Chapter 2.4, Problem 31E

To determine

To find: the coordinates of the car located at the loading platform, and then find its location at $Ro{t}_{90},Ro{t}_{180}\text{ and }Ro{t}_{270}$ .

Answer to Problem 31E

  $\left(0,-125\right),\left(125,0\right),\left(0,125\right),\left(-125,0\right)$

  $\begin{array}{l}\text{Location at }Ro{t}_{90}=\left[\begin{array}{cccc}\begin{array}{l}125\\ 0\end{array}& \begin{array}{l}-1\\ 125\end{array}& \begin{array}{l}-125\\ 0\end{array}& \begin{array}{l}-1\\ -125\end{array}\end{array}\right]\\ \text{Location at }Ro{t}_{180}=\left[\begin{array}{cccc}\begin{array}{l}0\\ 125\end{array}& \begin{array}{l}-125\\ 0\end{array}& \begin{array}{l}0\\ -125\end{array}& \begin{array}{l}125\\ 0\end{array}\end{array}\right]\\ \text{Location at }Ro{t}_{270}=\left[\begin{array}{cccc}\begin{array}{l}-125\\ 0\end{array}& \begin{array}{l}0\\ -125\end{array}& \begin{array}{l}125\\ 0\end{array}& \begin{array}{l}0\\ 125\end{array}\end{array}\right]\end{array}$

Explanation of Solution

Given information:

The diameter of wheel =250 ft

Axle length = 45 ft

Assume that axle is fitted at the origin of wheel.

  $R_{180}=\left[\begin{array}{cc}\begin{array}{l}-1\\ 0\end{array}& \begin{array}{l}0\\ -1\end{array}\end{array}\right]$

Calculation:

The radius of wheel can be evaluated as follows:

  $\begin{array}{c}\text{Radius}=\frac{\text{Diameter}}{2}\\ =\frac{250}{2}\text{ ft}\\ =125\text{ ft}\end{array}$

Since we have to assume that axle is fitted at the origin of wheel i.e. $\left(0,0\right)$ , so the coordinates of the car located at the loading platform will be $\left(0,-125\right),\left(125,0\right),\left(0,125\right),\left(-125,0\right)$ .

Now use the x and y coordinates to write the rows of the vertex matrix as follows:

  $\text{Vertex matrix}=\left[\begin{array}{cccc}\begin{array}{l}0\\ -125\end{array}& \begin{array}{l}125\\ 0\end{array}& \begin{array}{l}0\\ 125\end{array}& \begin{array}{l}-125\\ 0\end{array}\end{array}\right]$

Multiply the vertex matrix by each rotation matrix as follows:

  $\begin{array}{c}\left(R_{90}\right)\cdot \left(\text{Vertex matrix}\right)=\left[\begin{array}{cc}\begin{array}{l}0\\ 1\end{array}& \begin{array}{l}-1\\ 0\end{array}\end{array}\right]\cdot \left[\begin{array}{cccc}\begin{array}{l}0\\ -125\end{array}& \begin{array}{l}125\\ 0\end{array}& \begin{array}{l}0\\ 125\end{array}& \begin{array}{l}-125\\ 0\end{array}\end{array}\right]\\ =\left[\begin{array}{cccc}\begin{array}{l}125\\ 0\end{array}& \begin{array}{l}-1\\ 125\end{array}& \begin{array}{l}-125\\ 0\end{array}& \begin{array}{l}-1\\ -125\end{array}\end{array}\right]\end{array}$

  $\begin{array}{c}\left(R_{180}\right)\cdot \left(\text{Vertex matrix}\right)=\left[\begin{array}{cc}\begin{array}{l}-1\\ 0\end{array}& \begin{array}{l}0\\ -1\end{array}\end{array}\right]\cdot \left[\begin{array}{cccc}\begin{array}{l}0\\ -125\end{array}& \begin{array}{l}125\\ 0\end{array}& \begin{array}{l}0\\ 125\end{array}& \begin{array}{l}-125\\ 0\end{array}\end{array}\right]\\ =\left[\begin{array}{cccc}\begin{array}{l}0\\ 125\end{array}& \begin{array}{l}-125\\ 0\end{array}& \begin{array}{l}0\\ -125\end{array}& \begin{array}{l}125\\ 0\end{array}\end{array}\right]\end{array}$

  $\begin{array}{c}\left(R_{270}\right)\cdot \left(\text{Vertex matrix}\right)=\left[\begin{array}{cc}\begin{array}{l}0\\ -1\end{array}& \begin{array}{l}1\\ 0\end{array}\end{array}\right]\cdot \left[\begin{array}{cccc}\begin{array}{l}0\\ -125\end{array}& \begin{array}{l}125\\ 0\end{array}& \begin{array}{l}0\\ 125\end{array}& \begin{array}{l}-125\\ 0\end{array}\end{array}\right]\\ =\left[\begin{array}{cccc}\begin{array}{l}-125\\ 0\end{array}& \begin{array}{l}0\\ -125\end{array}& \begin{array}{l}125\\ 0\end{array}& \begin{array}{l}0\\ 125\end{array}\end{array}\right]\end{array}$

Thus, the location of the car at $Ro{t}_{90},Ro{t}_{180}\text{ and }Ro{t}_{270}$ is as follows:

  $\begin{array}{l}\text{Location at }Ro{t}_{90}=\left[\begin{array}{cccc}\begin{array}{l}125\\ 0\end{array}& \begin{array}{l}-1\\ 125\end{array}& \begin{array}{l}-125\\ 0\end{array}& \begin{array}{l}-1\\ -125\end{array}\end{array}\right]\\ \text{Location at }Ro{t}_{180}=\left[\begin{array}{cccc}\begin{array}{l}0\\ 125\end{array}& \begin{array}{l}-125\\ 0\end{array}& \begin{array}{l}0\\ -125\end{array}& \begin{array}{l}125\\ 0\end{array}\end{array}\right]\\ \text{Location at }Ro{t}_{270}=\left[\begin{array}{cccc}\begin{array}{l}-125\\ 0\end{array}& \begin{array}{l}0\\ -125\end{array}& \begin{array}{l}125\\ 0\end{array}& \begin{array}{l}0\\ 125\end{array}\end{array}\right]\end{array}$