Moving constraints and coupled motion

1. More than one moving part

The preceding examples had either a single coordinate or an immediately visible conservation law. With coupled motion, changing one coordinate can move several particles. The kinetic energy then contains cross terms. These are not optional corrections: they describe the simultaneous contribution of the coordinates to physical velocity.

For moving constraints, calculate velocities in an inertial frame before forming $T$. A coordinate attached to a support is allowed, but the support's motion must remain in the transformation. The following examples bring together coordinate choice, kinetic-energy structure, and conservation laws.

2. Applications

Example 1 · A pendulum carried by an accelerating support

A pivot has prescribed horizontal position $X(t)$. Choose $\theta$ from the downward vertical. From the previous chapter,

The derivative with respect to velocity is $ma^2\dot\theta+ma\dot X\cos\theta$. Its total derivative is

Meanwhile $\partial L/\partial\theta=-ma\dot X\sin\theta\dot\theta-mga\sin\theta$. Subtraction cancels the two terms containing $\dot X\dot\theta$, giving

Constant support velocity produces no extra angular acceleration. Constant rightward support acceleration $A$ gives a stationary tilt satisfying $g\sin\theta_0=-A\cos\theta_0$, so $\tan\theta_0=-A/g$. The bob leans opposite the acceleration. The result also follows in the accelerating support frame if the fictitious force $-mA\hat{\mathbf x}$ is included.

For $A=g$, the downward equilibrium branch is $\theta_0=-\pi/4$. The angle alone specifies relative position, but it does not erase energy exchange with the actuator moving the support.

Example 2 · Coupled spring motion

Two equal masses $m$ move on a horizontal line. Each is attached to a fixed wall by a spring of stiffness $k$, and the masses are joined by a spring of stiffness $\kappa$. Let $x_1,x_2$ be displacements from equilibrium. The middle spring extension is $x_2-x_1$.

Using $L=T-V$, differentiate the middle term carefully: $\partial V/\partial x_1=kx_1+\kappa(x_1-x_2)$ and $\partial V/\partial x_2=kx_2+\kappa(x_2-x_1)$. The equations are

Define $s=x_1+x_2$ and $d=x_1-x_2$. Adding and subtracting gives

The in-phase motion has frequency $\sqrt{k/m}$ because the middle spring does not change length. The opposite-phase motion has frequency $\sqrt{(k+2\kappa)/m}$ because it stretches that spring. General initial data are a sum of these two motions: $s=A\cos\omega_st+B\sin\omega_st$, $d=C\cos\omega_dt+D\sin\omega_dt$, with $x_1=(s+d)/2$, $x_2=(s-d)/2$.

This elementary decoupling is supporting material, not an invitation to a separate course on normal modes. It demonstrates why a change of generalized coordinates can simplify coupled equations.

Example 3 · The spherical pendulum

A bob on a massless rod of length $a$ can move in three dimensions. Use polar angle $\theta$ from the downward vertical and azimuth $\phi$. The transformations are $x=a\sin\theta\cos\phi$, $y=a\sin\theta\sin\phi$, $z=-a\cos\theta$. Differentiating and adding squares gives

The azimuth is cyclic, so $p_\phi=ma^2\sin^2\theta\dot\phi$ is conserved. It is the vertical component of angular momentum, not its full magnitude. The $\theta$ derivative of $L$ is $ma^2\sin\theta\cos\theta\dot\phi^2-mga\sin\theta$. Hence

For a conical pendulum, $\theta$ is constant and $\dot\phi=\Omega$. For $\sin\theta\ne0$ the equation gives $\Omega^2\cos\theta=g/a$. Therefore a nonzero conical angle with $0<\theta<\pi/2$ needs $\Omega^2>g/a$. The $\sin\theta=0$ cases must be examined separately before dividing by that factor.

Example 4 · A bead on a rotating vertical hoop

A smooth hoop of radius $a$ rotates at prescribed speed $\Omega$ about its vertical diameter. A bead of mass $m$ slides on it. Let $\theta$ measure position from the bottom. Its height is $-a\cos\theta$ and its distance from the rotation axis is $a\sin\theta$.

The velocity caused by sliding along the hoop is perpendicular to the azimuthal velocity caused by rotation. Thus

The Lagrange equation is

Equilibria satisfy $\sin\theta=0$ or $\cos\theta=g/(a\Omega^2)$. The bottom exists for every $\Omega$. For small $\theta$, $\ddot\theta\simeq(\Omega^2-g/a)\theta$, so it is stable for $\Omega^2<g/a$ and unstable for $\Omega^2>g/a$. At equality the linear term vanishes; expanding further gives $\ddot\theta\simeq-(g/2a)\theta^3$, so the linear test alone was inconclusive.

When $\Omega^2>g/a$, two off-bottom equilibria exist on opposite sides. Write the right-hand side as $f(\theta)$; at either equilibrium, $f'(\theta_0)=-\Omega^2\sin^2\theta_0<0$. Therefore small disturbances oscillate with frequency $|\Omega\sin\theta_0|$.

The conserved reduced energy is $\mathcal E=ma^2\dot\theta^2/2-ma^2\Omega^2\sin^2\theta/2+V$. Laboratory mechanical energy can change because the motor maintains $\Omega$. This example will be useful when interpreting the Hamiltonian.

3. Choosing what to solve

A full equation of motion need not be integrated to answer a physical question. A cyclic coordinate gives a first integral; setting an acceleration to zero can determine a steady motion; expansion around equilibrium determines its local stability. Each step uses the same Lagrangian but asks for different information.

When a problem requests a derivation, retain the coordinate convention, the expressions for $T$ and $V$, and the derivatives that produce the equation. When it requests a frequency or equilibrium, specify the approximation or branch used. A correct formula outside its stated range is not a correct physical answer.

4. From velocities to momenta

The kinetic-energy matrix converts generalized velocities into momenta. Lagrangian mechanics describes motion using $(q,\dot q)$; the next unit uses $(q,p)$ instead. This changes a set of second-order equations into a paired first-order evolution and gives a natural space in which to draw the mechanical state.

Exercises

Hints and solutions are collected separately.

1. 11.1 Why must a prescribed moving support appear in the coordinate transformation?

2. 11.2 A pivot has $X(t)=\tfrac12At^2$. Find the downward equilibrium tilt of its pendulum.

3. 11.3 Derive the moving-pivot pendulum equation from its Lagrangian.

4. 11.4 For the coupled springs with $m=1\,\mathrm{kg}$, $k=4\,\mathrm{N\,m^{-1}}$, and $\kappa=6\,\mathrm{N\,m^{-1}}$, find both mode frequencies.

5. 11.5 Why does the joining spring not affect the in-phase frequency of equal masses?

6. 11.6 Write the coordinate transformations and kinetic energy of a spherical pendulum.

7. 11.7 A conical pendulum has $a=1\,\mathrm m$ and $\theta=60^\circ$. Find its angular speed for $g=9.8\,\mathrm{m\,s^{-2}}$.

8. 11.8 Derive the equilibrium locations for a bead on a rotating vertical hoop.

9. 11.9 For the rotating hoop, determine the bottom equilibrium’s linear stability.

10. 11.10 Explain why constant pivot velocity has no effect on the pendulum equation but constant pivot acceleration does.

11. 11.11 For the coupled springs, find motion when $x_1(0)=A$, $x_2(0)=0$, and both velocities vanish.

12. 11.12 Why is laboratory energy not generally conserved for the rotating hoop?