91影视

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,

$\stackrel{鈫�}{p}=m\stackrel{鈫�}{v}$

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

The angular momentum can be expressed as,

$\stackrel{鈫�}{L}=\stackrel{鈫�}{r}脳\stackrel{鈫�}{p}$

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

The linear momentum of the child is calculated as follows,

$\stackrel{鈫�}{p}=m\stackrel{鈫�}{v}$

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

Therefore, the linear momentum is .

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

The vector ${\stackrel{鈫�}{r}}_{鈯�}$is the perpendicular component of position vector $\stackrel{鈫�}{r}$.

At location$A$ ,the angle between the momentum vector$\stackrel{鈫�}{p}$and velocity vector$\stackrel{鈫�}{v}$ is$胃=90掳$ .

The linear momentum of the child is,

Magnitude of the linear momentum,

$\left|\stackrel{鈫�}{p}\right|=300kg路m/s$

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

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

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

$\left|{\stackrel{鈫�}{L}}_A\right|=\left|\stackrel{鈫�}{r}\right|\left|\stackrel{鈫�}{p}\right|\sin 胃$

Substitute, $300kg路m/sfor\left|\stackrel{鈫�}{p}\right|for\left|\stackrel{鈫�}{r}\right|and90掳for胃$in the above relation,

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

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

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

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

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

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

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

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

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

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

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,

$\stackrel{鈫�}{p}=m\stackrel{鈫�}{v}$

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

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

${\stackrel{鈫�}{L}}_{AE}={\stackrel{鈫�}{r}}_E脳\stackrel{鈫�}{p}$

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

$7200kg路m^2/s$.