91影视

Here we have $v=1.25\text{鈥尘/蝉}$

Constant linear speed of scanning $r\left(胃\right)=r_0+尾胃$

$Theequationofaspiral{r}_0=theradiusofthespiralatthe胃=0$The radius of the spiral at the

$尾$is the constant

Here we have$dS=rd胃$,.

From above equation, we can determine the total distances $S=f\left(胃\right)$.

$\begin{array}{rcl}ds& =& rd胃\\ ds& =& \left(r_0+尾胃\right)d胃\\ {鈭�}_0^{s}ds& =& {鈭�}_0^{胃}\left(r_0+尾胃\right)d胃\\ S-0& =& {鈭�}_0^{胃}{r}_{0}d胃+{鈭�}_0^{胃}尾胃d胃\\ S& =& r_0+尾\frac{{胃}^2}{2}\end{array}$

In the case of uniform rectilinear motion, the distance traveled can be calculated as:

$S=vt$
Using the equation above, we can determine the form of function$胃=f\left(t\right)$.

$$\begin{gathered} S=vt \\ {r}_0胃+{尾}^{{胃}^2}=vt \\ \frac{尾}{2}{胃}^2+r_0胃鈭�vt=0 \end{gathered}$$

The equation above is a quadratic equation so, its solution can be written in the following form:

$\begin{array}{c}胃=\frac{鈭�r_0卤\sqrt{r_0^2+4\frac{尾}{2}vt}}{尾}\\ =\frac{鈭�r_0卤\sqrt{r_0^2+2尾vt}}{尾}\end{array}$

In the previous equation, we will consider only positive solutions. So,

$胃=\frac{鈭�r_0+\sqrt{r_0^2+2尾vt}}{尾}$

Here we have to consider $胃=f\left(t\right)$calculated in the part b), we can determine the change of the angular velocity over time ${蝇}_z=f\left(t\right)$, using the equation (1):

$\begin{array}{rcl}& & \begin{array}{c}{蝇}_z=\frac{d胃}{dt}\\ =\frac{d}{dt}\left(\frac{鈭�r_0+\sqrt{r_0^2+2尾vt}}{尾}\right)\end{array}\\ & & \begin{array}{c}=\frac{1}{尾}\frac{d}{dt}\left(\sqrt{r_0^2+2尾vt}\right)\\ =\frac{2尾v1}{尾}{\left(r_0^2+2尾vt\right)}^{鈭�\frac{1}{2}}\\ =\frac{v}{\sqrt{r_0^2+2尾vt}}\end{array}\end{array}$

Using the equation 2 we can calculate ${伪}_z=f\left(t\right)$ as

$\begin{array}{c}{伪}_z=\frac{d{蝇}_z}{dt}\\ =\frac{d}{dt}\left(\frac{v}{\sqrt{r_0^2+2尾vt}}\right)\\ =v\frac{d}{dt}{\left(r_0^2+2尾vt\right)}^{鈭�\frac{1}{2}}\\ =v\left(鈭�\frac{2尾v}{2}\right){\left(r_0^2+2尾vt\right)}^{鈭�\frac{3}{2}}\\ =鈭�尾v^2{\left(r_0^2+2尾vt\right)}^{鈭�\frac{3}{2}}\\ =鈭�\frac{尾v^2}{\sqrt{{\left(r_0^2+2尾vt\right)}^3}}\end{array}$

Using the parameters given from part d) we can determine the angle$胃$, Before, it is necessary to show $尾$in units of$\frac{\text{m}}{\text{rad}}$.

$\begin{array}{c}尾=1.55\frac{渭\text{m}}{\text{revolution}}\\ =1.55鈰�脳{10}^{鈭�6}\frac{\text{m}}{\text{revolution}}鈰�\frac{1\text{revoltuon}}{2蟺\text{rad}}\\ =0.247脳{10}^{鈭�6}\text{鈥尘/谤补诲}\end{array}$

Now, using the equation from part b) we can determine as follows.

$\begin{array}{c}胃=\frac{鈭�r_0+\sqrt{r_0^2+2尾vt}}{尾}\\ =\frac{鈭�25鈰�{10}^{鈭�3}\text{鈥尘}+\sqrt{{\left(25鈰�{10}^{鈭�3}\text{鈥尘}\right)}^2+2(0.247脳{10}^{鈭�6}\text{鈥尘/谤补诲})\left(1.25\text{鈥尘/蝉}\right)\left(4440\text{鈥塻}\right)}}{0.247鈰�{10}^{鈭�6}\text{鈥尘/谤补诲}}\\ =133697.45\text{鈥塺别惫辞濒耻迟颈辞苍}\end{array}$

Based on the laws calculated in the part under c ). And by applying the parameters from the part under d), it is possible to show graphs of the dependence of ${蝇}_z=f\left(t\right)$

So, we have ${蝇}_z=\frac{v}{\sqrt{r_0^2+2尾vt}}$

So, graph of above function is which is given by,

As well as ${伪}_z=f\left(t\right)$

Here we have,${伪}_z=鈭�\frac{尾v^2}{\sqrt{{\left(r_0^2+2尾vt\right)}^3}}$

So, graph of above function is given as