Table Of ContentFeynman Simplified
1D: Angular Momentum,
Sound, Waves,
Symmetry & Vision
Everyone’s Guide
to the
Feynman Lectures on Physics
by
Robert L. Piccioni, Ph.D.
Copyright © 2014
by
Robert L. Piccioni
Published by
Real Science Publishing
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Westlake Village, CA 91361, USA
Edited by Joan Piccioni
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Everyone’s Guide to the
Feynman Lectures on Physics
Feynman Simplified gives mere mortals access to the fabled Feynman Lectures on Physics.
This Book
Feynman Simplified: 1D covers about a quarter of Volume 1, the freshman course, of The Feynman
Lectures on Physics. The topics we explore include:
Angular Momentum
Moments of Inertia
Rotations in Three-Dimensions
Sound & Beat Frequencies
Modes & Harmonics
Fourier Series & Transforms
Complex Waves
Symmetry Properties of Natural Laws
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Table of Contents
Chapter 39: Rotation & Angular Momentum
Chapter 40: Centers & Moments of Rotation
Chapter 41: 3D Rotation & Review of Rotation
Chapter 42: Physics of Waves & Sound
Chapter 43: Theory of Beats
Chapter 44: Modes of Oscillation
Chapter 45: Harmonics & Fourier Analysis
Chapter 46: Complex Waves
Chapter 47: Review of Physics of Waves
Chapter 48: Physics of Vision
Chapter 49: Symmetry & Physical Laws
Chapter 39
Rotation &
Angular Momentum
We have so far explored Newton’s laws as they apply to point particles, and objects that we can
assume are equivalent to point particles. An example of the latter is the gravitational attraction of the
Earth and Sun, which Newton proved can be calculated assuming all of each body’s mass is
concentrated at its center.
We can now move up to more complex objects and motions.
As Feynman says in V1p18-1:
“When the world becomes more complicated, it also becomes more interesting… the phenomena
associated with the mechanics of a more complex object…are really quite striking.”
We will start with the simplest cases, preparing ourselves for the “really quite striking” ones to come
later. In this chapter, we will examine rotations of rigid bodies in two-dimensions moving at non-
relativistic velocities, so that we can employ Newtonian mechanics.
Parabolic path of center of mass
From our prior studies, we know how to calculate the motion of a football thrown by a quarterback to
a receiver. Assume the QB releases the ball at time t = 0, at location x=y=z=0, with velocity v=
(v,0,v), where +z is straight up and +x is downfield. Ignoring air resistance, if the ball were a single
x z
point, its trajectory would be:
x(t) = v t
x
y(t) = 0
z(t) = v t – 1/2 g t2
z
This arcing curve is called a parabola, as illustrated in Figure 39-1.
Figure 39-1 Parabolic Path Of Thrown Ball
The motion is much more complicated if a more complex object is thrown, such as a gaucho’s bolo,
three balls tied to a string. But even then, there is something that still moves in a parabolic arc —
some central essence moves in a perfect parabola, while all else flails around it. That central essence
is called the center of mass, a kind of average of the locations of all the particles within the complex
object. The center of mass, often-abbreviated CM, is a mathematically determined point that doesn’t
even have to be a material part of the object. For three balls attached to a loop of rope, the center of
mass is a point in-between the balls where nothing exists.
Let’s examine F, the sum of all forces on all particles within a complex object. Label the particles j =
1 to some enormous number N. The jth particle experiences a force F equal to its mass m times its
j j
acceleration, which equals the second derivative of its position r. Recall the linearity of
j
differentiation: d(A+B)/dt = dA/dt + dB/dt.
F = Σ F = Σ {d2 (mr) / dt2 }
j j j j j
F = d2 ( Σ {mr} ) / dt2
j j j
Define:
M = Σ {m}
j j
R = Σ {mr} / M
j j j
Rewriting the prior equations with these definitions yields:
F = M d2 (R) / dt2
We have recovered the equation for the motion of a single point body of mass M and position R. As
defined above, M equals the mass of the complex object and R is the position of its center of mass.
Everything we learned before about the motion of a single point body applies directly to the motion of
the CM of a complex body.
In V1p18-2, Feynman suggests mentally separating the forces on each particle into two groups:
internal and external. Newton’s third law (action begets reaction) ensures that the forces between
particle j and particle k in the body are equal and opposite; hence they cancel one another in the sum
F. Indeed all internal forces sum to zero in the calculation of F. This means the motion of the center
of mass is determined only by external forces, such as Earth’s gravity.
Imagine a rocket ship in outer space. If it is sufficiently isolated, we could assume it experiences no
external forces and is stationary in an appropriate reference frame. More precisely, its center of mass
is stationary. If the crew moves toward the front of the ship, the ship itself must move backward to
keep the center of mass stationary. Indeed, this must happen on the International Space Station,
although since the ISS is enormously more massive than its crew, this effect is imperceptible.
Newton’s third law prevents a rocket ship from moving its own center of mass, but it can propel part
of its mass (the ship) by expelling another part (the fuel). In whatever manner its rockets fire, the CM
of all the ship’s original mass remains stationary in an appropriate reference frame. In any other
reference frame, the ship’s CM moves at a constant velocity, regardless of how or whether its rockets
fire.
2D Rotation of Rigid Body V1p18-2
Since we know how to analyze the motion of the center of mass of a complex body, let’s now examine
the motion of the rest of the body. We’ll make things simpler by considering only rigid bodies, those
whose atoms are so strongly bound to one another that their relative orientations do not change. We
thus avoid dealing with bending, twisting, and vibrating. Having agreed on all these constraints, the
only motion left is for the body to rotate as one single object.
What does “rotating” mean? A three-dimensional object rotates around an axis, whereas a two-
dimensional object rotates about a single point. Figure 39-2 illustrates rotation in two-dimensions. On
the left side, the darker rectangle is rotated about the XY origin by a small angle, resulting in the
lighter rectangle. The right side of this figure focuses on the rectangle’s upper right corner, labeled P.
Initially, the radial line from the XY origin to P lies at angle ø relative to the x-axis. Rotation
increases that angle by dø, moving the rectangle’s upper right corner P to position Q.
Figure 39-2 Rotation in 2D
In the figure, the line u is perpendicular to r, hence the angle between u and the vertical dotted line is
also ø. Let the coordinates of P be (x,y) and of Q be (x+dx,y+dy). From trigonometry, we find these
relationships:
x = r cosø
y = r sinø
u = r tan(dø)
u = dx = –r sinø dø
x
u = dy = +r cosø dø
y
If we measure angles in radians, and take the limit of dø going to zero, we have:
u = r dø
u = –y dø
x
u = +x dø
y
If dø occurs during an infinitesimal time interval dt, and we define ω = dø/dt, the equations become:
v = dx/dt = –y ω
x
v = dydt = +x ω
y
v2 = v2+v2 = ω2 (y2+x2)
x y
v = ω r
The last equation jibes with P moving to Q, a distance u = r dø, in time dt, making v = r dø/dt = r ω.
A quick note about measuring angles in radians. As mentioned earlier, in equations and calculations
we always measure angles in radians rather than degrees. Radians are convenient: the arc length
subtended by angle θ equals rθ, for radius r. For example, for θ = π/2, arc length = π r/2, both
corresponding to 1/4 of a full circle. In the text, we sometimes use the more colloquial 90 degrees,
rather than π/2, but in equations it’s always radians.
To clarify the rotational parameters, Feynman stresses the analogy with linear motion. In linear
motion we have position r, velocity v = dr/dt, acceleration a = d2r/dt2, and force F = ma. In rotational
motion we have angular position ø, angular velocity ω = dø/dt, angular acceleration α = d2ø/dt2, and
perhaps an unknown X. Isn’t there some angular X analogous to force?
Torques V1p18-4
Ask and ye shall receive: X is torque, from the Latin word torquere, to twist. We’ll follow
Feynman’s lead and find quantitative expressions for torque by considering the work energy expended
in rotating an object. Recall that for linear motion, work equals force times the distance through
which the force acts. By analogy, work will equal torque times the angle through which the torque
acts.
Going back to our rectangle in Figure 39-2, pushing on the rectangle at point P and turning it by angle
dø requires work W, according to:
W = F • ds = (F dx + F dy)
x y
Using the expressions derived above for dx and dy, we derive the equation for torque τ in 2D:
W = (–y dø F + x dø F)
x y
W = τ dø
τ = + x F – y F
y x
You might recognize the combination xy – yx; it’s the vector cross product. So, we can rewrite this
equation in vector form, which we’ll discover later is also valid in 3D, as:
τ = r × F
If multiple torques act on a body, they simply add like vectors, as do forces in linear motion. Since an
