/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Write down the Lagrangian for a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write down the Lagrangian for a projectile (subject to no air resistance) in terms of its Cartesian coordinates \((x, y, z),\) with \(z\) measured vertically upward. Find the three Lagrange equations and show that they are exactly what you would expect for the equations of motion.

Short Answer

Expert verified
The Lagrangian is \( L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - mgz \). The equations: \( \ddot{x} = 0 \), \( \ddot{y} = 0 \), \( \ddot{z} = -g \).

Step by step solution

01

Understanding the Lagrangian Function

The Lagrangian function, defined as \( L = T - V \), where \( T \) is the kinetic energy and \( V \) is the potential energy, is used to derive the equations of motion. For a projectile in three-dimensional space, we'll express both the kinetic and potential energy in terms of the Cartesian coordinates \((x, y, z)\).
02

Express Kinetic Energy in Cartesian Coordinates

The kinetic energy \( T \) of a projectile with mass \( m \) moving with velocity components \( \dot{x}, \dot{y}, \dot{z} \) is given by \( T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) \). These are the squared velocity components in the Cartesian system.
03

Express Potential Energy in Cartesian Coordinates

The potential energy \( V \) of a projectile in a gravitational field (with constant gravity \( g \)) depends on the vertical position \( z \). It is expressed as \( V = mgz \), where \( z \) is the height above the reference point.
04

Write the Lagrangian

Substitute the expressions for kinetic and potential energy into the Lagrangian: \[ L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - mgz \] This function represents the Lagrangian of the system.
05

Derive the Lagrange Equations

Using the Euler-Lagrange equation, \( \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0 \), apply it to each coordinate \( x, y, \) and \( z \).
06

Equation for the x-direction

Applying the Lagrange equation to \( x \):\[ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{x}} \right) = m\ddot{x} \quad \text{and} \quad \frac{\partial L}{\partial x} = 0 \]Thus, the equation becomes: \[ m\ddot{x} = 0 \] indicating no acceleration in the \( x \)-direction.
07

Equation for the y-direction

Similarly, apply the Lagrange equation to \( y \):\[ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{y}} \right) = m\ddot{y} \quad \text{and} \quad \frac{\partial L}{\partial y} = 0 \]Thus, \[ m\ddot{y} = 0 \] indicating no acceleration in the \( y \)-direction as well.
08

Equation for the z-direction

For the \( z \) coordinate, we have:\[ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{z}} \right) = m\ddot{z} \quad \text{and} \quad \frac{\partial L}{\partial z} = -mg \]This results in:\[ m\ddot{z} = -mg \] which represents the acceleration due to gravity.
09

Summary of Equations of Motion

The Lagrange equations lead to the expected equations of motion: \[ \ddot{x} = 0, \quad \ddot{y} = 0, \quad \ddot{z} = -g \] These equations show that the projectile moves with constant velocity in the \( x \) and \( y \) directions, and with constant acceleration \( -g \) in the \( z \) direction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Projectile Motion
Projectile motion describes the path of an object thrown into space, subject to gravitational forces alone. In our exercise, we're analyzing this motion without air resistance, focusing on three-dimensional space with coordinates
  • \( (x, y, z) \)
Here, \( z \) is vertical. The key factors in projectile motion are:
  • Initial Velocity: Determines how the object starts its trajectory.
  • Gravity: Acts downward, affecting vertical motion.
In our case, the Lagrange equations reveal that horizontal movements (\( x \) and \( y \)) have no acceleration. The vertical motion (\( z \)) accelerates down due to gravity (-g). This analysis gives us an understanding of how an object like a ball thrown from a cliff behaves under gravity's influence.
Euler-Lagrange Equation
The Euler-Lagrange equation is a fundamental principle to derive equations of motion in Lagrangian mechanics. It connects the Lagrangian function to physical trajectories of systems. The equation is:\[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0\]Where \( L = T - V \), it represents the difference between kinetic and potential energy. Here’s how it applies:
  • For each coordinate (\( x, y, z \)): We take partial derivatives related to velocity and position.
  • Motion Derivation: This leads to equations describing how the system evolves over time.
In our projectile exercise, applying this to \( x \) and \( y \) showed no force acting horizontally thus no acceleration (\( \ddot{x} = 0 \)). But for \( z \), gravity force exists, leading to \( \ddot{z} = -g \), reflecting downward acceleration. Understanding the Euler-Lagrange equation lets us derive these natural laws from simple principles.
Kinetic and Potential Energy
Kinetic and potential energies are key to understanding Lagrangian mechanics. They are components of the Lagrangian function \( L = T - V \), representing how energy changes in motion.

Kinetic Energy (T)

Kinetic energy is associated with the motion of the projectile. Given by:\[ T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) \]
  • Mass \( m \): How heavy the object is.
  • Velocity Components (\( \dot{x}, \dot{y}, \dot{z} \)): Speed in the three axes.
These factors combine to show how fast an object is moving in the Cartesian plane.

Potential Energy (V)

Potential energy relates to the object's position in a gravitational field:\[ V = mgz\]
  • Gravitational Force (\( g \)): Constant pulling the object down.
  • Height (\( z \)): Position elevated above a reference point.
Thus, an object's potential energy changes as it moves vertically. Understanding these energies helps to reflect the trade-offs in motion, vital for predicting the projectile's path.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the components of \(\nabla f(r, \phi)\) in two-dimensional polar coordinates. [Hint: Remember that the change in the scalar \(f \text { as a result of an infinitesimal displacement } d \mathbf{r} \text { is } d f=\nabla f \cdot d \mathbf{r}.]\)

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 mass \(m_{1}\) rests on a frictionless horizontal table and is attached to a massless string. The string runs horizontally to the edge of the table, where it passes over a massless, frictionless pulley and then hangs vertically down. A second mass \(m_{2}\) is now attached to the bottom end of the string. Write down the Lagrangian for the system. Find the Lagrange equation of motion, and solve it for the acceleration of the blocks. For your generalized coordinate, use the distance \(x\) of the second mass below the tabletop.

Consider two particles moving unconstrained in three dimensions, with potential energy \(U\left(\mathbf{r}_{1}, \mathbf{r}_{2}\right) .\) (a) Write down the six equations of motion obtained by applying Newton's second law to each particle. (b) Write down the Lagrangian \(\mathcal{L}\left(\mathbf{r}_{1}, \mathbf{r}_{2}, \dot{\mathbf{r}}_{1}, \dot{\mathbf{r}}_{2}\right)=T-U\) and show that the six Lagrange equations are the same as the six Newtonian equations of part (a). This establishes the validity of Lagrange's equations in rectangular coordinates, which in turn establishes Hamilton's principle. since the latter is independent of coordinates, this proves Lagrange's equations in any coordinate system.

Write down the Lagrangian for a one-dimensional particle moving along the \(x\) axis and subject to a force \(F=-k x\) (with \(k\) positive). Find the Lagrange equation of motion and solve it.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.