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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

Write down the Lagrangian for a cylinder (mass \(m\), radius \(R\), and moment of inertia \(I\) ) that rolls without slipping straight down an inclined plane which is at an angle \(\alpha\) from the horizontal. Use as your generalized coordinate the cylinder's distance \(x\) measured down the plane from its starting point. Write down the Lagrange equation and solve it for the cylinder's acceleration \(\ddot{x}\). Remember that \(T=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2},\) where \(v\) is the velocity of the center of mass and \(\omega\) is the angular velocity.

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.]

Consider the well-known problem of a cart of mass \(m\) moving along the \(x\) axis attached to a spring (force constant \(k\) ), whose other end is held fixed (Figure 5.2 ). If we ignore the mass of the spring (as we almost always do) then we know that the cart executes simple harmonic motion with angular frequency \(\omega=\sqrt{k / m} .\) Using the Lagrangian approach, you can find the effect of the spring's mass \(M,\) as follows: (a) Assuming that the spring is uniform and stretches uniformly, show that its kinetic energy is \(\frac{1}{6} M \dot{x}^{2} .\) (As usual \(x\) is the extension of the spring from its equilibrium length.) Write down the Lagrangian for the system of cart plus spring. (Note: The potential energy is still \(\frac{1}{2} k x^{2}\).) (b) Write down the Lagrange equation and show that the cart still executes SHM but with angular frequency \(\omega=\sqrt{k /(m+M / 3)} ;\) that is, the effect of the spring's mass \(M\) is just to add \(M / 3\) to the mass of the cart.

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.

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.]

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