Goldstein, Classical Mechanics Second Edition Problem 2-4: Find the Euler-Lagrange equation describing the brachistochrone curve for a particle moving inside a spherical Earth of uniform mass density. Obtain a first integral for this differential equation by analogy to the Jacobi integral h. With the help of this integral show that the desired curve is a hypocycloid. Obtain an expression for the time of travel along the brachistochrone between two points on the surface of the Earth. How long would it take to go from New York to Los Angeles (assumed to be 3000 miles apart) along a brachistochrone tunnel (assuming (assuming no friction) and how far below the surface would the deepest point of the tunnel be? This problem is vaguely reminiscent of a standard introductory physics problem (e.g., Halliday & Resnick, p. 351) involving a a straight line path through the Earth connecting two points on the surface. This problem with an undetermined path is more general and correspondingly more difficult. Let R and M represent the radius and mass of the Earth, respectively. Let r and m represent the radial distance of the traveling object from the center of the Earth and the mass of the traveling object, respectively. From introductory physics we know that for r > R, U = –GMm/r –GMm/r 2
3
3
For 0 < r < R, F = –GMm/r –GMm/r (r /R ); that is, the force behaves as if all the mass of the Earth is concentrated at r = 0 3
2
U = – = – F dr = (GMm/2R )r – C' 3
2
Need to ensure continuity at r=R; therefore (GMm/2R )r – C' = – = –GMm/R, GMm/R, which tells us that C = 3GMm/2R 3
2
U = (GMm/2R )r – 3GMm/2R v=0r=R E = U(R) = – = –GMm/R GMm/R 2
3
2
-GMm/R = ½mv + (GMm/2R )r – 3GMm/2R 2
3
2
GMm/R = mv + (GMm/R )r
Note that as we would expect in this type of problem, the particle mass m drops out and is irrelevant to our solution. 2
2
2
2
2
v = (GM/R)(R /R – r /R ) 2
2
2
½
2
2
½
v = [(GM/R )(R – r )/R] = [(g/R)(R – r )] = ds/dt 2
2 ½
Integating, dt' = (R/g) ds'/ (R – r' )
Redefining dt' = (R/g) dt and ds' = Rds and r'=Rr 2 ½
t = ds / (1 – r )
2
2 ½
ds = [(x') + (y') ] dt (Goldstein, p. 40) 2
2
2
r =x +y 2
2 ½
2
2 ½
t = [(x') + (y') ] / (1 – x – y ) dt 2
2 ½
2
2 ½
f = [(x') + (y') ] / (1 – x – y ) There is no dependence on t. f – x' f/x' = c f – y' f/y' = c 2
2 ½
2
2 ½
2
2 ½
2
2 ½
2
2 ½
2
2 ½
2
2 ½
2
2 ½
2
2
2
2 ½
[(x') + (y') ] / (1 – x – y ) – x' {x' / [(x') + (y') ] } {1 / (1 – x – y ) } = C1 [(x') + (y') ] / (1 – x – y ) – y' {y' / [(x') + (y') ] } {1 / (1 – x – y ) } = C2 2
2
2 ½
2
2
2 ½
2 ½
2
2
2 ½
[(x') + (y') – (x') ] / {(1 – x – y ) [(x') + (y') ] } = y' / {(1 – x – y ) [(x') + (y') ] } = C1 2
2
2 ½
Similarly, x' / {(1 – x – y ) [(x') + (y') ] } = C2 Adding these two results together, we get 2
2
2
2 ½
2
2 ½
2
2 ½
2
2
2
2 ½
[(x') + (y') ] / {(1 – x – y ) [(x') + (y') ] } = C1 + C2 [(x') + (y') ] / (1 – x – y ) = C 2
2
2
[(x') + (y') ] = C (1 – x – y ) The solution to this equation can be found in any introductory calculus text (e.g., Anton) x(t) = (1-b) cos t + b cos [(1-b)/b t] y(t) = (1-b) sin t – b sin [(1-b)/b t] These are the parametric equations for a hypocycloid.
2
2
2
2
2
2
2
2
2
2
x = (1-b) cos t + b cos [(1-b)/b t] + 2b(1-b) cos t cos [(1-b)/b t] y = (1-b) sin t + b sin [(1-b)/b t] – 2b(1-b) sin t sin [(1-b)/b t] 2
2
2
2
2
2
x + y = (1-b) + b + 2b(1-b) cos (t/b) = 1 – 2b + b + b + 2b(1-b) cos (t/b)
2
2
x + y = 1 – 2b(1-b)(1 – cos t/b) 2
2
1 – x – y = 2b(1-b)(1 – cos t/b) x' = –(1-b) sin t + (1-b) sin [(1-b)/b t] 2
2
2
2
2
2
(x') = (1-b) sin t – 2(1-b) sin t sin [(1-b)/b t] + (1-b) sin [(1-b)/b t] y' = (1-b) cos t + (1-b) cos [(1-b)/b t] 2
2
2
2
2
2
(y') = (1-b) cos t – 2(1-b) cos t cos [(1-b)/b t] + (1-b) cos [(1-b)/b t] 2
2
2
2
2
2
(x') + (y') = (1-b) – 2(1-b) cos (t/b – 2t) + (1-b) = 2(1-b) (1 – cos t/b) 2
2
2
2
2
2
C = [(x') + (y') ] / (1 – x – y ) = 2(1-b) (1 – cos t/b) / 2b(1-b)(1 – cos t/b) = (1-b)/b
The starting and ending points are at r=0 and r=1 in our new u nits 2
2
2
1 = r = x + y = 1 – 2b(1-b)(1 – cos t/b) 1 – cos t/b = 0 t = 2b ½
Transit time = C dt = [(1-b)/b] t = 2[b(1-b)]
½
Based on the geometric properties of the hypocycloid = d/R
= 2b in general Therefore b = d/2R ½
Transit time = 2 [(d/2R)(1 – d/2R)] Reintroducing the factor of (R/g) ½
½
Transit time = 2 (R/g) [(d/2R)(1 – d/2R)]
½
And finally plugging in all the numbers including R 4000 km ½
½
Transit time = 2 (4000000/9.8) [(3000/8000)(1 – 3000/8000)] Transit time 1300 s 22 min
Based on the geometry the diameter = 2bR = d/ is the maximum depth, which is approximately equal to 955 km
Goldstein, Classical Mechanics Second Edition Problem 2-5: In the problem of the minimum surface of revolution examine the symmetric case y2 = y1, x2 = -x1 > 0, and express the condition for the parameter a as a transcendental
equation in terms of the dimensionless quantities k = x 2/a and = y2/x2. Show that for greater than a certain value 0 two values of k are possible, for = 0 only one value of k is possible, while if < 0 no real value of k (or a) can be found, so that no catenary solution exists in this region. Find the value of 0, numerically if necessary. (x1, y1) = (-x2, x2) and (x2, y2) = (x 2, x2) y = a cosh ((x-b)/a) In the symmetrical case y = a cosh (x/a)
x2 = a cosh (x2/a) = (a/x2) cosh (x2/a) = cosh k / k The minimum value occurs when k = 1.2, at which point = 0 ≈ 1.5089 as seen from the graph below: 1.550 1.545 1.540 1.535 1.530 1.525 1.520 1.515 1.510 1.505 1
1.05
1.1
1.15
1.2
1.25
1.3
1.35
1.4
1.45
1.5
Figure 1 - Graphical solution for alpha0 = 1.5089 at which value k = 1.2 For = 0 there is only one solution k = k0 = 1.2 For > 0 there are two solutions for k and for < 0 there are no real solutions, as seen in the graph below:
12
10
8
6
4
2
0 0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Figure 2 - illustrating that two solutions for k are possible whenever alpha > alpha0
Goldstein, Classical Mechanics Second Edition Problem 2-6: The broken segment solution described in the text (p. 42) in which the area of revolution is only that of the end circles of radii y 1 and y2 respectively is known as the Goldschmidt solution. For the symmetric situation described in exercise 2.5 obtain an expression for the ratio of the area generated by the catenary solutions to that given by the Goldschmidt solution. Your result should be a function only of the parameters k and . Show that for sufficiently large values of at least one of the catenaries gives an area below that of the Goldschmidt solution. On the other hand show that if = 0 the Goldschmidt solution gives a lower area than the catenary. 2
2
2
2
For the Goldschmidt solution, S = π(y2 + y2 ) = 2π x2 For the symmetrical catenary solution y = a cosh (x/a) y’ = sinh (x/a) 2 ½
*1 + y’ ] = cosh (x/a)
For = 0 ≈ 1.5089, k = 1.2, and the ratio = 3.279/3.933 ≈ 0.834 so the Goldschmidt solution yields a smaller area. α
k1
k2
Ratio1
Ratio2
1.5089
1.2
1.5168
1.1
1.305
0.836
0.836
1.5431
1.0
1.425
0.846
0.843
1.5923
0.9
1.559
0.866
0.856
1.6718
0.8
1.714
0.900
0.873
1.7931
0.7
1.893
0.954
0.893
1.8955
0.639
2.016
1.000
0.907
1.9758
0.6
2.102
1.037
0.916
2.2553
0.5
2.353
1.169
0.939
2.7027
0.4
2.661
1.385
0.960
3.4845
0.3
3.057
1.767
0.977
5.1003
0.2
3.604
2.567
0.990
10.0500
0.1
4.506
5.033
0.998
0.834
The values in the table show that for 0 ≤ < 1.8955 the Goldschmidt solution yields a smaller area. For 1.8955 ≤ exactly one catenary yields a smaller area. The larger value of k always yields a larger area than the Goldschmidt solution.
Goldstein 2nd Edition Problem 2-9 Reference: Goldstein, page 20
Goldstein says L includes all non-impulsive forces, so Q j is the generalized impulsive force
t 0 first term 0
Goldstein 2
nd
Edition Problem 2-13
(Wheatstone bridge circuit diagram obtained from Wikipedia) Using our knowledge of circuits from Physics II, we would obtain the equations
With some algebra this can be expressed as
However, the problem requires the use of t he method of Lagrange multipliers
The dissipative force is
Reference: Goldstein, pages 24 and 52
Balanced bridge means i1 = i2 = 0 and we get the usual network equations.
Goldstein 2
nd
Edition Problem 2-15
where {ri} forms a basis
If the forces are electromagnetic in character, then
Goldstein 2
nd
Edition Problem 2-16
The Lagrange equations are separate.
The problem can be reduced to quadratures.
nd
Goldstein 2 Edition Problem 2-20 Length of second spring << r 0 m can be considered to be at the end of first string T = ½mv = ½m r = ½m (r0+x1) 2
2 2
2
2
(a) T = ½m (r0+x1) + ½m + ½m 2
2
where in the last term we have again approximated because x 2 << r0 2
2
V = ½Kx1 + ½kx2
The first two terms are conserved, but energy E is not co nserved because the last term varies; this is because of the power expended in keeping constant.
(b) Laboratory system , r=r0+x1, x2
This is not conserved because the
term varies with x 1
(c) Rotating system x1, x2
This is conserved. The relation between the two Jacobi integrals is the energy related to keeping the system rotating at constant frequency that appears in b but not in c.
Goldstein 2
nd
Edition Problem 2-21
Reference: Equation 2.29
Force of constraint is always perpendicular, hence
L = T(p) – V(q)
In vector notation
E is conserved iff
, i.e. the surface doesn’t move.
The moving surface imparts energy to the system.
