Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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:\[\frac{d \hat{\mathbf{z}}}{dt} = 0.\]

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 Vaia!

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.

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

If you have some experience in electromagnetism and with vector calculus, prove that the magnetic forces, \(\mathbf{F}_{12}\) and \(\mathbf{F}_{21}\), between two steady current loops obey Newton's third law. [Hints: Let the two currents be \(I_{1}\) and \(I_{2}\) and let typical points on the two loops be \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\). If \(d \mathbf{r}_{1}\) and \(d \mathbf{r}_{2}\) are short segments of the loops, then according to the Biot- Savart law, the force on \(d \mathbf{r}_{1}\) due to \(d \mathbf{r}_{2}\) is $$\frac{\mu_{0}}{4 \pi} \frac{I_{1} I_{2}}{s^{2}} d \mathbf{r}_{1} \times\left(d \mathbf{r}_{2} \times \hat{\mathbf{s}}\right)$$ where \(\mathbf{s}=\mathbf{r}_{1}-\mathbf{r}_{2} .\) The force \(\mathbf{F}_{12}\) is found by integrating this around both loops. You will need to use the " \(B A C-C A B\) " rule to simplify the triple product.]

Verify by direct substitution that the function \(\phi(t)=A \sin (\omega t)+B \cos (\omega t)\) of (1.56) is a solution of the second-order differential equation \((1.55), \ddot{\phi}=-\omega^{2} \phi .\) (since this solution involves two arbitrary constants - the coefficients of the sine and cosine functions \(-\) it is in fact the general solution.)

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.

One of the many uses of the scalar product is to find the angle between two given vectors. Find the angle between the vectors \(\mathbf{b}=(1,2,4)\) and \(\mathbf{c}=(4,2,1)\) by evaluating their scalar product.

[Computer] The differential equation (1.51) for the skateboard of Example 1.2 cannot be solved in terms of elementary functions, but is easily solved numerically. (a) If you have access to software, such as Mathematica, Maple, or Matlab, that can solve differential equations numerically, solve the differential equation for the case that the board is released from \(\phi_{\mathrm{o}}=20\) degrees, using the values \(R=5 \mathrm{m}\) and \(g=9.8 \mathrm{m} / \mathrm{s}^{2} .\) Make a plot of \(\phi\) against time for two or three periods. ( \(\mathbf{b}\) ) On the same picture, plot the approximate solution (1.57) with the same \(\phi_{\mathrm{o}}=20^{\circ}\) Comment on your two graphs. Note: If you haven't used the numerical solver before, you will need to learn the necessary syntax. For example, in Mathematica you will need to learn the syntax for "NDSolve”and how to plot the solution that it provides. This takes a bit of time, but is something that is very well worth learning.

See all solutions

Recommended explanations on Physics Textbooks

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.

Sign-up for free