Problem 59 - Fundamentals of Physics

Question

A has the magnitude \(12.0 \mathrm{~m}\) and is angled \(60.0^{\circ}\) counterclockwise from the positive direction of the \(x\) axis of an \(x y\) coordinate system. Also, \(\vec{B}=(12.0 \mathrm{~m}) \hat{\mathrm{i}}+(8.00 \mathrm{~m}) \hat{\mathrm{j}}\) on that same coordinate system. We now rotate the system counterclockwise about the origin by \(20.0^{\circ}\) to form an \(x^{\prime} y^{\prime}\) system. On this new system, what are (a) \(\vec{A}\) and (b) \(\vec{B}\), both in unit-vector notation?

Step-by-Step Solution

Answer

\( \vec{A}' \approx 3.57 \hat{i}' + 11.0 \hat{j}' \) and \( \vec{B}' \approx 8.47 \hat{i}' + 11.38 \hat{j}' \).

Step 1: Express \( \vec{A} \) in Unit-Vector Notation

Given that the magnitude of \( \vec{A} \) is \( 12.0 \mathrm{~m} \) and it is at an angle of \( 60.0^{\circ} \) from the positive \( x \)-axis, we can find the components using:\[ A_x = A \cos(\theta) = 12.0 \cos(60.0^{\circ}) = 6.0 \mathrm{~m} \]\[ A_y = A \sin(\theta) = 12.0 \sin(60.0^{\circ}) = 10.39 \mathrm{~m} \]Thus, \( \vec{A} = 6.0 \hat{i} + 10.39 \hat{j} \).

Step 2: Rotation Transformation Matrix

The transformation matrix for rotating a vector counterclockwise by an angle \( \phi \) is:\[ \begin{bmatrix} \cos(\phi) & -\sin(\phi) \ \sin(\phi) & \cos(\phi) \end{bmatrix} \]For a rotation of \( 20.0^{\circ} \), the matrix becomes:\[ \begin{bmatrix} \cos(20.0^{\circ}) & -\sin(20.0^{\circ}) \ \sin(20.0^{\circ}) & \cos(20.0^{\circ}) \end{bmatrix} \approx \begin{bmatrix} 0.9397 & -0.3420 \ 0.3420 & 0.9397 \end{bmatrix} \]

Step 3: Rotate \( \vec{A} \) to New Coordinate System

Apply the rotation matrix to the components of \( \vec{A} \):\[ \begin{bmatrix} A_{x'} \ A_{y'} \end{bmatrix} = \begin{bmatrix} 0.9397 & -0.3420 \ 0.3420 & 0.9397 \end{bmatrix} \begin{bmatrix} 6.0 \ 10.39 \end{bmatrix} \]Calculate each component:\[ A_{x'} = 0.9397 \times 6.0 + (-0.3420) \times 10.39 \approx 3.57 \mathrm{~m} \]\[ A_{y'} = 0.3420 \times 6.0 + 0.9397 \times 10.39 \approx 11.0 \mathrm{~m} \]Thus, \( \vec{A}' \approx 3.57 \hat{i}' + 11.0 \hat{j}' \).

Step 4: Rotate \( \vec{B} \) to New Coordinate System

Apply the rotation matrix to the components of \( \vec{B} \):\[ \begin{bmatrix} B_{x'} \ B_{y'} \end{bmatrix} = \begin{bmatrix} 0.9397 & -0.3420 \ 0.3420 & 0.9397 \end{bmatrix} \begin{bmatrix} 12.0 \ 8.0 \end{bmatrix} \]Calculate each component:\[ B_{x'} = 0.9397 \times 12.0 + (-0.3420) \times 8.0 \approx 8.47 \mathrm{~m} \]\[ B_{y'} = 0.3420 \times 12.0 + 0.9397 \times 8.0 \approx 11.38 \mathrm{~m} \]Thus, \( \vec{B}' \approx 8.47 \hat{i}' + 11.38 \hat{j}' \).

Key Concepts

Rotation Matrix

A rotation matrix is a crucial tool in transforming vectors in the context of rotations. It allows you to turn a vector around a particular point without changing its length. For a two-dimensional rotation, such as the one found in this exercise, a rotation matrix is usually represented as a 2x2 matrix. The general form of a counterclockwise rotation by an angle \( \phi \) is given by:

• First row: \( [\cos(\phi), -\sin(\phi)] \)
• Second row: \( [\sin(\phi), \cos(\phi)] \)

This matrix effectively changes the orientation of a vector by \( \phi \) degrees. In the exercise, to rotate the vector by \( 20.0^{\circ} \), the specific rotation matrix becomes:

• \( \cos(20.0^{\circ}) \approx 0.9397 \) and \( \sin(20.0^{\circ}) \approx 0.3420 \).
• Thus, the matrix is: \[ \begin{bmatrix} 0.9397 & -0.3420 \ 0.3420 & 0.9397 \end{bmatrix} \]

This transformation rotates the vectors' coordinates, effectively switching them to a new reference frame. Utilizing the rotation matrix helps visually accomplish this seamless rotation in mathematical terms.

Coordinate System

Understanding the coordinate system is key to describing vectors and their transformations. A coordinate system consists of axes, usually denoted as the x-axis and y-axis in two-dimensional space. Each point or vector can be defined by its distance from the origin, measured along these axes.
In this exercise, the original coordinate system is simply the familiar Cartesian system with axes labeled \(x\) and \(y\). When we talk about the new \(x' y'\) system, that's the rotated version of our original axes. Each vector is initially given in terms of these axes, often starting from the origin and described in unit-vector notation.

• The new coordinate axes are derived by rotating the original axes by \(20.0^{\circ}\) counterclockwise.
• The vectors' coordinates in this system are found using the rotation matrix. their position relative to the newly defined x' and y' axes.

The rotation of this coordinate system results in vectors needing to be recomputed in terms of this new reference. This is important in physics and engineering for understanding motion and orientation changes.

Unit-Vector Notation

Unit-vector notation is a powerful way to express vectors using a combination of their magnitudes and direction components. This notation breaks down each vector into its individual components along the axes, often denoted as \( \hat{i} \) for the x-component and \( \hat{j} \) for the y-component.When a vector is expressed in unit-vector notation, such as \( \vec{B} = (12.0 \mathrm{~m}) \hat{i} + (8.0 \mathrm{~m}) \hat{j} \), the components show how far the vector stretches along each axis:

• \( \hat{i} \) indicates the unit vector in the direction of the x-axis.
• \( \hat{j} \) indicates the unit vector in the direction of the y-axis.

In this exercise, both vectors \( \vec{A} \) and \( \vec{B} \) are converted into this form before applying the rotation matrix. This allows the transformation to be made clearly as each component is multiplied and recalculated.
Unit-vector notation is very helpful for visualizing each vector's contribution along the axes, making it easier to understand and calculate rotations or translations.