Question

II You go to an amusement park with your friend Betty, who wants to ride the 30 -m-diameter Ferris wheel. She starts the ride at the lowest point of a wheel that, as you face it, rotates counter clockwise. What is her displacement vector when the wheel has rotated by an angle of ${60}^{°}$? Give your answer as a magnitude and direction.

Short answer

The magnitude of the displacement vector is$15\text{m}$ and the direction is${30}^{°}$ above the $x$axis.

Step-by-step solution

Step.1.

The equation for the displacement vector is as follows,

$\Delta \overrightarrow{r}={\overrightarrow{r}}_2-{\overrightarrow{r}}_1$

Here$\overrightarrow{r}$ and ${\overrightarrow{r}}_2$are the initial and final position vectors.

Step .2.

Consider the following diagram,

Step.3.

Origin of the coordinate system is taken as the center of the circle. Therefore, the position vector ${\overrightarrow{r}}_1$can be written as follows,

${\overrightarrow{r}}_{\text{1}}=(0.0\text{m})\widehat{i}+(15\text{m})\widehat{j}$

Also,

$\begin{array}{l}{\overrightarrow{r}}_2=\left((15\text{m})\sin {60}^{°}\right)\widehat{i}+\left((15\text{m})\cos {60}^{°}\right)\widehat{j}\\ =(13\text{m})\widehat{i}+(7.5\text{m})\widehat{j}\end{array}$

Therefore, the displacement vector is calculated as follow,

$\begin{array}{l}\Delta \overrightarrow{r}={\overrightarrow{r}}_2-{\overrightarrow{r}}_1\\ =(13\text{m})\widehat{i}+(7.5\text{m})\widehat{j}-(0.0\text{m})\widehat{i}-(15\text{m})\widehat{j}\\ =(13\text{m})\widehat{i}-(7.5\text{m})\widehat{j}\end{array}$

Find the magnitude of the displacement vector,

$\begin{array}{l}\left|\Delta \overrightarrow{r}\right|=\sqrt{{(13\text{m})}^2+{(-7.5\text{m})}^2}\\ =15\text{m}\end{array}$

Step .4.

The direction is calculated as follows,

$\begin{array}{l}\theta ={\tan }^{-1}\frac{|-7.5\text{m}|}{|13\text{m}|}\\ ={30}^{°}\end{array}$

Therefore, the magnitude of the displacement vector is$15\text{m}$ and the direction is ${30}^{°}$above the $x$axis.