Physics 31600 R. Wald Classical Mechanics Autumn, 2002

Problem Set V Solutions v1.1

1) A body of mass m is constrained to move only horizontally (i.e., in the xdirection) and is attached to a fixed support by a spring of spring constant k. A particle also of mass m is attached to the first body by a massless rod of length l as shown. A gravitational field is present, so the potential energy of the second particle is mgz. Suppose that, initially, the first body is at the equilibrium position of the spring, but the particle below it is displaced from the vertical by a small angle θ_{0} as shown. Solve for the subsequent motion of both bodies.

Solution: Let x be the position of the mass moving horizontally. Then the position of the pendulum mass is x_{p} = x + l sin θ, y_{p} = −l cos θ. Hence we can express the Lagrangian is

L = \frac{1}{2}mẋ^{2} + \frac{1}{2}m [ẋ^{2} + 2l cos θθ̇ẋ + l^{2}θ̇^{2}] − \frac{1}{2}kx^{2} + lg cos θ.

Since the disturbance is small we can expand the Lagrangian to second order in the coordinates and velocities. We further change variables to u = lθ so that both coordinates will have the same dimension, yielding

L = \frac{1}{2}m [2ẋ^{2} + 2u̇ẋ + u̇^{2}] − \frac{1}{2}kx^{2} − \frac{1}{2} \frac{mg}{l} u^{2}.

Thus, we have coupled oscillations with mass matrix A and potential matrix B given by:

\frac{A}{m} = [ \begin{matrix}2 & 1 \\ 1 & 1 \end{matrix} ] ; \frac{B}{k} = [ \begin{matrix}1 & 0 \\ 0 & r \end{matrix} ] with r = \frac{mg}{lk} .

In the above, we have divide B by k and A by m to simplify the arithmetic. The normal modes can be found by diagonalizing B with respect to A. Hence we must solve the equation

det (\frac{B}{k} − \frac{mω^{2}}{k} \frac{A}{m}) = | \begin{matrix}1 − 2\frac{mω^{2}}{k} & −\frac{mω^{2}}{k} \\ −\frac{mω^{2}}{k} & r − \frac{mω^{2}}{k} \end{matrix} | = 0

which gives the eigenfrequencies

ω_{±}^{2} = \frac{k}{m} (\frac{1}{2} + r ± \frac{1}{2}\sqrt{1 + 4r^{2}}) = \frac{k}{2m} + \frac{g}{l} ± \sqrt{\frac{k^{2}}{4m^{2}} + \frac{g^{2}}{l^{2}}} .

The normal modes are given by the kernels of B − λA, so we find

N_{±} = span [ \begin{matrix}1 \\ \frac{1}{2r} − 1 ∓ \frac{1}{2r} \sqrt{1 + 4r^{2}} \end{matrix} ] .

The general solution of each normal mode is of course X_{0} cos ωt + V_{0} sin ωt, but we are given that the initial conditions are pure displacement, i.e., the initial velocity is zero. Hence we know that we have pure cosine solutions:

[ \begin{matrix}x(t) \\ u(t) \end{matrix} ] = A [ \begin{matrix}1 \\ \frac{1}{2r} − 1 − \frac{\sqrt{1+4r^{2}}}{2r} \end{matrix} ] cos ω_{+}t + B [ \begin{matrix}1 \\ \frac{1}{2r} − 1 + \frac{\sqrt{1+4r^{2}}}{2r} \end{matrix} ] cos ω_{−}t.

Applying the initial condition

[ \begin{matrix}x(0) \\ u(0) \end{matrix} ] = [ \begin{matrix}0 \\ lθ_{0} \end{matrix} ] = [ \begin{matrix}A + B \\ (A + B) (\frac{1}{2r} − 1) + (B − A)\frac{\sqrt{1+4r^{2}}}{2r} \end{matrix} ] ,

we find that

B = −A = \frac{lθ_{0}r}{\sqrt{1 + 4r^{2}}} .

Switching back to the variable θ, the subsequent motion of the two bodies is

[ \begin{matrix}x(t) \\ θ(t) \end{matrix} ] = \frac{−θ_{0}}{\sqrt{\frac{l^{2}k^{2}}{m^{2}g^{2}} + 4}} \{[ \begin{matrix}l \\ \frac{lk}{2mg} − 1 − \sqrt{\frac{l^{2}k^{2}}{4m^{2}g^{2}} + 1} \end{matrix} ] cos (\frac{k}{2m} + \frac{g}{l} + \sqrt{\frac{k^{2}}{4m^{2}} + \frac{g^{2}}{l^{2}}} )^{\frac{1}{2}} t

− [ \begin{matrix}l \\ \frac{lk}{2mg} − 1 + \sqrt{\frac{l^{2}k^{2}}{4m^{2}g^{2}} + 1} \end{matrix} ] cos (\frac{k}{2m} + \frac{g}{l} − \sqrt{\frac{k^{2}}{4m^{2}} + \frac{g^{2}}{l^{2}}} )^{\frac{1}{2}} t.\}

2) Consider a “double pendulum’ with each pendulum of mass m and length l, and with a vertical gravitational force −mgẑ acting on each mass. At t = 0 the double pendulum is vertical, but the top mass is given a small “kick”, so that it has “velocity” θ̇_{10}. Solve for the subsequent motion of the masses.

Solution: Let θ be the angle of the top particle with respect to the vertical, and ψ the angle of bottom particle also with respect to the vertical. If we put the origin at the pivot, then the coordinates of the particles are

(x_{1}, z_{1}) = l(sin θ, − cos θ) ; (x_{2}, z_{2}) = l(sin θ + sin ψ, − cos θ − cos θ).

The Lagrangian T − V = T − mgz_{1} − mgz_{2} can then be written

L = \frac{1}{2}ml^{2} [2θ̇^{2} + ψ̇^{2} + 2 cos(θ − ψ)θ̇ψ̇] + 2mgl cos θ + mgl cos ψ.

Since the initial velocity is assumed small, we may expand the Lagrangian to second order in the generalized coordinates and velocities to obtain

L = \frac{1}{2}ml^{2} [2θ̇^{2} + ψ̇^{2} + 2θ̇ψ̇] − mgl [θ^{2} + \frac{1}{2}ψ^{2}] .

Thus, we have reduced the problem to couple oscillations with mass matrix A and potential matrix B of the form

A = ml^{2} [ \begin{matrix}2 & 1 \\ 1 & 1 \end{matrix} ] ; B = mgl [ \begin{matrix}2 & 0 \\ 0 & 1 \end{matrix} ]

The eigenfrequencies will obey the equation

det(B − ω^{2}A) = | \begin{matrix}2mgl − 2ml^{2}ω^{2} & −ml^{2}ω^{2} \\ −ml^{2}ω^{2} & mgl − gl^{2}ω^{2} \end{matrix} | = 0,

which after one does the arithmetic gives rise to

l^{2} (ω^{2})^{2} − 4glω^{2} + 2g^{2} = 0 ⇒ ω_{±}^{2} = (2 ± \sqrt{2}) \frac{g}{l}

The normal modes are the null vectors of B − ω^{2}A. If we divide B and A by mgl to make this equation dimensionless, we find

ker B−ω_{+}^{2}A = ker [ \begin{matrix}−2 − 2\sqrt{2} & −2 − \sqrt{2} \\ −2 − \sqrt{2} & −1 − \sqrt{2} \end{matrix} ] = ker [ \begin{matrix}\sqrt{2} & 1 \\ 0 & 0 \end{matrix} ] = span \{[ \begin{matrix}1 \\ −\sqrt{2} \end{matrix} ]\} .

To find the normal mode for ω_{−}, we can either repeat the above procedure, or exploit the fact that normal modes are orthogonal in the inner product defined by A. Following the latter procedure, we find

[ x y ] [ \begin{matrix}2 & 1 \\ 1 & 1 \end{matrix} ] [ \begin{matrix}1 \\ −\sqrt{2} \end{matrix} ] = (2 − \sqrt{2}) x + (1 − \sqrt{2}))y = 0,

which gives us that

ker B − ω_{−}^{2}A = span \{[ \begin{matrix}1 \\ √ & \overline{2} \end{matrix} ]\} .

The general solution for a normal mode is V_{0} sin ωt + X_{0} cos ωt. Since the initial condition is at zero displacement, we know that we have a pure sine solution:

[ \begin{matrix}θ(t) \\ ψ(t) \end{matrix} ] = V_{+} [ \begin{matrix}1 \\ −\sqrt{2} \end{matrix} ] sin ω_{+}t + V_{−} [ \begin{matrix}1 \\ √ & \overline{2} \end{matrix} ] sin ω_{−}t.

Applying the initial condition

[ \begin{matrix}θ′(t) \\ ψ′(t) \end{matrix} ] = ω_{+}V_{+} [ \begin{matrix}1 \\ −\sqrt{2} \end{matrix} ] + ω_{−}V_{−} [ \begin{matrix}1 \\ √ & \overline{2} \end{matrix} ] = [ θ̇_{10} ; 0 ] ,

solving for V_{+} and V_{−} (trivial, since the ψ′ component implies that ω_{+}V_{+} = ω_{−}V_{−}), and substituting in the positive roots of ω_{±}^{2} gives

\begin{aligned}[ \begin{matrix}θ(t) \\ ψ(t) \end{matrix} ] & = \frac{θ̇_{10}}{2} \sqrt{\frac{l}{(2 + \sqrt{2}) g}} [ \begin{matrix}1 \\ −\sqrt{2} \end{matrix} ] sin [t\sqrt{(2 + \sqrt{2}) \frac{g}{l}} ] \\ & +\frac{θ̇_{10}}{2} \sqrt{\frac{l}{(2 − \sqrt{2}) g}} [ \begin{matrix}1 \\ √ & \overline{2} \end{matrix} ] sin [t\sqrt{(2 − \sqrt{2}) \frac{g}{l}} ] . ■ \end{aligned}

3) Consider a particle of mass m moving in a 2-dimensional plane, with a potential, V = V (r), which depends only on the radius r.

(a) Use the symmetry under rotations to reduce the problem to one of onedimensional motion in r.

Solution: We begin by writing down the Lagrangian in polar coordinates:

L = T − V = \frac{1}{2}m [ṙ^{2} + r^{2}θ̇^{2}] − V (r).

This Lagrangian is cyclic in the angular variable θ, so the angular momentum

l = \frac{∂L}{∂θ̇} = mr^{2}θ̇

is conserved. Since the Lagrangian is time independent and of the standard form T − V, the total energy E = T + V is also conserved. Substituting the angular momentum into the expression for E, we obtain

E = \frac{1}{2}mṙ^{2} + \frac{l^{2}}{2mr^{2}} + V (r).

Hence, we have reduced the problem to an effective 1d problem. ■

(b) Explicitly write down the equation of motion in r and show that a circular orbit exists at every radius at which dV/dr > 0.

Solution: Differentiating the expression for E we obtain:

\frac{dE}{dt} = 0 = ṙ [mr̈ − \frac{l^{2}}{mr^{3}} + \frac{dV}{dr} ] .

For a general solution ṙ ≠ 0, so we identify the terms in brackets as the equation of motion. (We could also have obtained this by computing the EL equation for r, then substituting l into the resulting equation of motion.) At fixed r, l^{2} can be set to an arbitrary nonnegative value by suitable choice of θ̇(0). Hence the circular orbit

r(t) = r_{0} ; l_{0}^{2} = mr_{0}^{3} \frac{dV}{dr} |_{r_{0}}

satisfies the equations of motion. ■

(c) Obtain the general solution for the radial motion resulting from an arbitrary small perturbation about a circular orbit of radius r_{0}. (Allow the perturbation to change p_{ϕ}.) Determine the conditions on V such that this circular orbit is stable.

Solution: In parts (a) and (b) we reduced the problem to an effective one-dimensional problem with potential

U(r) = \frac{l^{2}}{2mr^{2}} + V (r)

and also found that the equations of motion for r are

mr̈ − \frac{l^{2}}{mr^{3}} + \frac{dV}{dr} = 0.

Either of these can be used as the starting point for our investigation of the effects of perturbations. However, in order to get the equations of motion correct to first order, the potential U must be expanded correctly to second order in the perturbations which is a bit of a mess. We therefore look directly at the equations of motion. Consider a change in the initial conditions from

r(0) = r_{0} ; ṙ(0) = 0 ; θ(0) = θ_{0} ; θ̇(0) = θ̇_{0} = \sqrt{\frac{1}{mr_{0}} \frac{dV}{dr} |_{r_{0}}}

by an arbitrary small perturbation:

r(0) = r_{0} + δr_{0} ; ṙ(0) = δṙ ; θ(0) = θ_{0} + δθ_{0} ; θ̇(0) = θ̇_{0} + δθ̇_{0}.

The new angular momentum to first order in the perturbations is

l^{2} = m^{2}(r_{0} + δr_{0})^{4}(θ + δθ̇)^{2} ≈ l_{0}^{2} (1 + \frac{4δr_{0}}{r_{0}} + \frac{2δθ̇_{0}}{θ̇_{0}} )

Expanding the force \frac{dV}{dr} to first order in the perturbations also we find that the equation of motion for δr is

m(r̈_{0} + δr̈) − \frac{l_{0}^{2}}{m(r_{0} + δr)^{3}} (1 + \frac{4δr_{0}}{r_{0}} + \frac{2δθ̇_{0}}{θ̇_{0}} ) + \frac{dV}{dr} |_{r_{0}} + \frac{d^{2}V}{dr^{2}} |_{r_{0}} δr = 0.

Expanding the (r_{0} + δr)^{−3} to first order and using the fact that l_{0}^{2} = mr_{0}^{3} \frac{dV}{dr} as in part (b), we find that to first order the equation is

mδr̈ + [\frac{3}{r_{0}} \frac{dV}{dr} |_{r_{0}} + \frac{d^{2}V}{dr^{2}} |_{r_{0}}] δr = \frac{dV}{dr} |_{r_{0}} [\frac{4δr_{0}}{r_{0}} + \frac{2δθ̇_{0}}{θ̇_{0}} ] .

The solution of this equation with the initial conditions given above is

δr(t) = A + (δr_{0} − A) cos ωt + \frac{δṙ_{0}}{ω} sin ωt

where A and ω are given by

ω = \sqrt{\frac{1}{m} [\frac{3}{r_{0}} \frac{dV}{dr} |_{r_{0}} + \frac{d^{2}V}{dr^{2}} |_{r_{0}}]} ; A = \frac{dV}{dr} |_{r_{0}} [\frac{4δr_{0}}{r_{0}} + \frac{2δθ̇_{0}}{θ̇_{0}} ] [\frac{3}{r_{0}} \frac{dV}{dr} |_{r_{0}} + \frac{d^{2}V}{dr^{2}} |_{r_{0}}]^{−1} .

The frequency ω will be real and the corresponding solution truly oscillatory (as opposed to exponentially growing) if

\frac{3}{r_{0}} \frac{dV}{dr} |_{r_{0}} + \frac{d^{2}V}{dr^{2}} |_{r_{0}} > 0,

hence this is the condition for stability. This makes perfect sense: the orbit is stable if r_{0} corresponds to a minimum of the effective potential U. Finally, we can (although the problem didn’t ask us to) also find the solution for θ to first order from conservation of angular momentum:

θ̇(t) = θ̇_{0} + δθ̇ = \frac{l}{m(r_{0} + δr)^{2}} = \frac{l_{0}}{mr_{0}^{2}} (1 + \frac{2δr_{0}}{r_{0}} + \frac{δθ̇_{0}}{θ̇_{0}} ) (1 − \frac{2δr}{r_{0}}) .

As l_{0} = mr_{0}^{2}θ̇_{0} the terms zeroth order in the perturbation cancel, and this equation is

δθ̇(t) = θ̇_{0} (\frac{2 (δr_{0} − δr(t))}{r_{0}} + \frac{δθ̇_{0}}{θ̇_{0}} ) = θ̇_{0} (\frac{δl}{l_{0}} − \frac{2δr(t)}{r_{0}}) .

Since we’ve already solved for δr(t), the right hand side is a well-defined function of time which gives δθ̇(t) explicitly. It can be integrated to give δθ(t). Since this expression doesn’t depend on δθ_{0}, the effect of the perturbation in initial angle is merely to change the offset in θ(t). The perturbation in the angular velocity has a constant offeset plus an oscillation oppositely signed from that of the radial perturbation. This is something which we could have guessed: since l is still conserved, the radial oscillations must be opposite to the the anglular velocity. ■