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| This Site : | Bottom-of-Page | Home | Part 1 | Part 2 | Other Topics |
Book and TV Program : The Mechanical Universe
Many of the ideas in this article were taken from the TV series called
"Mechanical Universe"
( hosted by Professor David Goodstein ).
There are several useful textbooks that accompany the series.
For example, "The Mechanical Universe Mechanics and
Heat Advanced Edition"
© 1986 Cambridge University Press
See especially chapters 16 and 17.
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| Symbol | Meaning | |
| t | time ( a scalar quantity ) | |
| Dt |
derivative with respect to t [ where t stands for time ] ( e.g. DtL = dL/dt = instantaneous rate of change of L ) |
|
| m | mass ( a scalar quantity ) | |
| v | velocity ( a vector quantity ) | |
| | a | | the length of vector a | |
| a X b | the cross product of vectors a and b | |
| F | Force ( a vector quantity ) | |
| r | radius vector ( its tail is at the origin of our coordinate system ) | |
| τ | torque ( vector cross product defined by τ = r X F ) | |
| θ | angle from positive X-axis to radius vector ( see Figure 1.0 ) | |
| ω | angular speed ( derivative of θ with respect to time ) | |
| ≈ | is approximately equal to ( e.g. 4.99999999 ≈ 5 ) | |
| << | is much less than ( e.g. 2 << 100,000 ) | |
| L | angular momentum ( L = radius vector cross linear momentum ) = r X ( mv ) |
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implies
( e.g. x = 6
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| Sometimes we put an arrow over a letter to indicate a vector | ||
|
For example,
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| A "'hat" ( ^ ) over a letter indicates a unit vector | ||
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For example,
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| A "'dot" ( . ) over a letter indicates derivative with respect to time | ||
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For example,
[ By definition, ω also equals the derivative of θ with respect to time. ] |
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We will use an X-Y Cartesian coordinate system.
A vector r from the origin to a point in space is called the
radius vector.
A Force F is a vector whose
twisting effect is called torque.
Torque is symbolized by the Greek letter tau ( τ ) and is defined to be the
vector cross product r X F.
If we cross the radius vector into the linear momentum we get angular momentum.
The angular momentum L is defined to be r X ( mv ).
If you recall that
| 1 ) | v = dr/dt by definition of velocity | |
| 2 ) | F = d/dt ( mv ) Newton's 2nd law of motion | |
| 3 ) | For any vector A, the cross product A X A is zero | |
| 4 ) | Dt ( A*B ) = [ ( dA/dt ) * B ] + [ A * ( dB/dt ) ] |
then you can easily see that torque is the rate of change of angular momentum:
| dL/dt | = | Dt [ r X (mv) ] | |
| = | [ dr/dt X mv ] + r X [ d/dt (mv) ] | ||
| = | [ v X mv ] + [ r X F ] | ||
| = | m [ v X v ] + τ | ||
| = | 0 + τ |
In any force field τ = dL/dt
In a central force field the force vector has the same line of action as
the radius vector, and the force magnitude is a scalar function
( call it k ) of the length of the radius vector.
| F | = |
k( r )
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| τ | = |
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| = |
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| = |
( r
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| = |
r k( r )
(
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| = | 0 | ||
But τ always equals dL/dt.
So in a central field dL/dt must be zero, and thus
L must be a constant.
L = length of
= | r X ( mv ) | = m | r X v | = a constant vector
In a central force field
the angular momentum L
is a constant vector
r X v = L / m = a constant vector
We will use this result later ( when deriving Kepler's 2nd Law ).
Since each of
and
is perpendicular to the ( fixed ) direction of
,
motion in a central force field will be confined to a single
plane
which we take to be the X-Y plane.
Whenever motion is confined to a single plane we can use
polar coordinates in addition to our X-Y coordinates.
Figure 1.0 shows the r-θ components of a radius vector.
The unit vector
is defined by
See Figure 2.0.
If the horizontal component of r has length x and
the vertical component
has length y then
we have the relationships shown in Figures 2.0 and 3.0.
Since r-hat dot theta-hat is zero, those two vectors must be perpendicular to each other, as shown in Figure 1.0. The sense of theta-hat ( i.e. which direction it points ) can be determined by the fact that the cross product of r-hat and theta-hat is k-hat, as shown in Figure 3.0.
We will keep a list of some of the key facts we derive that will be needed later.
Referring to Figure 2.0 above , we will now find
( see figure 5.0 below )
some derivatives of the r-hat and θ-hat unit vectors:
Next we will find an expression for the velocity vector.
For our purposes we won't need an expression for the
acceleration vector.
But just for curiosity you could
differentiate the velocity vector to verify that
the acceleration is given by :
To find the above result you can first extend the product rule
for differentiation (ab)' = ab' + a'b
to the less familiar rule for differentiating a triple product:
(abc)' = a(bc)' + a'(bc) = a( bc' + b'c ) + a'(bc)
= abc' + ab'c + a'bc
Thus, you could use this rule:
(abc)' = a'bc + ab'c + abc'
together with previously computed derivatives.
Recall the distributive law for vectors a, b, and c:
a X ( b + c ) = ( a X b ) + ( a X c )
Using this, we now compute ( see Figure 8.0 below )
the value of the cross product r X v.
( Here is where we use the top formula in Figure 6.0 :
)
But earlier we showed that in a central force field
the magnitude
of rXv is L / m. Therefore ....
( see Figure 9.0. )
We can regard an ellipse as the path traced out by a point that
moves
in the X-Y plane in such a way that the sum of
its distances from two
fixed points ( the foci ) remains constant.
See Figure 12.0 where the
foci F1 and F2
lie on the X-axis, and point A is one vertex of the ellipse.
The distance from the center of the ellipse to a focus is denoted by
c
,
and
a
is the distance from the center to point A.
In Figure 13.0, the major axis of the ellipse
is the line segment
from point A' to point A, and
the minor axis is the line segment
between
points B' and B.
| The length of the major axis is 2a. | ( a = length of the semi-major axis ) | |
| The length of the minor axis is 2b. | ( b = length of the semi-minor axis ) |
In Figure 12.0, the distance from A to F1 is a+c
and the distance from A to F2 is a-c
Since point A is on the ellipse, the constant
d1+d2
must equal (a+c) + (a-c) = 2a.
For any point on an ellipse, the sum of the
distances from that point to each focus is 2a.
In Figure 14.0 the point ( 0,b ) lies on the ellipse;
and so, w + w must equal 2a.
Thus w = a and
( see Figure 15.0 ) we see that a, b, and c
satisfy a
pythagorean relationship.
| a2 = b2 + c2 | |
|
|
Suppose that an ellipse with semi-major axis a and
semi-minor axis b has its center at the origin of an X-Y
coordinate system. If the foci are at ( -c,0 ) and ( c,0 ) then
| the standard ellipse equation is | x2 / a2 + y2 / b2 = 1 | |
| the right-side directrix equation is | x = a2 / c | |
| the left-side directrix equation is | x = - ( a2 / c ) |
If you want to see a derivation of these results, then click here
To develop useful formulas for the ellipse we use two coordinate
systems -- the usual X-Y rectangular system and also a polar ( r-θ )
coordinate system. We could place the origin of the polar system at
either focus. Let's first derive formulas for the case where the
r-θ
origin is placed at the right-side focus as shown in
Figure 16.0.
The point P will move along the ellipse provided that the
ratio r / PA remains constant.
The constant ratio is
denoted by ε and is called the
eccentricity of the ellipse.
Let ε = r / PA =
the eccentricity of the ellipse.
F is a focus and d is the distance
from F to the closer directrix.
d = r cos θ + PA
Since PA = r / ε,
| d | = | distance from focus to directrix | |
| = | PA + r cos θ | ||
| = | r / ε + r cos θ | ||
| = | r ( 1 / ε + cos θ ) | ||
| εd | = | εr ( 1 / ε + cos θ ) | |
| = | r ( 1 + ε cos θ ) |
| r = ε d / ( 1 + ε cos θ ) |
We can rewrite this in a more useful form by replacing
εd
with the value shown in the 3rd formula
below Figure 17.0 :
|
It's convenient to use special symbols for the radius vector
when θ is either zero or π radians.
| The value of r at θ = 0 is denoted by r0 and | |
| the value of r at θ = π is denoted by r1 [ See Figure 17.0 ] |
Assuming that
| a | = | the semi-major axis | |
| b | = | the semi-minor axis | |
| c | = | distance from the center of the ellipse to a focus |
we have the following "right-side focus" formulas :
| r0 = ε d / ( 1 + ε ) | |
| r1 = ε d / ( 1 - ε ) | |
| ε d = a ( 1 - ε2 ) | |
| r0 = a ( 1 - ε ) | |
| r1 = a ( 1 + ε ) | |
| r1 = a + c | |
| ε = c / a | |
| r0 = a - c | |
| b = a ( 1 - ε2 )1/2 | |
| b = ( r0r1 )1/2 | |
If you want to see how the above ten formulas were derived,
then
click here
An ellipse has two foci and two directrices.
If we had chosen the other focus ( see Figure 18.0 ) to be the origin of
our r-θ coordinate system then the equation for our ellipse would be
given by r = ε d / ( 1 - ε cos θ )
If you would like to see the formulas that apply to
an ellipse where
the origin of the polar coordinates is
placed at the left-side focus
instead of the right-side focus then
click here
| In our discussion of ellipses we define | ||
| a = length of the semi-major axis | ||
| b = length of the semi-minor axis = ( a2 - c2 )1/2 | ||
| c = distance from the Y-axis to either focus | ||
| ε = eccentricity = c / a | ||
| and we always require that | ||
| a > 0 | ||
| 0 <= c <= a | ||
| Usually we restrict ε to the range 0 <= ε < 1 | ||
| If ε = 1 then our "ellipse" degenerates to a straight line. | ||
| When we derived the standard ellipse equation | ||
| x2 / a2 + y2 / b2 = 1 | ||
| we could not allow ε = 1 because that would cause | ||
| b = a ( 1 - ε2 )1/2 = 0 | ||
Let's summarize what happens at the boundary endpoints.
| If | ε = 1 | then | |
| b = 0 ( the "ellipse" is totally flat ) | |||
| c = a ( foci are as far from the Y-axis as possible ) | |||
| The "ellipse" is a line segment of length 2a. | |||
| The directrices are as close as possible to the Y-axis. | |||
| The equation of the right-side directrix is x = a | |||
| The equation of the left-side directrix is x = -a | |||
| See Figure 19.0. | |||
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| If | ε = 0 | then | |
| b = a ( the ellipse is a circle of radius a ) | |||
| c = 0 ( both foci are at the center of the circle ) | |||
| The ellipse is a circle of radius a. | |||
| The directrices are as far away as possible from the Y-axis. | |||
|
The equation of the right-side directrix is
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The equation of the left-side directrix is
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| In Figure 20.0 the left-side directrix is not shown. | |||
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| The bigger the eccentricity, | the flatter the ellipse will be. | |
| The smaller the eccentricity, | the more circular the ellipse will be. |
We can regard any circle as an ellipse of eccentricity zero.
Consider a sequence of ellipses, all having the same fixed major axis,
but with foci that start far apart and then move closer together.
As the
values of c become smaller and the ellipses become more
circular, the
directrices move further away ( "toward infinity" ).
Figure 21.0 shows two of the ellipses in the sequence.
The outer ellipse has smaller eccentricity and is
therefore more circular.
The inner ellipse is flatter; and so,
b2 <
b1
Although the picture does not show any foci, it is also a fact that the
ellipse of smaller eccentricity would have to have its foci closer to
the Y-axis. ( To see a simple analytic proof that
ε1 < ε2
b2 < b1
and
c1 < c2
click here
)
|
Looking at limits gives us another way to see what happens to an
ellipse whose eccentricity is at or near zero. We can easily show
what happens to r as ε approaches zero.
To do this we first need
to recall that
| d | = | distance from right-side focus to right-side directrix | |
| = | distance from left-side focus to left-side directrix | ||
We also need to point out that in previous sections we developed
equations with the origin of the r-θ coordinate system
placed at
either the left or right focus, and in both cases we
found that
| ε d | = | a ( 1 - ε2 ) and | stmt 10 | |
| r | = | ε d / ( 1 ± ε cos θ ) | stmt 20 |
| r | = | a ( 1 - ε2 ) / ( 1 ± ε cos θ ) | stmt 30 |
|
To prove this statement we have to show that for any number
N > 0 ( no matter how large ), there must exist a number δ > 0
such that 0 < c < δ
a2 / c > N
The proof is very easy:
| Let N > 0 be given. | |
| Choose δ = a2 / N | |
| Then | |
| 0 < c < δ |
|
c < a2 / N | |
|
|
c N < a2 | ||
|
|
a2 / c > N | ||
From this it is also clear that
So the right-side directrix is at infinity when the eccentricity is zero.
Similarly, the left-side directrix is at minus infinity when the
eccentricity is zero.
Let A = area of the ellipse shown in Figures 13.0 and 14.0
Since the ellipse has equation
x2 / a2 + y2 / b2
= 1
we can remove the fractions and solve for y to get
| b2x2 + a2y2 | = | a2b2 | |
| a2y2 | = | a2b2 - b2x2 | |
| | y | | = | ( b/a ) ( a2 - x2 )1/2 |
The upper half of the ellipse has equation y = ( b/a ) ( a2 - x2 )1/2
In Figure 25.0 we use this equation and the fact that the ellipse is
symmetric about both axes to find A.
If you want to verify that the formula from the table of integrals
is correct
you will need this formula for the derivative of
the arcsine function:
Dxsin-1U
=
DxU / ( 1 - U2 )1/2
We will skip the verification.
A chord is a straight line that connects
two distinct points on an ellipse.
A chord that passes thru a focu