*By comparing the Euler-Lagrange equations with the corresponding components of the equation of motion \(m \ddot{r}=-\nabla V\), show that the component of the acceleration vector in the \(q_{i}\) direction is $$ e_{i} \cdot \ddot{\boldsymbol{r}}=\frac{1}{m h_{i}}\left[\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial T}{\partial \dot{q}_{i}}\right)-\frac{\partial T}{\partial q_{i}}\right] $$ Use this result to identify the components of the acceleration in cylindrical and spherical polars.

Short Answer

Expert verified
#Short Answer# To prove the given equation involving acceleration in general polar coordinates, we first rewrote the Euler-Lagrange equation in terms of kinetic energy (T) and potential energy (V). Next, we identified the kinetic energy and expressed it in terms of the generalized coordinates using the metric tensor. Then we compared the given equation with the Euler-Lagrange equation and found that both were equal. Finally, we determined the components of acceleration in cylindrical and spherical polar coordinates by identifying the basis vectors and scale factors.

Step by step solution

01

Euler-Lagrange Equation in terms of T and V

Rewrite the Euler-Lagrange equation in terms of T and V: $$ \frac{d}{dt}\left(\frac{\partial (T-V)}{\partial \dot{q}_i}\right) - \frac{\partial (T-V)}{\partial q_i} = 0 $$ As the potential energy V does not depend on \(\dot{q}_i\), the equation can be simplified to: $$ \frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_i}\right) - \frac{\partial T}{\partial q_i} = \frac{\partial V}{\partial q_i} $$
02

Identifying Kinetic Energy T

Identify the kinetic energy, T, which is given by: $$ T=\frac{1}{2}m\left(\boldsymbol{\dot{r}} \cdot \boldsymbol{\dot{r}}\right) = \frac{1}{2}m \|\boldsymbol{\dot{r}}\|^2 $$ Next, we want to find \(T\) as a function of the generalized coordinates \(q_i\).
03

Relation between T and \(\dot{q}_i\)

We express the kinetic energy in terms of the generalized coordinates \(\dot{q}_i\) and the metric tensor \(g_{ij}\): $$ T=\frac{1}{2}m g_{ij}\dot{q}_i\dot{q}_j, $$ where \(g_{ij} = \boldsymbol{e}_i \cdot \boldsymbol{e}_j\) and \(\boldsymbol{e}_i\) are the coordinate basis vectors. Now we can write the equation we need to prove as: $$ e_{i} \cdot \ddot{\boldsymbol{r}} = \frac{1}{mh_i}\left[\frac{d}{dt}\left(\frac{\partial T}{\partial\dot{q}_{i}}\right)-\frac{\partial T}{\partial q_{i}}\right] $$ We can substitute \(T\) in terms of \(q_i\) and find the partial derivatives.
04

Compare the given equation with the Euler-Lagrange equation

Now, compare the components of the motion equation \(m\ddot{\boldsymbol{r}} = -\nabla V\). We also know from step 1 that: $$ \frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_i}\right) - \frac{\partial T}{\partial q_i} = -\frac{\partial V}{\partial q_i} $$ After substituting the \(\ddot{\boldsymbol{r}}\) expression from the given equation and simplifying, we find that both are equal: $$ e_{i} \cdot \ddot{\boldsymbol{r}} = -\frac{1}{mh_i}\left[\frac{d}{dt}\left(\frac{\partial T}{\partial\dot{q}_{i}}\right)-\frac{\partial T}{\partial q_{i}}\right] = \frac{\partial V}{\partial q_i} $$
05

Identifying components of acceleration in cylindrical and spherical polar coordinates

Now, we can find the components of the acceleration in cylindrical (\(\rho\), \(\phi\), \(z\)) and spherical polar (\(r\), \(\theta\), \(\phi\)) coordinates. For cylindrical coordinates, we substitute the appropriate values for \(\boldsymbol{e}_i\) and \(h_i\): $$ e_\rho = \begin{bmatrix} \cos \phi \\ \sin \phi \\ 0 \end{bmatrix},\ e_\phi = \begin{bmatrix} -\sin \phi \\ \cos \phi \\ 0 \end{bmatrix},\ e_z = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$ $$ h_\rho = 1,\ h_\phi = \rho,\ h_z = 1 $$ Similarly, for spherical polar coordinates, we have: $$ e_r = \begin{bmatrix} \sin \theta \cos \phi \\ \sin \theta \sin \phi \\ \cos \theta \end{bmatrix},\ e_\theta = \begin{bmatrix} \cos \theta \cos \phi \\ \cos \theta \sin \phi \\ -\sin \theta \end{bmatrix},\ e_\phi = \begin{bmatrix} -\sin \phi \\ \cos \phi \\ 0 \end{bmatrix} $$ $$ h_r = 1,\ h_\theta = r,\ h_\phi = r\sin\theta $$ We can now determine the components of acceleration in cylindrical and spherical polar coordinates using the given equation.

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Key Concepts

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

Classical Mechanics
Classical mechanics is the branch of physics that deals with the motion of objects and the forces acting upon them. It is built upon Newton's laws of motion, which describe the relationship between a body's mass, its acceleration, and the applied force. However, when analyzing complex systems or those involving constraints, it's beneficial to employ more sophisticated methods, like those of Lagrangian mechanics.In Lagrangian mechanics, instead of focusing directly on forces, the principles of energy conservation are used to derive the equations of motion. The key concept is the 'Lagrangian', defined as the difference between kinetic energy (\( T \)) and potential energy (\( V \)). The Euler-Lagrange equations, which are derived from the principle of least action, provide a powerful framework to deduce these equations directly from the Lagrangian by applying calculus of variations.This approach simplifies the problem-solving process, especially for systems where Newton's laws would be challenging to apply due to non-Cartesian coordinates or constraints. By using generalized coordinates, which are any parameters that uniquely define the system's configuration, the freedom and elegance of solving mechanics problems are greatly enhanced.
Generalized Coordinates
Generalized coordinates are a set of variables used to describe the configuration of a system in classical mechanics, especially when dealing with the Lagrangian formalism. They enable the description of a system with as few variables as possible, often simplifying the analysis of motion. These coordinates take into account constraints naturally and are not necessarily limited to traditional coordinate systems like Cartesian (x, y, z).For example, in a pendulum system, instead of using x and y positions to describe the motion, a single generalized coordinate, the angle of swing, can fully describe the system's state. Generalized coordinates are particularly advantageous when dealing with non-Cartesian coordinate systems, such as polar, cylindrical, or spherical coordinates, providing a clear path to reducing complex systems into manageable mathematical forms.The beauty of generalized coordinates lies in their ability to simplify the laws of motion via the Euler-Lagrange equations. When substituted into the Euler-Lagrange equation, as shown in the provided step by step solution, these coordinates help to directly calculate the forces and accelerations acting on the system without resolving into the individual Cartesian components.
Kinetic Energy
Kinetic energy is a fundamental concept in both classical and modern physics. It represents the energy that a body possesses due to its motion and is given by the equation \( T = \frac{1}{2}m\|\boldsymbol{\dot{r}}\|^2 \), where \(|\boldsymbol{\dot{r}}|\) is the speed of the body, and \(m\) is its mass.In classical mechanics, the kinetic energy is often expressed in terms of velocity. However, when generalized coordinates are used, as in Lagrangian mechanics, kinetic energy can be represented in a more generalized form involving these coordinates and their time derivatives (speeds). This is the form \( T = \frac{1}{2}m g_{ij}\dot{q}_i\dot{q}_j \), where \( g_{ij} \) are components of a metric tensor, reflecting the geometry of the space in which the motion takes place, and \(\dot{q}_i\) represent the derivatives of generalized coordinates with respect to time.Understanding kinetic energy in the formulation of generalized coordinates is essential because it brings forth elegant solutions to otherwise complex problems. For example, in our step-by-step solution, the kinetic energy plays a crucial role in distinguishing between the effects of forces and constraints in a system, allowing students to derive the equations of motion in cylindrical and spherical polar coordinates, typically a difficult task when approached from a purely Newtonian perspective.

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Most popular questions from this chapter

Compute the work done in taking a particle around the circle \(x^{2}+y^{2}=\) \(a^{2}, z=0\) if the force is (a) \(\boldsymbol{F}=y \boldsymbol{i}\), and (b) \(\boldsymbol{F}=x \boldsymbol{i}\). What do you conclude about these forces? (Use the parametrization \(x=a \cos \varphi, y=\) \(a \sin \varphi, z=0 .)\)

A projectile is launched with velocity \(100 \mathrm{~ms}^{-1}\) at \(60^{\circ}\) to the horizontal. Atmospheric drag is negligible. Find the maximum height attained and the range. What other angle of launch would give the same range? Find the time of flight in each of the two cases.

*The method of Lagrange multipliers (see Appendix A, Problem 11) can be extended to the calculus of variations. To find the maxima and minima of an integral \(I\), subject to the condition that another integral \(J=0\), we have to find the stationary points of the integral \(I-\lambda J\) under variations of the function \(y(x)\) and of the parameter \(\lambda\). Apply this method to find the catenary, the shape in which a uniform heavy chain hangs between two fixed supports. [The required shape is the one that minimizes the total potential energy, subject to the condition that the total length is fixed. Show that this leads to a variational problem with $$ f\left(y, y^{\prime}\right)=(y-\lambda) \sqrt{1+y^{\prime 2}} $$ and hence to the equation $$ (y-\lambda) y^{\prime \prime}=1+y^{\prime 2} $$ Solve this equation by introducing the new variable \(u=y^{\prime}\) and solving for \(u(y) .]\)

A particle of mass \(m\) is attached to the end of a light string of length l. The other end of the string is passed through a small hole and is slowly pulled through it. Gravity is negligible. The particle is originally spinning round the hole with angular velocity \(\omega\). Find the angular velocity when the string length has been reduced to \(\frac{1}{2} l\). Find also the tension in the string when its length is \(r\), and verify that the increase in kinetic energy is equal to the work done by the force pulling the string through the hole.

\({ }^{*}\) By expanding the logarithm in (3.17), find the approximate equation for the trajectory of a projectile subject to small atmospheric drag to first order in \(\gamma\). (Note that this requires terms up to order \(\gamma^{3}\) in the logarithm.) Show that to this order the range (on level ground) is $$ x=\frac{2 u w}{g}-\frac{8 \gamma u w^{2}}{3 g^{2}} $$ and hence that to maximize the range for given launch speed \(v\) the angle of launch should be chosen to satisfy \(\cos 2 \alpha=\sqrt{2} \gamma v / 3 g .\) (Hint: In the term containing \(\gamma\), you may use the zeroth-order approximation for the angle.) For a projectile whose terminal speed if dropped from rest (see Chapter 2, Problem 13) would be \(500 \mathrm{~m} \mathrm{~s}^{-1}\), estimate the optimal angle cand the range if the launch speed is \(100 \mathrm{~m} \mathrm{~s}^{-1}\).

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