Chapter 11.5, Problem 11.164P

Some parasailing systems use a winch to pull the rider back to the boat. During the interval when θ is between 20° and 40° (where t = 0 at θ = 20°), the angle increases at the constant rate of 2°/s. During this time, the length of the rope is defined by the relationship $r=600-\frac{1}{8}{t}^{5/2}$, where r and t are expressed in feet and seconds, respectively. Knowing that the boat is traveling at a constant rate of 15 knots (where 1 knot = 1.15 mi/h), (a) plot the magnitude of the velocity of the parasailer as a function of time, (b) determine the magnitude of the acceleration of the parasailer when t = 5 s.

Fig. P11.163 and P11.164

To determine

Plot the magnitude of the velocity of the parasailer as a function of time.

Explanation of Solution

Given Information:

During the interval the $\left(\theta \right)$ is between $20°$ and $40°$.

The angle $\left(\dot{\theta }\right)$ increasing at a constant rate of $\text{2°/s}$.

The length of the rope is define by the relationship (r) of $600-\frac{1}{8}{t}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

The boat is travelling at a constant velocity $\left(v_B\right)$ of $15\text{knots}$.

Calculation:

Convert the knot to feet per second.

Consider $\left(v_B\right)$:

$\begin{array}{c}\left(v_B\right)=15\text{knots}\times \frac{1.15\text{mph}}{1\text{knots}}\times \frac{5280\text{ft}}{1\text{mi}}\frac{1\text{hr}}{3600\sec }\\ =25.3\text{ft/s}\end{array}$

Show the Free body diagram of parasailer and boat as in Figure (1).

Write the velocity $\left(v_B\right)$ of the boat in term of vector:

$v_B=25.3i\text{ft/s}$

The acceleration vector of the boat is as follows:

$a_B=0$

Differentiate angle $\left(\dot{\theta }\right)$ with respective to time (t),

$\ddot{\theta }=0$

Differentiate radius (r) with respective to time (t).

$\dot{r}=-\frac{5}{16}{t}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Differentiate $\left(\dot{r}\right)$ with respective to time (t).

$\ddot{r}=-\frac{15}{32}{t}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Write the expression for velocity vector $\left(v_P\right)$ of parasailer:

$v_P=v_B+v_{P/B}$ . (1)

Here, $v_{P/B}$ is relative velocity vector of parasailer with respect to boat.

Write the expression for acceleration vector $\left(a_P\right)$ of parasailer:

$a_P=a_B+a_{P/B}$ (2)

Here, $a_{P/B}$ is relative acceleration vector of parasailer with respect to boat.

Write the velocity vector ${\left(v_{P/B}\right)}_{\text{radial}}$ of parasailer with respect boat using radial and transverse component:

${\left(v_{P/B}\right)}_{\text{radial}}=\dot{r}{e}_r+r\dot{\theta }{e}_{\theta }$ (3)

Write the acceleration vector ${\left(a_{P/B}\right)}_{\text{radial}}$ of parasailer with respect boat using radial and transverse component:

${\left(a_{P/B}\right)}_{\text{radial}}=\left(\ddot{r}+r{\dot{\theta }}^2\right)e_r+\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)e_{\theta }$ (4)

Write the velocity vector $\left(v_{P/B}\right)$ of parasailer with respect boat in rectangular coordinates using equation (1):

$v_{P/B}=\dot{r}\left(-\cos \theta i+\sin \theta j\right)+r\dot{\theta }\left(\sin \theta i+\cos \theta j\right)$ (5)

Substitute $\dot{r}\left(-\cos \theta i+\sin \theta j\right)+r\dot{\theta }\left(\sin \theta i+\cos \theta j\right)$ for $v_{P/B}$ in equation (1).

$\begin{array}{c}{v}_P=v_B+\dot{r}\left(-\cos \theta i+\sin \theta j\right)+r\dot{\theta }\left(\sin \theta i+\cos \theta j\right)\\ =\left(v_B-\dot{r}\cos \theta +r\dot{\theta }\sin \theta \right)i+\left(\dot{r}\sin \theta +r\dot{\theta }\cos \theta \right)j\end{array}$ . (6)

Calculate velocity vector of parasailer at an angle $\left(\theta \right)$ of $20°$.

Substitute 0 for t, $25.3i\text{ft/s}$ for $v_B$, $-\frac{5}{16}{t}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ for $\dot{r}$, $\theta$ for $20°$, $600-\frac{1}{8}{t}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ for r and 2 for $\dot{\theta }$ in equation (6).

$\begin{array}{c}{v}_P=\left[\begin{array}{l}\left(25.3-\left(-\frac{5}{16}{\left(0\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\cos \left(20°\right)+\left(600-\frac{1}{8}{\left(0\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\left(2\text{°/s}\times \frac{\pi }{180°}\right)\sin \left(20°\right)\right)i\\ +\left(\left(-\frac{5}{16}{\left(0\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\sin \left(20°\right)+\left(600-\frac{1}{8}{\left(0\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\left(2\text{°/s}\times \frac{\pi }{180°}\right)\cos \left(20°\right)\right)j\end{array}\right]\\ =\left(25.3+7.163\right)i+\left(19.684\right)j\\ =32.463i+19.684j\end{array}$

Here, ${\left(v_p\right)}_x$ is $32.463\text{ft/s}$ and ${\left(v_p\right)}_y$ is $19.684\text{ft/s}$.

Calculate the velocity $\left(v_P\right)$ of parasailer at an angle $\left(\theta \right)$ of $20°$ using the relation:

$\left(v_p\right)=\sqrt{{\left(v_P\right)}_x^2+{\left(v_P\right)}_y^2}$

Substitute $32.463\text{m/s}$ for ${\left(v_p\right)}_x$ and $19.684\text{m/s}$ for ${\left(v_p\right)}_y$.

$\begin{array}{c}\left(v_p\right)=\sqrt{{\left(32.463\right)}^2+{\left(19.684\right)}^2}\\ =\sqrt{1441.306}\\ =37.965\text{ft/s}\end{array}$

The time (t) is increase with 1 sec for an angle of $2°$.

Similarly, calculate the velocity $\left(v_P\right)$ of parasailer for the angle between $20°$ and $40°$ .

Summarize the calculated values of velocity as in Table (1).

Time(t) (sec)	$\left(\theta \right)$(degree)	Radius (r)	$\dot{r}$	${\left(v_p\right)}_x$(ft/s)	${\left(v_p\right)}_y$ (ft/s)	$\left(v_P\right)$(ft/s)
0	20	600.000	0.000	32.463	19.681	37.963
1	22	599.875	-0.313	33.434	19.298	38.603
2	24	599.293	-0.884	34.616	18.751	39.369
3	26	598.051	-1.624	35.911	18.051	40.193
4	28	596.000	-2.500	37.274	17.195	41.050
5	30	593.012	-3.494	38.676	16.180	41.924
6	32	588.977	-4.593	40.090	15.001	42.804
7	34	583.795	-5.788	41.494	13.658	43.684
8	36	577.373	-7.071	42.867	12.149	44.555
9	38	569.625	-8.438	44.190	10.474	45.415
10	40	560.472	-9.882	45.446	8.635	46.259

Plot the magnitude of the velocity of the parasailer as a function of time as in Figure (1).

To determine

The magnitude of the acceleration $\left(a_p\right)$ of the parasailer when time (t) is 5 sec.

Answer to Problem 11.164P

The magnitude of the acceleration $\left(a_p\right)$ of the parasailer when time (t) is 5 sec is $\underset{\_}{1.787{\text{ft/s}}^2}$.

Explanation of Solution

Given Information:

During the interval the $\left(\theta \right)$ is between $20°$ and $40°$.

The angle $\left(\dot{\theta }\right)$ increasing at a constant rate of $\text{2°/s}$.

The length of the rope is define by the relationship (r) of $600-\frac{1}{8}{t}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

The boat is travelling at a constant velocity $\left(v_B\right)$ of $15\text{knots}$.

Calculation:

Write the expression for acceleration vector $\left(a_p\right)$:

Substitute $\left(\ddot{r}+r{\dot{\theta }}^2\right)e_r+\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)e_{\theta }$ for $a_{P/B}$ and 0 for $a_B$ in equation (2).

$\begin{array}{c}{a}_P=0+\left(\ddot{r}+r{\dot{\theta }}^2\right)e_r+\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)e_{\theta }\\ =\left(\ddot{r}-r{\dot{\theta }}^2\right)e_r+\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)e_{\theta }\end{array}$ . (7)

Calculate the acceleration vector $\left(a_P\right)$ of parasailer:

Substitute 5 sec for t, $-\frac{15}{32}{t}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ for $\ddot{r}$, $-\frac{5}{16}{t}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ for $\dot{r}$, $\theta$ for $20°$, $600-\frac{1}{8}{t}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ for r, 0 for $\ddot{\theta }$, and 2 for $\dot{\theta }$ in equation (7).

$\begin{array}{c}{a}_P=\left[\begin{array}{l}\left(\left(-\frac{15}{32}{\left(5\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)-\left(600-\frac{1}{8}{\left(5\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right){\left(2\text{°/s}\times \frac{\pi }{180°}\right)}^2\right)e_r\\ +\left(\left(600-\frac{1}{8}{t}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\left(0\right)+2\left(-\frac{5}{16}{5}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\left(2\text{°/s}\times \frac{\pi }{180°}\right)\right)e_{\theta }\end{array}\right]\\ =-1.771e_r-0.2439e_{\theta }\end{array}$

Here, $-1.771{\text{ft/s}}^2$ for ${\left(a_p\right)}_r$ and $-0.2439{\text{ft/s}}^{\text{2}}$ for ${\left(a_p\right)}_{\theta }$.

Calculate the acceleration $\left(a_P\right)$ of parasailer using the relation:

$\left(a_p\right)=\sqrt{{\left(a_P\right)}_r^2+{\left(a_P\right)}_{\theta }^2}$

Substitute $-1.771{\text{ft/s}}^2$ for ${\left(a_p\right)}_r$ and $-0.2439{\text{ft/s}}^{\text{2}}$ for ${\left(a_p\right)}_{\theta }$.

$\begin{array}{c}\left(a_p\right)=\sqrt{{\left(-1.771\right)}^2+{\left(-0.2439\right)}^2}\\ =\sqrt{3.1959}\\ =1.787{\text{ft/s}}^2\end{array}$

Therefore, the magnitude of the acceleration $\left(a_p\right)$ of the parasailer when time (t) is 5 sec is $\underset{\_}{1.787{\text{ft/s}}^2}$.