PHY6426/Fall 2007: CLASSICAL MECHANICS
HOMEWORK ASSIGNMENT #6
due by 9:35 a.m. Mon 10/8
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
1. A sphere rolls without slipping along a horizontal surface. Using Eulerian angles and nonholonomic rolling
constraint, show that the center of the sphere is moving along a straight line while the angular velocity remains
constant in time.
(Hint: use the projections of the angular velocity on the axes of the fixed system; see Goldstein, Derivation 14,
p. 182.)
Solution: Ω1,2,3 are the projections of the angular velocity on the axes of the fixed system.
a) Let’s first solve the problem in a simple way, without using the non-holonomic constraints.
Frictional force f acts in the plane. Equations of motion
mẍ1 = f1
mẍ2 = f2 .
This force also produces a torque about the c.o.m. of the sphere. The corresponding equations describing the
dynamics of the angular momentum are
N1 = f2 R = I Ω̇1
N2 = −f1 R = I Ω̇2 ,
N3 = 0.
where I is the moment of inertia.
Rolling constraint
ẋ1 = RΩ2 → ẍ1 = RΩ̇2
ẋ2 = −RΩ1 → ẍ2 = −RΩ̇1 .
Substituting expressions for ẍ1 into the equation of motion and using the equations for torque, we find that we
have two results for f1
f1 = mẍ1 = mRΩ̇2
I
f1 = − Ω̇2 →
R
I
Ω̇2 = 0.
mR +
R
As the coefficient in front of Ω̇2 is strictly positive, the last equations can be satisfied only if
Ω̇2 = 0.
Similarly, one shows that
Ω̇1 = 0.
Then, it follows that ẍ1 = ẍ2 = 0. The ratio of velocity components is obtained by dividing the first constraint
equation by the second one
Ω2
ẋ1
=− .
ẋ2
Ω1
2
Thus, we see that the sphere rolls along a straight line whose tangent is determined by the ratio of the angular
velocities
tan θ =
ẋ2
Ω1
=− .
ẋ1
Ω2
2) In order to apply the technique of non-holonomic constraints, we write the constraint equation as
α
Fα = A α
j q̇j = 0
and add Aα
j = ∂Fα /∂ q̇α multiplied by the Lagrange factors µα to the right-hand side of the Euler equation
∂L
d ∂L
−
= µ α Aα
j.
dt ∂ q̇j
∂qj
The Lagrangian is
3
1 X 2
1
2
Ω ,
L = m (ṙ) + I
2
2 i=1 i
where r is the position of the c.o.m. and Ωi are defined with respect to a stationary system (XY Z). Projecting
the angular velocities on the axes of XY Z, we find
Ω1 = θ̇ cos φ + ψ̇ sin θ sin φ
Ω2 = θ̇ sin φ − ψ̇ sin θ cos φ
Ω3 = φ̇ + ψ̇ cos θ.
Substituting these expressions into Ω21 + Ω22 + Ω23 , we find
Ω21 + Ω22 + Ω23 = θ̇2 + φ̇2 + ψ̇ 2 + 2φ̇ψ̇ cos θ.
As before, the rolling constraint reads
ẋ1 = Ω2 R
ẋ2 = −Ω1 R
or, using equations for Ω1 and Ω2 ,
ẋ1
=0
R
ẋ2
= 0.
= θ̇ cos φ + ψ̇ sin θ sin φ +
R
f1 = θ̇ sin φ − ψ̇ sin θ cos φ −
f2
Equations of motion
∂L
∂F1
∂F2
d ∂L
−
= µ1
+ µ2
,
dt ∂ q̇j
∂qj
∂ q̇j
∂ q̇j
where j labels x1 , x2 , φ, ψ, and θ. Performing differentiations, we find
mẍ1 = −µ1 /R
mẍ2
d I
φ̇ + ψ̇ cos θ
dt
d ψ̇ + φ̇ cos θ
I
dt
I θ̈ + I φ̇ψ̇ sin θ
= µ2 /R
d
= I Ω̇3 = 0
dt
= sin θ (−µ1 cos φ + µ2 sin φ)
= µ1 sin φ + µ2 cos φ
3
Note that Ω3 =const. Multiply the last equation by sin φ/ sin θ and the one before the last one by cos φ, and
add them up. This gives a relation for µ2
sin φ cos θ
sin φ
+ φ̈
− φ̈θ̈ sin φ .
µ2 = I θ̈ cos φ + φ̇ψ̇ sin θ cos φ + ψ̈
sin θ
sin θ
Noticing that
φ̈ + ψ̈ cos θ − ψ̇ θ̇ sin θ = 0,
the expression for µ2 is reduced to
i
d h
θ̇ cos φ + ψ̇ sin θ sin φ
dt
I
d
= I Ω1 = − ẍ2
dt
R
µ2 = I
Thus,
ẍ2 = −RΩ̇1 .
On the other hand,
mẍ2 =
I
µ2
= − ẍ2 .
R
R
It follows then that
ẍ2 = 0
Ω̇1 = 0.
Similarly,
ẍ1 = 0
Ω̇2 = 0.
The conclusions are made in the same way as in the “simple” solution.
2. Goldstein, Problem 5.7
Choose the the principal axes so that I1 < I2 < I3 . Focus on the axis x1 first. The angular velocity is almost
aligned with this axis, so that we can write
ω = ω1 x̂1 + ε2 x̂2 + ε3 x̂3 ,
where projections ε2 and ε3 are small.
The Euler equations for a torque-free motion
I1 ω̇1 = ω2 ω3 (I2 − I3 )
I2 ω̇2 = ω1 ω3 (I3 − I1 )
I3 ω̇3 = ω1 ω2 (I1 − I2 )
become
I1 ω̇1 = ε2 ε3 (I2 − I3 )
I2 ε̇2 = ω1 ε3 (I3 − I1 )
I3 ε̇3 = ω1 ε2 (I1 − I2 ) .
To lowest order in ε2 and ε3 , the RHS of the first equation is equal to zero, which means that ω1 =const.
Differentiating the second equation and using the third one gives
ε̈2 = −Ω21 ε2 ,
4
where
Ω21 = ω12
(I3 − I1 ) (I2 − I1 )
I2 I3
Likewise,
ε̈3 = −Ω21 ε3
The motion is stable if
Ω21 > 0
which is the case because of the ordering of I’s. Thus rotation about the x1 axis is stable. Components of ω,
transverse to this axis, rotate with the frequency
s
(I3 − I1 ) (I2 − I1 )
Ω1 = ω 1
.
I2 I3
Results for other axes can be simply obtained by a cyclic permutation of indices 1, 2, 3. This gives
Ω22 = ω22
(I1 − I2 ) (I3 − I2 )
.
I1 I3
Ω22 < 0 because I1 < I2 , and thus rotation about the x2 axis is unstable. Finally,
Ω23 = ω32
(I2 − I3 ) (I1 − I3 )
>0
I1 I3
and rotation about x3 is stable.
3. Goldtsein, Problem 4.23
Choose the frame as shown in the figure: the xy plane is tangent to the Earth surface, the z axis is along
the radius. The pivoting point of the pendulum is on the z axis. Consider first the motion of the pendulum
for stationary Earth. The pendulum swings in the plane normal to the xy plane. The motion in this plane is
described by
ẍ + ω02 x = 0
ÿ + ω0 y = 0,
p
where ω0 = g/l and l is the length of the pendulum. Earth’s rotation add extra terms to the RHS of these
equations. For slow rotation, one can neglect the centrifugal force (proportional to Ω 2 ) and keep only the Coriolis
force. Then the equations change to
ẍ + ω02 x = 2 (v × Ω)x = 2ẏΩz
ÿ + ω0 y = 2 (v × Ω)y = −2ẋΩz ,
where Ωz = Ω cos θ. Multiplying the second equation by i adding up the two equations, we obtain
ζ̈ + 2iΩz ζ̇ + ω02 ζ = 0,
where ζ = x + iy. A solution of the form ζ ∝ eiωt gives an equation for ω
ω 2 + 2Ωz ω − ω02 = 0
or
ω1,2 = −Ωz ±
As Ωz ω0 , one can expand the last result as
q
ω02 + Ω2z .
ω1,2 = −Ωz ± ω0 .
5
Ω
z
x
θ
y
Hence the general solution for ζ is
ζ = e−iΩz t Aeiω0 t + Be−iω0 t .
For Ωz = 0, ζ = x0 (t) + iy0 (t), where x0 (t) and y0 (t) are trajectories for Ωz = 0. Therefore,
ζ = x + iy = e−iΩz t (x0 + iy0 ) .
Separating real and imaginary parts, we find
x = x0 cos Ωz t + y0 sin Ωz t
y = −x0 sin Ωz t + y0 cos Ωz t.
The plane of the pendulum rotates about the z axis with angular frequency Ωz .