s equations,
˙q=∂H /∂p =p/mr2
, and
˙p=−∂H /∂q =−mGr sin q
. Thus,
¨q=
˙p/mr2=−Gsin q/r
. (Again, factors of
r
require adjustment.)
5
On a given trajectory,
H(q, p)
takes some constant value
E
. Since
cos q≤1
, I have
p2= 2mr2(E+mGr cos q)≤2mr2(E+mGr)
,
or
−r
È
2m(E+mGr)≤p≤r
È
2m(E+mGr)
.
Note that in order for this to make sense, I must have
E≥ −mGr
, so energies strtictly less than
−mGr
are inaccessible. (This makes sense physically, since
−mGr
is the potential energy when the mass is at
its lowest position.) Similarly, since
cos q≥ −1
, I have
p2≥2mr2(E−mGr)
. Thus,
p= 0
will never oc-
cur when
E > mGr
. Remembering that
E < −mGr
is impossible, then, there are
2
broad types of mo-
tion and
2
degenerate energies. When
|E|< mGr
, the motion will pass through
p= 0
but will never reach
cos q=−1
. Physically, this corresponds to an oscillation about the bottom position. When
E > mGr
, in
contrast, the motion will pass through
cos q=−1
but will never reach
p= 0
. Physically, this corresponds
to traversing the entire conguration space. One degenerate energy is
E=−mGr
, where the motion re-
mains at
cos q= 1
and
p= 0
, a stable equilibrium. Physically, this corresponds to remaining in the bot-
tom position. The other degenerate energy is
E=mGr
. In this case, we could have
p= 0
and
cos q=−1
,
another equilibrium. However, this is not required; we could have
p > 0
and
cos q > −1
, or alternatively
p < 0
and
cos q > −1
. This trajectory would approach the above equilibrium but never reach it. There-
fore, this equilibrium is unstable. Physically, the unstable equilibrium corresponds to balancing at the top
position, as indicated in digression 2, and the degenerate trajectory approaching this equilibrium corre-
sponds to ever more slowly climbing to the top. A graph illustrating this is on the next page.
6
As already noted,
p2= 2mr2(E+mGr cos q)
. Since
˙q=p/mr2
, it follows that
˙q2= 2(E+mGr cos q)/mr2
,
so
˙q=±
È
2(E+mGr cos q)/mr2
, which is equation (1) with the correct factors of
r
.
7
Of course, we really want
mGr = 1
, not so much
mG = 1
. (Note that the product
m2r2G
of these units
gives a scale for
p2
, as can be seen on the graph for problem 5, while the quotient
G/r
gives a scale for
t2
.) Then the previous problem gives
˙q2= 2(E+ cos q)
. Since
dt =dq/˙q
, equation (2) follows if I use the
principal square root. Note that this will break down when
p
changes sign, which can happen when
|E|<
1
in the new units (the oscillating case).
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