Question

Two equal masses, \(m_{1}=m_{2}=m,\) are joined by a massless string of length \(L\) that passes through a hole in a frictionless horizontal table. The first mass slides on the table while the second hangs below the table and moves up and down in a vertical line. (a) Assuming the string remains taut, write down the Lagrangian for the system in terms of the polar coordinates \((r, \phi)\) of the mass on the table. (b) Find the two Lagrange equations and interpret the \(\phi\) equation in terms of the angular momentum \(\ell\) of the first mass. (c) Express \(\dot{\phi}\) in terms of \(\ell\) and eliminate \(\dot{\phi}\) from the \(r\) equation. Now use the \(r\) equation to find the value \(r=r_{0}\) at which the first mass can move in a circular path. Interpret your answer in Newtonian terms. (d) Suppose the first mass is moving in this circular path and is given a small radial nudge. Write \(r(t)=r_{0}+\epsilon(t)\) and rewrite the \(r\) equation in terms of \(\epsilon(t)\) dropping all powers of \(\epsilon(t)\) higher than linear. Show that the circular path is stable and that \(r(t)\) oscillates sinusoidally about \(r_{\mathrm{o}}\) What is the frequency of its oscillations?

Short answer

The system is stable for \(r = r_0 = \left(\frac{\ell^2}{m^2g}\right)^{1/3}\), and \(r(t)\) oscillates with frequency \(\omega = \sqrt{\frac{3\ell^2}{m^2r_0^4}}\).

Step-by-step solution

Lagrangian Setup

For this system, we have two types of energy: kinetic and potential. The first mass moves on a table with polar coordinates \((r, \phi)\), having kinetic energy \(T_1 = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\phi}^2)\). The second mass has potential energy \(U = mg(L - r)\) and kinetic energy \(T_2 = \frac{1}{2}m\dot{r}^2\). The Lagrangian \(L\) is given by the sum of kinetic energies minus the potential energy: \(L = T_1 + T_2 - U = \frac{1}{2}m(2\dot{r}^2 + r^2\dot{\phi}^2) - mg(L - r)\).

Deriving Lagrange Equations

Using the Euler-Lagrange equation \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) = \frac{\partial L}{\partial q}\) for \(r\) and \(\phi\), we get: \(m\ddot{r} - mr\dot{\phi}^2 + mg = 0\) and \(\frac{d}{dt}(mr^2\dot{\phi}) = 0\). The \(\phi\) equation implies that angular momentum \(\ell = mr^2\dot{\phi}\) is conserved.

Express \(\dot{\phi}\) in Terms of \(\ell\)

From angular momentum conservation, \(\ell = mr^2\dot{\phi}\), we find \(\dot{\phi} = \frac{\ell}{mr^2}\). Substitute this into the \(r\)-equation: \(m\ddot{r} - m\left(\frac{\ell}{mr^2}\right)^2r + mg = 0\), resulting in \(\ddot{r} = \frac{\ell^2}{m^2r^3} - g\).

Circular Path Condition

For a stable circular path \(\ddot{r} = 0\) and \(r = r_0\). Solving \(\frac{\ell^2}{m^2r_0^3} = g\) gives \(r_0 = \left(\frac{\ell^2}{m^2g}\right)^{1/3}\). Newtonian interpretation: Radius is set by balancing centripetal force \(\frac{m\ell^2}{m^2r_0^3}\) and gravitational force \(mg\).

Perturbation and Oscillation Frequency

For small perturbations \(r(t) = r_0 + \epsilon(t)\), substitute into the modified \(r\) equation and linearize: \(m\ddot{\epsilon} + \frac{3\ell^2}{m^2r_0^4}\epsilon = 0\). This is a simple harmonic oscillator equation, \(\ddot{\epsilon} + \omega^2\epsilon = 0\), with \(\omega^2 = \frac{3\ell^2}{m^2r_0^4}\). The frequency is \(\omega = \sqrt{\frac{3\ell^2}{m^2r_0^4}}\).

Key concepts

Lagrange Equations

Lagrange equations form the mathematical foundation of Lagrangian mechanics. They offer a powerful method to analyze the dynamics of a system by considering both kinetic and potential energy.
Unlike Newton's laws, which deal directly with forces, Lagrange equations focus on energy transformations and symmetries in the system.
For any generalized coordinate \( q \), the Euler-Lagrange equation is given by:\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) = \frac{\partial L}{\partial q} \]Here, \( L \) is the Lagrangian, defined as the difference between kinetic energy \( T \) and potential energy \( U \):

• \( L = T - U \)

In our exercise, the setup involves two sets of coordinates, \( r \) and \( \phi \), reflecting the radial and angular movements of the mass on the table.
This setup allows us to write two distinct Lagrange equations, which lead to insights about the system's dynamics. The transformations and equations derived offer a strong perspective on the mechanics at play.

Angular Momentum Conservation

Angular momentum is a key concept in rotational dynamics, describing how much rotation a body has. It is conserved in isolated systems, meaning it remains constant if no external torque is acting.
In the exercise, we see this conservation principle in action through the relationship distinct for the angular coordinate \( \phi \):\[ \frac{d}{dt}(mr^2\dot{\phi}) = 0 \]This indicates that \( \ell = mr^2\dot{\phi} \), the angular momentum, remains constant over time. It's a central insight when interpreting the motions of the masses.
One consequence of this conservation is the ability to relate \( \dot{\phi} \) back to \( \ell \):

• \( \dot{\phi} = \frac{\ell}{mr^2} \)

This relationship simplifies the analysis, allowing us to eliminate \( \dot{\phi} \) from other equations. The understanding of this conservation law helps in analyzing circular or rotational movements.

Harmonic Oscillator

A harmonic oscillator is a system that experiences a force proportional to its displacement from equilibrium, often resulting in sinusoidal oscillations. It's a foundational concept in physics, representing the simplest type of periodic motion.
In our situation, the first mass executes circular motion and is described by an approximate linearized equation:\[ m\ddot{\epsilon} + \frac{3\ell^2}{m^2r_0^4}\epsilon = 0 \]Here, \( \epsilon(t) \) represents small deviations or perturbations from the stable circular path. This equation is characteristic of a simple harmonic oscillator, implying that \( r(t) \) undergoes sinusoidal oscillations around \( r_0 \).
The oscillation frequency \( \omega \) is determined by:

• \( \omega = \sqrt{\frac{3\ell^2}{m^2r_0^4}} \)

This behavior underscores the stability of the circular path and provides insights into potential oscillatory motions present in dynamic systems.

Centripetal Force

Centripetal force is essential for keeping an object moving in a circular path. It acts inward, towards the center of the circle, maintaining the curved trajectory.
For the given problem, the balance between centripetal and gravitational forces determines the steady state circular radius \( r_0 \). This balance is represented as:\[ \frac{m\ell^2}{m^2r_0^3} = g \]This equation arises from setting \( \ddot{r} = 0 \) in the radial Lagrange equation. Here, \( g \) is the gravitational acceleration affecting the second mass. From a Newtonian perspective, the centripetal force needed for equilibrium comes from the tension in the string, adjusted for hanging mass weight.
This concept illustrates the linkage between rotational dynamics and linear forces, providing a threshold for circular motion stability. It's a vivid example of how forces come together to produce a consistent rotational motion in mechanical systems.