Physics 31600 R. Wald Classical Mechanics Autumn, 2002

Problem Set IV-Solutions

1) Consider the “heavy symmetrical top” treated in class (and also discussed in Goldstein and Arnold). Suppose we impose the following additional constraint on the top: The symmetry axis is forced to rotate about the z-axis with uniform angular velocity ϕ̇ = Ω. Then the top has only two degrees of freedom, which may be “coordinatized” by the remaining Euler angles (θ, ψ).

(a) Write down the Lagrangian for this system.

(b) Write down the constants of motion and explicitly reduce the problem of finding the general motion “to quadratures”.

(c) Find the most general choice of initial conditions (θ_{0}, θ̇_{0}, ψ_{0}, ψ̇_{0}) for which the solution has θ̇ = 0 for all time (and hence θ(t) = θ_{0}). Are these solutions stable insofar as the θ-motion is concerned, i.e., if the initial conditions differ by a small amount from the ones yielding θ = 0, does θ(t) − θ_{0} remain small for all time?

The Lagrangian for the unconstrained system is:

L = \frac{I_{1}}{2} (θ̇^{2} + ϕ̇^{2} sin^{2} θ) + \frac{I_{3}}{2} (ψ̇ + ϕ̇ cos θ)^{2} − Mgl cos θ  (1)

where I_{1} and I_{3} are the moments of inertia, M is the mass, l is the height of the center of mass along the central principal axis, and ϕ, θ, ψ are the Euler angles. To constrain the system we add a ”non-dynamical” field β with the term −β(ϕ̇ − Ω) in the Lagrangian. This has the effect of replacing ϕ̇ with Ω in the above Lagrangian. The Lagrangian is , therefore:

L = \frac{I_{1}}{2} (θ̇^{2} + Ω^{2} sin^{2} θ) + \frac{I_{3}}{2} (ψ̇ + Ω cos θ)^{2} − Mgl cos θ  (2)

Since the Lagrangian does not depend on ψ, we can immediately write down one constant of the motion (from the equation for ψ):

\frac{d}{dt}[I_{3}(ψ̇ + Ω cos θ)] = \frac{d}{dt}[I_{3}ω_{3}] = 0  (3)

Since the system is conservative, the other constant of motion is the energy:

E = \frac{I_{1}}{2} (θ̇^{2} + Ω^{2} sin^{2} θ) + \frac{I_{3}}{2} ω_{3}^{2} + Mgl cos θ  (4)

Define E′ = E − \frac{I_{3}}{2} ω_{3}^{2} and u = cos θ, then we have:

E′ = \frac{I_{1}}{2} (\frac{u̇^{2}}{1 − u^{2}} + Ω^{2}(1 − u^{2})) + Mglu  (5)

Solving for u̇ we have:

u̇^{2} = (\frac{2(E′ − Mglu)}{I_{1}})(1 − u^{2}) − Ω^{2}(1 − u^{2})^{2}  (6)

Further define α = \frac{2E′}{I_{1}} and β = \frac{2Mgl}{I_{1}}. We can solve for u now with the following quadrature:

t = ∫_{u(0)}^{u(t)} \frac{du}{\sqrt{(α − βu)(1 − u^{2}) − Ω^{2}(1 − u^{2})^{2}}}  (7)

From the Lagrangian, the equation of motion for θ is

I_{1}θ̈ = I_{1}Ω^{2} sin θ cos θ − I_{3}(ψ̇ + Ω cos θ)(sin θ)Ω + Mgl sin θ  (8)

Setting θ̈ = 0 and noting that ψ̇ + Ω cos θ = ω_{3} is a constant we see that either sin θ = 0 or:

cos θ = \frac{I_{3}ω_{3}Ω − Mgl}{I_{1}Ω^{2}}  (9)

Here θ must be a constant, since the right hand side is a constant, thus θ̈ = 0 → θ̇ = 0. With sin θ(t) = 0, θ(t) = 0 or π, (although π is unphysical for a real top, we might be dealing with a gyroscope or similar system), then ψ̇ = ω_{3} ∓ Ω. So, our first sets of allowed initial conditions which produce θ̇(t) = 0 are:

θ̇ = 0, θ = 0, ψ̇ = ψ̇_{0}, ψ = ψ_{0}
θ̇ = 0, θ = π, ψ̇ = ψ̇_{0}, ψ = ψ_{0}

We use the equation of motion for θ to determine stability. Expanding about the equilibrium point, to first order in θ this equation is:

I_{1}δ̈θ = [I_{1}Ω^{2} − I_{3}(±ψ̇ + Ω)Ω ± Mgl]δθ  (10)

where the ± indicates expansion around θ = 0 or π, respectively. This solution is stable if the term in brackets in negative, otherwise it is unstable. The condition on ψ̇ for stability, then, is:

±ψ̇ > [\frac{I_{1}Ω^{2} ± Mgl}{I_{3}Ω} − Ω]  (11)

if Ω ≠ 0 and the stability is the same as that of a pendulum if Ω = 0 (as it should be; think about it, you’re forcing Ω = 0). The other solutions come from Equation (9), which is (substituting for ω_{3}):

cos θ_{0} = \frac{I_{3}(ψ̇ + Ω cos θ)Ω − Mgl}{I_{1}Ω^{2}}  (12)

which gives:

cos θ_{0} = \frac{I_{3}Ωψ̇ − Mgl}{(I_{1} − I_{3})Ω^{2}}  (13)

This has solutions only for the RHS being in the interval (−1, 1). So ψ̇ must satisfy:

\frac{(I_{3} − I_{1})Ω^{2} + Mgl}{I_{3}Ω} < ψ̇ < \frac{(I_{1} − I_{3})Ω^{2} + Mgl}{I_{3}Ω}  (14)

with cos θ_{0} determined by equation (13), θ̇_{0} = 0 and ψ = ψ_{0}. The reverse relationship holds if I_{3} > I_{1}. If we again expand the equation of motion for θ around this point we have:

I_{1}δ̈θ = [−I_{1}Ω^{2} sin^{2} θ_{0} + I_{3}Ω^{2} sin^{2} θ_{0}]δθ  (15)

so these points are stable if I_{1} > I_{3}, metastable if I_{1} = I_{3}, and unstable otherwise.

2) Consider a rigid body in “free motion’ (i.e. V = 0), with a point in the body fixed.

(a) Show that the motion of uniform rotation about an axis fixed in space is dynamically possible if and only if that axis coincides with a principal axis of the body.

(b) Suppose that the eigenvalues of the inertia tensor satisfy I_{1} < I_{2} < I_{3}. Show that the solutions of uniform rotation about the ê_{1} and ê_{3} axis are stable, but the solution of uniform rotation about the ê_{2} axis is unstable. [Note: This problem can be solved by finding the orbits in Ω⃗-space by intersecting the surface of constant energy with the surface of constant squared angular momentum, as done in Arnold and Goldstein. Rather than copy their solutions, please do the problem by writing down and solving the (approximate) Euler equations for a small departure from uniform rotation about a principal axis.]

(a) Uniform rotation means we have:

\frac{dL⃗}{dt})_{s} = 0  (16)

where the s denotes that the coordinate system is space-fixed. This implies:

\frac{dL⃗}{dt})_{s} = \frac{dL⃗}{dt})_{b} + ω⃗ × L⃗ = 0  (17)

Uniform motion in the space-fixed frame implies that \frac{dL⃗}{dt})_{b} = 0 so we are left with:

ω⃗ × L⃗ = 0  (18)

Since we have the relation:

L⃗ = I · ω⃗  (19)

Equation (18) is satisfied if and only if L⃗ is parallel to ω⃗. This will be true if and only if ω⃗ is an eigenvector of I. Q.E.D.

(b) The Euler equations are:

I_{i}ω̇_{i} − ω_{j}ω_{k}(I_{j} − I_{k}) = 0  (20)

for i, j, k a cyclic permutation of 1,2,3. Setting ω̇_{i} = 0 gives (since no 2 I’s equal each other):

ω_{j}ω_{k} = 0  (21)

which can only be solved if 2 of the 3 ω’s are 0. So uniform rotation corresponds to rotation about only 1 principal axis, as above. If we expand the Euler equations about one of these solutions (say for ω_{1} = ω_{0}, ω_{2} = ω_{3} = 0):

I_{i}δ̇ω_{i} = 0  (22)
I_{j}δ̇ω_{j} = (δω_{k})ω_{i}(I_{k} − I_{i})  (23)
I_{k}δ̇ω_{k} = (δω_{j})ω_{i}(I_{i} − I_{j})  (24)

Deferentiating the second two equations with respect to time and substituting the appropriate equations for δ̇ω_{i} we have:

I_{j}δ̈ω_{j} = ω_{i}^{2} \frac{(I_{k} − I_{i})(I_{i} − I_{j})}{I_{k}}δω_{j}  (25)
I_{k}δ̈ω_{k} = ω_{i}^{2} \frac{(I_{i} − I_{j})(I_{k} − I_{i})}{I_{j}}δω_{k}  (26)

If we take i = 1 or i = 3 then the right hand coefficients are negative, hence the direction of motion is stable. If i = 2, the coefficients are positive and the motion is unstable.

3) Three particles, each of mass m, are constrained to lie on a circle and are connected by identical springs lying on the circle, each of spring constant k, as shown. Find the general solution for the motion of these particles.

If the radius of the circle is a, the Lagrangian for this system is:

L = ∑_{i=1}^{3} \frac{1}{2}ma^{2}ϕ̇_{i}^{2} − \frac{1}{2}ka^{2}(ϕ_{i} − ϕ_{i+1})^{2}  (27)

We can rewrite this equation as:

L = \frac{1}{2}ma^{2} ˙ϕ⃗^{2} − \frac{1}{2}ka^{2}ϕ⃗ · Aϕ⃗  (28)

where:

A = ( \begin{matrix}2 & −1 & −1 \\ −1 & 2 & −1 \\ −1 & −1 & 2 \end{matrix} )  (29)

The equation of motion is:

ma^{2}¨ϕ⃗ = −ka^{2}Aϕ⃗  (30)

The solutions of this equation are ϕ⃗(t) = ϕ⃗_{i} exp(−iω_{i}t), where ϕ⃗_{i} is an eigenvector of A with eigenvalue λ_{i}. This yields:

ma^{2}ω_{i}^{2} = ka^{2}λ_{i}  (31)

so the frequency of oscillation of each mode is:

ω_{i} = \sqrt{\frac{kλ_{i}}{m}}  (32)

A quick calculation shows that the eigenvectors are:

ϕ⃗_{1} = ( 1 1 1 )  (33)
ϕ⃗_{2} = ( 1 0 −1 )  (34)
ϕ⃗_{3} = ( 1 −2 1 )  (35)

with eigenvalues

λ_{1} = 0  (36)
λ_{2} = 3  (37)
λ_{3} = 3  (38)

These have ω_{1} = 0 and ω_{2} = ω_{3} = \sqrt{\frac{3k}{m}}. Since ω_{1} = 0, we must mulitply ϕ⃗_{1} by a linear function of time (because this is also a solution). The general motion is then:

ϕ⃗(t) = (A_{1} + B_{1}t)ϕ⃗_{1} + (A_{2}ϕ⃗_{2} + A_{3}ϕ⃗_{3}) exp(−i\sqrt{\frac{3k}{m}} t)  (39)

with initial conditions determining the constants as:

ϕ⃗(0) = A_{1}ϕ⃗_{1} + A_{2}ϕ⃗_{2} + A_{3}ϕ⃗_{3}  (40)
˙ϕ⃗(0) = B_{1}ϕ⃗_{1} − i\sqrt{\frac{3k}{m}} (A_{2}ϕ⃗_{2} + A_{3}ϕ⃗_{3})  (41)