A common amusement park ride is a Ferris wheel (see figure, which is not drawn to scale). Riders sit in chairs that are on pivots so they remain level as the wheel turns at a constant rate. A particular Ferris wheel has a radius of 24 meters, and it make one complete revolution around its axle (at location $\mathit{A}$) in $\mathbf{20}\mathit{s}$In all of the following questions, consider location $\mathit{A}$(at the center of the axle) as the location around which we will calculate the angular momentum. At the instant shown in the diagram, a child of mass$\mathbf{40}\mathit{k}\mathit{g}$, sitting at location $\mathit{F}$, is traveling with velocity $\mathbf{<}\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{>}\mathit{m}\mathbf{/}\mathit{s}$.

(a.) What is the linear momentum of the child? (b) In the definition $\overrightarrow{\mathbf{L}}\mathbf{=}\overrightarrow{\mathbf{r}}\mathbf{\times }\overrightarrow{\mathbf{p}}\mathbf{,}$what is the vector $\overrightarrow{\mathbf{r}}$? (c) what is $\overrightarrow{\mathbf{r}}\mathbf{\perp }$? (d) what is the magnitude of the angular momentum of the child about location $\mathit{A}$? (e) What is the plane defined by $\overrightarrow{\mathbf{r}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\overrightarrow{\mathbf{p}}$(that is, the plane containing both of these vectors)? (f) Use the right-hand rule to determine thecomponent of the angular momentum of the child about location$\mathit{A}$. (g) You used the right-hand rule to determine the $\mathit{z}$component of the angular momentum, but as a check, calculate in terms of position and momentum: What is$\mathit{y}{\mathit{p}}_{\mathbf{x}}$? Therefore, what is$\mathit{z}$the component of the angular momentum of the child about location$\mathit{A}$? (h) The Ferris wheel keeps turning, and at a later time, the same child is at location$\mathit{E}$with coordinates$<16.971,-16.971,0>$m relative to location $\mathit{A}$, moving with velocity$<5.303,5.303,0>\mathit{m}\mathbf{/}\mathit{s}\mathbf{.}$Now what is the magnitude of the angular momentum of the child about location $\mathit{A}$?

In classical physics, the linear momentum of translation is a vector quantity equal to the product of the mass and the velocity of the centre of mass.

The formula to find the linear momentum is as follows,

$\overrightarrow{p}=m\overrightarrow{v}$

Here$m$ is the mass and$\overrightarrow{v}$is the velocity.

The angular momentum can be expressed as,

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

Here,$\overrightarrow{r}$is the position vector and$\overrightarrow{p}$ is the linear momentum vector.

The linear momentum of the child is calculated as follows,

$\overrightarrow{p}=m\overrightarrow{v}$

Substitute $40kg$for $m$and for $\overrightarrow{v}$in the above relation,

Therefore, the linear momentum is .

In the formula $\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p},$the vector$\overrightarrow{r}$ is the position vector with respect to the point at which the angular momentum is calculated from the center of the wheel.

The vector ${\overrightarrow{r}}_{\perp }$is the perpendicular component of position vector $\overrightarrow{r}$.

At location$A$ ,the angle between the momentum vector$\overrightarrow{p}$and velocity vector$\overrightarrow{v}$ is$\theta =90°$ .

The linear momentum of the child is,

Magnitude of the linear momentum,

$\left|\overrightarrow{p}\right|=300kg·m/s$

The distance of the child from the point$A$ is the radius of the wheel. That is,

$\begin{array}{rcl}\left|\overrightarrow{r}\right|& =& r\\ & =& 24\\ & & \end{array}$

The magnitude of the angular momentum of the child about location$A$ is calculated as follows,

$\left|{\overrightarrow{L}}_A\right|=\left|\overrightarrow{r}\right|\left|\overrightarrow{p}\right|\sin \theta$

Substitute, $300kg·m/sfor\left|\overrightarrow{p}\right|for\left|\overrightarrow{r}\right|and90°for\theta$in the above relation,

$\begin{array}{rcl}\left|{\overrightarrow{L}}_A\right|& =& \left(24m\right)\left(300kg.m/s\right)\left(\sin 90°\right)\\ & =& 7200kg·m^2/s\\ & & \end{array}$

The Ferris wheel is rotating in the anti-clockwise. So, according to the right-hand rule, the plane defined by$\overrightarrow{r}$and$\overrightarrow{p}$ is out of the page.

Position vector of the child at location$F$

Velocity of the child at point$F$ is,

The linear momentum of the child is calculated already.

Angular momentum of a particle location$A$ is,

The direction is given by the right-hand rule by using cross product rule,

So, thecomponent of the angular momentum can be calculated as follows,

$\begin{array}{rcl}{\overrightarrow{L}}_{A,z}& =& \left(x,p_y-y{p}_x\right)\hat{j}\\ & =& \left(0\right)\left(0\right)-\left(24m\right)\left(300kg·m/s\right)\hat{j}\\ & =& 7200\hat{j}kg·m^2/s\\ & & \end{array}$

Therefore, the $z$component of the angular momentum is$7200\hat{j}kg·m^2/s$.

The value of $x{p}_y$is calculated as follows,

$\begin{array}{rcl}x{p}_y& =& \left(0m\right)\left(0kg·m/s\right)\\ & =& 0kg·m^2/s\\ & & \end{array}$

The value of $y{p}_y$is calculated as follows,

$\begin{array}{rcl}y{p}_y& =& \left(24m\right)\left(300kg·m/s\right)\\ & =& 7200kg·m^2/s\end{array}$

Therefore, the$z$ component of the angular momentum is calculated as follows,

$\begin{array}{rcl}{\overrightarrow{L}}_{A,z}& =& \left(x{p}_y-y{p}_x\right)\hat{j}\\ & =& \left(0\right)\left(0\right)-\left(24m\right)\left(300kg·m/s\right)\hat{j}\\ & =& -7200\hat{j}kg·m^2/s\end{array}$

The position vector of the child at location$E$ is,

Velocity of the child is,

Linear momentum vector of the child is calculated as follows,

$\overrightarrow{p}=m\overrightarrow{v}$

Substitute,$40kg$for$m$ and for$\overrightarrow{v}$in the above relation,

The angular momentum of the child about$A$can be calculated as follows,

${\overrightarrow{L}}_{AE}={\overrightarrow{r}}_E\times \overrightarrow{p}$

Substitute,for and for in the above relation,

Use the following property of vectors cross product,

Hence,

Therefore, the magnitude of the angular momentum of the child about location$A$ is

$7200kg·m^2/s$.