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The "spherical pendulum" is just a simple pendulum that is free to move in any sideways direction. (By contrast a "simple pendulum"- unqualified - is confined to a single vertical plane.) The bob of a spherical pendulum moves on a sphere, centered on the point of support with radius \(r=R\) the length of the pendulum. A convenient choice of coordinates is spherical polars, \(r, \theta, \phi,\) with the origin at the point of support and the polar axis pointing straight down. The two variables \(\theta\) and \(\phi\) make a good choice of generalized coordinates. (a) Find the Lagrangian and the two Lagrange equations. (b) Explain what the \(\phi\) equation tells us about the \(z\) component of angular momentum \(\ell_{z^{*}}\) (c) For the special case that \(\phi=\) const, describe what the \(\theta\) equation tells us. (d) Use the \(\phi\) equation to replace \(\dot{\phi}\) by \(\ell_{z}\) in the \(\theta\) equation and discuss the existence of an angle \(\theta_{\mathrm{o}}\) at which \(\theta\) can remain constant. Why is this motion called a conical pendulum? (e) Show that if \(\theta=\theta_{0}+\epsilon,\) with \(\epsilon\) small, then \(\theta\) oscillates about \(\theta_{\mathrm{o}}\) in harmonic motion. Describe the motion of the pendulum's bob.

Short Answer

Expert verified
The pendulum's motion is governed by the Lagrangian mechanics, resulting in conical, harmonic oscillations about a stable angle.

Step by step solution

01

Define the Lagrangian

The Lagrangian, \( L \), is defined as the difference between the kinetic \( T \) and potential \( V \) energies of the system: \( L = T - V \). For a spherical pendulum, the kinetic energy in spherical coordinates is \( T = \frac{1}{2} m R^2 (\dot{\theta}^2 + \sin^2 \theta \dot{\phi}^2) \) and the potential energy is \( V = mgR(1 - \cos \theta) \). Thus, the Lagrangian is \( L = \frac{1}{2} m R^2 (\dot{\theta}^2 + \sin^2 \theta \dot{\phi}^2) - mgR(1 - \cos \theta) \).
02

Find the Euler-Lagrange Equations

The Lagrange equations for \( \theta \) and \( \phi \) are derived from the Euler-Lagrange equation \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0 \), where \( q \) is the generalized coordinate. Applying this to \( \theta \) and \( \phi \), we get:1. \[ \frac{d}{dt} (mR^2 \dot{\theta}) = mR^2 \sin \theta \cos \theta \dot{\phi}^2 - mgR \sin \theta \]2. \[ \frac{d}{dt} (mR^2 \sin^2 \theta \dot{\phi}) = 0 \]
03

Analyze the \( \phi \) Equation

The equation \( \frac{d}{dt} (mR^2 \sin^2 \theta \dot{\phi}) = 0 \) implies that the quantity \( mR^2 \sin^2 \theta \dot{\phi} \) is conserved. This represents the conservation of the \( z \) component of angular momentum \( \ell_{z^{*}} \).
04

Evaluate Special Case \( \phi = \) const

For the special case where \( \phi \) is constant, \( \dot{\phi} = 0 \) and consequently, the system simplifies to one-dimensional motion for \( \theta \). The \( \theta \) equation reduces to simple harmonic motion in \( \theta \). The pendulum oscillates in a single plane like a simple pendulum.
05

Replace \( \dot{\phi} \) Using \( \ell_{z} \)

Using the conservation of angular momentum, \( \dot{\phi} = \frac{\ell_{z}}{mR^2 \sin^2 \theta} \) can replace \( \dot{\phi} \) in the \( \theta \) equation. This substitution helps to find a stable equilibrium angle \( \theta_o \) where \( \theta \) can remain constant, leading to the concept of a conical pendulum where the pendulum swings in a conical motion.
06

Prove Harmonic Oscillation for Small \( \epsilon \)

Consider \( \theta = \theta_{0} + \epsilon \) with \( \epsilon \) small. Linearize the \( \theta \) equation around \( \theta_0 \) to find that \( \epsilon \) satisfies the equation for harmonic oscillation. The pendulum oscillates about \( \theta_0 \) with small oscillations, indicating it undergoes simple harmonic motion about the cone's axis.

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Key Concepts

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

Lagrangian mechanics
Understanding the movement of systems with Lagrangian mechanics is a fundamental aspect of physics studying motion. The Lagrangian, denoted by \( L \), plays a crucial role in formulating the equations of motion. In the context of the spherical pendulum, the Lagrangian is expressed as the difference between the system's kinetic and potential energies: \( L = T - V \). Here, \( T \) represents the kinetic energy, given by \( T = \frac{1}{2} m R^2 (\dot{\theta}^2 + \sin^2 \theta \dot{\phi}^2) \), while \( V \) denotes the potential energy, \( V = mgR(1 - \cos \theta) \). By setting up these equations, one can analyze the motion dynamics of the pendulum.
In simpler terms, the Lagrangian encompasses all the energy aspects of a system: how it moves, how fast it moves, and how high it might swing. This formulation allows us to derive important equations that govern the dynamics, specifically the Euler-Lagrange equations, bringing us a step closer to understanding the complex movements of a spherical pendulum.
Angular momentum
The conservation of angular momentum is a pivotal concept in understanding spherical pendulum motion. Angular momentum, particularly its component \( \ell_{z^{*}} \) (momentum related to the \( z \) direction), is conserved in isolated systems like our pendulum. This means that \( mR^2 \sin^2 \theta \dot{\phi} \), which represents the \( z \) component of angular momentum, remains constant over time. By analyzing this, we can deduce key characteristics of motion.
  • Angular momentum conservation tells us that as a pendulum swings and changes its angle \( \theta \), the speed at which it swings around the vertical axis \( \dot{\phi} \) varies to balance any changes in \( \theta \).
  • This conservation directly impacts how rotational motion is maintained, thus defining consistent, predictable motion patterns for the pendulum.
  • This is why angular dynamics are often more predictable than other types, given the lack of external forces disrupting this balance.
Euler-Lagrange equations
Euler-Lagrange equations are foundational tools in mechanics, especially for analyzing dynamic systems like the spherical pendulum. These equations are formulated as \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0 \) for a generalized coordinate \( q \). They allow us to derive the necessary equations of motion for the pendulum.
For a spherical pendulum, applying this form:
  • If \( q = \theta \), you derive an equation capturing linear dynamics affected by angles.
  • If \( q = \phi \), you gain insight into the rotational components, reflected in the conservation of angular momentum.
These insights enable physicists to explore the deeper mechanics involved in pendulum motion. They are essential for understanding momentum variations and rotational influences. By translating the Lagrangian into the Euler-Lagrange formalism, we distill its essence down to precise, solvable mathematical equations.
Harmonic motion
Harmonic motion becomes particularly significant in the analysis of a spherical pendulum. When the angle \( \theta \) slightly deviates from a mean position \( \theta_0 \), we see this motion come into play. Hence, if \( \theta = \theta_0 + \epsilon \) with \( \epsilon \) being small, the resulting motion can be approximated as simple harmonic. This means:
  • The pendulum bob undergoes minor oscillations about \( \theta_0 \), behaving as a harmonic oscillator.
  • This behavior is akin to how springs and waves operate, with periodic back-and-forth movements centered around a stable point.
  • In the context of the spherical pendulum, such oscillations mean gentle, predictable swagging within the conical path—a fundamental harmonic trait.
Thus, recognizing harmonic motion in these oscillations simplifies the complexity of pendulum motion, giving us clear insights into its predictable patterns of repeated movement.

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Most popular questions from this chapter

Prove that the potential energy of a central force \(\mathbf{F}=-k r^{n} \hat{\mathbf{r}}(\text { with } n \neq-1)\) is \(U=k r^{n+1} /(n+1)\). In particular, if \(n=1,\) then \(\mathbf{F}=-k \mathbf{r}\) and \(U=\frac{1}{2} k r^{2}\).

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.)

The center of a long frictionless rod is pivoted at the origin, and the rod is forced to rotate in a horizontal plane with constant angular velocity \(\omega\). Write down the Lagrangian for a bead threaded on the rod, using \(r\) as your generalized coordinate, where \(r, \phi\) are the polar coordinates of the bead. (Notice that \(\phi\) is not an independent variable since it is fixed by the rotation of the rod to be \(\phi=\omega t\).) Solve Lagrange's equation for \(r(t) .\) What happens if the bead is initially at rest at the origin? If it is released from any point \(r_{\mathrm{o}}>0,\) show that \(r(t)\) eventually grows exponentially. Explain your results in terms of the centrifugal force \(m \omega^{2} r\).

Consider a mass \(m\) moving in two dimensions with potential energy \(U(x, y)=\frac{1}{2} k r^{2},\) where \(r^{2}=x^{2}+y^{2} .\) Write down the Lagrangian, using coordinates \(x\) and \(y,\) and find the two Lagrange equations of motion. Describe their solutions. [This is the potential energy of an ion in an "ion trap," which can be used to study the properties of individual atomic ions.]

A particle is confined to move on the surface of a circular cone with its axis on the \(z\) axis, vertex at the origin (pointing down), and half-angle \(\alpha\). The particle's position can be specified by two generalized coordinates, which you can choose to be the coordinates \((\rho, \phi)\) of cylindrical polar coordinates. Write down the equations that give the three Cartesian coordinates of the particle in terms of the generalized coordinates ( \(\rho, \phi\) ) and vice versa.

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