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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 \(\phi\):\[ \phi = \tan^{-1}\left(\frac{y}{x}\right) \]

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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.

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Most popular questions from this chapter

A small cart (mass \(m\) ) is mounted on rails inside a large cart. The two are attached by a spring (force constant \(k\) ) in such a way that the small cart is in equilibrium at the midpoint of the large. The distance of the small cart from its equilibrium is denoted \(x\) and that of the large one from a fixed point on the ground is \(X,\) as shown in Figure \(7.13 .\) The large cart is now forced to oscillate such that \(X=A \cos \omega t,\) with both \(A\) and \(\omega\) fixed. Set up the Lagrangian for the motion of the small cart and show that the Lagrange equation has the form $$\ddot{x}+\omega_{0}^{2} x=B \cos \omega t$$ where \(\omega_{\mathrm{o}}\) is the natural frequency \(\omega_{\mathrm{o}}=\sqrt{k / m}\) and \(B\) is a constant. This is the form assumed in Section 5.5, Equation (5.57), for driven oscillations (except that we are here ignoring damping). Thus the system described here would be one way to realize the motion discussed there. (We could fill the large cart with molasses to provide some damping.)

(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.

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?

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

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.
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