Transcript Solution - Physics at Oregon State University

Classical Mechanics Homework 5
Due October 31, 2014
1. Problem 1: Taylor 6.22
You are given a string of fixed length l with one end fastened at the origin
O, and you are to place the string in the x-y plane with its other end
on the x-axis in such a way as to enclose the maximum area between the
string and the x axis. Show
∫ that the required shape is a semicircle. The
are enclosed is of course ydx, but show that you can write this in the
∫l
form 0 f ds, where s denotes the distance measured along the string from
√
O, where f = y 1 − y ′ 2 , and y ′ denotes dy
ds . Since f does not involve
the independent variable s explicitly, you can exploit the ”first integral”
(6.43) of Problem 6.20.
If there are loops in the string, we can always enlarge the area by unfolding
the loop and stretching the string further. So we assume that there are
no loops, and that there is therefore a unique correspondence between the
distance s along the string and the distance x along the x-axis. The area
is found by integrating over x from zero to the unknown point xf ,
∫ xf
A=
ydx
0
which we can convert to an integral over s via
∫ l
dx
A=
y ds
ds
0
We also have for the length along the curve
ds2 = dx2 + dy 2
which with y ′ =
dy
ds
gives
ds2 = dx2 + y ′ ds2
2
or
dx2 = (1 − y ′ )ds2
2
which gives
∫
l
A=
y
√
1 − y ′ 2 ds
0
as we were supposed to show. In other words, we have
∫ l
A=
L(y, y ′ , s)ds
0
with
′
L(y, y , s) = y
√
1 − y′ 2
which assumes that we want to have y as a function of s. The EulerLagrange equation involves:
√
∂L
= 1 − y′ 2
∂y
and
∂L
yy ′
= −√
′
∂y
1 − y′ 2
√
d
yy ′
√
=0
1 − y′ 2 +
ds 1 − y ′ 2
and hence
√
2
2
yy ′
y ′ + yy ′′
+
1 − y′ 2 + √
3 = 0
(1 − y ′ 2 ) 2
1 − y′ 2
Multiply by (1 − y ′ ) 2 to get
2
3
(1 − y ′ )2 + (1 − y ′ )(y ′ + yy ′′ ) + yy ′ = 0
2
2
2
2
1 − y ′ + (1 − y ′ )yy ′′ + yy ′ = 0
2
2
2
yy ′′ = −1 −
yy ′
1 − y′ 2
2
This turns out to be harder, so the hint in the problem might be useful.
Using 6.43 we get
√
yy ′
y 1 − y′ 2 + y′ √
=C
1 − y′ 2
√
2
2
y(1 − y ′ ) + yy ′ = C 1 − y ′ 2
√
y = C 1 − y′ 2
y 2 = C 2 (1 − y ′ )
2
Does this describe a semicircle? For a semicircle we can use the angle θ
and relate the angle to the distance along the string,
s=l
we have
θ
π
s
y = R sin(θ) = R sin( π)
l
R
s
y ′ = π cos( π)
l
l
and hence
s
R
s
R2 sin2 ( π) = C 2 (1 − ( π)2 cos2 ( π))
l
l
l
This can be solved only if l = Rπ, which is the correct distance along the
string for a given radius, and C = R. Therefore, the string forms a semicircle.
Evaluation In order to apply the principle of the extremum of the action,
we need to integrate over a fixed length. Therefore, we need to integrate
along the string. This is possible, because we can argue that there will
be no loops in the string. If we know y(s) we know everything about the
curve, the only thing is that we need to transform dx into ds. We have
done all these steps. This is just a description of what we have done. The
result makes sense, because we know that a circle has the optimal ratio of
circumference to area. Therefore, we expect our result to be a semi-circle.
2. Problem 2: Taylor 6.23
An aircraft whose speed is v0 has to fly from town O (at the origin) to
town P , which is a distance D due east. There is a steady gentle wind
shear, such that ⃗vwind = V y x
ˆ, where x and y are measured east and north
respectively. Find the path, y = y(x), which the plane should follow to
minimize its flight time, as follows: (a) Find the plane’s ground speed in
terms of v0 , V , ϕ (the angle by which the plane heads to the north of east),
and the plane’s position. (b) Write down the time of flight as an integral
∫D
of the form 0 f dx. Show that if we assume that y ′ and ϕ both remain
small (as is certainly reasonable if the wind speed is not too large), then
2
the integrand f takes the approximate form f = (1+ 21 y ′ )/(1+ky) (times
an uninteresting constant) where k = V /v0 . (c) Write down the EulerLagrange equation that determines the best path. To solve it, make the
intelligent guess that y(x) = λx(D − x), which clearly passes through the
two towns.
Show that it satisfies the Euler-Lagrange equation, provided
√
λ = ( 4 + 2k 2 D2 −2)/(kD2 ). How far north does this path take the plane,
if D=2000 miles, v0 =500 mph, and the wind shear is V =0.5 mph/mi? How
much time does the plane save by following this path? [You’ll probably
want to use a computer to do this integral.]
The speed of the plane is given by
vx = v0 cos(ϕ) + V y , vy = v0 sin(ϕ)
The time of flight follows from the integration along the path
∫ D
∫ D
dt
dt ds
T =
dx =
dx
dx
ds
dx
0
0
The speed along the path is v =
(
ds
dt ,
which follows from
ds 2
) = vx2 + vy2 = v02 + 2V v0 y cos(ϕ) + V 2 y 2
dt
We also have
ds2 = dx2 + dy 2 = (1 + y ′ )dx2
2
and hence
∫
D
T =
0
1
v
√
1 + y ′ 2 dx
When the path is close to the direct path, we can write
v 2 ≈ vx2 + vy2 = v02 + 2V v0 y = v02 (1 + 2ky)
In this case the derivative is also small and we have
√
1 2
1 + y′ 2 ≈ 1 + y′
2
and
v ≈ v0 (1 + ky)
which leads to the required result
1
T =
v0
∫
D
0
We set
1 + 12 y ′
dx
1 + ky
2
1 + 12 y ′
1 + ky
2
F (y, y ′ ) =
and need to solve
∂F
d ∂F
−
=0
∂y
dx ∂y ′
which leads to
1 + 12 y ′
d
y′
−
=0
2
(1 + ky)
dx 1 + ky
2
−k
2
1 + 12 y ′
y′
y ′′
+
k
−
=0
(1 + ky)2
(1 + ky)2
1 + ky
2
−k
1 2
−k + ky ′ − (1 + ky)y ′′ = 0
2
Now we try y(x) = λx(D − x), which gives y ′ = λ(D − 2x) and y ′′ = −2λ.
Plugging in gives
1
−k + kλ2 (D − 2x)2 − (1 + kλx(D − x)(−2λ) = 0
2
which needs to be true for all values of x!
1
−k + kλ2 D2 + 2λ + x(−2Dkλ2 + 2kλ2 D) + x2 (2kλ2 − 2kλ2 ) = 0
2
This is good, the terms in x and x2 have zero coefficients, so we just need
to require that
1 2 2
kλ D + 2λ − k = 0
2
which has solutions
√
−2 ± 4 + 2k 2 D2
λ=
kD2
and since we need λ > 0 we take the plus sign. More on this later. The
furthest north we get is half way, so
ymax =
1
λD2
4
Note that for negative values of λ we slow down, and at the furthest point
we have
1
1
v = v0 + V ymax = v0 + V λD2 = v0 (1 + kλD2 )
4
4
and for the negative solution of λ this is negative. We have past the distance south for which the speed is zero and the time has become infinitely
large! This part of the solution has no longer any physical meaning.
0.5
In our case we have D=2000 miles, k= 500
=0.001 per mile, kD=2, and
√
4+8−2
hence λD= 2 =0.73 which means that we went north 41 (0.73)D =
0.18D=360 miles. This number is a bit large for the small deviation approximation, but it is not too bad.
The normal time for the direct flight is 4 hours. The time along the path
is
∫ D
1 + 21 λ2 (D − 2x)2
1
T =
dx
v0 0
1 + kλx(D − x)
Use x = sD and we get
T =
T =
D
v0
D
v0
∫
1
0
∫
0
1
1 + 12 λ2 D2 (1 − 2s)2
ds
1 + kD2 λs(1 − s)
1 + 12 (0.73)2 (1 − 2s)2
ds
1 + 2(0.73)s(1 − s)
Using Maple we find
T = 0.93
D
v0
which gives T equals 3 hours and 43 minutes. We saved 17 minutes.
Evaluation We need to make sure that all units are correct. From the
definitions we see that V is not a speed, but a speed change per length, so
the dimensions of V is 1/T. This gives that the dimensions of k is 1/L.
For the path we have that the dimensions of λ is also 1/L. So both kD and
λD are dimensionless numbers. This is consistent with our final integral
for the time, the dimensions are all in the prefactor vD0 . The average angle
of the first half of the flight follows from ymax = 12 D sin(ϕav ) which gives
ϕav about 0.36 radians, which is not small. Error terms go like the square
of this number, so we expect answers to be wrong by about 10 per cent.
3. Problem 3: Taylor 6.24
Consider a medium in which the refractive index n is inversely proportional
to r2 ; that is, n = a/r2 , where r is the distance from the origin. Use
Fermat’s principle, that the integral (6.3) is stationary, to find the path
of a ray of light travelling in a plane containing the origin. [Hint: Use
two-dimensional polar coordinates and∫write the path as ϕ = ϕ(r). The
Fermat integral should have the form f (ϕ, ϕ′ , r)dr, where f (ϕ, ϕ′ , r) is
actually independent of ϕ.The Euler-Lagrange equation therefore reduces
to ∂f /∂ϕ′ = const. You can solve this for ϕ′ and then integrate to give ϕ
as a function of r. Rewrite this to give r as a function of ϕ and show that
the resulting path is a circle through the origin. Discuss the progress of
the light around the circle.]
∫2
Fermat’s principle states that 1 n(r, ϕ)ds is a minimum. In polar coordinates we have
ds2 = dr2 + r2 dϕ2
which leads to
ds2 = (1 + r2 ϕ′ )dr2
2
. Hence we need to minimize
∫ 2
1
a
r2
√
1 + r2 ϕ′ 2 dr
which gives
√
a
f (ϕ, ϕ , r) = 2 1 + r2 ϕ′ 2
r
This is indeed independent of ϕ. Euler-Lagrange then simplifies to ∂f /∂ϕ′ =
const or
a
r 2 ϕ′
√
=C
r2 1 + r2 ϕ′ 2
′
This gives
a2 ϕ′ = C 2 (1 + r2 ϕ′ )
2
2
C2
− C 2 r2
C
ϕ′ = ± √
2
a − C 2 r2
Take a path in which ϕ increases with r. Integration gives
ϕ′ =
2
a2
ϕ = arcsin(
Cr
)+D
a
where D is another constant.Inverting this equation gives
Cr
= sin(ϕ − D)
a
a
sin(ϕ − D)
C
What is the equation for a circle through the origin in polar coordinates?
Supposes the origin of the circle is at (0, R) and the angle θ measures the
angle along the circle, starting at the bottom. We then have
r=
x = R sin(θ)
y = R − R cos(θ)
This gives
r2 = x2 + y 2 = 2R2 − 2R2 cos(θ)
r2 = 2R2 (1 − cos(θ))
θ
r2 = 2R2 2 sin2 ( )
2
θ
r = 2R cos( )
2
which has the form we have with ϕ = 12 θ. This is indeed the relation
between ϕ and θ. So our path is part of a circle through the origin.
Evaluation The first thing to notice is that the actual path of the light
will never go through the origin, because that would take an infinite time.
At the origin the integrand diverges like 1/r2 , which cannot be integrated.
So if we take two points, we construct a circle through these two points
and the origin, and the path of light takes the part of the circle not going
through the origin. Second, if light goes from a region with a lower index
of refraction to one with a higher one, it curves towards the normal. In
our example the normals are lines coming out of the origin, and a circle
through the origin has indeed the correct form, bending towards the origin
at very small distance. The curvature of the path is correct.