bradley j. nartowt Saturday, July 06, 2013, 09:32:04
PHYS 6246 – 6246 – classical classical mechanics Dr. Whiting Whiting
The Foucault pendulum experiment consists in setting a long pendulum in motion at a point on the surface of the rotating earth with its momentum originally in the vertical plane containing the pendulum bob and the point of suspension. Show that the pendulum’s subsequent motion may be described by saying that the pla ne of oscillation rotates uniformly co-latitude. 2 cos radians per day, where is the co-latitude. The equations of motion are the same as for a projectile, except with the constraint that the pendulum moves in the z = 0 F T mbob
plane, and and that you you have x and y components components of pendulum-a pendulum-arm-tension rm-tension acting acting as
a ( x, y ) g ( x, y) . This
effectively is a consideration of small oscillations. r
x, y ,
d 2 ( z 0 ) dt 2
d ( z 0) dt
sin y cos
g
x, x cos
g
y , 0
x y cos g x y x cos g y
[I.1]
Here, is the co-latitude (the effect disappears and reverses direction when one nears and then crosses the equator). This is the whole point of designing the focault pendulum with, say, a 67-meter arm — arm — to to get motion in the xy-plane free enough such that such a tiny force as the Coriolis force can be detected, and exaggerate the motion by large swings. Isotropy: Now, we have a coupled system [I.1]; to decouple it, write x , and put in y ; then compute x , and put in y ,
x cos y g x cos ( x cos g y )
g
x g
x x
x cos ( y x cos ) x cos ( x cos ( cos )) x ( cos 2 ) x g
g
g
Notice that that the equations equations of motion [I.1] are symmetric symmetric under x y and g y cos x g y cos ( y co c os x )
y ( cos ) y 2
2
g
g cos
g
2
2
g
2
g 2 2
[I.2]
x
; therefore, we should have,
y
x ( cos ) y 2
g
g
g cos
g
(
y y cos
) ( cos 2 ) y ( ) y 2
2
g
g 2
[I.3]
This is the exact same equation of motion, so any motion in y-space will be true in x-space as well. However , we will not solve these equations. We only want to establish homogeneity of the motion from these equations. Isotropic hooke’s law from homogeneity: Homogeneity motivates us to switch to xy-plane polar coordinates. Compute
the combination xx yy from the coupled [I.1], and use the identity
yx xy x2 y 2
d dt
tan1 ( xy )
d dt dt
; also note that from
2 2 ( x, y ) r (cos , si sin ) we can get xx yy rr r . Incorporating all this,
xx yy cos ( xy xy) g ( x2 y 2 ) cos ( x2 y 2 ) dtd tan1 xy g r2 ( cos g )r 2 rr r 2 2 [I.4] Dividing out by r, we now have an isotropic Hooke’s law,
for this inhomogeneity in ( cos g )r r r 2 , except
. Only appears, and going to a rotating frame: Since only first derivatives of appear, we infer that
now go to a rotating frame of reference where
0,
we have just Hooke’s Law alone,
g
2 g g r r , a stand-alone-true-statem stand-alone-true-statement, ent, from ( cos )r r , we are left with
0.
If we
r r . If we subtract this
r cos r 2 cos , a promising-looking milestone. Meanwhile, the solution to the isotropic hooke’s law is r (t ) A cos(t ) B sin(t ) for g /
. Averaging this over 1 day (note that 1 day corresponds to T
r cos cos r cos
1
2
sin2 cos2 dt 2 cos
0
1
[I.5]
An alternate solution is in many textbooks, but the treatment is cumbersome. One source instructs us to multiply the second equation in [I.1] by i and add them together to get rotational ansatz. Guessing Ae
i t
2i cos
g
0 , where x iy , a sort of
then gives an equation for the guessed frequency , which is the frequency of
x iy ; this leads to a quadratic with important roots,
2 2 cos
g
0 cos
This gives us an equation of motion (t ) e
i cos t
( Aei ( g /
)t
g
2 cos2 cos g O( 2 )
[I.6]
Be ( g / ) t ) . Real and imaginary parts of this equation, by i
construction, yield x(t ), y (t ) , meaning, r(t )
A cos( cos t) B sin( cos t) x B cos( cos t) A sin( cos t) y ˆ
ˆ
[I.7]
What is the direction of rotation? By looking at [I.7] (a rotation matrix for the + direction) or my derived cos , one can see that you are rotating counterclockwise in the xy-plane. Since the coordinate system matches that for which we fired projectiles in (see problem 4.23) refer to there in order to interpret which cartesian directions are north, south, east west.
