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Motion of a Particle in Three Dimensions
V. Lindberg
June 17, 2010
1
General Principles
What changes as we move to 2D or 3D motion?
We must deal with vectors and components. In Cartesian coordinates Newton’s Second
Law becomes
Fx = mẍ
(1)
Fy = mÿ
(2)
Fz = mz̈
(3)
Each of the force components can be functions, explicitly or implicitly, of position, x, y, z,
velocity, ẋ, ẏ, ż and time, t. The differential equations can quickly become much more complicated than those we saw in 1D, and in general will be partial differential equations.
We will begin with a constant force. Cases of force depending solely on position will be
treated next, followed by simple cases where the force depends solely on velocity or solely
on time. Finally a few words will be said about forces that depend on more than one type
of variable.
For the moment we will not worry about angular momentum.
1.1
Work in 3D
Consider the dot product of Newton’s Second Law with velocity,
d~
p
d(m~v )
F~ · ~v =
· ~v =
· ~v
dt
dt
Assuming constant mass, we can recognize that ~v · ~v̇ =
dT
F~ · ~v =
dt
1
1 d(~v ·~v )
2 dt
(4)
and hence
(5)
where T is the kinetic energy of the particle.
Now ~v = d~r/dt so we can write
d~r
dT
F~ ·
=
dt
dt
~
F · ~r = dT
and integrating using a line integral between points A and B,
Z B
W =
F~ · ~r = TB − TA = ∆T
(6)
(7)
(8)
A
Keep in mind that F~ is the net force on the particle.
Example Suppose that the net force on a particle is F~ = 3xyz î + 2xĵ + 2xy k̂ N.
(a) Suppose we start at (0,0,0) move in a straight line to (0,2,0), then in a straight
line to (3,2,0), with all the distances in meters. Find the work done in moving
along this path.
(b) Instead suppose we start at (0,0,0) move in a straight line to (3,0,0), then in a
straight line to (3,2,0), with all the distances in meters. Find the work done in
moving along this path.
(c) Could this force be conservative? Why or why not?
1.2
Force Fields and Conservative Forces
In University Physics you discussed potential energy, perhaps using one of the definitions
“The net work done by a conservative force on a particle moving around any closed path is
zero.” or “The work done by a conservative force on a particle moving between two points
does not depend on the path taken by the particle1 .”
We want to discuss the details of this statement and find a reliable method to decide
whether a force is conservative.
A pictorial method of showing a force (at least in 2D) is to take a coordinate grid, and at
regularly spaced points in the grid to draw short vectors that have the properties that the
length of the vector is proportional to the force at that point, and the direction indicates
the direction of the force. Examples are shown in the text, Figures 4.1.2 and 4.1.3.
Suppose we have a force defined by Fx = −by and Fy = +bx where b is a positive constant.
Figure 4.1.2 shows this. Consider going around a closed rectangular path from (x, y) to
1
Fundamentals of Physics, Halliday, Resnick & Walker, 8th edition.
2
(x + ∆x, y) to (x + ∆x, y + ∆y) to (x, y + ∆y) and back to (x, y), and computing the work.
(The circle on the integral sign indicates a closed path.)
I
W =
F~ · d~r
Z x+∆x
Z y+∆y
Z x
Z y
=
Fx (y)dx +
Fy (x + ∆x)dy +
Fx (y + ∆y)dx +
Fy (x)dy
x
Z
y
x+∆x
x+∆x
[Fx (y) − Fx (y + ∆y)]dx +
=
x
y+∆y
y+∆y
Z
[Fy (x + ∆x) − Fy (x)]dy
y
= [b(y + ∆y) − by]∆x + [b(x + ∆x) − bx]∆y
= 2b∆x∆y
(9)
For this force, the work done around a closed path is non-zero, therefore this force is
non-conservative and has no potential energy.
Try a different force, defined by Fx = +by and Fy = +bx where b is a positive constant.
The force field is shown in Figure 4.1.3. Evaluating the closed-loop work integral for the
same path results in W = 0, meaning that the force is conservative, and that a potential
energy does exist.
This is still a tedious method (and we need to check that the integral is zero for all closed
paths) so we want something better. Stokes’ Theorem tells us that in general
I
Z
~
F · d~r =
(∇ × F~ ) · n̂ da
(10)
surf
The closed path of the line integral defines a surface, and the dot product between the curl
of the force and the normal to the surface is evaluated over the area.
The criterion that we have a conservative force when the closed line integral is zero is thus
equivalent to saying that the curl of the force must be zero:
For a conservative force, one that has a potential energy defined,
∇ × F~ = 0.
1.3
ConservativeForce
Computing curl
In Cartesian coordinates we can easily write the curl using a determinant.


î
ĵ
k̂
∂
∂ 
∂
∇ × F~ = det  ∂x
∂y
∂z
Fx Fy Fz
3
(11)
= î
∂Fy
∂Fz
−
∂y
∂z
+ ĵ
∂Fx ∂Fz
−
∂z
∂x
+ k̂
∂Fy
∂Fx
−
∂x
∂y
(12)
Think about this in terms of cyclic permutations: The first term has î, y in the denominator,
z in the numerator: think of this as (xyz), x-component is partial with y of Fz . The curl
terms consist of [(xyz) − (xzy)] + [(yzx) − (yxz)] + [(zxy) − (zyx)]. Getting used to thinking
in cyclic permutations takes a little effort, but is very valuable.
We may want to find curls in other coordinate systems. Appendix F gives some details,
and I will summarize it here.
Suppose we have a general orthogonal coordinate system of coordinates (u, v, w) and associated unit vectors (ê1 , ê2 , ê3 ): for example, spherical polar coordinates (r, θ, φ). It is right
handed, meaning ê1 × ê2 = ê3 .
There are associated with the coordinates Lamé factors (or scale factors) (h1 , h2 , h3 ) that
allow us to write volume elements, line elements, gradients, divergences, and curls2 . For
spherical coordinates they are (1, r, r sin θ).
The volume element in general is
dV = h1 h2 h3 du dv dw
(13)
d~r = ê1 h1 du + ê2 h2 dv + ê3 h3 dw
(14)
The line element is
Gradient is
∇f =
ê1 ∂f
ê2 ∂f
ê3 ∂f
+
+
h1 ∂u h2 ∂v
h3 ∂w
(15)
Divergence is (note the cyclic nature!)
1
∂(h2 h3 Q1 ) ∂(h3 h1 Q2 ) ∂(h1 h2 Q3 )
~
∇·Q=
+
+
h1 h2 h3
∂u
∂v
∂w
and curl is
h1 ê1 h2 ê2 h3 ê3
∂
1
∂
∂
~
∇×Q=
det ∂u
∂v
∂w
h1 h2 h3
h1 Q1 h2 Q2 h3 Q3
(16)
(17)
For Cartesian coordinates, h1 = h2 = h3 = 1, for spherical polar, h1 = 1, h2 = r, h3 =
r sin θ,and for cylindrical coordinates (R, φ, z), h1 = h3 = 1, h2 = R.
Example 1 Suppose F~ = Ax2 y î − Axy 2 ĵ + Cz k̂ with A a positive constant. Is this a
conservative force?
2
For a description of how to find the Lamè factor, try http://www.luc.edu/faculty/dslavsk/courses/
phys301/classnotes/scalefactorscomplete.pdf
4
Example 2 Suppose F~ = Ar2 ê1 in spherical polar coordinates with A a positive constant.
Is this a conservative force?
Example 3 Suppose F~ = Ar2 cos θê1 + Br2 sin θê2 + C ê3 in spherical polar coordinates.
Could this be a conservative force for the proper choice of constants A, B, C? If so
what are the values of these constants?
Example 4 Suppose F~ = Ar cos2 θê1 + Br2 sin θ cos θê2 + C ê3 in spherical polar coordinates. Could this be a conservative force for the proper choice of constants A, B, C?
If so what are the values of these constants?
2
Gradients: Relating Force to Potential Energy
We now have an easy way to determine if a force is conservative. When the force is
conservative we want to find its potential energy function, V (x, y, z) (or other orthogonal
coordinates.) Conversely, if we know the potential energy function, we want to find the
vector force.
The second task is easier, so we start there. Given V , we can find the force by
∂V
∂V
∂V
F~ = −∇V = −
−
−
∂x
∂y
∂z
(18)
where the gradient is given in Cartesian coordinates.
It is a simple matter to show that ∇×∇V = 0 verifying that the force is conservative.
Example 1 What force is associated with V = 12 k(x2 + y 2 + z 2 )?
Example 2 What force is associated with V = 21 kr2 in spherical polar coordinates?
Example 3 What force is associated with V = − A3 r2 cos θ in spherical polar coordinates?
So now let’s do the reverse, find a potential energy related to a force. Suppose that
F~ = −k1 xî − k2 y ĵ − k3 z k̂. This is an anisotropic spring force.
1
2
We know that ∂V
∂x = −Fx , so we integrate dV = −Fx dx to get V = 2 k1 x +A(y, z). Likewise
1
2
from the other force components we get V = 2 k2 y + B(x, z) and V = 12 k3 z 2 + C(x, y).
Combining these we get
1
V (x, y, z) = (k1 x2 + k2 y 2 + k3 z 2 ) + V0
2
(19)
As another example, suppose that in spherical polar coordinates F~ = Ar2 cos θê1 −(Ar2 /3) sin θê2 ,
with A being a positive constant. Referring back to Equation 15 and using h1 = 1, h2 =
5
r, h3 = r sin θ we get the equations
Fr = Ar2 cos θ = −
1 ∂V
1 ∂r
Fθ = −(Ar2 /3) sin θ = −
Fφ = 0 =
1 ∂V
r ∂θ
1 ∂V
r sin θ ∂φ
(20)
Doing the integrals results in
V
V
V
Ar3
cos θ + A(θ, φ)
3
Ar3
cos θ + B(r, φ)
= −
3
= C(r, θ)
= −
(21)
Combining
Ar3
cos θ + V0
3
where V0 is a constant (positive or negative).
V =−
3
(22)
Separable Forces
A separable force can be written in Cartesian Coordinates as
F~ = Fx (x)î + Fy (y)ĵ + Fz (z)k̂
(23)
Almost by inspection you can see that this is a conservative force. The equations of motion
are independent,
Fx (x) = mẍ
Fy (y) = mÿ
Fz (z) = mz̈
(24)
Solving these equations by the methods of Chapter 2 gives us ~r = x(t)î+y(t)ĵ +z(t)k̂.
In chapter 2 we also looked at forces that were functions of velocity as well. In the case
where each component depends on the corresponding position and velocity components,
the equations of motion look like mẍ = Fx (x, ẋ, t) and these can in principle be solved
fairly easily.
6
3.1
Constant Force: Projectile Motion with No Drag
Consider projectile motion in a uniform and constant3 gravitational field ~g in a non-rotating
coordinate system.. We choose the field to define the z-direction, with up as positive. This
is the usual case of two dimensional projectile motion, and we look to see how our formalism
gives the well known results. The equation of motion is
m
d2~r
d~v
=m
= −mg k̂
2
dt
dt
(25)
If we choose the coordinates so that the initial velocity ~v0 is in the x-z plane, then y = const
and the motion is two dimensional. For simplicity we choose y = 0 so that the location at
arbitrary time can be written
~r = xî + z k̂
(26)
Integrating the equation of motion once gives
~v = −k̂gt + ~v0
(27)
Integrating again, with the initial position being ~r0 gives
1
~r = −k̂ gt2 + ~v0 t + ~r0
2
(28)
The force is conservative with a potential energy V = mgz + V0 and for convenience we
define V0 = 0 at z = 0.
The energy equation, T0 + V0 = T + V thus becomes
1
1
1
m(ẋ20 + ż02 ) + mgz0 = m(ẋ2 + ż 2 ) + mgz = mv 2 + mgz
2
2
2
(29)
Frequently we can choose z = 0 at t = 0 so that the energy equation just becomes
1
1
mv02 =
mv 2 + mgz
2
2
v02 − 2gz = v 2
(30)
We can write the initial velocity using magnitude and direction as ~v0 = (v0 cos α0 )î +
(v0 sin α0 )k̂ where α0 is the angle from the x axis. (The text uses α.)
3
Recall the meaning of these two terms: uniform means having the same value for all locations in
space, but possibly changing in time; constant means unchanging in time, but possibly varying with spatial
location.
7
Then the solutions to the equation of motion are
~v = î(v0 cos α0 ) + k̂(v0 sin α0 − gt)
1 2
~r = î(v0 cos α0 )t + k̂ (v0 sin α0 )t − gt
2
(31)
(32)
The position components are
x = (v0 cos α0 )t
(33)
y = 0
(34)
1
z = (v0 sin α0 )t − gt2
2
and eliminating time we can write
g
z = (tan α0 )x −
x2
2v02 cos2 α0
(35)
(36)
which is the equation of a parabola.
These results should all be familiar to you, and it is easy to use them to get quantities such
as the text does: Maximum height where ż = 0, Time to reach maximum height, Time
to return to original height, Range = horizontal distance when particle returns to original
height.
Equation 4.3.14 in the text has an error. It uses the trig identity 2 sin α0 cos α0 =
sin 2α0 (See Appendix B). The correct result is
R=x=
3.2
v02 sin 2α0
g
(37)
Velocity Dependent Force: Projectile Motion With Linear Drag
We have already discussed air drag and found that for most objects the drag is quadratic.
With quadratic drag in 2D, the solution is found numerically. We will assume that the
drag is linear because then we can solve the result analytically, even if does not apply to
most projectiles in air.
In Chapter 2 we used F = −c1 v with c1 = 1.55 × 10−4 D in SI units. Choose the same orientation of coordinate system with initial velocity in the x-z plane and positive z upwards.
For convenience let c1 = mγ so that the equation of motion becomes
m
d2~r
dt2
d2~r
dt2
= −mγ~v − k̂mg
= −γ~v − k̂g
8
(38)
In components
ẍ = −γ ẋ
(39)
ÿ = −γ ẏ = 0
(40)
z̈ = −γ ż − g
(41)
This shows that the motion is separable and the solutions are variations of those seen in
Chapter 2, namely
ẋ = ẋ0 e−γt
(42)
ẏ = 0
ż = ż0 e
or
(43)
g
− (1 − e−γt )
γ
−γt
(44)
g
~v = ~v0 e−γt − k̂ (1 − e−γt )
γ
(45)
and integrating again,
ẋ0
(1 − e−γt )
γ
y = 0
ż0
g
gt
+ 2 (1 − e−γt ) −
z =
γ
γ
γ
x =
or
~r =
~v0 k̂g
+ 2
γ
γ
!
(1 − e−γt ) − k̂
gt
γ
(46)
(47)
(48)
(49)
Notice that as t → ∞, ẋ → 0 meaning that asymptotically the particle falls vertically,
unlike parabolic motion, at a final horizontal location x → ẋ0 /γ.
We have seen that the motion is not parabolic. We can again eliminate t to see what the
motion actually is, resulting in
ż0
g γx
g
γx
+ 2
+ 2 ln 1 −
(50)
z=
γ
γ
ẋ0
γ
ẋ0
If we want to find the horizontal range, R, that is the distance travelled until the particle
returns to its original height, we set z = 0 and solve the resulting transcendental equation.
9
Approximation methods must be used to solve this equation, starting with a Taylor series
expansion of the natural logarithm. The result is
R =
=
2ẋ0 ż0 8ẋ0 ż02
−
γ + ···
g
3g 2
v02 sin 2α0 4v03 sin 2α0 sin α0
−
γ + ···
g
3g 2
(51)
(52)
with the first term being the result with no drag, and the second term being the first order
(in γ) correction for drag.
Getting this result from Equation 50 is not a trivial project, and is well worth doing.
3.3
Quadratic Drag
For projectiles like baseballs or ping-pong balls, the drag in air will be quadratic. For a
one dimensional drop we used F = −c2 v|v| with c2 = 0.22D2 for spheres of diameter D in
air (all in SI units). In two dimensions we can write
p
(53)
F~drag = −c2~v |v| = −c2 vx2 + vz2 (îvx + k̂vz )
This force is not separable, so the solution is much more difficult than that for linear drag,
and we will not attempt it analytically. If we have values for all the quantities we can solve
it numerically using, for example, EJS. You will do this as an assignment.
3.4
Other Forces on Projectiles
In addition to gravitational and drag forces, real balls are subject to buoyant forces and, if
they are spinning, Magnus forces. The buoyant force is a constant for motion in a uniform
medium.
The Magnus force4 is actually an offshoot of drag and Bernoulli forces. Consider a baseball
moving with a velocity ~v through air. It spins with an angular velocity ω
~ , and at the surface
of the ball, air is dragged with the ball.
Imagine being in the frame of reference of the ball. The air moves with a translational
velocity −~v , and in addition there is a velocity due to the rotation, of magnitude ωR. The
net velocity of the air is the combination of these two, resulting in air having a higher
velocity on one side of the ball than on the other, and from Bernoulli’s Principle, the
pressure is higher when the velocity is lower, resulting in a Magnus force proportional to
ω
~ × ~v .
4
See http://en.wikipedia.org/wiki/Magnus_effect.
10
4
Position Dependent Forces: Harmonic Oscillator in 3D
Suppose we have a particle subject to a linear restoring force
F~ = −k~r
(54)
Producing such an isotropic restoring force is NOT easy. The text suggests a set of springs
in Figure 4.4.1, but showing whether it really does produce the desired force is not simple5 .
This is an example of a central force that, in spherical polar coordinates, depends only the
magnitude of the radius vector and not its angular orientation.
4.1
Two Dimensional Case
The equation of motion using the x-y plane is separable and is
m
d2~r
= −k~r
dt2
mẍ = −kx
(55)
mÿ = −ky
(56)
The differential equation should be an old friend and defining ω =
is
p
k/m, a solution
x = A cos(ωt + α)
(57)
y = B cos(ωt + β)
(58)
The four constants A, B, α, β are found from the initial conditions.
To find the equation for the path we eliminate time from the equations. Defining ∆ =
β − α,
x2
2 cos ∆
y2
−
xy
+
= sin2 ∆
(59)
A2
AB
B2
4.2
A Primer on Conic Sections
Both here, the two-dimensional harmonic oscillator, and later, planetary orbits, we will
need to discuss conic sections: ellipses, parabolas, and hyperbolas6 . The term conic section
5
And I suspect that to make it really work that all the springs must initially be slightly stretched. You
will solve a related problem in the second quarter of mechanics.
6
Those preferring older spellings will note that the plurals are parabolae and hyperbolae.
11
derives from the intersection of a plane surface with a cone, as is illustrated in http:
//en.wikipedia.org/wiki/Conic_section.
There are several ways to describe the conic sections. First consider the geometric view.
Two cones meet at an apex, and the z-axis is defined as the symmetry axis. If a plane
surface is in the x-y plane, the intersection of the plane with the cone will be a circle, or
a point. If the plane is parallel to the edge of the cones, the intersection will be either a
parabola or a straight line.
As we tilt the plane from the case of the circle to the case of the parabola, we get an ellipse.
When the plane is tilted beyond the point of the parabola, we get a pair of hyperbolas.
A second method describes the conic sections in terms of a fixed point, the focus, F, a
line not through the focus called the directrix, L, and a non-negative real number, the
eccentricity. Let us call the separation of focus and directrix q. These are illustrated in
Figure 1 and Figure C.1a in Appendix C of the text.
d
dc .
The eccentricity is defined as the ratio of lengths = F
C/CL
A circle has eccentricity = 0, and q = ∞. The focus serves as the center of the circle. All
points on the circle are equidistant from F.
An ellipse has eccentricity 0 < < 1, and the distance from a point on the ellipse to F is times the distance from the point on the ellipse to L. This is shown in Appendix C of the
text. Figure 1 has q = 3 and = 0.5.
A parabola has = 1. Hyperbolas have > 1.
Ellipses have a second symmetrically placed focus and directrix. Likewise hyperbolas have
a second symmetrically placed focus and directrix.
We can write a general equation for conic sections in terms of and q using the focal point
as the origin.
q
1
1 + cos θ
r=
=
(60)
1 + cos θ
r
r0 (1 + )
where we have introduced r0 = q/(1 + ). You can see what r0 means on Figure 1.
The first form of Equation 60 can be written in Cartesian coordinates as7
(1 − 2 )x2 + 22 qx + y 2 = 2 q 2
7
(61)
This is easy to derive if you start down the correct path. From the first of Eq. 60 cross multiply to get
r(1 + cos θ) = q, use x = r cos θ, rearrange to get r on one side, and proceed.
12
Figure 1: A conic section symmetric about the x-axis with the origin at the focus, F. The
directrix, L, is separated from the focus by a distance q. For a point on a conic section
d
dc where Lc is a point on the directrix opposite C. In polar
C, the eccentricity = F
C/CL
coordinates, C is at (r, θ), and r0 is the value at θ = 0.
4.2.1
The Circle
Using the second form in Equation 60 we see that for a circle with = 0, r = r0 for all
angles, or equivalently x2 + y 2 = r02 .
4.2.2
The Ellipse
Now consider the ellipse with 0 < < 1. The latus rectum, α, is the y coordinate when
x = 0, θ = 90◦ , and it is easy to show that it is α = q.
It is also helpful to know the height, 2b and width, 2a, of the ellipse, called the minor
and major axes. Half of these are called the semi-minor and semi-major axes. To find the
semi-minor axis, write the expression for y = r sin θ and find its maximum value. You will
13
find the maximum value occurs at cos θ = −, and has a value
b= √
q
1 − 2
(62)
The minor axis occurs at a location x = −c relative to the focus where
c=
2 q
1 − 2
(63)
The semi-major axis is then
q
(64)
1 − 2
It is more common for ellipses to use the center of the ellipse as the origin. Thus we make
a shift of coordinates y 0 ← y, x0 ← x + c. I will now drop the primes. With a bit of work we
can show that the equation of the ellipse relative to coordinates located at its center
is
x2 y 2
+ 2 =1
(65)
a2
b
a = c + r0 =
Foci at x = ±c
p
b = a 1 − 2
a2 − b2 = c2
q = ±b2 /c
= c/a
(66)
r0 = (1 − )a
(67)
The ellipse thus written is symmetrical about both x- and y- axes.
4.2.3
The Parabola
For the parabola, = 1. I find it easier to consider a parabola opening to the left, similar
to Figure 1. Calling r0 ≡ c we quickly see that q = 2c. Beginning with the polar expression
for the parabola using the focus as the origin, r = 2c/(1 + cos θ), with a bit of work we can
get to the relation between the cartesian coordinates, y 2 = −4c(c − x0 ).
It is more useful to shift the origin to the apex of the parabola, x = x0 − c and then find the
Cartesian equation for a parabola with the origin at the apex, opening to the left
y 2 = −4cx
4.2.4
(68)
The Hyperbola
Figure 2 shows an hyperbola8 . Now > 1. (Note that the figure in Appendix C of the
text, page A10, has an error. Can you see what it is?) If we use the center of the the two
8
Some authors write a hyperbola.
14
Figure 2: Hyperbola (both branches) with important parameters.
branches for our origin, the hyperbola can be written
x2 y 2
− 2 =1
a2
b
(69)
The apex of the right branch is at x = a. At large magnitudes of x the branches are
asymptotic to the lines y = ±bx/a, so b is the distance from the horizontal line
√ to the
asymptote at the apex. The triangle formed by a and b has a hypotenuse c = a2 + b2 ,
and c is also the distance from the center to the focus. At the focus the latus rectum is
α = b2 /a. Other relations are
a2 + b2 = c2
4.2.5
=
c
a
b=a
p
2 − 1
q=±
b2
c
(70)
Effect of Translation of the Origin
Suppose we translate the origin both in x and y. It is easy to show that this introduces
linear terms in x and y so that
Ellipse b2 x2 + Bx + a2 y 2 + Dy + E = 0
15
(71)
y 2 + By + E = −4cx
Parabola
Hyperbola
4.2.6
2 2
2 2
b x + Bx − a y + Dy + E = 0
(72)
(73)
Translation and Rotation
Suppose the coordinate sytem is rotated and translated, the resulting conic function will
be, in general,
Ax2 + Bxy + Cy 2 + Dx + Ey = F
(74)
To determine which conic section this is, compute the discriminant disc = B 2 − 4AC.
Negative discriminants indicate ellipses, positive discriminants indicate hyperbolas, and
zero indicates a parabola.
It is a bit of work to show this in the general case, but if we consider just rotation about
the origin, and use the rotation matrix from Chapter 1, it is relatively straight forward to
show that the discriminant condition is valid.
4.3
Completing the Two Dimensional Isotropic Oscillator
In Section 4.1 we obtained the equation of the particles motion,
x2
2 cos ∆
y2
−
xy
+
= sin2 ∆
A2
AB
B2
(75)
where ∆ = β − α is the phase difference
p between the motion along orthogonal axes, (x =
A cos(ωt+α), y = B cos(ωt+β), ω = k/m.) It is quick work to calculate the discriminant
and determine that the motion is an ellipse, and a lot more work to show that the major
axis is rotated from the x-axis by an amount ψ,
tan 2ψ =
2AB cos ∆
A2 − B 2
(76)
Recall that trig functions are double valued: there are two values of 2ψ that give the same
result. You will do an activity to try to clear this up.
4.4
3D Isotropic Oscillator
The equations of motions are
mẍ = −kx
mÿ = −ky
16
mz̈ = −kz
(77)
with solutions
x = A1 sin ωt + B1 cos ωt
y = A2 sin ωt + B2 cos ωt
z = A3 sin ωt + B3 cos ωt (78)
or in vector notation
~ sin ωt + B
~ cos ωt
~r = A
(79)
~ and B
~ define a plane, and the motion is an ellipse located in that plane. Planar
Vectors A
motion is characteristic of central forces as we shall see in Chapter 6.
4.5
Non-Isotropic Oscillator
With different elastic constants in three orthogonal directions, we have equations of motion
mẍ = −k1 x
mÿ = −k2 y
mz̈ = −k3 z
(80)
and solutions
x = A cos(ω1 t + α)
y = B cos(ω2 t + β)
z = A cos(ω3 t + γ)
(81)
The path of the object is constrained to be in a box of dimensions 2A, 2B, 2C centered
on the origin. If A, B and C are commensurate, that is any ratio of the constants is a
rario of integers, the path is a closed Lissajous figure. Otherwise the path does not close
and will eventually fill the box. An example of a non-isotropic oscillator is an atom in a
crystal.
5
Charged Particles in Electric and Magnetic Fields
Consider a charged particle of mass m and charge q moving with velocity ~v in electric and
magnetic fields. The Lorentz force is
~ + ~v × B)
~
F~ = q(E
5.1
(82)
Electric Field Only
The equation of motion is
mẍ = qEx
mÿ = qEy
mz̈ = qEz
17
or
m
d2~r
~
= qE
dt2
(83)
In general the electric field can be a function of location and time (although that will
introduce a magnetic field). Consider the simplest case of a constant electric field with
axes chosen so that the field is along the z axis. The equations of motion reduce to
mẍ = 0
mÿ = 0
z̈ = qE/m = constant
(84)
and we once again have the equations of motion under a constant acceleration.
~ = 0 and therefore
In E & M you will see that when static charges produce a field, ∇ × E
the static electric field can be written in terms of an electric potential
~ = −∇Φ
E
(85)
with the potential energy being V = qΦ.9
5.2
Static Magnetic Field Only
The equation of motion is
d2~r
~
= q~v × B
(86)
dt2
We are here concerned only about a static field (unchanging in time) but it could depend on
location. For simplicity, consider the case of a constant magnetic field that we choose to be
~ = B k̂. The equation of motion is not separable, but becomes
along the z direction, B
m
î ĵ k̂ d2~r
m 2 = q~v × B k̂ = qB ẋ ẏ ż dt
0 0 1 (87)
m(ẍî + ÿ ĵ + z̈ k̂) = qB(ẏ î − ẋĵ)
(88)
Equating components, letting ω = qB/m we get the following differential equations that
we can integrate once with respect to time
ẍ = ω ẏ
ẋ = ωy + C1
ÿ = −ω ẋ
ẏ = −ωx + C2
z̈ = 0
ż = constant = ż0
(89)
(90)
Putting the expression for ẏ into the first equation in (89), and defining a = C2 /ω, we can
get
ẍ + ω 2 x = ω 2 a
(91)
9
Yup notation change here. In Physics III you used U for potential energy, V for electric potential. In
this course we use V for potential energy and Φ for electric potential.
18
with solution
x = a + A cos(ωt + θ0 )
(92)
It is then a relatively simple matter to find
y = b − A sin(ωt + θ0 )
(93)
and
qB 2
= ẋ + ẏ = A ω = A
(94)
m
This tells us that the speed of the particle is constant, and you should recognize that the
motion is circular in the x − y plane with motion along z making a helix. Notice that the
size of the helix, A depends on the speed in the x − y plane, but that the frequency of
rotation is constant. This is the basis for the cyclotron, an early (non-relativistic) particle
accelerator.
2
vxy
2
2
2 2
2
Suggested tasks: What is the pitch of the helix? In cylindrical polar coordinates, what are
the components of the motion?
Motion of charged particles in combined electric and magnetic fields is common. One
example is a magnetron used in deposition of material. A non-uniform magnetic field
is produced near a target material, and an electric field is perpendicular to the target
material. Charged particles (electrons and positive ions) are concentrated in the vicinity
of the target and the light electrons have a large mean-free path between collisions.
6
Constraint Forces
There are many situations where a particle is constrained in its motion. Consider a bead
on a wire: it is constrained to move along the wire and cannot move off the wire. Consider
a piece of ice in a bowl: it is constrained to either move on the surface of the bowl or into
space above the bowl, but it cannot penetrate the bowl.
These are examples of complete and one-sided constraints respectively. Next quarter when
we talk about Lagrangians we will term these holonomic and non-holonomic constraints. It
is sometimes useful to describe the bead on a wire as a one-dimensional problem, meaning
one variable (length along wire) is sufficient to uniquely locate the bead. The ice in the
bowl, providing the ice stays on the surface, is two-dimensional problem. This will be
useful in Lagrangians, Chapter 10.
~ is normal to the surface, and
In the simplest case, the force introduced by the constraint, R,
the velocity of the particle, ~v , is tangent to the surface. If the other forces are a combined
value F~ we write
d~v
~
m
= F~ + R
(95)
dt
19
d~v
~ · ~v
· ~v = F~ · ~v + R
dt
d 1
m~v · ~v
= F~ · ~v
dt 2
m
(96)
(97)
If the other forces are conservative, we can still write conservation of mechanical energy.
Eg Period of Ice slipping in a spherical bowl. Consider a frictionless spherical bowl
of radius R. A small object of mass m is released from rest at an initial angle θ0
measured from the vertical. Find an expression for the period of the oscillation.
We can write the energy equation
1
E = mv 2 − mgR cos θ = −mgR cos θ0
2
Solving for speed as a function of angle,
p
v = 2gR(cos θ − cos θ0 )
(98)
(99)
We can find the time to move from an angle θ to θ + dθ: the distance travelled is
Rdθ at a speed v, so
Rdθ
dt = p
(100)
2gR(cos θ − cos θ0 )
and we integrate to find the period,
T
=
4
Z
0
θ0
Rdθ
p
2gR(cos θ − cos θ0 )
(101)
This integral cannot be done in closed form: it is an elliptic integral, and we will discuss
this further in Chapter 8.
20