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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?

Short Answer

Expert verified
In the limit \( R \rightarrow 0 \), \( \ddot{z} = -g \). This is consistent with free fall.

Step by step solution

01

Identify the Generalized Coordinate

The problem specifies using \( z \) as the generalized coordinate. Note that \( z = \lambda \phi \), meaning as the bead moves along the wire, \( z \) changes with \( \phi \). This will relate the radial and vertical motion.
02

Express the Kinetic Energy

The bead can move in both the radial and vertical directions. In cylindrical coordinates, the velocity components are \( \dot{\rho} = 0 \), \( \rho \dot{\phi} = R \dot{\phi} \), and \( \dot{z} \). The kinetic energy \( T \) is given by:\[T = \frac{1}{2} m \left((R \dot{\phi})^2 + (\dot{z})^2 \right)\]Now, since \( z = \lambda \phi \), we have \( \dot{z} = \lambda \dot{\phi} \). Substitute this into the kinetic energy expression:\[T = \frac{1}{2} m \left( R^2 \dot{\phi}^2 + (\lambda \dot{\phi})^2 \right) = \frac{1}{2} m \dot{\phi}^2 (R^2 + \lambda^2)\]Using \( \dot{\phi} = \frac{\dot{z}}{\lambda} \), the kinetic energy becomes \( T = \frac{m \dot{z}^2}{2 \lambda^2}(R^2 + \lambda^2) \).
03

Express the Potential Energy

The potential energy \( V \) is due to gravity, which is dependent on the vertical height. Therefore, \( V = mgz \).
04

Write the Lagrangian

The Lagrangian \( \mathcal{L} \) is defined as the difference between the kinetic and potential energies:\[\mathcal{L} = T - V = \frac{m}{2 \lambda^2} (R^2 + \lambda^2) \dot{z}^2 - mgz\]
05

Derive the Euler-Lagrange Equation

Using the Euler-Lagrange equation \( \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{z}}\right) - \frac{\partial \mathcal{L}}{\partial z} = 0 \), find the expression:1. Calculate \( \frac{\partial \mathcal{L}}{\partial \dot{z}} = \frac{m}{\lambda^2}(R^2 + \lambda^2) \dot{z} \)2. Then compute \( \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{z}}\right) = \frac{m}{\lambda^2}(R^2 + \lambda^2) \ddot{z} \)3. Finally, calculate \( \frac{\partial \mathcal{L}}{\partial z} = -mg \)Substitute into the Euler-Lagrange equation:\[\frac{m}{\lambda^2}(R^2 + \lambda^2) \ddot{z} + mg = 0\]
06

Solve for Vertical Acceleration \(\ddot{z}\)

From the Euler-Lagrange equation, isolate \( \ddot{z} \):\[\ddot{z} = -\frac{\lambda^2 g}{R^2 + \lambda^2}\]
07

Evaluate the Limit \( R \rightarrow 0 \)

Substitute \( R = 0 \) into the expression for \( \ddot{z} \):\[\ddot{z} = -g\]This matches the free-fall acceleration due to gravity, indicating that in this limit, the bead would fall freely as if there were no constraints (the wire becomes a vertical line).
08

Conclusion

When \( R \rightarrow 0 \), \( \ddot{z} = -g \) describes the bead's acceleration under gravity alone, with no radial constraint. This makes physical sense as the helix collapses into a vertical line.

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Key Concepts

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

Helix
A helix is a fascinating geometric shape that resembles a spiral or coil. Imagine a wire twisted in a perfect spring-like form. In our exercise, the helix is not just any helix; it has a constant radius and is defined in cylindrical polar coordinates. This type of helix has a fixed cylindrical radius, denoted as \( \rho = R \), which determines how far it is from the central axis at any point. The vertical position \( z \) relates to the angular displacement \( \phi \) through the equation \( z = \lambda \phi \). This means that for every complete circle around the helix, there is a fixed vertical rise \( \lambda \). This constant rise-to-turn ratio gives the helix its uniformly spiraling nature. Understanding these properties helps in solving problems involving motion along helices, where both radial and vertical motions are intricately linked.
Cylindrical Polar Coordinates
Cylindrical polar coordinates are a three-dimensional coordinate system useful for describing positions in space where rotational symmetry exists. In this system, any point in space is described using three values: the radial distance \( \rho \), the azimuthal angle \( \phi \), and the height \( z \). Try to envision it as a combination of polar coordinates, which you might use on a flat circle, with an added vertical component. This makes it particularly useful for problems involving round, cylindrical, or spiral shapes, such as the helix in our problem.
  • \( \rho \) indicates how far from the cylinder's central axis the point is.
  • \( \phi \) describes the angle around that axis, much like longitude on Earth.
  • \( z \) is simply the vertical height, indicating how high or low a point is.
By understanding this coordinate system, you can better describe and analyze movement in spirals and rotations.
Vertical Acceleration
Vertical acceleration refers to the change in velocity along the vertical axis over time. In this exercise, we consider the motion of a bead along a helix, which involves both radial and vertical components. To find the bead's vertical acceleration, we employ Lagrangian mechanics, which simplifies the process of dealing with complex motion by focusing on energy rather than forces. Using the Lagrangian, which is derived from the difference between kinetic and potential energy, we apply the Euler-Lagrange equation to determine \( \ddot{z} \), the vertical acceleration.
The result \( \ddot{z} = -\frac{\lambda^2 g}{R^2 + \lambda^2} \) shows how both the geometry of the helix and gravitational force influence the bead’s acceleration. Notably, when \( R \rightarrow 0 \), the vertical acceleration simplifies to \( -g \), signifying free fall under gravity without radial constraint. This insight highlights the beauty of how theoretical physics matches our expectations based on everyday experiences.

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Most popular questions from this chapter

Consider a bead that is threaded on a rigid circular hoop of radius \(R\) lying in the \(x y\) plane with its center at \(O,\) and use the angle \(\phi\) of two- dimensional polar coordinates as the one generalized coordinate to describe the bead's position. Write down the equations that give the Cartesian coordinates \((x, y)\) in terms of \(\phi\) and the equation that gives the generalized coordinate \(\phi\) in terms of \((x, y)\).

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.

Consider a particle of mass \(m\) and charge \(q\) moving in a uniform constant magnetic field \(\mathbf{B}\) in the \(z\) direction. (a) Prove that \(\mathbf{B}\) can be written as \(\mathbf{B}=\nabla \times \mathbf{A}\) with \(\mathbf{A}=\frac{1}{2} \mathbf{B} \times \mathbf{r} .\) Prove equivalently that in cylindrical polar coordinates, \(\mathbf{A}=\frac{1}{2} B \rho \hat{\phi}\). (b) Write the Lagrangian (7.103) in cylindrical polar coordinates and find the three corresponding Lagrange equations. (c) Describe in detail those solutions of the Lagrange equations in which \(\rho\) is a constant.

Lagrange's equations in the form discussed in this chapter hold only if the forces (at least the nonconstraint forces) are derivable from a potential energy. To get an idea how they can be modified to include forces like friction, consider the following: A single particle in one dimension is subject to various conservative forces (net conservative force \(=F=-\partial U / \partial x)\) and a nonconservative force (let's call it \(F_{\text {fric }}\) ). Define the Lagrangian as \(\mathcal{L}=T-U\) and show that the appropriate modification is $$\frac{\partial \mathcal{L}}{\partial x}+F_{\mathrm{fric}}=\frac{d}{d t} \frac{\partial \mathcal{L}}{\partial \dot{x}}.$$

(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.

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