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Chapter 7: Problem 12

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 $$\frac{\partial \mathcal{L}}{\partial x}+F_{\mathrm{fric}}=\frac{d}{d t} \frac{\partial \mathcal{L}}{\partial \dot{x}}.$$

Short Answer

Expert verified
Adding the nonconservative force as \(F_{\text{fric}}\), the equation becomes \(\frac{\partial \mathcal{L}}{\partial x} + F_{\text{fric}} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}}\)."

Step by step solution

01

Write Lagrange's Equation

Lagrange's equation for a single coordinate, say \(x\), in the presence of conservative forces is given by \( \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}} - \frac{\partial \mathcal{L}}{\partial x} = 0 \), where \(\mathcal{L}\) is the Lagrangian, \(T\) is kinetic energy, and \(U\) is the potential energy resulting in \( \mathcal{L} = T - U \).
02

Express Conservative Force

For a conservative force, we have \( F = -\frac{\partial U}{\partial x} \). Therefore, the term involving the partial derivative with respect to \(x\) is: \( -\frac{\partial U}{\partial x} = F \).
03

Consider Nonconservative Force

Introduce the nonconservative force \( F_{\text{fric}} \). We want the equation to be modified to include this force, so on the left side of the Lagrange's equation, we add \( F_{\text{fric}} \) to account for this force.
04

Write the Modified Equation

The equation now becomes \( \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}} - \frac{\partial \mathcal{L}}{\partial x} + F_{\text{fric}} = 0 \). Rearrange the equation to have \( \frac{\partial \mathcal{L}}{\partial x} + F_{\text{fric}} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}} \).

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Key Concepts

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

Conservative Forces
Conservative forces are those that conserve mechanical energy in a system. This means that the work done by a conservative force like gravity or spring force is stored as potential energy and can be fully recovered. A classic example is gravity, which allows energy to be converted between potential and kinetic forms without any loss.
  • The force is path-independent, meaning the total work done depends only on the initial and final positions, not on the specific path taken.
  • It can be expressed as the negative gradient of potential energy, typically formulated as \( F = -abla U \).
  • In systems governed primarily by conservative forces, Lagrange's equations can be applied directly to analyze motion.
Conservative forces provide a foundation in physics for deriving the equations of motion through potential energy functions, facilitating the use of analytical mechanics.
Nonconservative Forces
Nonconservative forces, unlike conservative forces, do not conserve mechanical energy. This means that they cause a loss of energy in the form of heat, sound, or other forms. Common examples include friction and air resistance.
  • These forces depend on the path taken, meaning the work done is not solely a function of the starting and ending points but varies with the path.
  • They are often considered a source of dissipation in mechanical systems, leading to a gradual loss of energy over time.
  • When nonconservative forces act on a system, Lagrange's equations are modified by including additional terms to account for these forces.
Understanding nonconservative forces is crucial in real-world applications where friction and resistance play significant roles, such as in automotive dynamics and machinery.
Potential Energy
Potential energy is the stored energy in a system due to its position or configuration. It is a crucial concept in mechanics that enables us to understand how energy is conserved and transformed.
  • Common forms of potential energy are gravitational potential energy \( U = mgh \) and elastic potential energy \( U = \frac{1}{2}kx^2 \) from a spring.
  • Potential energy can be converted to kinetic energy, and vice versa, allowing the total mechanical energy to remain constant in conservative systems.
  • The potential energy function \( U \) plays a central role in establishing the Lagrangian \( \mathcal{L} = T - U \), which is used to derive equations of motion in a system.
By analyzing potential energy, physicists can predict system behavior, such as the oscillation of a pendulum or orbit of planets.
Lagrangian Mechanics
Lagrangian mechanics provides a powerful framework for analyzing the motion of complex systems. It utilizes the concept of the Lagrangian \( \mathcal{L} = T - U \) to derive equations of motion.
  • Here, \( T \) represents kinetic energy and \( U \) the potential energy, providing a difference that reflects the energy state of the system.
  • This approach simplifies many aspects of dynamics, especially when dealing with problems that involve multiple constraints or generalized coordinates.
  • The principle of least action forms the basis of Lagrangian mechanics, where the actual path taken by a system is the one that minimizes the action \( S = \int \mathcal{L} \, dt \).
The power of Lagrangian mechanics lies in its ability to handle a broad range of problems, from the simple pendulum to the motion of particles in a field, making it a cornerstone of modern theoretical physics.

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Most popular questions from this chapter

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?

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.]

A mass \(m\) is suspended from a massless string, the other end of which is wrapped several times around a horizontal cylinder of radius \(R\) and moment of inertia \(I\), which is free to rotate about a fixed horizontal axle. Using a suitable coordinate, set up the Lagrangian and the Lagrange equation of motion, and find the acceleration of the mass \(m\). [The kinetic energy of the rotating cylinder is \(\frac{1}{2} I \omega^{2} .\)]

(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.

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.
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