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Energy
Suppose the body on our track is acted upon by a
force which depends on the coordinate x in a
known way:
Consider the quantity
x
V(x) = − I F(x´) dx´
F = F(x) ,
where F(x) is a specified function of x. (It
doesn’t depend on the velocity or on the time.)
The differential equation which determines x(t) is
m
d 2 x(t)
= F(x(t)) .
dt2
Maybe there’s a quantity that is always the same
as time goes on. Such a quantity is called a
constant of the motion, and gives us powerful
information even in the absence of an explicit
solution. The most important constant of the
motion, and the one we will talk about the most,
is called the total energy. Let’s see how it is
defined and how it is used.
.
This is the negative of the area under the graph of
F(x´), from an arbitrary initial position x´=x0 to
x´=x :
Force
F(x')
0101010101
Now, the dependence of F on x could be very
complicated, perhaps so complicated that the
only way we can solve the relevant differential
equation is by using some elaborate numerical
procedure. Is there an easy way to get at least
some information about the motion, without resorting to a computer?
x0
AREA = −V(x)
x0
x
x'
Note that x´ is just a dummy variable. The
function V(x) does not depend on x´. Notice also
that the force is just the negative of the slope of
V(x):
dV(x)
F(x) = −
dx
(by the Fundamental Theorem of Calculus.)
Example: the harmonic oscillator
This all might seem pretty abstract, until you see
a specific example. For the harmonic oscillator
F(x) = – kx .
Then V(x) is just quadratic in x:
x
V(x) = − I ( − kx´) dx´
x0
=
1
1
k x 2 − k x 20
2
2
=
1
k x 2 + arbitrary constant .
2
V(x) change, when x changes this way?
d
dV(x(t)) dx(t)
V(x(t)) =
dt
dt
dx
= − F(x(t))
dx(t)
.
dt
(The first step uses the chain rule.) This may be
more compactly written as
dV
= − Fv ,
dt
where it is understood that everything is evaluated at x(t).
Next, let’s consider a completely different
quantity, which we will symbolize by T (not to
be confused with the period of periodic motion,
which is sometimes also symbolized by T !):
T=
1
mv2
2
.
The above example highlights the fact that the
potential energy is only defined to within an
arbitrary additive constant. This reflects the arbitrariness of the choice of starting point for the
integral. Another way of saying this is that we
may choose to make the potential energy equal
zero at any desired point.
where the last step uses Newton’s law ma=F.
Now, as the body moves, the value of x changes
in time according to x(t). How does the value of
We notice that V and T change in equal but
opposite ways over time. If V loses a bit, then T
How does it change in time?
dT
dv
= mv
= mva = Fv ,
dt
dt
gains the same amount, and vice versa. Another
way to say this is that their sum, T + V , is a
constant. This constant is called the total
energy:
E=T+V
.
When the external force is not a function of
velocity or time, the total energy is conserved:
dE
=0
dt
.
The quantity T is called the kinetic energy - it is
the energy associated with the motion of the
body. The quantity V is called the potential
energy - it is the amount of energy stored up in
whatever is causing the force on the body.
We note that the kinetic energy is not an intrinsic
property of a body. It depends on the frame of
reference. For example, we can always view a
body from a frame in which the body is at rest.
Its kinetic energy is then zero, which is not true
in other frames. Similarly, the potential energy is
not an intrinsic property, because we are free to
add any constant value to it.
Despite these shortcomings, the concept of
energy is extremely useful because it is con-
served in any inertial frame, as long as the force
depends only on position. The energy provides
us with one of our main tools (sometimes our
only tool) to get information about motion.
To find the dimension of energy, we look at an
equation which defines an energy in terms of
more fundamental quantities. A good choice is,
T=
1 2
mv .
2
Then it’s easy to see that the dimension of energy
is
[E] = mass (length)2 (time)− 2 .
In the system of units we are using, the unit of
energy is called the Joule (sounds like “jewel”):
1 Joule = 1 kg m2 s-2 .
Example: the harmonic oscillator
A very instructive example is the simple harmonic oscillator. We worked out the potential
energy above; we are free to take the constant to
be zero, corresponding to zero potential energy
when the spring is not stretched. Then the
amount of energy stored in the spring when it is
stretched by a distance x is
V(x) = kx2/2.
The farther the spring is stretched, the more
energy is stored in the spring and the less is
available for kinetic energy; the velocity therefore decreases. Eventually, all the energy is
stored in the spring and the body comes to rest;
this occurs at one of the turning points of the
oscillator.
The next plot shows the kinetic, potential and
total energies versus distance, for the oscillator.
The resources for this section contain a movie
which displays the exchange of potential and
kinetic energies during the motion. It is worth
studying closely.
Well, right away we know the total energy of the
body, because at the beginning the kinetic energy
was zero. The total energy is equal to the amount
of potential energy at the beginning:
E=
1 2
kx0 .
2
(Recall that we worked out the potential energy
earlier, as an example.) The body reaches its
maximum velocity when the potential energy is a
minimum; that is, at x=0. At that point, all of its
energy is kinetic. Hence,
1
1
m v 2max = k x 20 ,
2
2
Energy
total
or
kinetic
v max =
potential
x
Sub-example: what is the maximum velocity?
Here is an example of the kind of problem that is
easily solved using energy. Suppose we start off
the oscillator by pulling it to some initial point x0
and releasing it with zero initial velocity. What
will be its maximum velocity?
k
x .
m 0
You can easily verify this answer from the
detailed treatment of the harmonic oscillator presented earlier.
Example: gravitational potential near earth
Let’s analyze the gravitational force near the
surface of the earth, from the point of view of
energy. If we orient our coordinates with positive values pointing upwards, then the force due
to gravity is
F = – mg ,
This agrees with an earlier result.
the minus sign indicating that the force is
downwards.
In summary, the relation
E=
The potential energy is
x
V(x) = − I ( − mg) dx´
x0
= mgx + constant .
We see that the potential energy grows linearly
with x. The higher the body goes, the less energy
is available for kinetic energy, so the body slows
down.
Sub-example: how high does the body go?
relates the velocity v and the position x, allowing
you to solve for one in terms of the other if you
wish. And you can do this without needing to
know x(t).
Application: periodic motion
What property is shared by every periodic
motion? The body stops and turns around at both
ends. Here is a typical case:
Suppose we start it at x=0, with velocity v0. How
high does it go?
We may choose V to be zero at any particular
value of x we desire, by adjusting the constant. It
is convenient to let V=0 at x=0; then the constant
is zero. The initial potential energy of the body
is therefore zero, and its total energy is E=mv02/2.
At the top, its kinetic energy is zero, and all the
energy is potential. Hence
1 2
mv 0 = mg xmax, or
2
xmax
v 20
=
.
2g
1
mv 2 + V(x)
2
Energy
V(x)
E
total
kinetic
potential
x min
x
x max
Using energy methods, we can arrive at a closed
expression for the period of such a motion. Sol-
ving the above expression for v, we obtain
dx
=
dt
2
á E − V(x) é .
m
Here we took the positive square root; we’ll
multiply by two for the return journey below. Rearranging, we get
dt =
m
2
dx
.
E − V(x)
Let’s integrate both sides of this expression, from
one turning point to the other. The integral of the
left-hand side will give half the period, and
therefore the full period is
T=
xmax
2m I
xmin
dx
E − V(x)
.
Dissipation
In previous sections, we considered motion in
the presence of a damping force –bv, where b is a
positive constant. For example, we considered
the damped harmonic oscillator, and saw that its
oscillations become smaller and smaller as time
goes on. Evidently, energy is lost by the moving
body. This energy loss is called dissipation.
What is the rate of energy loss?
Let’s not make any assumptions about the total
force acting on the body, except that it is of the
form
Ftot = F(x) − bv .
As before, we define the total energy of the
moving body as the sum of its kinetic plus
potential energy:
E=
Exercise
The potential energy for the simple harmonic
oscillator is
V(x) =
1
m ω20 x 2 .
2
Using the above formula, verify our earlier
expression T=2π/ω 0 for the period.
1
mv 2 + V(x) ,
2
where V is the potential energy due to F :
F(x) = −
dV(x)
.
dx
Differentiating the total energy, we find
dE
dv dx dV
= mv +
= v(ma − F) .
dt
dt dt dx
The differential equation governing the motion is
ma = F(x) − bv ,
where b is the positive damping constant. Hence,
the rate of change of total energy is
dE
= − bv 2 .
dt
This is negative, indicating that energy is lost
when the velocity is nonzero.
Where does the energy go? It is transferred to
whatever is causing the damping force. For
example, if the damping is due to friction with a
track, then the lost energy is transferred to the
material of the track. It heats up as a result. The
total energy of body plus track is conserved, of
course.
Advanced example of use of energy method:
gravity including variation with height
Here is another interesting example of the use of
energy methods. Let’s analyze the motion of a
body falling in the gravitational field of the earth,
taking into account the fact that the force varies
with distance. (In an earlier section, we took the
gravitational force to be constant, which is a
good approximation near the earth’s surface.)
Although we have not yet studied gravity, you
may know the result for the force law already. If
we measure x upwards from the center of the
earth, the force exerted by the earth on a body at
or above the earth’s surface is
GMm
F= − 2 .
x
The minus sign indicates that the force points
downwards. G is Newton’s constant, M is the
mass of the earth, and m is the mass of the body.
The potential energy is
V(x) = −
GMm
+ constant ,
x
and we choose the constant to be zero by fixing
V=0 as x becomes infinite. The total energy of
the body is therefore
E=
1 2 GMm
mv −
.
x
2
Now, suppose the body starts at x=x0 with zero
initial velocity. Then its total energy is E =
–GMm/x0. Solving for the square of the velocity,
we find
åäåå dx
ã dt
ä
ëì 2
ë
ìì = 2GM åååå 1 − 1 ìììì .
í
ã x x0 í
This has an interesting solution, one that you
probably couldn’t derive without a lot of work
and inspiration. It involves a new variable η
(read “eta”), which is related to the time by
t=
force is accounted for, because the force is
stronger, the further down the body gets.
x0 x 0
(η + sin η) .
2GM 2
In terms of this new variable, you can easily
verify (try it!) that the solution is given by
x=
x0
(1 + cos η) .
2
x(t) is the curve traced out by a point on an
imaginary circle of diameter x0 “rolling” through
angle η along the t axis. Such a curve is called a
cycloid.
x
0101010101010101
1001010101
x0
η
x0 η
2
These facts are illustrated in the following figure,
which compares the paths taken with constant
gravitational force and with the true force which
varies with distance:
x0
2
x0 η
sin
2
Second, according to the above figure the body
apparently reaches infinite velocity at the center
of the earth! However, this is not true because
this force law ceases to be valid once the surface
of the earth is reached. (We will explain the
latter point when we talk about the gravitational
interaction in more detail later.) If you bored a
hole into the earth so the body could continue
falling even after it reached the earth’s surface,
the force law that we used to derive the solution
would no longer apply.
x(t)
2GM
t
x0
The resources for this section contain a movie
illustrating the trajectory.
There are two things to note. First, the body falls
faster when the variation of the gravitational
constant force
varying force
surface of earth
t