PHY4222/Spring 08: CLASSICAL MECHANICS II
HOMEWORK ASSIGNMENT #10: Solutions
due by 12:50 p.m. Wed 04/16
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.
1. Thornton & Marion, Problem 12.5 [33 points]
The Lagrangian
L=
1
1
k
k
k
m1 (ẋ1 )2 + m1 (ẋ2 )2 − x21 − (x1 − x2 )2 − x22
2
2
2
2
2
Equations of motion
m1 ẍ1 = −2kx1 + kx2
m2 ẍ2 = − − 2kx2 + kx1
Normal modes
x1 = a1 eiωt
x2 = a2 eiωt
Secular equation
m1 ω 2 − 2k a1 + ka2 = 0
ka1 + m2 ω 2 − 2k a2 = 0.
Equation for the eigenfrequencies
Det
m1 ω 2 − 2k
k
k
m1 ω 2 − 2k
= 0→
k
3k 2
ω4 − 2 ω2 +
= 0,
µ
m1 m2
where µ = m1 m2 / (m1 + m2 ) is the reduced mass. Roots
r
k
3µ
2
ω1,2
=
1± 1−
.
µ
m1 + m 2
Substituting either ω12 or ω22 into the system for a1 , a2 , we obtain the relations
r
m1
3µ
(2)
(1)
1+ 1−
a1 = a 1
2−
µ
m1 + m 2
r
3µ
m1
(2)
(2)
,
1− 1−
a1 = a 2
2−
µ
m1 + m 2
where the upper index refers to the index of the frequency. The orthonormality condition
!
1
1
q
(1)
a
= √
3µ
m
D1 2 − µ1 1 + 1 − m1 +m2
q
!
3µ
m1
1
2
−
1
−
1
−
µ
m1 +m2
,
a(2) = √
D2
1
P
jk
Tjk aj ak = 1 gives
2
where
r
m21
2
m1 m2
1−3
+
m21 − m22
2
m2
m2
(m1 + m2 )
r
m2
m2
m2
m1 m2
= 2m1 1 + 21 + m2 1 − 3 21 + 2m1 1 − 21
1−3
2.
m2
m2
m2
(m1 + m2 )
D1 = 2 (m2 − m1 ) + 2
D2
2. Thornton & Marion, Problem 12.16 [33 points]
From Fig. 1, the x and y coordinates of mass M at point Q are
x = R (sin θ + sin φ)
y = −R (cos θ + cos φ) .
Inertia of the hoop about point O
IO0 = IO + M R2 = 2M R2 ,
where IO = M R2 is the inertia of the hoop about its center of mass (O).
Potential energy, expanded for small angles,
U = −M gR cos θ − M gR (cos θ + cos φ) = −M gR (2 cos θ + cos φ)
≈ −M gR 2 − θ2 + 1 − φ2 /2 → (1/2) M gR 2θ2 + φ2 ,
where I discared the constant term in U.
Kinetic energy
1
1
IO0 θ̇2 + M ẋ2 + ẏ 2
2
2
1
2 2
= M R θ̇ + M R2 θ̇2 + φ̇2 + 2 cos θ cos φθ̇φ̇ + 2 sin θ sin φθ̇ φ̇
2
1
2 2
≈ M R θ̇ + M R2 3θ̇2 + φ̇2 + 2θ̇φ̇ .
2
T =
Matrix of masses
T̂ = M R2
3 1
1 1
V̂ = M gR
2 0
0 1
Matrix of spring constants
Secular matrix
V̂ − ω 2 T̂ = M R2
2g/R − 3ω 2
−ω 2
2
−ω
g/R − ω 2
.
Secular equation
Det V̂ − ω 2 T̂ = 0 →
2g/R − 3ω 2 g/R − ω 2 − ω 4 = 0 →
√ p
ω1 = 2 g/R
√
p
ω2 =
2/2
g/R
Substituting the frequencies into the equations for the amplitudes, we obtain the relations
(1)
a2
(1)
= −2a1
(2)
a21 = a2
3
y
O
x
R
θ
P
Q
φ
FIG. 1: Problem 2
Expanding θ and φ over the normal modes
(1)
(2)
(1)
(2)
θ (t) = C1 a1 cos ω1 t + C2 a1 cos ω2 t
φ (t) = C1 a2 cos ω1 t + C2 a2 cos ω2 t.
At t = 0 :
(1)
(2)
(1)
(2)
(1)
(2)
θ (0) = C1 a1 + C2 a1 = C1 a1 + C2 a1
(1)
(2)
φ (0) = C1 a2 + C2 a2 = −2C1 a1 + C2 a1 .
Solving for C1 and C2
C1 =
θ (0) − φ (0)
(1)
3a1
, C2 =
2θ (0) + φ (0)
(2)
3a1
.
The mode in which C1 = 0 and C2 6= 0 is excited if θ (0) = φ (0) (symmetrical mode). The mode in which
C2 = 0 and C1 6= 0 is excited if φ (0) = −2θ (0) (antisymmetrical mode).
3. Thornton & Marion, Problem 12.27 [34 points]
x1 = L1 sin θ1
y1 = L1 cos θ1
x2 = L1 sin θ1 + L2 sin θ2
y2 = L1 cos θ1 + L2 cos θ2
4
h
i 1
1
2
2
2
m1 (ẋ1 ) + (ẏ1 ) = m1 L21 (θ1 )
2
2
= −m1 gL1 sin θ1
T1 =
U1
i 1
h
h
i
1
2
2
2
2
m2 (ẋ2 ) + (ẏ2 ) = m2 L21 (θ1 ) + L22 (θ2 ) + 2L1 L2 θ1 θ2 cos (θ1 − θ2 )
2
2
= −m2 gL2 sin θ2
T2 =
U2
Notice that the Solution Manual has an incorrect sign in front of the 2L 1 L2 θ1 θ2 cos (θ1 − θ2 ) term.
L = T 1 + T2 − U 1 − U 2
m1 + m2 2 2 m2 2 2
=
L1 θ̇1 +
L θ̇2 + m2 L1 L2 θ̇1 θ̇2 cos (θ1 − θ2 )
2
2 2
+ (m1 + m2 ) gL1 cos θ1 + m2 gL2 cos θ2 .
Approximating sin θ ≈ θ and cos θ ≈ 1, we obtain
(m1 + m2 ) L1 θ̈1 + m2 L2 θ̈2 + (m1 + m2 ) gθ1 = 0
L2 θ̈2 + L1 θ̈1 + gθ2 = 0
Substituting
θ1 = a1 eiωt
θ2 = a2 eiωt ,
we obtain
a1 (m1 + m2 ) g − ω 2 L1 − m2 L2 ω 2 a2 = 0
−a1 L1 ω 2 + a2 m2 g − L2 ω 2 = 0
The condition that the determinant is equal to zero yields
v
(
)
u
r
h
i
u
g
2
t
(m1 + m2 ) (L1 + L2 ) ± (m1 + m2 ) (m1 + m2 ) (L1 + L2 ) − 4m1 L1 L2 .
ω=
2m1 L1 L2
The relations between the amplitudes of the normal modes are obtained by susbstituting the found frequencies
back into the equations for a1,2


(1)
a2
(2)
a2
=
=
m1 + m 2 L 1 
1 −
m1
L2 
m1 + m2 ) (L1 + L2 ) +
r
m1 + m2 ) (L1 + L2 ) −
r

m1 + m 2 L 1 
1 −
m1
L2 
 (1)
2m1 L2

h
i  a1
2
2
(m1 + m2 ) m1 (L1 − L2 ) + m2 (L1 + L2 )

 (2)
2m1 L2

i  a1
h
2
2
(m1 + m2 ) m1 (L1 − L2 ) + m2 (L1 + L2 )
4. Bonus: A small metallic ball rolls on the rail under the force of gravity. Determine the shape of the rail that
provides for the frequency of oscillations being independent of the amplitude even if the amplitude is not small.
[30 points]
The period does not depend on the amplitude if the potential energy is exactly a parabolic function of the
distance separating the ball from the minimum, i.e.,
U = (1/2) mω 2 s2 ,
5
where s is the arclength from the minimum to the ball. In a gravitational field, U = mgy which we need to
equate to (1/2) mω 2 s2 to obtain the curve. Since ds2 = dx2 + dy 2 , we have
q
2
dx =
(ds/dy) − 1
Z
q
2
x =
dy (ds/dy) − 1
p
mgy = (1/2) mω 2 s2 → s = 2gy/ω 2
p
g/2ω 2y
ds/dy =
r
Z
g
−1
x =
dy
2mω 2 y
The integral is solved introducing a new variable
g
(1 − cos ξ) ,
4ω 2
g
dy =
sin ξdξ
4ω 2
y =
√
Z
Z
Z
1 + cos ξ
g
g
ξ cos ξ/2
g
cos ξ/2
ξ
√
=
=
dξ
sin
ξ
dξ
sin
ξ
dξ sin cos
4ω 2
4ω 2
sin ξ/2
2ω 2
2
2 sin ξ/2
1 − cos ξ
Z
Z
g
g
g
2 ξ
=
=
(ξ + sin ξ) .
dξ
cos
dξ (1 + cos ξ) =
2
2
2ω
2
4ω
4ω 2
x =
The set of equations
g
(ξ + sin ξ)
4ω 2
g
(1 − cos ξ)
y =
4ω 2
x =
give a parametric description of the curve called cycloid. This is the same curve that gives the solution to the
brachistochrone problem. To visualize this curve, think of ξ being “time” and watch how x and y develop. It is
also convenient to rearrange the equations for x and y in the following manner
g
g
ξ =
sin ξ
4ω 2
4ω 2
g
g
y−
=
cos ξ
2
4ω
4ω 2
2
g
g 2
g 2
x−
ξ
=
.
+
y
−
4ω 2
4ω 2
4ω 2
x−
Then it is obvious that x and y are points on a circle of radius g/4ω 2 . The circle is rolling in the x direction.
See the plot of cycloid for −3π ≤ ξ ≤ 3π. Oscillatory motion within any of the ”parabolic” part has the required
property that the frequency of oscillations does not depend on the amplitude.
2
1.5
FIG. 2: Cycloid
1
0.5
–8
–6
–4
–2
2
4
6
8
6