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Find expressions for the unit vectors \(\hat{\boldsymbol{\rho}}, \hat{\boldsymbol{\phi}},\) and \(\hat{\mathbf{z}}\) of cylindrical polar coordinates (Problem 1.47) in terms of the Cartesian \(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}}\). Differentiate these expressions with respect to time to find \(d \hat{\boldsymbol{\rho}} / d t, d \hat{\boldsymbol{\phi}} / d t,\) and \(d \hat{\mathbf{z}} / d t\)

Short Answer

Expert verified
\(d\hat{\boldsymbol{\rho}}/dt = \frac{d\phi}{dt} \hat{\boldsymbol{\phi}}; \) \( d\hat{\boldsymbol{\phi}}/dt = -\frac{d\phi}{dt} \hat{\boldsymbol{\rho}}; \) \(d\hat{\mathbf{z}}/dt = 0\).

Step by step solution

01

Identify relations between coordinates

In cylindrical coordinates, the position vector \(\mathbf{r}\) can be expressed as \((\rho, \phi, z)\). In Cartesian coordinates, it is \((x, y, z)\). The relationships are: \(x = \rho \cos \phi\) and \(y = \rho \sin \phi\).
02

Express cylindrical unit vectors in Cartesian coordinates

The unit vectors for cylindrical coordinates are \( \hat{\boldsymbol{\rho}}, \hat{\boldsymbol{\phi}}, \) and \( \hat{\mathbf{z}} \). We have:- \( \hat{\boldsymbol{\rho}} = \cos \phi \hat{\mathbf{x}} + \sin \phi \hat{\mathbf{y}} \) - \( \hat{\boldsymbol{\phi}} = -\sin \phi \hat{\mathbf{x}} + \cos \phi \hat{\mathbf{y}} \) - \( \hat{\mathbf{z}} = \hat{\mathbf{z}} \).
03

Differentiate \(\hat{\boldsymbol{\rho}}\) with respect to time

Differentiate \( \hat{\boldsymbol{\rho}} = \cos \phi \hat{\mathbf{x}} + \sin \phi \hat{\mathbf{y}} \) with respect to \(t\):\[\frac{d \hat{\boldsymbol{\rho}}}{dt} = -\sin \phi \frac{d\phi}{dt} \hat{\mathbf{x}} + \cos \phi \frac{d\phi}{dt} \hat{\mathbf{y}} = \frac{d\phi}{dt} \hat{\boldsymbol{\phi}}.\]
04

Differentiate \(\hat{\boldsymbol{\phi}}\) with respect to time

Differentiate \( \hat{\boldsymbol{\phi}} = -\sin \phi \hat{\mathbf{x}} + \cos \phi \hat{\mathbf{y}} \) with respect to \(t\):\[\frac{d \hat{\boldsymbol{\phi}}}{dt} = -\cos \phi \frac{d\phi}{dt} \hat{\mathbf{x}} - \sin \phi \frac{d\phi}{dt} \hat{\mathbf{y}} = -\frac{d\phi}{dt} \hat{\boldsymbol{\rho}}.\]
05

Differentiate \(\hat{\mathbf{z}}\) with respect to time

Since \(\hat{\mathbf{z}}\) is constant, its derivative with respect to time is zero:\[\frac{d \hat{\mathbf{z}}}{dt} = 0.\]

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Key Concepts

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

Differentiation of Unit Vectors
When working with cylindrical coordinates, we often need to differentiate unit vectors to understand motion and dynamics in a 3D space. Unit vectors in the cylindrical system include \(\hat{\boldsymbol{\rho}}\), \(\hat{\boldsymbol{\phi}}\), and \(\hat{\mathbf{z}}\). These vectors describe directions in space within the cylindrical framework. To determine how these directions change over time, we calculate their time derivatives. Let's take for example the differentiation of \(\hat{\boldsymbol{\rho}}\), expressed in terms of Cartesian components as \(\cos \phi \hat{\mathbf{x}} + \sin \phi \hat{\mathbf{y}}\). Differentiating each term with respect to time gives:
  • \( -\sin \phi \frac{d\phi}{dt} \hat{\mathbf{x}} \)
  • \( + \cos \phi \frac{d\phi}{dt} \hat{\mathbf{y}} \)
This eventual interaction results in the derivative being: \( \frac{d \hat{\boldsymbol{\rho}}}{dt} = \frac{d\phi}{dt} \hat{\boldsymbol{\phi}} \).Similarly, differentiating \(\hat{\boldsymbol{\phi}}\), which is \(-\sin \phi \hat{\mathbf{x}} + \cos \phi \hat{\mathbf{y}}\), over time leads to:
  • \( -\cos \phi \frac{d\phi}{dt} \hat{\mathbf{x}} \)
  • \( -\sin \phi \frac{d\phi}{dt} \hat{\mathbf{y}} \)
The result here is \( \frac{d \hat{\boldsymbol{\phi}}}{dt} = -\frac{d\phi}{dt} \hat{\boldsymbol{\rho}} \).Finally, since \(\hat{\mathbf{z}}\) remains constant, its derivative with respect to time \( \frac{d \hat{\mathbf{z}}}{dt} \) is simply 0. Understanding these derivatives helps in analyzing rotations and transformations within cylindrical systems.
Cartesian Coordinates
Cartesian coordinates are a fundamental system used in geometry to specify points in a plane or space. Using two (2D) or three (3D) perpendicular axes, commonly labeled as \(x\), \(y\), and \(z\), they provide a straightforward way to describe the location of points.In a 2D Cartesian coordinate system:
  • \(x\)-axis represents the horizontal direction.
  • \(y\)-axis represents the vertical direction.
For a 3D system, we include:
  • \(z\)-axis, signifying depth or height.
Each axis in the Cartesian system has its unit vector, denoted as \(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \) and \(\hat{\mathbf{z}}\). These indicate the direction along each axis.

Cartesian coordinates serve as the basis for expressing cylindrical coordinates in terms of the more familiar linear framework. For instance, when we have a cylindrical coordinate point \((\rho, \phi, z)\), it can be expressed in Cartesian as:
  • \(x = \rho \cos \phi \)
  • \(y = \rho \sin \phi \)
  • \(z = z \)
Using these transformations enables us to work seamlessly between the two coordinate systems and is particularly useful in physics and engineering when analyzing motion or forces.
Coordinate Transformation
Coordinate transformation is a powerful mathematical tool essential for converting one type of coordinate system to another. This process allows for solving problems in one coordinate system when they are more naturally described in another. For students learning about physics or engineering, this ability is crucial. A common transformation is from cylindrical to Cartesian coordinates. The equations for this transformation are as follows:
  • \(x = \rho \cos \phi \)
  • \(y = \rho \sin \phi \)
  • \(z = z \)
These allow us to convert a point given in cylindrical coordinates \((\rho, \phi, z)\) into Cartesian coordinates. Similarly, the unit vectors transformation is
  • \(\hat{\boldsymbol{\rho}} = \cos \phi \hat{\mathbf{x}} + \sin \phi \hat{\mathbf{y}}\)
  • \(\hat{\boldsymbol{\phi}} = -\sin \phi \hat{\mathbf{x}} + \cos \phi \hat{\mathbf{y}}\)
  • \(\hat{\mathbf{z}} = \hat{\mathbf{z}}\)
These transformations allow one to not only work with points but also vectors in different systems, facilitating calculations of dynamics and motions that would be complex in a non-intuitive frame.Understanding coordinate transformations extends your capability to view problems from various perspectives, optimizing the analysis and comprehension of spatial relationships.

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Most popular questions from this chapter

In Section 1.5 we proved that Newton's third law implies the conservation of momentum. Prove the converse, that if the law of conservation of momentum applies to every possible group of particles, then the interparticle forces must obey the third law. [Hint: However many particles your system contains, you can focus your attention on just two of them. (Call them 1 and 2.) The law of conservation of momentum says that if there are no external forces on this pair of particles, then their total momentum must be constant. Use this to prove that \(\mathbf{F}_{12}=-\mathbf{F}_{21}\).]

Conservation laws, such as conservation of momentum, often give a surprising amount of information about the possible outcome of an experiment. Here is perhaps the simplest example: Two objects of masses \(m_{1}\) and \(m_{2}\) are subject to no external forces. Object 1 is traveling with velocity \(\mathbf{v}\) when it collides with the stationary object \(2 .\) The two objects stick together and move off with common velocity \(\mathbf{v}^{\prime}\). Use conservation of momentum to find \(\mathbf{v}^{\prime}\) in terms of \(\mathbf{v}, m_{1},\) and \(m_{2}\)

If you have some experience in electromagnetism, you could do the following problem concerning the curious situation illustrated in Figure \(1.8 .\) The electric and magnetic fields at a point \(\mathbf{r}_{1}\) due to a charge \(q_{2}\) at \(\mathbf{r}_{2}\) moving with constant velocity \(\mathbf{v}_{2}\) (with \(v_{2} \ll c\) ) are \(^{15}\) $$\mathbf{E}\left(\mathbf{r}_{1}\right)=\frac{1}{4 \pi \epsilon_{\mathrm{o}}} \frac{q_{2}}{s^{2}} \hat{\mathbf{s}} \quad \text { and } \quad \mathbf{B}\left(\mathbf{r}_{1}\right)=\frac{\mu_{0}}{4 \pi} \frac{q_{2}}{s^{2}} \mathbf{v}_{2} \times \hat{\mathbf{s}}$$ where \(\mathbf{s}=\mathbf{r}_{1}-\mathbf{r}_{2}\) is the vector pointing from \(\mathbf{r}_{2}\) to \(\mathbf{r}_{1}\). (The first of these you should recognize as Coulomb's law.) If \(\mathbf{F}_{12}^{\mathrm{el}}\) and \(\mathbf{F}_{12}^{\text {mag }}\) denote the electric and magnetic forces on a charge \(q_{1}\) at \(\mathbf{r}_{1}\) with velocity \(\mathbf{v}_{1},\) show that \(F_{12}^{\operatorname{mag}} \leq\left(v_{1} v_{2} / c^{2}\right) F_{12}^{\mathrm{el}} .\) This shows that in the non-relativistic domain it is legitimate to ignore the magnetic force between two moving charges.

In case you haven't studied any differential equations before, I shall be introducing the necessary ideas as needed. Here is a simple excercise to get you started: Find the general solution of the firstorder equation \(d f / d t=f\) for an unknown function \(f(t) .\) [There are several ways to do this. One is to rewrite the equation as \(d f / f=d t\) and then integrate both sides.] How many arbitrary constants does the general solution contain? [Your answer should illustrate the important general theorem that the solution to any \(n\) th-order differential equation (in a very large class of "reasonable" equations) contains \(n\) arbitrary constants.]

Let \(\mathbf{u}\) be an arbitrary fixed unit vector and show that any vector \(\mathbf{b}\) satisfies $$b^{2}=(\mathbf{u} \cdot \mathbf{b})^{2}+(\mathbf{u} \times \mathbf{b})^{2}$$ Explain this result in words, with the help of a picture.

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