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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

Find the angle between a body diagonal of a cube and any one of its face diagonals. [Hint: Choose a cube with side 1 and with one corner at \(O\) and the opposite corner at the point (1,1,1) . Write down the vector that represents a body diagonal and another that represents a face diagonal, and then find the angle between them as in Problem 1.4.]

Two vectors are given as \(\mathbf{b}=(1,2,3)\) and \(\mathbf{c}=(3,2,1)\). (Remember that these statements are just a compact way of giving you the components of the vectors.) Find \(\mathbf{b}+\mathbf{c}, 5 \mathbf{b}-2 \mathbf{c}, \mathbf{b} \cdot \mathbf{c},\) and \(\mathbf{b} \times \mathbf{c}\).

Imagine two concentric cylinders, centered on the vertical \(z\) axis, with radii \(R \pm \epsilon,\) where \(\epsilon\) is very small. A small frictionless puck of thickness \(2 \epsilon\) is inserted between the two cylinders, so that it can be considered a point mass that can move freely at a fixed distance from the vertical axis. If we use cylindrical polar coordinates \((\rho, \phi, z)\) for its position (Problem 1.47 ), then \(\rho\) is fixed at \(\rho=R,\) while \(\phi\) and \(z\) can vary at will. Write down and solve Newton's second law for the general motion of the puck, including the effects of gravity. Describe the puck's motion.

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.

A ball is thrown with initial speed \(v_{\mathrm{o}}\) up an inclined plane. The plane is inclined at an angle \(\phi\) above the horizontal, and the ball's initial velocity is at an angle \(\theta\) above the plane. Choose axes with \(x\) measured up the slope, \(y\) normal to the slope, and \(z\) across it. Write down Newton's second law using these axes and find the ball's position as a function of time. Show that the ball lands a distance \(R=2 v_{\mathrm{o}}^{2} \sin \theta \cos (\theta+\phi) /\left(g \cos ^{2} \phi\right)\) from its launch point. Show that for given \(v_{\mathrm{o}}\) and \(\phi,\) the maximum possible range up the inclined plane is \(R_{\max }=v_{\mathrm{o}}^{2} /[g(1+\sin \phi)]\)

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