Rigid Bodies: The Rotational Inertia Tensor

If all position vectors ri, for the particles of mass mi within a rigid body, are measured withrespect to the center of mass, then the angular momentum vector is

L =∑i

r× p =∑i

r×miri =∑i

miri × (ω × ri)

You can reduce this expression with some tedium. (See K&K.) You’ll find, for example,

Lz =

[∑i

mi(x2i + y2

i )

]ωz −

[∑i

mixizi

]ωx −

[∑i

miyizi

]ωy or L = I · ω (92)

where I is the rotational inertia tensor, a 3× 3 matrix of numbers. This is a general formu-lation of the angular momentum of a rigid body. Note that if ω = ωk then Lz = Iωz whereI =

∑mr2 just as we had before. Also, as we have seen, if the “off diagonal elements” of I

are nonzero, then there are other terms in L. (How much do you guys know about matrices?Can I tell you that the axes which diagonalize I are called the principle axes?)

Worksheet #3: The Rotational Inertia Tensor for the Skew Rod

Collect the worksheets, and work the problem out on the board, if there is time.

27

Name:


Worksheet #1: Angular velocity for motion in a plane

Consider a particle which moves in the (x, y) plane. Answer the following questions:

A. In one sentence, why does r(t) = x(t)i + y(t)j tell us the position of the particle?

B. The particle moves in a circle with angular velocity ω. It starts at (x, y) = (r, 0). Findthe functions x(t) and y(t) in terms of r, ω, and t.

C. Find the velocity vector v(t) = r(t) in terms of r, ω, and t, and unit vectors i and j.

D. The angular velocity is ω = ωk. Take the cross product ω× r(t) and show that it is thesame as v(t). Refer to Sec. 1.4 in your book for the cross product of the unit vectors.

28

Name:


Worksheet #2: Conservation of Angular Momentum for Rigid Bodies

A student sits in a swiveling chair that allows her to rotate about a vertical axis. Herrotational inertia is I0. She is holding a wheel on an axle. The wheel has rotational inertiaIW . The axle is oriented vertically. Initially, the student is not moving, but the wheel isspinning with an angular velocity ω, counter clockwise as viewed from the top.

A. What is the magnitude and direction of the initial angular momentum vector Li?

B. The student turns the wheel and axle over by 180 so that it again is oriented vertically,but now the wheel rotates clockwise as seen from the top. If the student’s angular velocityis now Ω, write the final angular momentum Lf in terms of I0, IW , ω, and Ω. Be careful ofsigns. You can ignore the mass of the wheel as compared to the mass of the student.

C. Find Ω in terms of I0, IW , and ω.

29

Name:


Worksheet #3: The Rotational Inertia Tensor for the Skew Rod

Use Eq. 92 to calculate the angular momentum of the skew rodwhich we studied last class. The angular velocity is ω = ωk.

A. Using ρ as defined at the left, derive Lz in terms of m, ρ, and ω.

B. By rearranging x, y, and z in Eq. 92, show that

Lx = −

[∑i

mixizi

]ω and Ly = −

[∑i

miyizi

]ω

C. Complete the following table for the coordinates of the two masses in terms of ρ, ω, and h:

Particle 1 Particle 2x1 = ρ cosωt x2 =y1 = y2 = −ρ sinωtz1 = z2 = h

D. Derive Lx and Ly in terms of m, ρ, h, and ω. Compare to our result from last class.

30

Physics I Honors Tuesday 14 October Fall 2008

Today we finally (!?) leave rotations of rigid bodies, but we will use those topics later.

Accelerating Reference Frames

Newton’s first law of motion says something like “An object at rest stays at rest unless someforce acts on it. Also, an object in motion doesn’t change its direction or speed unless someforce acts on it.” These statements are fundamental assumptions and cannot be proven. Infact, they aren’t always true. We say that an “inertial reference frame” is one in whichNewton’s first law, sometimes called the “law of intertia”, holds.

A reference frame in which Newton’s law doesn’t hold is an “accelerating reference frame.”The second law really embodies the first law. The second law says that∑

F = mA (93)

so that if there are no forces (∑

F = 0) then A = 0, i.e. the first law. If you are “inside” theframe that is accelerating, then you feel something that’s like a force, but isn’t really there.So, invent a “fictitious force” Ffict = −mA and everything is fine. Here’s a simple example:

Analyze this two ways, for an observer standing on thestreet, and for someone inside the car.

Outside (“inertial”) Inside (“accelerating”)

Ffict 0 −mAi∑Fx T sin θ T sin θ + Ffict∑Fy T cos θ −mg T cos θ −mg

So, the person outside the car says that T sin θ = mA, while the person inside the car saysthat T sin θ + Ffict = T sin θ −mA = 0. These are the same equations. This is really prettytrivial, but more complex situations are more complicated.

Practice Exercise

Here’s a slightly more complex situation, from Example 8.2 in your textbook.

The plank accelerates to the right with acceleration A. Thecylinder rolls without slipping on the plank. To an observerin an inertial frame, the cylinder accelerates to the right dueto the frictional force f . To an observer on the plank, afictional force Ffict = MA acts on the center of mass andaccelerates the cylinder to the left. If the acceleration of thecylinder is a′ to the observer on the plank, then the torquefR accelerates the cylinder by α′ via fR = I0α

′ = I0a′/R

where I0 = MR2/2.

So, first, write x = x′ + X to locate the cylinder in the inertial frame relative to the accel-erating frame, where X locates the plank. Use this to explain why the acceleration in theinertial frame is a = a′ + A. Then, use the relationships between rolling and translationalmotion above, to find the accelerations a and a′ in terms of A.

31

Galilean Relativity

No let’s be a little more formal about this idea of “reference frames.” In particular, we willshow that Newton’s laws of motion are the same in different “inertial” reference frames, thatis, two frames that are moving with constant velocity with respect to each other.

For this discussion, I am using the notation and figures from Sec. 11.4 in your textbook, asopposed to Sec. 8.2. Both are called “The Galiliean Transformations”, but the latter is amore common formulation.

Two observers follow motion of the same particle. It is lo-cated at r in the coordinate system of one of them, and atr′ in the coordinate system of the other. The origin of the“primed” system is located at R in the unprimed system.We therefore have

r′ = r−R where R = vt (94)

That is, the primed coordinate system moves with a (constant) velocity v relative to theunprimed system. We let v be along the x (and x′) axis.

Obviously r′ = r − v and r′ = r. Therefore, even though the velocity of the particle isshifted by v in one coordinate system relative to the other, the accelerations are the same.This means that

∑F = ma in both systems, so Newton’s second law is intact. In fact, you

cannot tell which frame you are in by studying the particle’s motion. Notice also that, fortwo particles a and b, r′a−r′b = ra−rb. This means that the distances and directions betweenthe two particles are the same in both frames, and so are two-body forces like gravity.

(This all makes some plausible, albeit ultimately wrong, assumptions. For example, it as-sumes that lengths are measured the same by both observers, but this ends up being incorrect.Leave this for now, but remember it when you come to study special relativity.)

Now assume that some sort of pulse is emitted at t = 0 fromthe origin in the (x, y) system. Its position along the x axisin the (x, y) system is

x = ct

where I’ve assumed that its velocity is c. Its position alongthe x′ axis in the (x′, y′) system is therefore, from Eq. 94,

x′ = x− vt = ct− vt = (c− v)t

In other words, if you follow the pulse in the (x′, y′) system, then it appears to move in the x′

direction with a velocity dx′/dt = c− v. Of course, this is just what you’d expect. However,as you will learn next semester, the laws of electromagnetism (aka Maxwell’s Equations)say that the speed of light is c in any inertial frame. This inconsistency is what led to thedevelopment of special relativity, and the eventual recognition that Newton’s laws are onlyan approximation in a world where all velocities are small compared to c.

Practice Exercise

Show that the force of gravity is the same between two objects a and b, regardless of whetherthey are viewed in a primed or unprimed system, related by the transformation Eq. 94.

32


Thanks to Brian for covering class today (and lab yesterday!)

Remember: Exam #2 is two weeks from today.

Note: Section 8.4 in book on Equivalence Principle is interesting, but we’re skipping it.

Rotating Coordinate Systems

Recall from last class: Frame accelerating with acceleration A doesn’t obey Newton’s laws,but you can fix things by inventing a fictitious force Ffict = −mA.

Familiar example: For a turntable rotating at constant angular velocity Ω, there is an inwardacceleration Ω2R where R is the distance from the rotation axis. So, if you are on theturntable you feel a fictitious force mΩ2R outward. This is called “centrifugal force.”

Our goal today is to be more formal with this concept of “rotating coordinate systems.”

See the figure. One set or coordinates rotatesabout the other with angular velocity Ω. Theorigins coincide, that is, we are only concernedwith rotation, not translation. The (x, y, z)system is inertial. At time t all three axescoincide. Rotation is about the z (z′) axis.

Our job is to relate the derivatives of the vector r(t) in the inertial (unprimed) and rotating(primed) frames. Examine the change in this vector between times t and t+ ∆t, below:

In short, the rotating observer is “fooled.” After the time ∆t has elapsed, he measures achange ∆r′ based on the correct location of the final position, but a “wrong” starting positionbecause his axes have rotated. The figure makes it clear that

∆r(t) = [r′(t) + ∆r′(t)]− r(t) (95)

We know how to relate r′(t) and r(t). They are connected through the angular velocityvector Ω = Ωk according to Equation 86, that is

r′(t)− r(t) = (Ω× r)∆t (96)

If we combine equations 95 and 96 we can write

∆r(t)

∆t=

∆r′(t)

∆t+ Ω× r (97)

33

This is essential. If we just let ∆t→ 0, we can relate the velocities in the two frames as

vinertial = vrotating + Ω× r (98)

This proof relies only on the geometric properties of r, and really relates the derivative of thisvector when viewed by two different coordinate systems. So, we could just as well write(

dr

dt

)in

=

(dr

dt

)rot

+ Ω× r or, generally,

(dB

dt

)in

=

(dB

dt

)rot

+ Ω×B (99)

for any vector B. (See also Note 8.2.) For example, the acceleration vector becomes

ain =

(dvin

dt

)in

=

(dvin

dt

)rot

+ Ω× vin

=

[d

dt(vrot + Ω× r)

]rot

+ Ω× (vrot + Ω× r)

= arot + 2Ω× vrot + Ω× (Ω× r) (100)

where we have assumed that the rotation vector Ω does not change with time. Thus, thefictitious force in the rotating coordinate system

Ffict = −2mΩ× vrot −mΩ× (Ω× r) (101)

has two terms. The second term has magnitude mΩ2r for a position r in the rotation plane,so is just our old friend the centrifugal force. The first term is new. It depends on thevelocity vrot in the rotating frame, and is called the coriolis force.

Example 8.7. Bead sliding without friction on a radial rod,rotating with angular velocity ω. Use the rotating coordinatesystem. The vector ω× r is in the horizontal plane, upwardin the lower figure. So, Fcent = −mω × (ω × r) is outward,and Fcor = −2mω×vrot is as shown for vrot radially outward.

Radial∑

F = ma so mω2r = mr

Transverse∑

F = ma so N − 2mωr = 0

The radial equation gives r(t) = Aeωt + Be−ωt where A and B are determined from initialconditions. Thus N = 2mω2(Aeωt−Be−ωt) for the normal force of the rod on the bead.

Practice Exercise

Get some practice with the directions of coriolis and centrifugal forces on the Earth’s surface.What is the direction of Ω? What is the magnitude and direction of Ω × r for a particleat the equator? What if the particle is at the North Pole? Do the same for Ω × (Ω × r).Now calculate, in terms of Ω and the radius R of the Earth, the magnitude of the centrifugalforce on a particle at the equator and at the pole.

Next, instead of being stationary, let the particle fall from some height near the Earth’ssurface. It therefore has a velocity vrot downward in the rotating frame. Go through asimilar process, and determine the magnitude and direction of the coriolis force on thefalling particle, in terms of vrot, Ω, and R.

34


Motion under Central Forces

Recall our discussion of two-body motion and center of mass. We found that we can treat atwo body problem as a one-body problem with a “reduced mass.” Let’s review that.

First, write Newton’s Second Law for each of the two bodies:

m1r1 = f(r)r and m2r2 = −f(r)r

Add these two equations to see what happens to the center of mass:

m1r1 +m2r2 = MR = 0

where M = m1 +m2 and R = (m1r1 +m2r2)/M is the position of the center of mass. On theother hand, the two equations of motion can also be combined by subtracting, hence

r = r1 − r2 =

(1

m1

+1

m2

)f(r)r so µr = f(r)r where µ =

m1m2

M

is called the reduced mass. We now can deal with the one-body “central force” problem.

Note that if m2 m1 (like Sun & planet) then M ≈ m2 and µ ≈ m1, so we ignore it.

Recall that angular momentum L = µr × r is conserved, in both magnitude and direction,for central forces. One obvious conclusion is that the motion must be in a plane.

We can go further by using polar coordinates. The mag-nitude of the angular momentum is ` = µr2θ. The totalmechanical energy is

E =1

2µv2 + U(r) =

1

2µ(r2 + r2θ2) + U(r) (102)

where U(r) is the potential energy with f(r) = −dU/dr. So,

E =1

2µ

(r2 +

`2

µ2r2

)+ U(r) =

1

2µr2 + Ueff(r) where Ueff(r) =

`2

2µr2+ U(r) (103)

is called the effective potential energy. It is useful for analyzing central force motion in termsof the radial coordinate. The coordinate θ obeys dθ/dt = `/µr2, where ` is constant.

Effective potential for the potential energy of two gravitatingbodies. The solid curve is

Ueff(r) =`2

2µr2− Gm1m2

r(104)

For E = Emin, the radial coordinate doesn’t change. ForE < 0, the radial coordinate oscillates between min andmax values. For E ≥ 0, there is no maximum. Later!

Practice Exercise

Put m1 = mM = m2 in Eq. 104. Show that ` = mvR is the correct circular orbit.

35

Dark Matter in Galaxies

We can learn a lot about one of the most important problems in physics today, just bycombining what we know now about circular orbits, Newtonian gravity, and observations ofthe Milky Way and some distant galaxies.

A circular orbit of some small mass m about some large mass M obeys Newton’s second lawin the form

GmM

r2= m

v2

rwhich gives v =

√GM

r

for the “orbital velocity”. Note that the period of revolution is T = 2πr/v so we have

T 2 =4π2

GMr3

which is called Kepler’s Law of Periods. (We’ll look at this more on Thursday.) Note that thisis a way to determine the mass of the sun from the revolution periods of the planets.

One very basic result from this is that the tangential rotation speed v should be proportionalto 1/

√r as soon as the mass is concentrated at distances smaller than r. Galaxies should

look like this, since there is generally a big “bulge” of bright stars near the center. However,the “galactic rotation curves” look very different! (See slides.) This is one of the strongpieces of evidence for “dark matter” in the universe.

For some interesting reading, see Alternatives to dark matter and dark energy. Philip D.Mannheim (UConn). Published in Prog.Part.Nucl.Phys.56:340-445,2006. You can also visithttp://arxiv.org/ and get e-Print: astro-ph/0505266.

Practice Exercise

Use the Milky Way galactic rotation curve to estimate the mass of the galaxy containedwithin a distance of 16 kpc. (Note that 1 kpc=103 pc=3.1× 1019 m.) The mass of the sun is2×1030 kg. How many sun-like stars do you need to get this mass? Note that the brightnessof a typical spiral galaxy (like the Milky Way?) seems to be about 1010 times that of thesun. (Table 21-1 in Introductory Astronomy and Astrophysics by Zeilik and Gregory.)

36

The Milky Way Galaxy

Visible Light

Infrared Light

37

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39


Remember: Exam #2 is one weeks from today. Same ground rules as before.

Motion under Newtonian Gravity

We want to find the motion under Newtonian gravity in three dimensions. Assume an object(say, a planet) with mass m orbiting a (massive) body M . (Don’t worry about reduced massµ distinction now.) The potential energy is (See Example 5.3 in your textbook)

U(r) = −GmMr

= −Cr

where C ≡ GmM (105)

We use C because that makes it easy to generalize to other central force laws.

First let’s review from last class. Motion is in a plane since L = mr× r is conserved forcentral forces. This is easy to show using Newton’s second law:

dL

dt= mr× r +mr× r = 0 + r×

(∑F)

= r× f(r)r = 0 (106)

where the sum of forces takes the form that defines a “central” force.

Central forces are conservative. (See Example 4.8 in your textbook.) So write f(r) = −dU/drfor some U(r). We know that this means that the energy

E =1

2mr2 + U(r) =

1

2m(r2 + r2θ2

)+ U(r) (107)

doesn’t change with time. It is a “constant of the motion.” Note that we’ve written this interms of polar coordinates r and θ. See Section 1.9 of your textbook for a review.

The magnitude ` = |L| of the angular momentum is also a constant of the motion. Thecomponent of velocity perpendicular to the moment arm r is vθ = rθ by definition. So

` = mr2θ (108)

Once again, for as long as there is motion, the value of ` does not change, even though rand θ are of course functions of time. They just always combine so that ` is constant.

Now we can find the shape of the orbits. We can combine Equations 107 and 108 toform two differential equations describing the motion as r(t) and θ(t), namely

dθ

dt=

`

mr2and

dr

dt=

√2

m[E − Ueff(r)] where Ueff(r) ≡ U(r) +

`2

2mr2(109)

However, the shape of the orbit is some function r(θ). So, divide the two to find

dθ

dr=

`

mr2

1√(2/m)[E − Ueff(r)]

=`

r(2mEr2 + 2mCr − `2)1/2(110)

The integral looks messy, but in fact it can be done analytically. (You can find it in tables.)Integrating both sides, and putting the constant of integration, −θ0, on the left, we have

θ − θ0 = sin−1

(mCr − `2

r√m2C2 + 2mE`2

)(111)

40

Divide through my mC in the parentheses, take the sine of both sides, and solve for r:

r = r(θ) =`2/mC

1−√

1 + (2E`2/mC2) sin(θ − θ0)(112)

Let’s clean this up. Pick θ0 = −π/2 (it’s arbitrary) and define

r0 ≡ `2/mC and ε ≡√

1 +2E`2

mC2> 0 so r = r(θ) =

r0

1− ε cos θ(113)

The parameter ε is called the eccentricity. The curves r(θ) are called conic sections. Theforms become familiar if we work in cartesian coordinates x = r cos θ and y = r sin θ, so√

x2 + y2 − εx = r0 =⇒ (1− ε2)x2 − 2εr0x+ y2 = r20 (114)

Case ε = 0. The curve is r = r0, a circle. Setting mv2/r = p2/mr = `2/mr3 = C/r2 givesr = `2/mC = r0. This is precisely what we know we should get for a circular orbit.

Case ε = 1, i.e. E = 0. In cartesian form, we see that the “orbit” is a parabola,

x =y2

2r0

− r0

2(115)

Case ε < 1, i.e. 1− ε2 > 0. Your book works this out in polar coordinates, but let’s try itin cartesian. First find the “ends” of the orbit by setting y = 0. This leads to

xmin = − r0

1 + εand xmax =

r0

1− ε=⇒ a ≡ 1

2(xmax − xmin) =

r0

1− ε2(116)

Now translate to x′ = x−x0 where x0 = a+xmin = r0ε/(1− ε2) is the halfway point betweenthe two ends. Substituting for x in Equation 114, and simplifying, we find

(1− ε2)x′2

+ y2 =r2

0

1− ε2≡ b2 (117)

Clearly y = ±b are the “top” and “bottom” of the orbit. Since a2 = b2/(1−ε2) we have

x′2

a2+y2

b2= 1 (118)

which is the familiar equation of an ellipse. The parameters a and b are called the semimajorand semiminor axes. The position x = 0 is called the focus; it is not the “center”!

Case ε > 1, i.e. ε2 − 1 > 0. Multiply Equation 117 through by −1, but keep the sign infront of b2 so that it is still positive. Equation 118 becomes

x′2

a2− y2

b2= 1 (119)

This is the equation of a hyperbola. Note that x0 is now a negative number.

This is enough for today. You should read through your book on Kepler’s laws, but we won’tspecifically cover them in this course. They will be important, though, if you should take acourse on astrophysics.

41

Name:


Worksheet: The following shows a circular orbit of a planet around the Sun.

a. Draw on this graph a circular orbit with larger total energy.

b. Draw on this graph an elliptical orbit with the same total energy as the original circularorbit. Make the eccentricity e of the orbit equal to 4/5. (Hint: Use the definitions of r0, ε,and a to show that the energy E only depends on C and a.) Calculations:

c. At a point where the elliptical and (original) circular orbit cross, draw the velocity vectorsfor each of the two orbits at that point. Indicate which is which.

d. Explain briefly why the magnitudes of the two velocity vectors must be the same.

e. Of the original circular orbit and your elliptical orbit, which has the largest angularmomentum? (Assuming the same orbiting mass in each case.) Explain your answer.

42


Second exam is on Thursday. Everything the same as for Exam #1.

The Simple Harmonic Oscillator (Again)

First, start with a review of old stuff. Pay attention to the notation, e.g. ω0.Newton’s Second Law says

∑F = ma = mx. For this

example,∑F = Fspring = −kx. So we need to solve

−kx = mx =⇒ x(t) = −ω20x(t) where ω0 ≡

√k

m(120)

Any function x(t) proportional to cosω0t or sinω0t solves this. So, general solution is

x(t) = B sinω0t+ C cosω0t (121)

We need more information to determineB and C. Typically, we use “initial conditions.”

Everybody knows that x and x(t) mean exactly the same thing, right?

Solution using complex numbers. We solved Eq. 120 by “recognition.” A better way isto use an ansatz, namely x(t) = Aeiωt. Substitute into Eq. 120 and solve for ω:

−ω2Aeiωt = −ω20Ae

iωt =⇒ ω = ±ω0 (122)

Once again, there are two possible solutions, and so once again the general solutions is:

x(t) = A1eiω0t + A2e

−iω0t (123)

Initial conditions. Work with solution 121. Obviously x(0) = C. We also have

v(t) = x(t) = ω0B cosω0t− ω0C sinω0t =⇒ v(0) = ω0B or B = v(0)/ω0 (124)

A trig identity gives us a different, more useful way, to write the solution. We write

x(t) = A cos(ω0t+ φ) = A cosφ cosω0t− A sinφ sinω0t (125)

which we compare to 121 to see that

B = −A sinφ and C = A cosφ =⇒ A =

√x(0)2 +

v(0)2

ω20

and tanφ = − v(0)

ω0x(0)(126)

Energy. Have U(x) = kx2/2 since −dU/dx = Fspring. Also K = mv2/2 = mx2/2, so

E = K + U =1

2mω2

0A2 sin2(ω0t+ φ) +

1

2kA2 cos2(ω0t+ φ) =

1

2kA2 (127)

So, the total energy is the potential energy when the position x = A, the “amplitude.”

Practice Exercise

Apply the initial conditions to solution 123. Solve for A1 and A2 and then show that thecomplete solution is the same as (121) with B and C in terms of x(0) and v(0). RecallEuler’s formula, namely

eiθ = cos θ + i sin θ

43

The Damped Harmonic Oscillator

Now∑F = Fspring +Fdamp = −kx− bv = −kx− bx. The equation of motion becomes

mx = −kx− bx =⇒ x+ γx+ ω20x = 0 where γ ≡ b

mand ω0 ≡

√k

m(128)

(The definition of ω0 is just as before.) Solving this by “recognition” is too hard, so weimmediately move on to our ansatz x(t) = Aeiωt. Inserting this into Eq. 128 gives

−ω2 + iγω + ω20 = 0 =⇒ ω = i

γ

2±√ω2

0 −γ2

4(129)

Case 1: γ < 2ω0. Define ω1 ≡√ω2

0 − γ2/4 and the solution to Eq. 128 becomes

x(t) = Ae−(γ/2)t cos(ω1t+ φ) (130)

Some comments are in order. First, the oscillation frequency is ω1 < ω0. Secondly, theoscillation maximum is not a fixed amplitude, but decays away with a 1/e “time constant”(γ/2)−1 = 2m/b. Finally, the exponential factor in front makes the association of A andφ with x(0) and v(0) kind of messy. This isn’t so interesting, so we won’t worry about it.

0 2 4 6 8 10−8

−6

−4

−2

0

2

4

6

8

10

Time t

x(t)

The red plot shows a damped oscillator with γ = ω0/4, withthe dashed line showing the exponential factor alone. It isclear that the energy decreases with time. In fact, in thecase where γ ω0 (a situation called “lightly damped”) itis not hard to show that

E(t) = E0e−γt where E0 =

1

2kA2 (131)

So, 1/γ ≡ τ is the “energy decay time constant.”

Case 2: γ > 2ω0. In this case, we write the solutions to Eq. 129 as ω = iα where

α =γ

2± γ

2

√1− 4ω2

0

γ2= α1, α2 =⇒ x(t) = C1e

−α1t + C2e−α2t (132)

where both α1 and α2 are positive real numbers, so both terms of the solution are decayingexponential functions. The blue curve in the figure is for γ = 3ω0, and the same initialconditions as for the underdamped case. This situation is called “overdamped.”

Case 3: γ = 2ω0. This case is called “critical damping.” Something is weird, though. Itlooks like our ansatz has broken down, because we only have one solution, α = γ/2, whenwe know there should be two. But don’t worry, math is good. See your textbook, or a bookon differential equations. The second solution in this case is proportional to te−αt, so

x(t) = D1e−αt +D2te

−αt (133)

is the general solution. This is the green line plotted in the figure.

Practice Exercise

Derive Eq. 131 for the decay time of a lightly damped harmonic oscillator. Follow the discus-sion in your textbook, pages 416 and 417. Our Eq. 131 is the book’s equation 10.16.

44

Physics I Honors Monday 3 November Fall 2008

Hand back exam? If so, give grade statistics. Some Notes:

• Wednesday is “make-up lab” day; need email to me or Scott by the end of today

• Preliminary lab notebooks due in class next Monday!

• Stay tuned: Testing something new, perhaps, in lab on coupled oscillations (Nov.19)

Review: Simple and Damped Harmonic Oscillations

The simple harmonic oscillator is based on∑F = −kx and has the equation of motion

x(t) = −ω20x(t) (134)

where ω0 ≡√k/m. We solve this with the “ansatz” x(t) = Aeiωt. Substituting,

−ω2 = −ω20 =⇒ ω = ±ω0 (135)

So, both x(t) = A1eiω0t and x(t) = A2e

−iω0t are valid solutions. In fact, any linear combina-tion of the two is a valid solution. We find A1 and A2 by applying the initial conditions.

The damped harmonic oscillator is based on∑F = −kx − bx and has the equation of

motionx(t) = −ω2

0x(t)− γx(t) (136)

where γ ≡ b/m. We again solve this with x(t) = Aeiωt. Substituting, we find a quadratic“characteristic equation” for ω. The solutions to this equation are

ω = iγ

2± ω1 where ω1 ≡

√ω2

0 −γ2

4(137)

(This is the “underdamped” solution, where γ < 2ω0, which is the only case which yieldsharmonic oscillations.) So, both x(t) = A1e

−γt/2eiω1t and x(t) = A2e−γt/2e−iω1t are valid

solutions. Again, any linear combination of the two is a valid solution, and we find A1 andA2 by applying the initial conditions. For future reference,

x(t) = A1e−γt/2eiω1t + A2e

−γt/2e−iω1t (138)

is the general solution to the equation of motion for a mass m attached to a spring with forceconstant k and being damped out with a force proportional to velocity, with coefficient b; theparameters ω1 and γ are defined in terms of m, k, and b by the relations given above.

Note that we could rewrite Eq. 138 simply by using e±iω1t = cosω1t ± sinω1t and thendefining coefficients B1 and B2 appropriately in terms of A1 and A2 so that

x(t) = B1e−γt/2 cosω1t+B2e

−γt/2 sinω1t (139)

= Ae−γt/2 cos(ω1t+ φ) (140)

This is exactly the same result which we got in Eq. 130 from last class.

45

The Forced Harmonic Oscillator

Now imagine that we are “forcing” the damped harmonic oscillator by applying a sinusoidallyvarying force as a function of time. That is

∑F = −kx − bx + F0 cosωt where ω is now a

frequency that we can control independently. The equation of motion becomes

x+ γx+ ω20x =

F0

mcosωt (141)

We can solve this equation using something of a trick. Pick another variable y(t) so that

y + γy + ω20y =

F0

msinωt (142)

and define z(t) = x(t) + iy(t). The differential equation that governs z(t) is therefore

z + γz + ω20z =

F0

meiωt (143)

where we just need to recall that x(t) = <z(t) to get the motion of our oscillator.

Applying our ansatz z(t) = Aeiωt yields something a little different than before. Instead offinding a characteristic equation to solve for ω, we instead find an equation for the (complex)amplitude A as a function of ω. That is

A[−ω2 + iγω + ω2

0

]=F0

m=⇒ A =

F0

m

1

ω20 − ω2 + iγω

(144)

(Note: We can always add in a solution like Eq. 139 since it just yields zero in the differentialequation. This is where we get the constants we need to apply the initial conditions. Thispart of the solution, though, dies out with time, so we usually just ignore it; it is called a“transient.”) This is easy to interpret if we write

A = Aeiφ where A =F0

m

1

[(ω20 − ω2)2 + ω2γ2]1/2

and φ = tan−1

(γω

ω2 − ω20

)(145)

In other words, the motion of the forced damped oscillatoris given by x(t) = A cos(ωt + φ) where A and φ are strongfunctions of the driving frequency ω. These functionsare shown on the left, for the lightly camped case whereγ ω0. The motion has an amplitude A that becomes verylarge when the driving frequency gets close to the naturalfrequency ω0. This phenomenon is called resonance.

Note the phase. The motion is in phase φ = 0 for frequencieswell below ω0, and precisely out of phase (φ = 180) far aboveω0. Also, it passes through φ = (−)90 as ω moves throughthe resonance.

The resonance has a width that depends directly on γ. In fact, if we define the width ∆ωas the distance between the two ω values for which the curve drops to half its value, then∆ω = γ. The value Q ≡ ω0/γ is a dimensionless number which characterizes the width,and can be interpreted as the amount of energy lost in one oscillation. Large Q means lowdamping, and is an important figure of merit for resonating systems.

46

Physics I Honors Thursday 6 November Fall 2008

Hand back exam, discuss grade statistics. (Average on test was 76.73.)

Waves on a Stretched String: The Wave Equation

These notes follow Chap.18 of “Physics”, by Resnick, Halliday, and Krane, 5th Ed.

What is the “equation of motion” for a string that is stretched under some tension F?

The motion is up and down, so analyze the forcesin the vertical direction on a small piece of stringat fixed x. We assume small amplitude waves soδx can represent the length of the small piece. Seethe figure. If F is the tension in the string, then∑

Fy = F sin θ2 − F sin θ1 = (δm)ay (146)

Put δm = µδx for linear mass density µ. Obviously ay = y = ∂2y/∂t2 since we are workingat fixed x. Also, for small deflections, θ 1 so sin θ ≈ θ ≈ tan θ = slope = ∂y/∂x. So,

F (sin θ2 − sin θ1) ≈ F

(∂y

∂x

∣∣∣∣2

− ∂y

∂x

∣∣∣∣1

)= µδx

∂2y

∂t2=⇒ 1

δx

(∂y

∂x

∣∣∣∣2

− ∂y

∂x

∣∣∣∣1

)=µ

F

∂2y

∂t2(147)

The left side Eq. 147 is just ∂2y/∂x2 as δx→ 0. This gives

∂2y

∂x2=

1

v2

∂2y

∂t2(148)

where 1/v2 ≡ µ/F . Eq. 148 is called the wave equation. Itdescribes a shape of the string, function y of x, which changesin time, hence y(x, t). The solution is any function like

y(x, t) = f(x′) = f(x± vt) i.e. x′ ≡ x± vt (149)

This means that the change in time is rather peculiar. Theshape f(x) moves to the right or left with speed v. This isthe definition of a (non-dispersive) wave.

Practice Exercise

Which of the following solves the wave equation, Eq. 148, for v = 1? Take derivatives, orshow which are of the form Eq. 149. Note that the sum of two waves is also a wave, i.e. the“principle of superposition;” This is because Eq. 148 is a linear differential equation.

(A) y(x, t) = x2 − t2; then try y(x, t) = x2 − t2

(B) y(x, t) = sinx2 sin t

(C) y(x, t) = log(x2 − t2)− log(x− t)

(D) y(x, t) = ex sin t

47

Sine Waves

We almost always describe waves as if the shape f(x± vt) is a sine function.

It is handy to describe a sine wave as follows:

y(x, t) = ym sin

[2π

λ(x− vt)

](150)

We call λ the wavelength, for obvious reasons. Note that for fixed x (say x = 0) we havey(0, t) = ym sin [−2πvt/λ] which means that the transverse oscillations are sinusoidal withperiod T = λ/v = 1/f where f (or, maybe, ν) is the frequency.

The more common way to write the equation of a sine wave is

y(x, t) = ym sin(kx− ωt− φ) (151)

where k = 2π/λ, ω = 2πf and φ is called the phase constant.

Standing Waves

Add two sine waves of the same speed and frequency, but moving in opposite directions:

y1(x, t) = ym sin(kx− ωt) and y2(x, t) = ym sin(kx+ ωt)

⇒ y(x, t) = y1(x, t) + y2(x, t) = [2ym cos(ωt)] sin(kx) (152)

where the result follows from a trig identity. This wave has zero velocity (i.e. kx = k(x−0t))but an amplitude that varies like cosωt. It is called a “standing wave.” Watch:

Something important happens when we set up standingwaves on a stretched string that is fixed at both ends, likea guitar string. As shown in the figure, we can only havestanding waves that “fit” into the distance L between thetwo fixed ends. That is

L = nλ

2⇒ λn =

2L

n⇒ fn = n

v

2L(153)

That is, frequencies have to be integer multiples of a “fun-damental” frequency v/2L. This is the basis of the har-monic sound of nearly all string and wind instruments.

48


The topic we call “coupled oscillations” has far reaching implications. The formalism endsup being appropriate for many different applications, some of which bear only a passingresemblance to classical oscillation phenomena. This includes the mathematics of eigenvaluesand eigenvectors, for example.

These notes describe the elementary features of coupled mechanical oscillations. We use onespecific example, and describe the method of solution and the physical implications impliedby that solution. More complicated examples, and their solutions, are easy to come by, forexample on the web.1

The following figure describes the problem we will solve:

x1 x2

m mk kkc

Two equal masses m slide horizontally on a frictionless surface. Each is attached to a fixedpoint by a spring of spring constant k. They are “coupled” to each other by a spring withspring constant kc. The positions of the two masses, relative to their equilibrium position,are given by x1 and x2 respectively.

Now realize an important point. We have two masses, each described by their own positioncoordinate. That means we will have two equations of motion, one in terms of x1 and theother in terms x2. Furthermore, since the motion of one of the masses determines the extentto which the spring kc is stretched, and therefore affects the motion of the other mass, thesetwo equations will be “coupled” as well. We will have to develop some new mathematics inorder to solve these coupled differential equations.

We get the equations of motion from “F = ma”, so let’s do that first for the mass on theleft, i.e., the one whose position is specified by x1. It is acted on by two forces, the spring kon its left and the spring kc on its right. The force from the spring on the left is easy. It isjust −kx1.

The spring on the right is a little trickier. It will be proportional to (x2 − x1) since that isthe extent to which the spring is stretched. (In other words, if x1 = x2, then the length ofspring kc is not changed from its equilibrium value.) It will also be multiplied by kc, but weneed to get the sign right. Note that if x2 > x1, then the string is stretched, and the force onthe mass will be to the right, i.e. positive. On the other hand, if x2 < x1, then the spring iscompressed and the force on the mass will be negative. This makes it clear that we shouldwrite the force on the mass as +kc(x2 − x1).

1See http://math.fullerton.edu/mathews/n2003/SpringMassMod.html.

49

The equation of motion for the first mass is therefore

−kx1 + kc(x2 − x1) = mx1 (154a)

We get some extra reassurance that we got the sign right on the second force, because ifk = kc, then the term proportional to x1 does not cancel out.

The equation of motion for the second mass is now easy to get. Once again, from the springk on the right, if x2 is positive, then the spring pushes back so the force is −kx2. The forcefrom the “coupling” spring is the same magnitude as for the first mass, but in the oppositedirection, so −kc(x2 − x1). This equation of motion is therefore

−kx2 − kc(x2 − x1) = mx2 (154b)

So now, we can write these two equations together, with a little bit of rearrangement:

(k + kc)x1 − kcx2 = −mx1 (155a)

−kcx1 + (k + kc)x2 = −mx2 (155b)

To use some jargon, these are coupled, linear, differential equations. To “solve” these equa-tions is to find functions x1(t) and x2(t) which simultaneous satisfy both of them. We cando that pretty easily using the exponential form of sines and cosines, which we discussedearlier when we did oscillations with one mass and one spring.

Following our discussions about single mass oscillating systems, we assume the same ansatzand write

x1(t) = a1eiωt x2(t) = a2e

iωt (156)

where the task is now to see if we can find expressions for a1, a2, and ω which satisfy the dif-ferential equations, including the initial conditions. You’ll recall that taking the derivativetwice of functions like this, brings down a factor of iω twice, so a factor of −ω2. There-fore, plugging these functions into our coupled differential equations gives us the algebraicequations2

(k + kc)a1 − kca2 = mω2a1 (158a)

−kca1 + (k + kc)a2 = mω2a2 (158b)

This is good. Algebraic equations are a lot easier to solve than differential equations. Tomake things even simpler, let’s divide through by m, do a little more rearranging, and definetwo new quantities ω2

0 ≡ k/m and ω2c ≡ kc/m. Our equations now become

(ω20 + ω2

c − ω2)a1 − ω2ca2 = 0 (159a)

−ω2ca1 + (ω2

0 + ω2c − ω2)a2 = 0 (159b)

2A mathematician would call collectively call these two equations an eigenvalue equation. The reasonbecomes clearer when you write this using matrices. In this case, Eqs. 158 become[

k + kc −kc

−kc k + kc

] [a1

a2

]= mω2

[a1

a2

](157)

50

So, what have we accomplished? We think that some kind of conditions on a1, a2, and ωwill solve these equations. Actually, we can see a solution right away. In mathematician’slanguage, these are two coupled homogeneous (i.e. = 0) equations in the two unknowns a1

and a2. The solution has to be a1 = a2 = 0. Yes, that is a solution, but it is a very boringone. All it means as that the two masses don’t ever move.

The way out of this dilemma is to turn these two equations into one equation. In otherwords, if the left hand side of one equation was a multiple of the other, then both equationswould be saying the same thing, and we could solve for a relationship between a1 and a2,but not a1 and a2 separately. Mathematically, this condition is just that the ratio of thecoefficients of a2 and a1 for one of the equations is the same as for the other, namely3

−ω2c

ω20 + ω2

c − ω2=

ω20 + ω2

c − ω2

−ω2c

(160)

ω4c = (ω2

0 + ω2c − ω2)2

±ω2c = ω2

0 + ω2c − ω2

ω2 = ω20 + ω2

c ∓ ω2c (161)

In other words, these two equations are really one equation if

ω2 = ω20 ≡ ω2

A (162)

or, instead, ifω2 = ω2

0 + 2ω2c ≡ ω2

B (163)

Borrowing some language from the mathematicians, the physicist refers to ω2A and ω2

B as“eigenvalues.” We will also refer to A and B as “eigenmodes.”

What is the physical interpretation of the eigenmodes? To answer this, we go back toEq. 159a or, equivalently, Eq. 159b. (Remember, they are the same equation now.) If wesubstitute ω2 = ω2

A into Eq. 159a we find that

aA1 = aA2 (164)

where the superscript just marks the eigenmode that we’re talking about. In other words,the two masses move together in “lock step”, with the same motion. This happens whenthe frequency is ω = ±ω0 = ±

√k/m, and that is just what you expect. It is as if the two

masses are actually one, with mass 2m. The effective spring constant is 2k, as you learnedin your laboratory exercise, where you had one cart (with variable mass) and a spring oneach side. Therefore, we expect this double-mass to oscillate with ω2 = (2k)/(2m) = k/m.Good. This all makes sense.

Now consider eigemode B. Substituting ω2 = ω20 + 2ω2

c into Eq. 159a, we find that

aB1 = −aB2 (165)

In other words, the two masses oscillate “against” each other, and with a somewhat higherfrequency. In this case, it is interesting to see the difference between ωc ω0 (in otherwords, a very weak coupling spring) and ωc ω0 (strong coupling spring). For the weak

3A mathematician would say that we are setting the determinant equal to zero.

51

coupling spring, the frequency is just the same as ω0, and that makes sense. If the twomasses aren’t coupled to each other very strongly, then they act as two independent massesm each with their own spring constant k. On the other hand, for a strong coupling betweenthe masses, the frequency is ω = ωc

√2, which gets arbitrarily large. Sometimes, this “high

frequency mode” can be hard to observe.

A simple experimental test of this result is to set up two identical masses with three identicalsprings. In other words, ωc = ω0. In that case ω2

B = 3ω2A and therefore the frequency for

mode B should be√

3 times larger than the frequency for mode A. We should be able tomake this test as a demonstration in class.

We have been talking about general properties of the two eigenmodes. Of course, thisdoesn’t tell you how the system behaves given certain starting values, that is, specific initialconditions.

For this, we need to understand that the two eigenmodes can be combined. In fact, the gen-eral motion of each of the masses really needs to be written as a sum of the two eigenmodes.Each of these comes with a positive and negative frequency, just as it did when we appliedall this to the motion of a single mass and spring. (Recall that the sum and difference ofa positive and negative frequency exponential, are equivalent to a cosine and sine of thatfrequency.) Incorporating what we’ve learned about the relative motions of modes A and B,we can write the motions as follows:

x1(t) = aeiωAt + be−iωAt + ceiωBt + de−iωBt (166a)

x2(t) = aeiωAt + be−iωAt − ceiωBt − de−iωBt (166b)

where a, b, c, and d are constants which are determined from the initial conditions. Notethat these four constants appear as they do in Eqs. 166 because each frequency term mustmatch up between the equation for x1(t) and x2(t), with the correct relative sign, in orderto solve the coupled differential equations.

An obvious set of initial conditions is starting mass #1 from rest at x = x0, and with massx2 from rest at its equilibrium position. This leads to the solution

x1(t) = (x0/2) [cos(ωAt) + cos(ωBt)] (167)

x2(t) = (x0/2) [cos(ωAt)− cos(ωBt)] (168)

The last laboratory experiment for this course will investigate this motion. Some numberof setups will be using new electronics that allows each mass to be followed separately andsimultaneously.

On to the plots. Two cases are shown, one with ωc = omega0, and the other with ωc = ω0/4.For each case, we plot x1(t) with x2(t), and also x1(t) + x2(t) with x1(t) − x2(t). The firstcase shows clearly that the motion is somewhat complicated for either mass, but clearlysplits into two eigenmodes with rather different frequencies for the sum and difference. Thesecond case shows something much closer to a beat pattern, with the energy shifting fromone mass to the other and then back again.

52

0 5 10 15 20 25 30−20

−15

−10

−5

0

5

10

15

20

ω0=1 ω

c=1

x1

x2

0 5 10 15 20 25 30−20

−15

−10

−5

0

5

10

15

20

ω0=1 ω

c=1

x1+x

2x

1−x

2

53

0 20 40 60 80 100 120−20

−15

−10

−5

0

5

10

15

20

ω0=1 ω

c=0.25

x1

x2

0 20 40 60 80 100 120−20

−15

−10

−5

0

5

10

15

20

ω0=1 ω

c=0.25

x1+x

2x

1−x

2

54


Today we start a new general topic, namely Thermodynamics. I will introduce it to youby discussing the related subject of Statistical Mechanics. This is a way of simplifyingcomplicated systems by averaging over their properties, an approach that should work ifthey are complicated enough.

The Ideal Gas

Let’s imagine a box filled with a very large number of particles. The particles are very smalland don’t interact with each other. There are so many particles, and they are moving soquickly, that all you feel is a steady pressure on the walls.

See the figure. Put N gas molecules each of mass m in acubic box of side length L. Molecule has x velocity vx. Itbounces off the wall and incurs momentum change 2mvx.The time between bounces is 2L/vx so Fx(2L/vx) = 2mvx.The pressure on the wall at x = L is the sum of the forcesof all the molecules, divided by the area of the wall. That is

p =1

L2

∑Fx =

1

L2

∑ mv2x

L=Nm

V

(1

N

∑v2x

)(169)

where we recognize that the volume of the cube is V = L3. Now the quantity in parenthesesis the average value of v2

x, written 〈v2x〉. There is nothing special about the x direction, and

the number N is very large, so we expect that 〈v2x〉 = 〈v2

y〉 = 〈v2z〉 = 1

3〈v2〉 where we call 〈v2〉

the “mean square velocity.” Since all molecules have the same mass, 〈K〉 = 12m〈v2〉 is the

average kinetic energy of the molecules. So, we can now write

pV = N2

3〈K〉 (170)

Now we invent something called temperature which measures the amount of energy perparticle in the gas. For historical reasons, the temperature T is measured in different unitscalled Kelvin (K). The conversion constant from temperature to energy is called k, knownas Boltzmann’s constant, and has the value k = 1.38 × 10−23J/K. We assign half a unit ofkT to each of the three directions in which the particle can move. In other words

〈K〉 =3

2kT =⇒ pV = NkT (171)

This is known as the ideal gas law. You may have already seen this in a chemistry course,but chemists count particles in a different way. They prefer to count moles where a mole isAvogadro’s number NA = 6.02× 1023 of particles. That is, N = nNA where n is the numberof moles. Chemist’s also write R ≡ NAk, that is pV = nRT .

Practice Exercise

A one liter cubical box is 10 cm on a side. Assume it is filled with molecules of oxygen (atomicmass 32 amu) at room temperature (T = 300K) and under a pressure of one atmosphere(105 Pa, where a pascal is one newton per square meter). Calculate the number N of oxygenmolecules in the box, the average speed, i.e 〈v2〉1/2, and the average number of collisionswith the box walls in one second.

55

Practical Aspects of Temperature

The kind of temperature we’ve just defined is called absolute temperature. Sometimes it iscalled the Kelvin temperature scale. As far as physics is concerned, it is the only kind oftemperature that matters.

There are other common temperature scales used in everyday life, although neither is usedmuch by physicists. The two ordinary scales are Fahrenheit (TF) and Celsius (TC), i.e.

TC = T − 273.15 and TF =9

5TC + 32 (172)

This is no historical introduction to temperature. Initially people wanted to measure how“hot” or “cold” something was. They found some material with a property that was sensitivethis quantity. Then they defined temperature scales by pinning two values to the behaviorof the particular material property, that is, boiling and freezing of pure water in the case ofCelsius, and body temperature and freezing point of brine in the case of Fahrenheit.

Then we came across the more physical meaning of temperature, based on internal energy,and the absolute temperature scale was born. It was chosen to have the same size of “degree”as Celsius, but put the zero of the scale at the place where there would be no internal kineticenergy. Sometimes we call that point “absolute zero.”

Thermal Expansion: A Material Property that Used to Define Temperature

Most solids expand when they get hot. The reason is subtle.

Higher temp means fastermolecular motion. Solidsare like atoms connected by“anharmonic” springs, andaverage position shifts tolarger separations as the en-ergy increases. So, solids ex-pand.

The effect is rather linear and not large for typical temperatures. For solids, we write

∆L = αL ∆T (173)

where α is called the coefficient of linear expansion and expresses the fractional change inlength (∆L/L) per degree of temperature change, and varies from solid to solid. Since thefractional change is small, we expect ∆A = 2αA ∆T for expansion of surface area, and∆V = 3αV ∆T for volume. (Remember your Taylor expansions.) Note that fluids (likewater!) are more complicated than solids.

Practice Exercises

(1) Determine the temperature at which the Fahrenheit and Celsius scales coincide, that is,give the same value for the temperature. Have you ever been outside on a day or night whenthis was the air temperature?

(2) Explain why our definition 〈K〉 ≡ 32kT for the temperature of an ideal gas, would not

make sense if we wrote it in terms of TF or TC, instead of absolute temperature T .

56


Energy, Work, and Heat

We are going to start talking about systems like the oneshown on the right. It is a cylinder of gas with a pistonthat moves without friction. Weight on the top of thepiston keeps the gas in an “equilibrium” state. That is,all changes are slow ones. The gas has some internal en-ergy that we will call U . (For an ideal gas, U = 3NkT/2,the average kinetic energy of a molecule times the num-ber of molecules.) Work can be done on the piston bythe gas, and work can be done on the gas by the piston.

This figure plots pressure versus volume for the gas in thecylinder. (It is called a “pV diagram.”) Both quantitiescan be changed in different ways, like adding or subtract-ing gas molecules or changing the mass M sitting on topof the piston. (There is also something called a “thermalreservoir” but we’ll get to that in a minute.) We writethat work W is done on the gas by the piston. We have

dW = Fdh = −pAdh = −pdV (174)

where A is the area of the piston and F = −pA is the force acting on the gas, in the directionopposite to that which increases h.

Now consider taking the gas from “state” B (p = pf and V = Vi) to state D (p = pf andV = Vf ) along path #1. The pressure p = pf , a constant, along this path. Therefore thework done on the gas is

Wpath #1 = −∫

path #1

pdV = −pf (Vf − Vi) = −pf∆V (175)

In contrast, along path #2, the volume does not change, so the work done along this path,connecting state C to state D, is zero. From there, it is simple to see that if we go from Bto D instead of along path #1, but rather down to A, then across to C and then back up toD, the work done on the gas is

Wpath BACD = −pi(Vf − Vi) = −pi∆V 6= Wpath #1 (176)

So, the work done in going from one state to another will depend on the “path” taken onthe pV diagram. We refer to U , p, V , and also the temperature T , as “state variables”because they only depend on the properties of the gas at any one time. Work W is not astate variable; it depends on the path taken to get between different states.

Next let’s think about what happened to the gas in going from B to D along path #1. Forsimplicity, assume it is an ideal gas. Also assume that the gas supply valve is turned off, so

57

no gas molecules are added or taken away. Then since pV = NkT , the temperature musthave increased, since p = pf has not changed but the volume has increased from Vi to Vf .Since the temperature has gone up, so has the internal energy U . The work done on thegas, Eq. 175, is negative, so that would have decreased the internal energy. Where did theextra internal energy come from? It came from the thermal reservoir, by transferring energyin the form of heat. Heat (Q) is a new concept. It is a different way of transferring energy,other than work (W ), in a thermodynamic system.

Energy is conserved. Always. In thermodynamic systems as well as mechanical ones. Ourexpression of energy conservation for the gas in the cylinder can be written as

Q+W = ∆U (177)

This is called the First Law of Thermodynamics. Note the signs. If heat is added to asystem, then Q is positive. If the gas expands against the piston, then W is negative.

Specific Heat Capacity

We pause for a moment and talk about heat in a general sense. This will be useful later.

Add heat and the internal energy goes up, so the temperature increases. For solids andliquids, the volume doesn’t change, so work is not an issue. For all materials, the amountby which the temperature increases depends on the mass and the specific properties of thematerial. This is all collected into the definition of the specific heat capacity c, namely

c =1

m

Q

∆T(178)

where m is the mass of the object, Q is the heat added or lost, and ∆T is the change in tem-perature. The specific heat capacity is tabulated in many places for lots of substances.

We write C = mc for “heat capacity.” Clearly, C depends on how much matter is present.

Specific heat capacity depends on temperature, but it is not a strong function of temperaturefor most substances for common temperatures. Under these conditions

Q = m

∫ Tf

Ti

cdT = mc(Tf − Ti) (179)

For an ideal gas under constant volume, Q = ∆U = (3/2)Nk∆T . Therefore C = (3/2)Nk =(3/2)(N/NA)NAk = (3/2)nR. We write CV = 3R/2 for the “molar heat capacity” atconstant volume. Real, monatomic gases show molar specific heat capacities very close tothis value. Diatomic or polyatomic gases show higher values, reflecting the additional degreesof freedom in those molecules.

For an ideal gas that expands under constant pressure, W = −p∆V = −Nk∆T . ThereforeQ = ∆U − W = (5/2)Nk∆T and the molar heat capacity at constant pressure is Cp =5R/2. Again, real monatomic gases agree with this very well. In addition, one expects thatCp − CV = R = 8.31 J/mol-K. We commonly write

γ ≡ CpCV

=5

3(180)

for a monatomic gas. A table is included which demonstrates these concepts.

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Cyclic Processes: An Introduction

The world makes a lot of use of so-called cyclic processes in thermodynamics. These areprocesses that operate in a cycle, ending up at the same place from which they start. We’lldo more with these next class, but let’s start with a simple one and see what we can learnfrom it.

The figure shows a cyclic process for an ideal gas. It goesfirst fromA toB by reducing the pressure at constant vol-ume, taking out heat Q1 and lowering the temperature;then to C by increasing the volume at constant pressure,in which case heat Q2 is added to raise the temperatureand work W2 < 0 is done on the gas; and then back toA by reducing the volume and increasing the pressure insuch a way that the temperature remains constant.

Since the process is cyclic, the change in internal energy around the cycle must be zero.Therefore, by the first law of thermodynamics, Q + W = 0 where Q and W represent thenet work around the cycle. As drawn in the figure, W2 < 0. Recall Eq. 175. On the otherhand, W3 > 0 since the volume decreases. Furthermore, W3 > −W2 since the area under thecurve defined by path #3 is larger than that for path #2. The work done along path #1 iszero.

Therefore, the net work is W = W2 +W3 > 0, and so the net heat Q < 0. That is, net workis done on the gas and heat leaves it. We call such a thing a refrigerator. If instead the cycleoperated clockwise, then heat is added and the gas does work on the piston. We call thisan engine. These conclusions will always be the case for cyclic processes, regardless of thecurves that bound them.

The constant volume path #3 is called an isotherm. It follows the curve p = NkT3/V , whereT3 is the temperature at both C and A. Therefore, the work done along this path is

W3 = −∫ VA

VC

pdV = NkT3

∫ VC

VA

dV

V= NkT3 log

(VCVA

)> 0 (181)

as advertised. Recall that we know that Q3 = −W3 since the path is an isotherm.

Adiabatic Processes in an Ideal Gas Left to Next Class

A process carried out in “thermal isolation” is called adiabatic. It is characterized by Q = 0,so W = ∆U = (3/2)Nk∆T = (3/2)nR∆T = nCV ∆T . If the temperature change isinfinitesimal, then we can write dW = −pdV = nCV dT . The ideal gas law gives d(pV ) =pdV + V dp = nRdT or V dp = nCV dT + nRdT = nCpdT . Dividing these gives us

V dp

pdV= − nCpdT

nCV dT= −Cp

CV= −γ =⇒ dp

p= −γ dV

V(182)

Integrating this equation is simple. For limits called A and B, it just says that

log

(pBpA

)= −γ log

(VBVA

)= log

(VAVB

)γor pV γ = constant (183)

This is the equation that describes an adiabatic path on a pV diagram.

59

Physics I Honors Fall 2008

Useful Tables for the Study of Thermodynamics

Taken from Physics, by Halliday, Resnick, and Krane, Fifth Edition

Coefficients of Linear ExpansionSubstance α (10−6/K)Ice 51Lead 29Aluminum 23Brass 19Copper 17Steel 11Ordinary glass 9Pyrex glass 3.2Invar alloy 0.7Fused quartz 0.5

Heat Capacities of Some SubstancesSubstance Specific Heat Capacity (J/kg-K)Lead 129Tungsten 135Silver 236Copper 387Carbon 502Aluminum 900Brass 380Granite 790Glass 840Mercury 139Water 4190

Molar Heat Capacities of Ideal and Real GasesCp CV Cp − CV

Gas (J/mol-K) (J/mol-K) (J/mol-K) γMonatomic

Ideal 20.8 12.5 8.3 1.67Helium 20.8 12.5 8.3 1.67Argon 20.8 12.5 8.3 1.67

DiatomicIdeal 29.1 20.8 8.3 1.40H2 28.8 20.4 8.4 1.41N2 29.1 20.8 8.3 1.40O2 29.4 21.1 8.3 1.40

PolyatomicIdeal 33.3 24.9 8.3 1.33CO2 37.0 28.5 8.5 1.30NH3 36.8 27.8 9.0 1.31

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Homework Assignment Due Thursday 20 November

(1) The following diagram shows the problem in coupled oscillators that we solved in class:

x1 x2

m mk kkc

Using the parameters ω20 ≡ k/m and ω2

c ≡ kc/m, find the motions x1(t) and x2(t) for thefollowing sets of initial conditions:

a) x1(0) = x2(0) = A and x1(0) = x2(0) = 0

b) x1(0) = x2(0) = 0 and x1(0) = −x2(0) = V

c) x1(0) = x2(0) = x2(0) = 0 and x1(0) = V

(2) As a result of a temperature rise of 32 C, a barwith a crack at its center buckles upward, as shownin the figure on the right. The distance L0 = 3.77 mbetween the bar supports is fixed, and the coefficientof linear expansion of the bar is 25× 10−6/K. Find x,the distance to which the center rises.

(3) In an experiment, 1.35 moles of oxygen (O2, a diatomic molecule) are heated at con-stant pressure, starting at 11C. How much heat must be added to the gas to double itsvolume?

(4) Gas within a chamber undergoes the process shown in the pV diagram below:

Find the net heat (in Joules) added to the system during one complete cycle.

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Review: Statistical Mechanics and Thermodynamics

Concentrate for now on gases. For a monatomic ideal gas, the “equation of state” is

pV =2

3〈K〉NkT = nRT (184)

which relates the state variables p, V , and T to the number of molecules N through afundamental constant k. The internal energy U = N(3/2)kT = N〈K〉 where 〈K〉 is theaverage molecular kinetic energy. The “3” counts the number of degrees of freedom,i.e.three because all a monatomic gas molecule can do is move in three directions. If, forexample, the molecule is diatomic and can rotate, there are five degrees of freedom.

If the gas container has movable walls, then work W can be done on the gas. If it is incontact with some kind of thermal bath, then heat Q can be transferred into the gas. Thesechange the state of the gas by an amount the depends on the “path” taken between thestates. The first law of thermodynamics says that

∆U = W +Q (185)

where by convention, W is the work done on the gas. In effect, this equation defines Q.

Entropy: A New State Variable

Now we define a new state variable S called entropy. Actually, rather than defining entropyitself, we’ll define the change in entropy ∆S between two states. We have

∆S ≡∫dQ

T

∣∣∣∣reversible

(186)

The notation implies that the path for the integral is “reversible”, that is, connected by aseries of equilibrium states. For example, ∆S = ∆Q/T for a reversible isothermal path.

Here’s a trickier example, a class problem of “free expan-sion.” The walls are perfectly insulated, so no heat comesin to the gas. Opening the stopcock lets gas molecules moveinto the evacuated chamber, until there is equal densityeverywhere. What is the change in entropy?

You are tempted to say ∆S = 0 since ∆Q = 0. However, theprocess shown is not reversible. We need to compute ∆S =∫dQ/T along some reversible path. Since no work is done

and no heat is transferred, ∆U = 0. So, choose an isothermalexpansion from Vi to Vf . Therefore dQ = −dW = pdV and

∆S =

∫ Vf

Vi

pdV

T= Nk

∫ Vf

Vi

dV

V= Nk ln

(VfVi

)> 0 (187)

62

The Second Law of Thermodynamics

Note that free expansion of a gas (Eq. 187) always yields an increase in entropy. This is asystem completely isolated from its surroundings, and the expansion process is irreversible.In a reversible process, like isothermal expansion or compression, heat is transferred througha common thermal reservoir, so we don’t expect any change in net entropy if we include boththe gas and the thermal bath with which it is in contact.

So, now we state the Second Law of Thermodynamics, namely “When changes occur withina closed system, its entropy either increases (for irreversible processes) or remains constant(for reversible processes). It never decreases.” In other words ∆S ≥ 0.

Adiabatic Processes in an Ideal Gas

A process carried out in “thermal isolation” is called adiabatic. It is characterized by Q = 0,so W = ∆U = (3/2)Nk∆T = (3/2)nR∆T = nCV ∆T . If the temperature change isinfinitesimal, then we can write dW = −pdV = nCV dT . The ideal gas law gives d(pV ) =pdV + V dp = nRdT or V dp = nCV dT + nRdT = nCpdT . Dividing these gives us

V dp

pdV= − nCpdT

nCV dT= −Cp

CV= −γ =⇒ dp

p= −γ dV

V(188)

Integrating this equation is simple. For limits called A and B, it just says that

log

(pBpA

)= −γ log

(VBVA

)= log

(VAVB

)γor pV γ = constant (189)

This is the equation that describes an adiabatic path on a pV diagram.

The Carnot Engine

The figure on the right shows a cyclic processwhich is bounded by two isotherms (at twotemperatures TL < TH) and two adiabats whichconnect those two temperatures. This is called aCarnot cycle. The cycle proceeds clockwise, sonegative work is done on the gas from A to B andheat QH enters the gas. During the isothermalcontraction C to D, the work is positive and soheat QL leaves the gas. (Remember that ∆U = 0along an isotherm, and Q = 0 along an adiabat.)

Since ∆U = 0 around the cycle, we must have

WDA +WH +QH +WBC +WL +QL = 0 (190)

Now WH < 0 and WL > 0. Futhermore, since paths H and L are isotherms, WH +QH = 0 =WL +QL. Therefore WDA +WBC = 0, and also QH = −WH > 0 (heat enters along H) andQL = −WL < 0 (heat exits along L). Since (by areas) |WL| < |WH |, we have |QL| < |QH |,so net heat enters the gas. Furthermore, the work done around the cycle is

W = WDA +WH +WBC +WL = WH +WL < 0 (191)

and the gas does work on the piston. Heat in, work out. This is an engine.

63

How efficient is this engine? In other words, how much work do you get out, for heat thatyou need to put in? Let us define an engine efficiency just that way, namely

ε ≡ |W ||QH |

=|WH +WL||QH |

=|QH | − |QL||QH |

= 1− |QL||QH |

(192)

We can go further by realizing that entropy change is zero around the cycle. That is

∆S =QH

TH+QL

TL=|QH |TH− |QL|

TL= 0 (193)

Therefore |QL|/|QH | = TL/TH and the efficiency of a Carnot engine is

ε = 1− TLTH

(Carnot engine efficiency) (194)

Of course, operated counter-clockwise around the cycle, a Carnot engine becomes a Carnotrefrigerator. Instead of efficiency we speak of coefficient of performance. This is the ratioof what you “want”, i.e. the heat QL taken out of the low temperature reservoir, and whatyou pay for, namely the work W you have to do on the gas. We write

K =|QL||W |

=|QL|

|QH | − |QL|=

TLTH − TL

(Carnot refrig coeff of performance) (195)

A very large coefficient of performance is needed for a refrigerator that operates between twotemperatures that are close to each other.

There is no such thing as a perfect engine, i.e. one with ε = 1. This requires TL = 0for a cold reservoir, that is one with no internal energy. There is also no such thing as aperfect refrigerator, i.e. one with an infinite coefficient of performance. This would meanthat |QH | = |QL| ≡ Q and ∆S = Q/TH −Q/TL < 0, which violates the second law.

These statements can be put together to show that “No real engine can have an efficiencygreater than that of a Carnot engine working between the same two temperatures.” Thatis, Eq. 194 is the most efficient an engine can ever be. In other words, by Eq. 192, any realengine must always discharge some heat QL to its surroundings.

Don’t Forget: Short homework assignment due in class on Monday.

64


Homework Assignment Due Monday 24 November

1) Two identical blocks, each of mass M and made from a material with specific heat capacityc, are kept thermally isolated from each other and from their surroundings. One block is attemperature T1 and the other is at temperature T2 > T1. The blocks are then put in thermalcontact with each other, but still thermally isolated from their surroundings.

a. Use the relationship between heat transfer and specific heat capacity to prove that thefinal equilibrium temperature of the two blocks together is T = (T1 + T2)/2.

b. Find the change in entropy for the system consisting of the two blocks. (You canimagine a reversible process by which one block is taken slowly to its final temperature,and then the other block is taken to this same temperature, but contact with a slowlychanging thermal bath.)

2) The figure on the right shows a cyclicprocess, plotted as a TS diagram, ratherthan a pV diagram.

a. Explain why this is a Carnot cycle.b. Calculate the heat that enters.c. Calculate the work done on the gas.

65

Ludwig Boltzmann is one of the few people who “understood” en-tropy. He came up with a way to understand it in terms of disor-der, which he wrote as W . His result, in the form of the equationS = k logW , is etched on his grave stone.

66


Remember that homework and final lab notebooks are due on Monday 1 Dec.

How to Build a Star: Hydrostatics of really big ball of gas

Start with the balance of pressure against gravitation. Then use ideal gas law to relate totemperature. (Will assume that stars like the sun have uniform density.) Finish with thebehavior of white dwarf stars, aka “degenerate Fermi gas.”

More reading (in order of increasing sophistication):

• Introductory Astronomy and Astrophysics, 4th Edition, Zeilik and Gregory, esp. Chap.10

• Order-of-magnitude “theory” of stellar structure, George Greenstein, Am. J. Phys. 55,804 (1987); Erratum Am. J. Phys. 56, 94 (1988)

• Stars and statistical physics: A teaching experience, Roger Balian and Jean-PaulBlaizot Am. J. Phys. 67, 1189 (1999)

Begin. First, set up the notation. Build our star with shells of radius r. Total mass of staris M and radius is R. Density is ρ = M/(4πR3/3) assumed to be independent of r. WriteM(r) be the mass enclosed at radius r, i.e. M(R) = ρ(4πr3/3) = M(r/R)3.

Now, balance pressure p(r) against gravity. Take a small piece of area A, thickness dr andmass dm = ρAdr, at radius r. Outward force from difference in pressure must equal thegravitational force from mass inside radius r. (Recall “shell theorems.”) Write it out:

pA− (p+ dp)A = GM(r)dm

r2= G

M(r)ρAdr

r2or,

dp

dr= −GM(r)ρ(r)

r2(196)

This equation expresses “hydrostatic equilibrium” for a large, spherical self-gravitating objectlike a star (or a planet, or a moon). It is written here in a way that holds for any densityfunction ρ(r), but we will be taking ρ to be a constant.

Let’s use this to determine the pressure pC at the center of a star, i.e. r = 0. Make thereplacements for constant density in Eq. 196 to get

dp

dr= −GM

( rR

)3 M

4πR3/3

1

r2= −3GM2

4πR6r =⇒ p(r) = pC −

3GM2

8πR6r2

At the surface of the star, the pressure is zero “by definition”, i.e. p(R) = 0. Therefore,

pc =3GM2

8πR4(197)

Next let us try to estimate the temperature at the center of a star. This requires us to knowthe “equation of state” which relates the gas pressure and density to the temperature, forthe matter that makes up the star. Understanding the equation of state is a big deal, and

67

an area of active research interest. For now, we will use an equation of state based on anideal gas, that is Eq. 184, namely pV = NkT .

If we assume that the star is made of N identical particles with mass m, then ρ = Nm/V andp = ρkT/m = (3/4π)(M/m)kT/R3. In fact, stars are hot, so hot that matter is a “plasma”of free nuclei and electrons. The nuclei are from mostly hydrogen, about 25% helium, anda few percent of heavier elements. For now, though, it is close enough to assume the star is100% hydrogen atoms. Therefore, the temperature at the center of a star is

TC =1

k

4πR3

3

m

MpC =

1

k

GmM

2R(198)

Can you see why we would not be able to get this answer from dimensional analysis?

The White Dwarf Star

We found some properties of stars assuming they consisted of an ideal gas, but the as-sumptions used to get Eq. 184 don’t always hold. Instead, we need another equation ofstate.

In deriving Eq. 184 we wrote Fx = ∆p/∆t = (2px)/(2L/vx), and that is still fine. For Nparticles, this gives a pressure P = NFx/L

2 = Npv/V . (We temporarily switch to P toavoid confusion with momentum. Also, we are making a rough estimate, so put px and vxto p and v.) Our concern will be with electrons, so write an electron density ne = N/V inwhich case P = nepv.

At very high densities, the ideal gas law assumptions are violated. The electrons are so closetogether, they start to violate the Pauli Exclusion Principle. The distance between them isd ≈ 1/n

1/3e and the Heisenberg Uncertainty Principle says that pd ≈ h so that p ≈ hn

1/3e .

Putting v = p/me and switching back to p for pressure, the equation of state becomes

p = ne(hn1/3

e

) (hn1/3

e /me

)=h2

me

n5/3e ∼ h2

me

n5/3 =h2

me

( ρm

)5/3

(199)

where n is the number density of (what would have been) atoms, and we make the (somewhatincorrect) assumption that the star is still made up of hydrogen atoms with mass m. Thisis an odd sort of equation of state. Combined with Eq 197 it leads to a star whose radiusdecreases with mass like 1/M1/3.

As the mass increases, and the radius decreases, the electrons are more and more confineduntil they are moving as fast as they can, i.e. v = c. The equation of state becomes

p = ne(hn1/3

e

)c = hc n4/3

e ∼ hc n4/3 = hc( ρm

)4/3

(200)

Combined this with Eq 197 gives an equation where the radius drops out! One solves for amass that is the highest possible mass of a white dwarf star. Doing a more careful job leadsto a value of 1.44 solar masses. This is the “Chandrasekhar limit” after the physicist whofirst derived it. What do you suppose would happen for a star of mass larger than this?

68


Homework Assignment Due Monday 1 December

(1) This problem asks you to carry through some numerical calculations that lead to anunderstanding of what it is like in the center of the Sun. The following web sites containsuseful data:

http://pdg.lbl.gov/2008/reviews/consrpp.pdfhttp://pdg.lbl.gov/2008/reviews/astrorpp.pdf

a. Use Eq. 197 to estimate the central pressure of the Sun. Express it in units of atmo-spheres, where one atmosphere is the air pressure at the surface of the Earth.

b. Use Eq. 198 to estimate the central temperature. Express it in Kelvin (K).

c. Recall that temperature is related to the average kinetic energy of the gas particles.What is the average kinetic energy of the hydrogen atoms in the center of the Sun?Express the answer in electron volts (eV). What does this tell you about the kind ofmatter in the center of the Sun? (A typical “binding energy” of an electron in an atomis around 10 eV.)

(2) Show that Eq. 199 leads to a “star” with radius R and mass M where R ∝ 1/M1/3.

69

Physics I Honors Monday 1 December Fall 2008

Last class! I will miss all of you, but don’t be a stranger.

Hand in lab books. Third midterm exam is this Thursday. Final exam is Wednesday, 3 Dec3-6pm, this room.

Conservation Laws and the Continuity Equation

Our last class will concentrate on how to write down a general conservation principle. Then,we’ll apply this to fluid dynamics.

The basic idea is simple. Imagine some substance that is conserved. In other words, itcannot appear or disappear unless it is somehow supplied. If we talk about the amount Qof this substance in some region R of space, then the only way to change this amount is tolet some flow into (or out of) the region.

Make this statement mathematical. The rate of change of Q is dQ/dt. Define some quantityJ = J(r, t) to be the “flux density” of the substance. That is, it is the amount of thesubstance flowing (perpendicularly) through some small area per unit time. So, throughsome small surface dA = dAn, an amount J · dA = JdA cosφ flows per unit time. Here, nis a unit vector perpendicular to the surface, and φ is the angle between n and J.

If we let S denote the surface of R, we then write our conservation condition as

dQ

dt= −

∮S

J · dA (201)

The negative sign is there because of the convention that elements of area dA point outwardon S. Therefore, if the integral is positive, there is a net flux out of R so Q decreases.

Now let ρ = ρ(r, t) be the substance density. Then Q is the integral of ρ over R, and

dQ

dt=

d

dt

∫Rρ(r, t)dV =

∫R

∂ρ

∂tdV (202)

The derivative becomes a partial derivative inside the integral, because all we care about isthe explicit dependence on time. That’s all that would be left after we carry out the integralover space.

For the right side of Eq. 201 we will make use of a result we first mentioned in Eq. 70.Imagine that R is small and rectangular, with one corner at (x, y, z) and with sides of lengthdx, dy, and dz. In this case∮

SJ · dA = Jx(x+ dx, y, z)dydz − Jx(x, y, z)dydz

+ Jy(x, y + dy, z)dxdz − Jy(x, y, z)dxdz

+ Jz(x, y, z + dz)dxdy − Jz(x, y, z)dxdy

=

[Jx(x+ dx, y, z)− Jx(x, y, z)

dx+ · · ·

]dxdydz

=

[∂Jx∂x

+∂Jy∂y

+∂Jz∂z

]dxdydz (203)

70

Now we can divide any arbitrary region R into little rectangular boxes that butt up againsteach other. The flux out of one box goes into the one next to it, so in the end, all that mattersare the sides of the boxes along the surface S. We also have dV = dxdydz, and

∇ · J =∂Jx∂x

+∂Jy∂y

+∂Jz∂z

(204)

which is called the “divergence” of J(r, t). Therefore, for any region R,∮S

J · dA =

∫R∇ · J dV (205)

which is known at the Divergence Theorem or Gauss’ Theorem. So, combining Eq. 202 withEq. 205, we turn Eq. 201 into ∫

R

[∂ρ

∂t+∇ · J

]dV = 0 (206)

which holds for any region R. In particular, if we make R very tiny, then there is noappreciable variation over the volume, and the therefore the integrand must be zero. Inother words, our statement of conservation becomes

∂ρ

∂t+∇ · J = 0 (207)

This is called the Continuity Equation. It finds application in all areas of physics, includingthe conservation of material in fluid mechanics; conservation of electric charge in electro-magnetism; conservation of energy and momentum in general relativity; and conservation ofprobability in quantum mechanics.

Let’s make this physical with the specific example of fluid flow. In this case, the conservedsubstance is the fluid mass, and density ρ(r, t) is just ordinary mass density. To identifyJ(r, t), consider the tiny fluid mass flowing through a small element of area dA in a timedt. Let the velocity of this tiny mass be v(r, t), and n be a unit vector perpendicular to thesurface of dA. Then the size of the volume element of mass swept out is v(r, t)dt · ndA, andthe mass which passes through dA is ρv(r, t) · ndtdA = J ·dAdt. In other words, J(r, t) = ρvis the amount of mass flowing (perpendicularly) through some small area per unit time.

The figure shows flow from P to Q in a closedregion. Specialize to ρ = ρ(r), independent oftime. Then Eq. 201 says that the integratedflux over the surface is zero. Assume the areasA1 and A2 on the right are small, so that thevelocities v1 and v2 do not vary much overthem, and are perpendicular to them. Then

ρ1v1A1 = ρ2v2A2 (208)

If the fluid is “incompressible” (liquids), thatis, the density ρ(r, t) =constant everywhere,then the continuity equation becomes

v1A1 = v2A2 (209)

I.e., water flows faster in constricted pipes.

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Practice Exercises on the Continuity Equation

(1) A pipe of diameter 34.5 cm carries water moving at 2.62 m/sec. How long will it take todischarge 1600 m3 of water? Answer: 1h 49 min.

(2) The figure below shows the flow of an incompressible liquid through a straight pipe witha constriction midway between the input and output:

Let z measure the distance along the axis of the pipe, with x and y perpendicular to z.What are Jx and Jy at values of z near points P and Q? For a value of z near the midpointbetween P and Q (i.e., in the middle of the constriction), make a sketch of Jx as a functionof x. Use your sketch, and Eq. 207, to explain why |v2| > |v1|.

(3) The figure shows the confluence of two streamsto form a river. One stream has a width of 8.2 m,depth of 3.4 m, and current speed of 2.3 m/sec. Theother stream is 6.8 m wide, 3.2 m deep, and flowsat 2.6 m/sec. The width of the river is 10.7 m andthe current speed is 2.9 m/sec. What is its depth?Answer: 3.9 m

(4) The figure shows a familiar effect, namely the “necking down”of a stream of water when it falls under gravity. (Water adheresto itself, so that no pipe is necessary to keep the mass continuous.)Use the continuity equation and conservation of mechanical energyto explain this effect. If the cross sectional area A1 is 1.2 cm2 andthat of A2 is 0.35 cm2, and their separation h = 45 mm, at whatrate R does water flow from the tap? Answer: R = 34 cm3/sec.

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