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\(\vec{A}\) 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?

Short Answer

Expert verified
For \(\vec{A}\), the unit vector notation is \(-0.4033 \hat{\imath} + 11.795 \hat{\jmath}\). For \(\vec{B}\), it's \(8.5404 \hat{\imath} + 11.6216 \hat{\jmath}\).

Step by step solution

01

- Convert \(\vec{A}\) to Cartesian Coordinates

First, convert \(\vec{A}\) from polar to Cartesian coordinates. Use the formulas \(A_x = A \cos(\theta)\) and \(A_y = A \sin(\theta)\). Given \(A = 12.0 \text{ m}\) and \(\theta = 60.0^\circ\), calculate \(A_x\) and \(A_y\). \[ A_x = 12.0 \cos(60.0^\circ) = 12.0 \left( \frac{1}{2} \right) = 6.0 \text{ m} \] \[ A_y = 12.0 \sin(60.0^\circ) = 12.0 \left( \frac{\sqrt{3}}{2} \right) = 6.0 \sqrt{3} \text{ m} \] So, \(\vec{A} = 6.0 \hat{\imath} + 6.0 \sqrt{3} \hat{\jmath}\)
02

- Rotation Matrix

Use the rotation matrix to rotate the coordinates by 20.0° counterclockwise. The rotation matrix is \[ R = \begin{bmatrix} \cos(20.0^\circ) & -\sin(20.0^\circ) \ \sin(20.0^\circ) & \cos(20.0^\circ) \end{bmatrix} \] The angle 20.0° in radians is \( \theta = \frac{20.0 \pi}{180} = \frac{\pi}{9} \).
03

- Rotate \(\vec{A}\)

Apply the rotation matrix to \(\vec{A}\) \[ \begin{bmatrix} A'_x \ A'_y \end{bmatrix} = \begin{bmatrix} \cos(20.0^\circ) & -\sin(20.0^\circ) \ \sin(20.0^\circ) & \cos(20.0^\circ) \end{bmatrix} \begin{bmatrix} 6.0 \ 6.0 \sqrt{3} \end{bmatrix} \] Calculate each component as follows: \[ A'_x = 6.0 \cos(20.0^\circ) - 6.0 \sqrt{3} \sin(20.0^\circ) \] \[ A'_y = 6.0 \sin(20.0^\circ) + 6.0 \sqrt{3} \cos(20.0^\circ) \]
04

- Rotate \(\vec{B}\)

Similar to \(\vec{A}\), rotate \(\vec{B}\). From \(\vec{B} = 12.0 \hat{\imath} + 8.0 \hat{\jmath}\) \[ \begin{bmatrix} B'_x \ B'_y \end{bmatrix} = \begin{bmatrix} \cos(20.0^\circ) & -\sin(20.0^\circ) \ \sin(20.0^\circ) & \cos(20.0^\circ) \end{bmatrix} \begin{bmatrix} 12.0 \ 8.0 \end{bmatrix} \] Calculate each component as follows: \[ B'_x = 12.0 \cos(20.0^\circ) - 8.0 \sin(20.0^\circ) \] \[ B'_y = 12.0 \sin(20.0^\circ) + 8.0 \cos(20.0^\circ) \]
05

- Numerical Values

Evaluate the numerical values using \( \cos(20.0^\circ) \approx 0.9397 \) and \( \sin(20.0^\circ) \approx 0.3420 \) For \(\vec{A}\): \[ A'_x = 6.0 \times 0.9397 - 6.0 \sqrt{3} \times 0.3420 \approx 6.0(0.9397) - 6.0(1.732)(0.3420) \approx 5.6382 - 6.0415 \approx -0.4033 \] \[ A'_y = 6.0 \times 0.3420 + 6.0 \sqrt{3} \times 0.9397 \approx 6.0(0.3420) + 6.0(1.732)(0.9397) \approx 2.052 + 9.743 \approx 11.795 \] \(\vec{A'} = -0.4033 \hat{\imath} + 11.795 \hat{\jmath}\) For \(\vec{B}\): \[ B'_x = 12.0 \times 0.9397 - 8.0 \times 0.3420 \approx 12.0(0.9397) - 8.0(0.3420) \approx 11.2764 - 2.736 \approx 8.5404 \] \[ B'_y = 12.0 \times 0.3420 + 8.0 \times 0.9397 \approx 12.0(0.3420) + 8.0(0.9397) \approx 4.104 + 7.5176 \approx 11.6216 \] \(\vec{B'} = 8.5404 \hat{\imath} + 11.6216 \hat{\jmath}\)

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Key Concepts

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

Cartesian coordinates
Cartesian coordinates represent points in a plane using two values, typically referred to as the x (horizontal) and y (vertical) coordinates. Imagine a grid where each point is determined by how far it is from the x-axis and y-axis.
For instance, if we have a point \( P(3, 4) \), it means it's 3 units to the right of the y-axis and 4 units up from the x-axis.

In the original exercise, vector \( \vec{A} \) is first given in polar coordinates with a magnitude (or length) and an angle. To work with it in equations, we convert it to Cartesian coordinates using trigonometric functions:
    \( A_x = A \cos(\theta) \)

    \( A_y = A \sin(\theta) \)
With \( A = 12 \text{ m} \) and \( \theta = 60^\text{°} \), this gives us: \( A_x = 6 \text{ m} \) and \( A_y = 12 \times \sin(60^\text{°}) = 6\text{ m}\times\ticksqrt{3} \), resulting in the Cartesian coordinates \( 6 \hat{\text i} + 6 \sqrt{3} \hat{\text j} \).
polar coordinates
Polar coordinates represent points in a plane using a distance from a reference point (usually the origin) and an angle from a reference direction (usually the positive x-axis).
Imagine a clock: the center is the origin, and each point on the clock can be described by how far it is from the center (magnitude) and its angle (position around the clock).
For vector \( \vec{A} \) in polar coordinates, the magnitude is 12 meters, and the angle is 60 degrees counterclockwise from the positive x-axis. It effectively tells us how 'long' the vector is and in what direction it points.

This system is very useful for problems involving rotational movement, as you can easily describe circular paths and rotations using just a distance and an angle.
rotation matrix
A rotation matrix is a mathematical tool used to rotate points or vectors in a plane around the origin. It helps us change the orientation of a system while keeping the same relative positions of points.

For a counterclockwise rotation by an angle \( \theta \), the rotation matrix is:
\[ \begin{bmatrix} \cos(\theta) & -\sin(\theta) \ \ \sin(\theta) & \cos(\theta) \end{bmatrix} \]
When we multiply this matrix by a vector, it rotates the vector counterclockwise by \( \theta \) degrees.
In the exercise, we rotate the vector \( \vec{A} \) by 20 degrees counterclockwise. This means using:

    \[ \begin{bmatrix} \cos(20^\text{°}) & -\sin(20^\text{°}) \ \ \sin(20^\text{°}) & \cos(20^\text{°}) \end{bmatrix} \]
then multiplying that by the Cartesian coordinates of \( \vec{A} \) and \( \vec{B} \).
trigonometry
Trigonometry is a branch of mathematics that deals with the relationships between the angles and sides of triangles. It's especially useful for converting between polar and Cartesian coordinates.
Key trigonometric functions: \( \cos \) (cosine) and \( \sin \) (sine) help us relate the horizontal and vertical components of a vector to its magnitude and direction (angle).

In the exercise, we use these functions to convert vector \( \vec{A} \) from polar to Cartesian coordinates:

    \( A_x = A \cos(60^\text{°}) = 12 \left(\tfrac{1}{2}\right) = 6 \text{ m} \)

    \( A_y = A \sin(60^\text{°}) = 12 \left(\tfrac{\ticksqrt{3}}{2}\right) = 6 \sqrt{3} \)

Furthermore, when we rotate the vectors, we again rely on trigonometric functions within the rotation matrix to determine the new coordinates in the rotated system.

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Most popular questions from this chapter

A pea leaves a pea shooter at a speed of \(5.4 \mathrm{~m} / \mathrm{s}\). It makes an angle of \(+30^{\circ}\) with respect to the horizontal. (a) Calculate the \(x\) -component of the pea's initial velocity. (b) Calculate the \(y\) -component of the pea's initial velocity. (c) Write an expression for the pea's velocity, \(\vec{v}\), using unit vectors for the \(x\) direction and the \(y\) direction.

A car is driven east for a distance of \(50 \mathrm{~km}\), then north for \(30 \mathrm{~km}\), and then in a direction \(30^{\circ}\) east of north for \(25 \mathrm{~km}\). Sketch the vector diagram and determine (a) the magnitude and (b) the angle of the car's total displacement from its starting point.

Two vectors are given by $$\begin{array}{l} \vec{a}=(4.0 \mathrm{~m}) \hat{\mathrm{i}}-(3.0 \mathrm{~m}) \hat{\mathrm{j}}+(1.0 \mathrm{~m}) \hat{\mathrm{k}} \\ \vec{b}=(-1.0 \mathrm{~m}) \hat{\mathrm{i}}+(1.0 \mathrm{~m}) \hat{\mathrm{j}}+(4.0 \mathrm{~m}) \hat{\mathrm{k}} . \end{array}$$ In unit-vector notation, find (a) \(\vec{a}+\vec{b}\), (b) \(\vec{a}-\vec{b}\), and \((\mathrm{c})\) a third vector \(\vec{c}\) such that \(\vec{a}-\vec{b}+\vec{c}=0\).

If \(\vec{d}_{1}+\vec{d}_{2}=5 \vec{d}_{3}, \vec{d}_{1}-\vec{d}_{2}=3 \vec{d}_{3}\), and \(\vec{d}_{3}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}\), then what are (a) \(\vec{d}_{1}\) and (b) \(\vec{d}_{2}\) ?

A golfer takes three putts to get the ball into the hole. The first putt displaces the ball \(3.66 \mathrm{~m}\) north, the second \(1.83 \mathrm{~m}\) southeast, and the third \(0.91 \mathrm{~m}\) southwest. What are (a) the magnitude and (b) the direction of the displacement needed to get the ball into the hole on the first putt?

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