91Ó°ÊÓ

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,

$\stackrel{→}{p}=m\stackrel{→}{v}$

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

The angular momentum can be expressed as,

$\stackrel{→}{L}=\stackrel{→}{r}×\stackrel{→}{p}$

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

The linear momentum of the child is calculated as follows,

$\stackrel{→}{p}=m\stackrel{→}{v}$

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

Therefore, the linear momentum is .

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

The vector ${\stackrel{→}{r}}_{âŠ\yen }$is the perpendicular component of position vector $\stackrel{→}{r}$.

At location$A$ ,the angle between the momentum vector$\stackrel{→}{p}$and velocity vector$\stackrel{→}{v}$ is$\mathrm{θ}=90°$ .

The linear momentum of the child is,

Magnitude of the linear momentum,

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

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

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

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

$\left|{\stackrel{→}{L}}_A\right|=\left|\stackrel{→}{r}\right|\left|\stackrel{→}{p}\right|\sin \mathrm{θ}$

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

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

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

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

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,

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

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,

$\stackrel{→}{p}=m\stackrel{→}{v}$

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

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

${\stackrel{→}{L}}_{AE}={\stackrel{→}{r}}_E×\stackrel{→}{p}$

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

$7200kg·m^2/s$.