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Consider a bead that is threaded on a rigid circular hoop of radius \(R\) lying in the \(x y\) plane with its center at \(O,\) and use the angle \(\phi\) of two- dimensional polar coordinates as the one generalized coordinate to describe the bead's position. Write down the equations that give the Cartesian coordinates \((x, y)\) in terms of \(\phi\) and the equation that gives the generalized coordinate \(\phi\) in terms of \((x, y)\).

Short Answer

Expert verified
\(x = R \cos \phi\), \(y = R \sin \phi\); \(\phi = \tan^{-1}(y/x)\).

Step by step solution

01

Define the relationship between polar and Cartesian coordinates

In polar coordinates, a point on the plane is defined by the radius \(r\) and angle \(\phi\). For a point on a circle of radius \(R\), the radius is constant and equal to \(R\). The angle \(\phi\) defines the position on the circle.
02

Relate Cartesian coordinates to polar coordinates

The Cartesian coordinates \((x, y)\) are related to polar coordinates \((r, \phi)\) by the equations: \(x = r \cos \phi\) and \(y = r \sin \phi\). Since the bead is on a hoop of radius \(R\), substitute \(r = R\).
03

Write equations for Cartesian coordinates

Substituting \(r = R\) in the equations, we get:\[ x = R \cos \phi \]\[ y = R \sin \phi \]These equations give the Cartesian coordinates \((x, y)\) in terms of the angle \(\phi\).
04

Write equation for angle in terms of Cartesian coordinates

To find \(\phi\) in terms of \(x\) and \(y\), use the inverse tangent function which gives the angle from the origin to the point \((x, y)\):\[ \phi = \tan^{-1}\left(\frac{y}{x}\right) \] provided that the bead is not at the origin. Since the bead is always on the hoop, \(x\) and \(y\) are non-zero as long as \(\phi\) is defined.

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Key Concepts

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

Cartesian coordinates
Cartesian coordinates are a system that defines a point's position on a plane using two numbers, called coordinates. These coordinates are the horizontal (x) and vertical (y) distances from a fixed point, which is usually referred to as the origin. The way you can visualize this is like finding a specific point on a piece of graph paper where the x-axis marks left and right movements, and the y-axis marks up and down movements.

For example, if you have a point \(5, 3\), it means you have moved 5 units to the right (along the x-axis) and 3 units up (along the y-axis) from the origin. Cartesian coordinates are incredibly useful for various calculations and transformations in geometry and physics. They help in translating complex shapes and curves into manageable equations and numbers that can be more easily manipulated.
  • Used for plotting points on a 2D plane
  • Essential for graphing equations
  • Commonly apply in geometry, physics, engineering
generalized coordinate
A generalized coordinate is a parameter that can describe the configuration of a mechanical system uniquely. In the context of the bead on the hoop, the angle \(\phi\) serves as a generalized coordinate. This is because it uniquely specifies the position of the bead on the circle regardless of how the coordinate system is rotated or translated.

This parameter is 'generalized' because it might not always be a typical distance or angle but rather any variable that can fully describe a system's state. Utilizing a generalized coordinate often simplifies problem-solving by reducing complex systems into simpler forms with fewer variables.
  • Helps describe complex systems with fewer variables
  • Not always traditional distances or angles
  • Simplifies solving mechanics problems
inverse tangent function
The inverse tangent function, also known as arctan or \(\tan^{-1}\), allows us to determine the angle whose tangent is a given number. In the bead exercise, this is used to find the angle \(\phi\) when the Cartesian coordinates (x, y) are known. Essentially, you're asking "what angle would have a tangent of y/x?".

For practical purposes, inverse tangent is crucial because it helps translate real-world coordinates back into a comprehensible angle. This function is particularly useful in cases where you need to convert positions defined by Cartesian coordinates into their equivalent angles in polar coordinates.
  • Used to find angles from known tangent values
  • Crucial for converting from Cartesian to polar coordinates
  • Ensures precision in interpreting angle positions

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Most popular questions from this chapter

Consider a mass \(m\) moving in two dimensions with potential energy \(U(x, y)=\frac{1}{2} k r^{2},\) where \(r^{2}=x^{2}+y^{2} .\) Write down the Lagrangian, using coordinates \(x\) and \(y,\) and find the two Lagrange equations of motion. Describe their solutions. [This is the potential energy of an ion in an "ion trap," which can be used to study the properties of individual atomic ions.]

[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 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.

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.

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