/*! 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 10 A particle is confined to move o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 10

A particle is confined to move on the surface of a circular cone with its axis on the \(z\) axis, vertex at the origin (pointing down), and half-angle \(\alpha\). The particle's position can be specified by two generalized coordinates, which you can choose to be the coordinates \((\rho, \phi)\) of cylindrical polar coordinates. Write down the equations that give the three Cartesian coordinates of the particle in terms of the generalized coordinates ( \(\rho, \phi\) ) and vice versa.

Short Answer

Expert verified
The Cartesian coordinates are: \( x = \rho \cos(\phi) \), \( y = \rho \sin(\phi) \), \( z = \frac{\rho}{\tan(\alpha)} \). The generalized coordinates are \( \rho = \sqrt{x^2 + y^2} \), \( \phi = \tan^{-1}(y/x) \).

Step by step solution

01

Understand the Geometrical Setup

The particle is on the surface of a cone with its vertex at the origin and its axis along the z-axis. The cone's half-angle is \(\alpha\). The generalized coordinates given are \( \rho \) (the radial distance from the z-axis) and \( \phi \) (the azimuthal angle around the z-axis).
02

Express the Height of the Particle

For a cone with half-angle \(\alpha\), any point on the surface of the cone satisfies the relation between \(z\) and \(\rho\):\[ \rho = z \tan(\alpha) \]Therefore, the height \(z\) of the particle is given by:\[ z = \frac{\rho}{\tan(\alpha)} \]
03

Write Cartesian Coordinates in Terms of Generalized Coordinates

In cylindrical coordinates, the radius \(\rho\) and angle \(\phi\) translate to Cartesian coordinates (x, y) as:- \( x = \rho \cos(\phi) \)- \( y = \rho \sin(\phi) \)The \(z\)-coordinate is expressed in terms of \(\rho\) from Step 2 as:- \( z = \frac{\rho}{\tan(\alpha)} \)Thus, the particle鈥檚 Cartesian coordinates are:\[ x = \rho \cos(\phi) \]\[ y = \rho \sin(\phi) \]\[ z = \frac{\rho}{\tan(\alpha)} \]
04

Express Generalized Coordinates in Terms of Cartesian Coordinates

Given the Cartesian coordinates \(x\), \(y\), and \(z\), the generalized coordinates \(\rho\) and \(\phi\) can be expressed as:- Calculating \(\rho\) as the radial distance in the \(xy\)-plane:\[ \rho = \sqrt{x^2 + y^2} \]- Calculating the azimuthal angle \(\phi\):\[ \phi = \tan^{-1}\left(\frac{y}{x}\right) \]

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 91影视!

Key Concepts

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

generalized coordinates
When dealing with complex systems, it can be beneficial to use generalized coordinates. These are coordinates that simplify the mathematical description of the system.
In this scenario, the movement of a particle on the surface of a cone is analyzed using generalized coordinates. Instead of using standard Cartesian coordinates, the problem can be simplified using
  • Radial distance (\( \rho \) ) from the z-axis
  • Azimuthal angle (\( \phi \) ) around the z-axis.
These coordinates are particularly useful because they naturally follow the cone's shape. They provide a compact way to describe the particle's location with just two parameters, even though the particle is moving in three-dimensional space.
By understanding and using generalized coordinates, we simplify the problem and make it easier to calculate other necessary values.
cylindrical polar coordinates
Cylindrical polar coordinates are a type of generalized coordinate system that expands upon the idea of two-dimensional polar coordinates to three dimensions.
This system consists of
  • Radial distance (\( \rho \) ) from the origin to the point
  • Azimuthal angle (\( \phi \) ) around an axis (typically the z-axis)
  • Height (\( z\) ) along this axis.
In the context of the cone surface, these coordinates effectively describe points by their horizontal and rotational position rather than their linear position in a Cartesian grid.
This is advantageous as it aligns more closely with the natural geometric properties of the cone. For our problem, this system helps by giving a straightforward description of the particle's position on the cone surface, making calculations less complex and more intuitive.
Cartesian coordinates
Cartesian coordinates describe the position of points in space using three values:
  • A horizontal distance along the x-axis
  • A vertical distance along the y-axis
  • A height along the z-axis
They are very familiar and intuitive, often used in basic geometry and algebra to express positions in space.
To convert generalized (cylindrical) coordinates into Cartesian coordinates, we use the following relations:
  • \( x = \rho \cos(\phi) \)
  • \( y = \rho \sin(\phi) \)
  • \( z = \frac{\rho}{\tan(\alpha)} \)
These equations translate positions from cylindrical to Cartesian, allowing us to describe the particle's location in rectangular space.
This is particularly useful when the task requires interactions with systems or equations that are best analyzed in a Cartesian framework.
half-angle of a cone
The half-angle of a cone is a crucial geometric parameter. It is defined as the angle between the cone's side and its axis.
For the exercise at hand, this half-angle is denoted as \(\alpha\). It directly influences the relationship between the radial distance \(\rho\) and the height \(z\) of a point on the cone's surface.
The relation is described by the equation \( \rho = z \tan(\alpha) \). This means that for a fixed angle, increasing \(\rho\) increases \(z\) linearly, following a slope determined by \(\tan(\alpha)\).
Understanding the half-angle helps visualize and predict how the dimensions of the cone will alter as we change the particle's position, making it a vital aspect of the cone dynamics.

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

[Computer] Consider a massless wheel of radius \(R\) mounted on a frictionless horizontal axis. A point mass \(M\) is glued to the edge, and a massless string is wrapped several times around the perimeter and hangs vertically down with a mass \(m\) suspended from its bottom end. (See Figure 4.28.) Initially I am holding the wheel with \(M\) vertically below the axle. At \(t=0,\) I release the wheel, and \(m\) starts to fall vertically down. (a) Write down the Lagrangian \(\mathcal{L}=T-U\) as a function of the angle \(\phi\) through which the wheel has turned. Find the equation of motion and show that, provided \(m

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 a frictionless plane that slopes at an angle \(\alpha\) with the horizontal. Write down the Lagrangian in terms of coordinates \(x,\) measured horizontally across the slope, and \(y\), measured down the slope. (Treat the system as two-dimensional, but include the gravitational potential energy.) Find the two Lagrange equations and show that they are what you should have expected.

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) Write down the Lagrangian \(\mathcal{L}\left(x_{1}, x_{2}, \dot{x}_{1}, \dot{x}_{2}\right)\) for two particles of equal masses, \(m_{1}=m_{2}=m,\) confined to the \(x\) axis and connected by a spring with potential energy \(U=\frac{1}{2} k x^{2} .\) [Here \(x\) is the extension of the spring, \(x=\left(x_{1}-x_{2}-l\right),\) where \(l\) is the spring's unstretched length, and I assume that mass 1 remains to the right of mass 2 at all times.] (b) Rewrite \(\mathcal{L}\) in terms of the new variables \(X=\frac{1}{2}\left(x_{1}+x_{2}\right)\) (the CM position) and \(x\) (the extension), and write down the two Lagrange equations for \(X\) and \(x\). (c) Solve for \(X(t)\) and \(x(t)\) and describe the motion.
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Physical Quantities And Units

Read Explanation

Electricity and Magnetism

Read Explanation

Turning Points in Physics

Read Explanation

Fluids

Read Explanation
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.