PHY 3003 SPRING 2015
Solutions 2
(1) A particle of mass m moves subject to a one dimensional potential
x2 x20
U (x) = U0
x20 + x2
2
(here U0 is a constant with dimension of energy and x0 is a constant with dimension
of length).
(a) Sketch U (x), identify all of the equilibrium points and indicate whether they
are stable or unstable. Be sure your sketch shows proper behavior as x → ±∞).
(b) Sketch the phase space orbits, indicating the bound, unbound and separatrix
orbits and giving the energy of the separatrices.
Unstable equilibria
U
U0
0.25
0.20
0.15
0.10
0.05
x
Stable equilibrium
-4
2
-2
4
x0
Unbound orbits
0.6
0.4
0.2
Bound orbits
0.0
-0.2
-0.4
separatrix
-0.6
-4
-2
0
2
4
(c) Sketch all of the orbits with energy equal to U0 /4.
The separatrix orbit has energy U0 /4 and is shown as the dashed line
(2) For the situation described in Taylor 4.36, please answer questions a-c asked in
˙
problem 1 of this set. Note that the coordinate is θ and the velocity is θ.
Here is the potential as a function of angle This problem is a little bit tricky. We
1
2
PHY 3003 SPRING 2015
U
Mgb
1.4
1.2
1.0
0.8
0.6
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
angle
must have M > m for there to be a minimum. If θ gets too small, the mass M
collides with the pulley. We cant have θ > π/2 because in this case the mass would
collide with the support. So it only makes sense to talk about phase space orbits for
energies less than the less restrictive of the two conditions. For this case, the phase
space orbits are closed loops: all orbits are bound.
(3) A particle of mass m moves in one dimension in the half-plane x > 0 subject to the
potential U (x) = A/x2 . If at time t = 0 the particle is at position x = x0 > 0 with
velocity v = 0 please find the subsequent motion x(t).
At the initial time the velocity is 0 so the energy is U (x0 ) so
r s
2
A
dx
A
=
v(x) =
− 2
2
dt
m x0 x
so that
Z
x
t=
x0
dx0 q q
2
m
1
A
x20
−
A
(x0 )2
or
r
t=
m
2A
Z
x
x0
dx0 q
x0
(x0 )2
x20
−1
Doing the integral gives
r
t=
m 2
x
2A 0
s
x2
−1
x20
PHY 3003 SPRING 2015
3
Solving for x gives
s
x = x0
2t2 A
+1
mx40
Sketch the potential and the phase portrait
1.0
Ux02/A
0.5
0.0
-0.5
-1.0
0
1
2
3
4
5
x/x0
(4) 4.32 (a)
s
|~v | =
=
dx
dt
2
+
dy
dt
2
+
dz
dt
2
p
dx2 + dy 2 + dz 2
dt
p
but dx2 + dy 2 + dz 2 is just the line element ds
(b) We have
d|v|2
d|v|
d
= 2|v|
= 2|v| s˙ = 2|v|¨
s
dt
dt
dt
but also
d (~v · ~v )
d~v
F~
= 2~v ·
= ~v ·
dt
dt
m
~
F
and ~v · m
is 1/m times the tangential component of F~ . Putting these together gives
the desired result.
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PHY 3003 SPRING 2015
(c) The force from the conservative part of the system is
~
F~ = −∇U
The tangential component of this is the negative of the change in U for an infinitesimal displacement along the wire which is −dU/ds.
(5) 4.36 The potential energy is (note the H and h are measured downwards from
support plane.
U = −M gH − mgh
we have that h = b cotθ while the diagonal portion of the string has length b/sinθ.
Given that H + b/sinθ = l is constant we have H = l − b/sinθ so
M gb
− mgbcotθ
U = −M gl +
sinθ
Equilibrium requires dU/dθ = 0 or
dU
M gbcosθ
mgb
=−
+
dθ
sin2 θ
sin2 θ
m
which is possible if m < M .
which can vanish if cosθ = M
The second derivative is
d2 U
M gb 2M gbcos2 θ 2mgbcosθ
=
+
−
dθ2
sinθ
sin3 θ
sin3 θ
Evaluating this at the minimum position gives qM gbm2 > 0 so the minimum is
1−
M2
stable.
(6) 4.52
We have a system with total KE given by
m1~v12 m2~v22 m3~v32 m4~v42
+
+
+
2
2
2
2
The change in kinetic energy during an infinitesimal time interval dT is then
d~v1
d~v2
d~v3
d~v4
d(KE) = m1~v1 ·
+ m2~v2 ·
+ m3~v3 ·
+ m4~v4 ·
dT
dt
dt
dt
dt
or, using Newton’s equation
d(KE) = ~v1 · F~1 + ~v2 · F~2 + ~v3 · ·F~3 + ~v4 · F~4 dT
KE =
which is just the work.
Next, the change in the potential energy is just
dU d~r2
dU d~r3
dU d~r4
dU d~r1
·
+
·
+
·
+
·
dU (~r1 , ~r2 , ~r3 , ~r4 ) =
d~r1 dt
d~r2 dt
d~r3 dt
d~r4 dt
Noting that the force is the negative of the gradient of the potential while the derivative of position with respect to time is the velocity gives the desired result.