PHY4222: CLASSICAL MECHANICS
FINAL EXAM: SOLUTIONS
30 April, 2007
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
Every (algebraic) final result must be supplemented by a check of units. Without such a check, no
more than 75% of the credit will be given even for an otherwise correct solution. On the other hand,
an answer obtained just by the dimensional analysis may give you up to 25% of the credit.
1. A bead of mass m slides without friction on a wire which connects the origin (O) with the point (P ). The bead
starts at origin with an initial speed v0 .
a) Obtain an integral equation for the time of descent. [15 points]
Energy conservation
q
v (x) = v02 + 2gx
Arclength
ds =
Time of descent
t=
Z
xp
0
p
p
dx2 + dy 2
1 + y 02
1
dx p 2
=√
2g
v0 + 2gx
Z
xp
0
p
dx √
1 + y 02
,
x0 + x
where x0 = v02 /2gx. Notice that by a shift of variable ξ = x0 + x, the equation for the time reduces to
Z xp +x0 p
1 + y 02
1
t= √
dξ √
,
2g x0
ξ
where now y 0 = dy/dξ. Therefore the problem reduces to the classic brachistochrone problem with zero
initial speed.
b) Use the calculus of variations to minimize the time of descent and obtain the differential equation for the
shape of the wire. [15 points]
We need to find a minimum of the integral (a constant prefactor does not matter)
Z xp +x0 p
1 + y 02
,
dx √
I=
ξ
x0
so that
F (y 0 , y; x) =
p
1 + y 02
√
.
ξ
Variational equation
∂F
d ∂F
=
=0
dξ ∂y 0
∂y
1
∂F
= √
∂y 0
2a
y0
1
1
p
√ = √
2a
1 + y 02 ξ
2
s
ξ
2a − ξ
s
Z
ξ
y =
dξ
2a − ξ
0
y =
c) Find the shape of the wire in a parametric form x = x (θ) , y = y (θ). Sketch this shape. [20 points]
Substitution
ξ = a(1 − cos θ) = 2a sin2 (θ/2)
dξ = a sin θdθ
2a − ξ = a(1 + cos θ) = 2a cos2 (θ/2)
√
Z
Z
2a sin (θ/2)
= a dξ sin θ = a (θ − sin θ) + y0 ,
y = a dξ sin θ √
2a cos (θ/2)
where y0 is a constant. Restoring the original x = ξ − x0 , we find the parametric equations of the wire
v02
+ a (1 − cos θ)
2g
y = y0 + a (θ − sin θ) .
x = −
Parameters a and y0 are to be found from the boundary conditions. The wire is a shifted cycloid.
2. A point of support of a simple pendulum of mass m and length ` moves uniformly in a vertical circle of radius
R with a constant angular velocity ω.
a) Construct the Lagrangian for this system. [15 points]
x = ` sin φ + R cos ωt
y = ` cos φ + R sin ωt
ẋ = lφ̇ cos φ − Rω sin ωt
ẏ = −`φ̇ sin φ + Rω cos ωt
Kinetic energy
1
m 2
1
ẋ + ẏ 2 = m`2 φ̇2 + mω 2 R2 − mR`φ̇ω [cos φ sin ωt + sin φ cos ωt]
2
2
2
1 2 2 1
2 2
= m` φ̇ + mω R − mR`φ̇ω sin (φ + ωt) .
2
2
T =
Potential energy
U = −mgy = −mg (` cos φ + R sin ωt)
Lagrangian
L=T −U =
1 2 2 1
m` φ̇ + mω 2 R2 − mR`φ̇ω sin (φ + ωt) + mg` cos φ + mgR sin ωt
2
2
The second term is a constant and the second term is the total time derivative. Those can be safely
dropped. The third term can be re-written via a total time derivative
φ̇ sin (φ + ωt) = −
Dropping
d
dt
d
cos (φ + ωt) − ω sin (φ + ωt) .
dt
cos (φ + ωt) , we reduce the Lagrangian to
L=
1 2 2
m` φ̇ + mR`ω 2 sin (φ + ωt) + mg` cos φ
2
3
b) Derive the equation of motion. [15 points]
∂L
d ∂L
=
dt ∂ φ̇
∂φ
m`2 φ̈ = mR`ω 2 cos (φ + ωt) − mg` sin φ
R
g
φ̈ = − sin φ + ω 2 cos (φ + ωt)
`
`
g
R 2
= − sin φ + ω [cos φ cos ωt − sin φ sin ωt]
`
`
R 2
R
g
= − 1 + ω sin ωt sin φ + ω 2 cos φ cos ωt
`
g
`
b) Linearize the equation of motion keeping
pthe system exhibits
p only the leading term in φ, and show that
g/` and a parametric resonance at ω = 2 g/`. Hint: expand
both an ordinary resonance at ω =
cos (φ + ωt) = cos φ cos ωt − sin φ sin ωt in the equation of motion, obtained in part b), and then linearize
with respect to φ. [20 points]
Using the hint,
g
R
φ̈ = − sin φ + ω 2 cos (φ + ωt)
`
`
g
R 2
= − sin φ + ω [cos φ cos ωt − sin φ sin ωt]
`
`
R 2
g
R
= − 1 + ω sin ωt sin φ + ω 2 cos φ cos ωt
`
g
`
Expand in to the first order in φ
R 2
R
g
φ̈ = − 1 + ω sin ωt φ + ω 2 cos ωt.
`
g
`
p
The second term is an external force which leads to an ordinary resonance at ω = g/`. The first term
leads to a parametric resonance.
p The second term can be neglected near the parametric resonance because
it is important only for ω ≈ g/`. Then
g
R 2
φ̈ = − 1 + ω sin ωt φ.
`
g
For small φ, this equation can be solved iteratively
φ = φ0 + φ1 ,
where φ0 satisfies
g
φ̈0 = − φ0 .
`
The solution for φ0 can be chosen as φ0 = A sin ω0 t, where ω0 =
p
g/`. Correction φ1 satisfies
R
g
R
g
φ̈1 = − φ1 − ω 2 sin ωtφ0 = − φ1 − A ω 2 sin ωt sin ω0 t
`
`
`
`
R 2
g
= − φ1 − A ω [cos ((ω − ω0 ) t) − cos ((ω + ω0 ) t)]
`
2`
The resonance occurs if ω − ω0 = ω0 or ω = 2ω0 .
4
3. A comet of mass m in moving in a central force potential U (r) = −k/r with angular momentum L. Neglect the
difference between m and the reduced mass µ.
a) Find the radius of a circular orbit in terms of k, m, and L. [15 points]
For a circular orbit, the minimum of the effective potential energy
Ueff (r) = −
k
L2
+
r
2mr2
coincides with the total energy. The radius of the circular orbit corresponds to the minimum of U eff (r) .
L2
k
−
=0
2
r
mr3
L2
=
.
mk
0
Ueff
(r) =
r0
b) Now consider a parabolic orbit in the same potential with the same m and L. Show that the distance of
the closest approach on this parabolic orbit is half the radius of the circular orbit. [15 points]
Parabolic orbit corresponds to E = 0. The distance of the closest approach is the zero of U eff (r)
Ueff (r) = 0
rmin =
1.
r0
L2
= .
2mk
2
c) Show that at the points
√ where the parabolic orbit intersects the circular one, the speed of the comet on
the parabolic orbit is 2 times the speed on the circular orbit.[20 points]
Parabolic orbit
mvp2
k
E = 0=− +
r0
2
√ p
vp = 2 k/mr0
Circular orbit
E=−
mv02
k
+
r0
2
On the other hand
E = Ueff (r0 ) = −
Using the equation for r0 =
L2
mk ,
L2
k
+
r0
2mr02
this reduces to
E=−
k
k
k
+
=−
r0
2r0
2r0
k
k
mv02
= − +
2r0
r0
2
mv02
k
=
2
2r
p0
k/mr0
v0 =
−
vp =
√
2v0
R
O
l
P
m
Problem 2
Problem 1
Problem 3