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A pendulum is made from a massless spring (force constant \(k\) and unstretched length \(l_{\mathrm{o}}\)) that is suspended at one end from a fixed pivot \(O\) and has a mass \(m\) attached to its other end. The spring can stretch and compress but cannot bend, and the whole system is confined to a single vertical plane. (a) Write down the Lagrangian for the pendulum, using as generalized coordinates the usual angle \(\phi\) and the length \(r\) of the spring. (b) Find the two Lagrange equations of the system and interpret them in terms of Newton's second law, as given in Equation (1.48). (c) The equations of part (b) cannot be solved analytically in general. However, they can be solved for small oscillations. Do this and describe the motion. [Hint: Let \(l\) denote the equilibrium length of the spring with the mass hanging from it and write \(r=l+\epsilon .\) "Small oscillations" involve only small values of \(\epsilon\) and \(\phi,\) so you can use the small-angle approximations and drop from your equations all terms that involve powers of \(\epsilon\) or \(\phi\) (or their derivatives) higher than the first power (also products of \(\epsilon\) and \(\phi\) or their derivatives). This dramatically simplifies and uncouples the equations.]

Short Answer

Expert verified
The Lagrangian is \( \mathcal{L} = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\phi}^2 + mgr\cos\phi - \frac{1}{2} k (r - l_{\mathrm{o}})^2 \). For small oscillations, \( \epsilon \) and \( \phi \) oscillate independently.

Step by step solution

01

Kinetic Energy of the Pendulum

The kinetic energy consists of two parts: one due to the radial motion and the other due to angular motion. The radial part is \\[ T_r = \frac{1}{2} m \dot{r}^2 \] \and the angular part is \\[ T_\phi = \frac{1}{2} m r^2 \dot{\phi}^2. \] \Thus, the total kinetic energy is \\[ T = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\phi}^2. \]
02

Potential Energy of the System

The potential energy of the system is due to the gravitational potential energy and the elastic potential energy of the spring. The gravitational potential energy is \\[ V_g = -mgr\cos\phi. \] \The elastic energy stored in the spring is \\[ V_s = \frac{1}{2} k (r - l_{\mathrm{o}})^2. \] \Therefore, the total potential energy is \\[ V = -mgr\cos\phi + \frac{1}{2} k (r - l_{\mathrm{o}})^2. \]
03

Lagrangian of the System

The Lagrangian \( \mathcal{L} \) is defined as the difference between the kinetic and potential energies. Thus, \\[ \mathcal{L} = T - V = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\phi}^2 + mgr\cos\phi - \frac{1}{2} k (r - l_{\mathrm{o}})^2. \]
04

Lagrange's Equations

The Lagrangian's equations are derived by applying the Euler-Lagrange equation for each generalized coordinate \( q \), given by \\[ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}} \right) - \frac{\partial \mathcal{L}}{\partial q} = 0. \] \Let's first apply it to \( r \): \\[ m \ddot{r} - mr\dot{\phi}^2 + mg\cos\phi - k(r - l_{\mathrm{o}}) = 0. \] \Now apply it to \( \phi \): \\[ \frac{d}{dt}(mr^2\dot{\phi}) + mgr\sin\phi = 0. \]
05

Simplification for Small Oscillations

Considering small oscillations where \( r = l + \epsilon \) and adopting small-angle approximations \( \sin\phi \approx \phi \) and \( \cos\phi \approx 1 \), the Lagrangian equations simplify. For the radial equation: \\[ m \ddot{\epsilon} - ml\dot{\phi}^2 + (k - \frac{mg}{l})\epsilon = 0. \] \For the angular motion: \\[ \frac{d}{dt}(ml^2\dot{\phi}) + mgl\phi \approx mgl\phi = 0. \] \This implies harmonic oscillations in both \( r \) and \( \phi \).
06

Interpretation of Small Oscillation Motion

The radial equation suggests that \( \epsilon \) oscillates harmonically with a modified spring constant \( k - \frac{mg}{l} \). The angular equation forms a simple harmonic motion in \( \phi \), suggesting a simple pendulum motion. Both \( \epsilon \) and \( \phi \) oscillate independently when small values are assumed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Small Oscillations
In the context of physics, small oscillations refer to the slight deviations from an equilibrium position that a system undergoes. When a system oscillates, it moves back and forth. For small oscillations, these movements are minimal, allowing certain approximations to make complex calculations easier. In the case of our pendulum exercise, when the spring's length changes slightly around its equilibrium length, we describe this change as small oscillations, represented by \(\epsilon\). Likewise, small deviations in angle \(\phi\) are considered small oscillations.

For small oscillations, we use the small-angle approximation:
  • \(\sin\phi \approx \phi\)
  • \(\cos\phi \approx 1\)
These approximations are valid when the angles are nearly zero, allowing us to simplify complex trigonometric expressions into linear expressions. This becomes useful in uncoupling the equations of motion for the pendulum. By assuming \(\epsilon\) and \(\phi\) are small, we can eliminate higher order terms in those variables, simplifying our calculations.

This approach is not only applicable to pendulums but to any oscillating system that stays close to its equilibrium. It is a foundational concept that finds utility in various physics and engineering applications.
Simple Harmonic Motion
Simple harmonic motion (SHM) is a specific type of oscillatory motion where the restoring force is directly proportional to displacement, and acts in the opposite direction. In this exercise, the pendulum undergoes simple harmonic motion both in its radial and angular components.

For the radial motion, the equation derived:
  • \(m \ddot{\epsilon} + (k - \frac{mg}{l})\epsilon = 0\)
resembles the standard format of simple harmonic motion: \(m \ddot{x} + kx = 0\). Here \(x\) is replaced by \(\epsilon\), and the modified spring constant is \(k - \frac{mg}{l}\). This implies that the spring stretches and compresses symmetrically about its equilibrium position, demonstrating SHM.

Meanwhile, for the angular motion, the simplified equation:
  • \(mgl\phi = 0\)
indicates that \(\phi\) follows a pendulum-like SHM. The angle oscillates around its resting position, much like a conventional simple pendulum does under the influence of gravity.

Simple harmonic motion is characterized by:
  • Sinusoidal patterns (like sine or cosine waves)
  • A constant frequency and period
  • The system's total energy swaying between potential and kinetic forms
Studying SHM provides essential insights into more complex oscillating systems and underpins many physical processes from waves to oscillating circuits.
Euler-Lagrange Equation
The Euler-Lagrange equation is a central tool in Lagrangian mechanics, providing a formal way to derive equations of motion for a system. It is based on the principle of least action and is applicable in both classical and modern physics.

In this exercise, the Euler-Lagrange equation is used to find the equations of motion of the pendulum system. This is accomplished by treating \(\phi\) and \(r\) as generalized coordinates in the Lagrangian \(\mathcal{L}\), which is the difference between kinetic and potential energy:
\[\mathcal{L} = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\phi}^2 + mgr\cos\phi - \frac{1}{2} k (r - l_{\mathrm{o}})^2\]

By applying the equation:
  • \(\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}} \right) - \frac{\partial \mathcal{L}}{\partial q} = 0\)
separately to both \(r\) and \(\phi\), we derive the system's equations of motion.

The genius of this method lies in its ability to handle complex systems in a systemic, clean approach, bypassing the need to directly apply Newton's second law. The Euler-Lagrange equation elegantly uses energy principles to yield motion descriptions. This abstraction is particularly powerful for multi-coordinate systems and forms the backbone of modern theoretical physics.

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Most popular questions from this chapter

Consider a double Atwood machine constructed as follows: A mass 4 \(m\) is suspended from a string that passes over a massless pulley on frictionless bearings. The other end of this string supports a second similar pulley, over which passes a second string supporting a mass of \(3 m\) at one end and \(m\) at the other. Using two suitable generalized coordinates, set up the Lagrangian and use the Lagrange equations to find the acceleration of the mass \(4 m\) when the system is released. Explain why the top pulley rotates even though it carries equal weights on each side.

(a) Write down the Lagrangian for a particle moving in three dimensions under the influence of a conservative central force with potential energy \(U(r),\) using spherical polar coordinates \((r, \theta, \phi)\). (b) Write down the three Lagrange equations and explain their significance in terms of radial acceleration, angular momentum, and so forth. (The \(\theta\) equation is the tricky one, since you will find it implies that the \(\phi\) component of \(\ell\) varies with time, which seems to contradict conservation of angular momentum. Remember, however, that \(\ell_{\phi}\) is the component of \(\ell\) in a variable direction.) (c) Suppose that initially the motion is in the equatorial plane (that is, \(\theta_{0}=\pi / 2\) and \(\dot{\theta}_{0}=0\) ). Describe the subsequent motion. (d) Suppose instead that the initial motion is along a line of longitude (that is, \(\dot{\phi}_{0}=0\) ). Describe the subsequent motion.

A smooth wire is bent into the shape of a helix, with cylindrical polar coordinates \(\rho=R\) and \(z=\lambda \phi,\) where \(R\) and \(\lambda\) are constants and the \(z\) axis is vertically up (and gravity vertically down). Using \(z\) as your generalized coordinate, write down the Lagrangian for a bead of mass \(m\) threaded on the wire. Find the Lagrange equation and hence the bead's vertical acceleration \(\ddot{z}\). In the limit that \(R \rightarrow 0\), what is \(\ddot{z} ?\) Does this make sense?

Noether's theorem asserts a connection between invariance principles and conservation laws. In Section 7.8 we saw that translational invariance of the Lagrangian implies conservation of total linear momentum. Here you will prove that rotational invariance of \(\mathcal{L}\) implies conservation of total angular momentum. Suppose that the Lagrangian of an \(N\) -particle system is unchanged by rotations about a certain symmetry axis. (a) Without loss of generality, take this axis to be the \(z\) axis, and show that the Lagrangian is unchanged when all of the particles are simultaneously moved from \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}\right)\) to \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}+\epsilon\right)\) (same \(\epsilon\) for all particles). Hence show that $$\sum_{\alpha=1}^{N} \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}}=0.$$ (b) Use Lagrange's equations to show that this implies that the total angular momentum \(L_{z}\) about the symmetry axis is constant. In particular, if the Lagrangian is invariant under rotations about all axes, then all components of \(\mathbf{L}\) are conserved.

A small cart (mass \(m\) ) is mounted on rails inside a large cart. The two are attached by a spring (force constant \(k\) ) in such a way that the small cart is in equilibrium at the midpoint of the large. The distance of the small cart from its equilibrium is denoted \(x\) and that of the large one from a fixed point on the ground is \(X,\) as shown in Figure \(7.13 .\) The large cart is now forced to oscillate such that \(X=A \cos \omega t,\) with both \(A\) and \(\omega\) fixed. Set up the Lagrangian for the motion of the small cart and show that the Lagrange equation has the form $$\ddot{x}+\omega_{0}^{2} x=B \cos \omega t$$ where \(\omega_{\mathrm{o}}\) is the natural frequency \(\omega_{\mathrm{o}}=\sqrt{k / m}\) and \(B\) is a constant. This is the form assumed in Section 5.5, Equation (5.57), for driven oscillations (except that we are here ignoring damping). Thus the system described here would be one way to realize the motion discussed there. (We could fill the large cart with molasses to provide some damping.)

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