Question

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 $\mathit{A}$?

Short answer

The linear momentum of the child is .

The vector, $\overrightarrow{r}$is the position vector.

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

The magnitude of the angular momentum of the child about location $A$is $7200kg·m^2/s$.

The plane defined by $\overrightarrow{r}$and$\overrightarrow{p}$ is out of the page.

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

The value of$y{p}_y$ is $7200kg·m^2/s$.

The magnitude of the angular momentum of the child about location$A$ is$7200kg·m^2/s$.

.

Step-by-step solution

Definition of Linear Momentum.

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.

Find the Linear Momentum of the Child.

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 .

Define the vector r→and r→⊥.

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 $\overrightarrow{r}$.

Find the magnitude of the Angular Momentum of the child about the location A.

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,

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

Find the Z component of the Angular Momentum.

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$7200\hat{j}kg·m^2/s$.

Find the value of xpy

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,

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

Find the magnitude of the angular momentum of the child about location A.

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

$7200kg·m^2/s$.