Chapter 11.5, Problem 11.164P

To determine

(a)

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

Answer to Problem 11.164P

Explanation of Solution

Given information:

Length of rope is defined as,

$r=600-\frac{1}{8}{t}^{5/2}$

Constant velocity of the boat is $15kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=v_r{e}_r+v_{\theta }{e}_{\theta }=\dot{r}{e}_r+r\dot{\theta }{e}_{\theta }$

The acceleration in radial and transverse components,

$a=a_r{e}_r+a_{\theta }{e}_{\theta }=\left(\ddot{r}-r{\dot{\theta }}^2\right)e_r+\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)e_{\theta }$

Calculation:

According to given information,

Convert,

$\begin{array}{l}{v}_B=15kn=25.3ft/s\\ a_B=0\end{array}$

We know that,

$r=600-\frac{1}{8}{t}^{5/2}$

$\dot{\theta }=2°s=0.0349rad/s$

Therefore,

$\begin{array}{l}\dot{r}=-\frac{5}{16}{t}^{3/2}\\ \ddot{r}=-\frac{15}{32}{t}^{1/2}\end{array}$

$\ddot{\theta }=0$

According to the explanation, the relative velocity of parasailer with respect to boat is,

$v_{P/B}=\dot{r}{e}_r+r\dot{\theta }{e}_{\theta }$

The relative co-ordinates of the relative velocity is equal to,

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

Therefore the velocity of parasailer is equal to,

$\begin{array}{l}{v}_P=v_B+v_{P/B}\\ v_P=v_B+\dot{r}\left(-\cos \theta i+\sin \theta j\right)+r\dot{\theta }\left(\sin \theta i+\cos \theta j\right)\\ v_P=\left(v_B-\dot{r}\cos \theta +r\dot{\theta }\sin \theta \right)i+\left(\dot{r}\sin \theta +r\dot{\theta }\cos \theta \right)j\end{array}$

We can rewrite this as,

$v_P=v_{P_x}i+v_{P_y}j$

Where,

$\begin{array}{l}{v}_{P_x}=v_B-\dot{r}\cos \theta +r\dot{\theta }\sin \theta \\ v_{P_y}=\dot{r}\sin \theta +r\dot{\theta }\cos \theta \end{array}$

To find the magnitude,

$v_P=\sqrt{v_{P_x}{}^2+v_{P_y}{}^2}$

Now, plot the graph,

Conclusion:

The magnitude of velocity of the parasailer is equal to,

$v_P=\sqrt{v_{P_x}{}^2+v_{P_y}{}^2}$

The relevant graph is shown above.

To determine

(b)

The magnitude of acceleration of parasailer at $t=5s$

Answer to Problem 11.164P

$a_P=1.7871m/s^2$

Explanation of Solution

Given information:

Length of rope is defined as,

$r=600-\frac{1}{8}{t}^{5/2}$

Constant velocity of the boat is $15kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=v_r{e}_r+v_{\theta }{e}_{\theta }=\dot{r}{e}_r+r\dot{\theta }{e}_{\theta }$

The acceleration in radial and transverse components,

$a=a_r{e}_r+a_{\theta }{e}_{\theta }=\left(\ddot{r}-r{\dot{\theta }}^2\right)e_r+\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)e_{\theta }$

Calculation:

According to given information,

Convert,

$\begin{array}{l}{v}_B=15kn=25.3ft/s\\ a_B=0\end{array}$

We know that,

$r=600-\frac{1}{8}{t}^{5/2}$

$\dot{\theta }=2°s=0.0349rad/s$

Therefore,

$\begin{array}{l}\dot{r}=-\frac{5}{16}{t}^{3/2}\\ \ddot{r}=-\frac{15}{32}{t}^{1/2}\end{array}$

$\ddot{\theta }=0$

According to the explanation, the relative acceleration of parasailer with respect to boat is,

$\begin{array}{l}{a}_{P/B}=\left(\ddot{r}-r{\dot{\theta }}^2\right)e_r+\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)e_{\theta }\\ a_{P/B}=\left[-\frac{15}{32}{t}^{1/2}-\left(600-\frac{1}{8}{t}^{5/2}\right){\left(0.0349rad/s\right)}^2\right]e_r+\left[2\left(-\frac{5}{16}{t}^{3/2}\right)\left(0.0349rad/s\right)\right]e_{\theta }\end{array}$

The acceleration of parasailer is equal to,

$\begin{array}{l}{a}_P=a_B+a_{P/B}\\ a_P=0+\left[\left(-\frac{15}{32}{t}^{1/2}-\left(600-\frac{1}{8}{t}^{5/2}\right){\left(0.0349rad/s\right)}^2\right)e_r+\left(2\left(-\frac{5}{16}{t}^{3/2}\right)\left(0.0349rad/s\right)\right)e_{\theta }\right]\end{array}$

At, $t=5s$

$\begin{array}{l}{a}_P=\left[\left(-\frac{15}{32}{\left(5\right)}^{1/2}-\left(600-\frac{1}{8}{\left(5\right)}^{5/2}\right){\left(0.0349rad/s\right)}^2\right)e_r+\left(2\left(-\frac{5}{16}{\left(5\right)}^{3/2}\right)\left(0.0349rad/s\right)\right)e_{\theta }\right]\\ a_P=-1.7704e_r-0.2439e_{\theta }\end{array}$

Therefore the magnitude is equal to,

$\begin{array}{l}{a}_P=\sqrt{{\left(-1.7704\right)}^2+{\left(-0.2439\right)}^2}\\ a_P=1.7871m/s^2\end{array}$

Conclusion:

The magnitude of acceleration of the parasailer is equal to,

$a_P=1.7871m/s^2$