It is not clear whether one can consider the centrifugal force (a fictious force existent only
in non-inertial frames) a conservative force. The more rigorous solution, if one were to be
careful, would need to show this by calculating the work done by the
In the second diagram, the potential falls off to . Thus at r = 0 the particle has infinite
kinetic energy. Classically we allow this, and the particle still comes back from the abyss.
Observe the addition of the extra term to the Kepler potential. It has
This looks awkward. Conceptually speaking, the elliptical trajectory is because of the fact
that the effective force is a vector parallel to the relative position and linearly dependant on
the radius. To go deeper into abstract significances, this is a co
7.52 a: As the string unwinds, it is clear that :1: : Rq, so the constraint equation is
The Lagrangian is E = érririt2 + £1952 + mga: and the two modied Lagrange equations are
81: 3f _ 0! BE _ .. ..
55+ABEE-5E => mg+Am1:t (xxu)
6C 8 f d
In part (e), we are essentially Taylor expanding f() to first order around 0:
= 0 + 1! 0 0 +
We can check that f '(0) is indeed negative.
3cos 2 0
In the 2 = 2gk case, it may not be obvious why t
Note: The first substitution is done as follows.
Using cos( 2) = cos(2) = 2sin2 1 = 1 2cos2 and 0 = ( 0) we write:
[1 cos 2 ]
2 = 2
g[cos 20 cos 2 ]
1 1 + 2 cos 2
2 sin2 0 12sin2 +1
Comment: We simply verified the cycloidal solution in this p
Comment: It may seem like this solution does not quite use any calculus of variations. But
one must remember that in calculating the total time above, were effectively integrating the
inverse speed of light along the path traversed by the ray. To do this
The Perpendicular Axis Theorem states that if a 2-dimensional object lies in the x-y plane,
its moment of inertia around the z-axis (perpendicular to the plane of the object) is given by
Izz = Ixx + Iyy. The proof is by definition, since for all points on