A Bead Sliding on a Parabola-Shaped Smooth Wire

Monday, February 18, 2008

I present here a beautiful solution (as many, including myself, would find) to a classical problem in mechanics.

A smooth massless wire is bent into a shape of a parabola y = Ax^2, where A is a positive constant. The wire is oriented vertically, opening upward, and is subjected into a uniform gravitational field of strength \mathbf{g}, directed downward. The wire rotates with a constant angular velocity \omega about the y axis, while a bead of mass m is free to slide on the wire without friction. We want to analyse this system for stability.

First, I will solve the problem, and then I will make a few remarks about the solution.

As the bed is confined to the wire, which rotates in \mathbb{R}^3, the problem is reduced to a one-dimensional motion of the bead, with rotation and wire’s shape playing the role of holonomic constraints.

As the gravitation field \mathbf{g} is acting only along the y axis, we choose the zero of the potential energy at y=0 (id est, at the bottom of the parabola), in which case the potential energy of the system can be written:

U = mgy = mgAx^2.

Note that the potential energy of the wire plays no role, since the wire is massless and hence it is a mere constraint for the bead’s motion, as has been mentioned previously. The kinetic energy of the system is written as

K = \frac{1}{2}m \left( \dot{y}^2 + \dot{x}^2 + \omega^2 x^2 \right) = 2mA^2x^2\dot{x}^2 + \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega^2x^2.

We can now write down the Lagrangian for the system:

\mathcal{L} = 2mA^2x^2\dot{x}^2 + \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega^2x^2 - mgAx^2.

After computing the derivatives \displaystyle\frac{\partial \mathcal{L}}{\partial \dot{x}}, \displaystyle\frac{d}{dt} \left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right), \displaystyle\frac{\partial \mathcal{L}}{\partial x}, we can write down the Lagrange equation of motion for the system:

4A^2 x \dot{x}^2 + 4A^2 x^2 \ddot{x} + \ddot{x} = \omega^2 x - 2gAx.

An equilibrium can occur only at a fixed point — a point where \dot{x} = \ddot{x} = 0;that is, if a material point is placed at a fixed point at rest, it will stay at rest at that point. Such points are found from the Lagrange equation by simply setting \dot{x} = \ddot{x} = 0:

\omega^2 x - 2gAx = 0 \iff \left( x = 0 ~ \vee ~ \omega^2 = 2gA \right).

Let us first analyse the first solution: x = 0. Consider small perturbations x = \delta x around x = 0. In this case, the Lagrange equation becomes

4A^2 \delta x (\delta \dot{x})^2 + 4A^2 (\delta x)^2 \delta \ddot{x} + \delta \ddot{x} = \omega^2 \delta x - 2gA\delta x,

or

\delta \ddot{x} = - \left( 2gA - \omega^2 \right) \delta x,

where we have neglected all higher order terms (such as \delta x (\delta \dot{x})^2). The resulting equation has the same form as that describing the simple harmonic motion; hence we can conclude that the stability of the system about point x=0 depends solely on the sign of the factor \left( 2gA - \omega^2 \right). Thus, there are three cases:

  1. 2gA > \omega^2 \Rightarrow the fixed point x = 0 is stable with a frequency of oscillation \frac{1}{2\pi} \sqrt{ 2gA - \omega^2};
  2. 2gA = \omega^2 \Rightarrow the fixed point x = 0 is a separatrix; the period of oscillation is infinitely long (id est, the frequency is zero);
  3. 2gA < \omega^2 \Rightarrow the fixed point x = 0 is unstable.

Let us now discuss the second solution, namely \omega^2 = 2gA. The solution implies that every point x \in \mathbb{R} is a fixed point, when this condition is satisfied. It can easily be verified, though, that all points x are unstable (or, more correctly, each one is a separatrix). Indeed, suppose we consider the stability of the system at a point x = x_0. We consider x = x_0 + \delta x with small perturbations \delta x about point x. The Lagrange equation of motion rewrites:

\delta \ddot{x} (4A^2 x^2_0 + 1 ) = \delta x (\omega^2 - 2gA) + (\omega^2 - 2gA) x_0,

\delta \ddot{x} (4a^2 x_0^2 + 1) = 0,

(note that \forall a, x_0 \in \mathbb{R} : 4a^2 x_0^2 + 1 >0)

\delta \ddot{x} = 0,

\delta \dot{x} = \mathrm{const},

\Rightarrow motion is unstable, unless \delta \dot{x} = 0 to begin with.

Remarks: the solution involves Lagrangian mechanics. This particular problem can be solved using simple Newtonian formulation. Such solution would, in principle, be quite easier, for there would be no need to look for auxiliary tools like the Lagrangian and hence no need to take three derivatives of it. However, the Newtonian formulation would be a perfect choice for finding the second solution to the Lagrange equation for fixed points (namely, \omega^2 = 2gA.) Such arguments are presented here. Indeed, we can see that we here have arrived at the fact that the bead will rotate in a circle of a given radius if and only if

A = \displaystyle\frac{\omega^2}{2g},

precisely the result found here.

On the other hand, the Newtonian formulation does not reveal the subtle nature of stability of the system in various other configurations, discussed above. This is in opposition to the Lagrangian formulation, which naturally leads to the analysis of these configuration.

One further advantage of the Lagrange formulation is that it possesses a by far greater generality than the Newton’s solution. For example, the above solution can easily be generalised to the case when the wire’s shape is bent into a biquadratic parabola, y = Ax^4 (where, perhaps surprisingly, one finds two equilibrium points, none of which is located at the origin), or any other analytic smooth function for that matter.

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Posted by Nikita