Chapter 12.1, Problem 12.66P

An advanced spatial disorientation trainer allows the cab to rotate around multiple axes, as well as to extend inward and outward. It can be used to simulate driving, fixed-wing aircraft flying, and helicopter maneuvering. In one training scenario, the trainer rotates and translates in the horizontal plane, where the location of the pilot is defined by the relationships $r=10+2\cos \left(\frac{\pi }{3}t\right)$ and $\theta =0.1\left(2t^2-t\right)$ , where $r,\theta ,$ and t are expressed in feet, radians: and seconds, respectively. Knowing that the pilot has a weight of 175 Ibs, (a) determine the magnitude of the resulting force acting on the pilot at $t=5$ s, (b) plot the magnitudes of the radial and transverse components of the force exerted on the pilot from 0 to 10 seconds.

  

To determine

(a)

The magnitude of the resulting force on pilot.

Answer to Problem 12.66P

We got force $F=230.113\text{ lb}$

Explanation of Solution

Given information:

Time $t=5\text{ sec}$

$r=10+2\cos \left(\frac{\pi }{3}t\right)$

$\theta =0.1\left(2t^2-t\right)$

Concept used:

$F_r=m(\ddot{r}-r{\dot{\theta }}^2)$

$F_{\theta }=m(r\ddot{\theta }+2\dot{r}\dot{\theta })$

Calculation:

Derivatives,

$\begin{array}{l}\dot{r}=\frac{d}{dt}\left(r\right)\\ \dot{r}=\frac{d}{dt}\left(10+2\cos \left(\frac{\pi }{3}t\right)\right)\\ \dot{r}=-\frac{2\pi }{3}\sin \left(\frac{\pi }{3}t\right)\end{array}$

$\begin{array}{l}\ddot{r}=\frac{d}{dt}\left(\dot{r}\right)\\ \ddot{r}=\frac{d}{dt}\left(-\frac{2\pi }{3}\sin \left(\frac{\pi }{3}t\right)\right)\\ \ddot{r}=-\frac{2{\pi }^2}{9}\cos \left(\frac{\pi }{3}t\right)\end{array}$

$\begin{array}{l}\dot{\theta }=\frac{d\theta }{dt}\\ \dot{\theta }=\frac{d}{dt}\left(0.1\left(2t^2-t\right)\right)\\ \dot{\theta }=0.1\left(4t-1\right)\end{array}$

$\begin{array}{l}\ddot{\theta }=\frac{d\dot{\theta }}{dt}\\ \ddot{\theta }=\frac{d}{dt}\left(0.1\left(4t-1\right)\right)\\ \ddot{\theta }=0.4\end{array}$

Force components,

$\begin{array}{l}{F}_r=m(\ddot{r}-r{\dot{\theta }}^2)\\ F_r=175\left[-\frac{2{\pi }^2}{9}\cos \left(\frac{\pi }{3}t\right)-\left(10+2\cos \left(\frac{\pi }{3}t\right)\right)\times {\left(0.1\left(4t-1\right)\right)}^2\right]\\ F_r=175\left[-\frac{2{\pi }^2}{9}\cos \left(\frac{\pi }{3}\times 5\right)-\left(10+2\cos \left(\frac{\pi }{3}\times 5\right)\right)\times {\left(0.1\left(4\times 5-1\right)\right)}^2\right]\\ F_r=-7141.2087{\text{ lb-ft/s}}^{\text{2}}\end{array}$

$\begin{array}{l}{F}_{\theta }=m(r\ddot{\theta }+2\dot{r}\dot{\theta })\\ F_{\theta }=175\left[0.4\left(10+2\cos \left(\frac{\pi }{3}t\right)\right)+2\left(-\frac{2\pi }{3}\sin \left(\frac{\pi }{3}t\right)\right)\left(0.1\left(4t-1\right)\right)\right]\\ F_{\theta }=175\left[0.4\left(10+2\cos \left(\frac{\pi }{3}\times 5\right)\right)+2\left(-\frac{2\pi }{3}\sin \left(\frac{\pi }{3}\times 5\right)\right)\left(0.1\left(4\times 5-1\right)\right)\right]\\ F_{\theta }=1976.333{\text{ lb-ft/s}}^{\text{2}}\end{array}$

Resultant force,

$\begin{array}{l}F=\sqrt{F_r{}^2+F_{\theta }{}^2}\\ F=\sqrt{{\left(-7141.2087\right)}^2+{1976.333}^2}\\ F=7409.64{\text{ lb-ft/s}}^{\text{2}}\\ F=\frac{7409.64}{32.2}\text{ lb}\\ F=230.113\text{ lb}\end{array}$

Conclusion:

We got force $F=230.113\text{ lb}$

To determine

(b)

Plot the radial and transverse components of force.

Answer to Problem 12.66P

Plot is in explanation part.

Explanation of Solution

Given information:

Time $t=5\text{ sec}$

$r=10+2\cos \left(\frac{\pi }{3}t\right)$

$\theta =0.1\left(2t^2-t\right)$

Concept used:

$F_r=m(\ddot{r}-r{\dot{\theta }}^2)$

$F_{\theta }=m(r\ddot{\theta }+2\dot{r}\dot{\theta })$

Calculation:

Derivatives,

$\begin{array}{l}\dot{r}=\frac{d}{dt}\left(r\right)\\ \dot{r}=\frac{d}{dt}\left(10+2\cos \left(\frac{\pi }{3}t\right)\right)\\ \dot{r}=-\frac{2\pi }{3}\sin \left(\frac{\pi }{3}t\right)\end{array}$

$\begin{array}{l}\ddot{r}=\frac{d}{dt}\left(\dot{r}\right)\\ \ddot{r}=\frac{d}{dt}\left(-\frac{2\pi }{3}\sin \left(\frac{\pi }{3}t\right)\right)\\ \ddot{r}=-\frac{2{\pi }^2}{9}\cos \left(\frac{\pi }{3}t\right)\end{array}$

$\begin{array}{l}\dot{\theta }=\frac{d\theta }{dt}\\ \dot{\theta }=\frac{d}{dt}\left(0.1\left(2t^2-t\right)\right)\\ \dot{\theta }=0.1\left(4t-1\right)\end{array}$

$\begin{array}{l}\ddot{\theta }=\frac{d\dot{\theta }}{dt}\\ \ddot{\theta }=\frac{d}{dt}\left(0.1\left(4t-1\right)\right)\\ \ddot{\theta }=0.4\end{array}$

Force components,

$\begin{array}{l}{F}_r=m(\ddot{r}-r{\dot{\theta }}^2)\\ F_r=175\left[-\frac{2{\pi }^2}{9}\cos \left(\frac{\pi }{3}t\right)-\left(10+2\cos \left(\frac{\pi }{3}t\right)\right)\times {\left(0.1\left(4t-1\right)\right)}^2\right]\end{array}$

$\begin{array}{l}{F}_{\theta }=m(r\ddot{\theta }+2\dot{r}\dot{\theta })\\ F_{\theta }=175\left[0.4\left(10+2\cos \left(\frac{\pi }{3}t\right)\right)+2\left(-\frac{2\pi }{3}\sin \left(\frac{\pi }{3}t\right)\right)\left(0.1\left(4t-1\right)\right)\right]\end{array}$

Table of $F_r$, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
Fr (lb)	-12.6	-11.0	-11.3	-13.6	-18.0	-26.1	-40.7	-65.6	-104.1	-157.1	-221.8	-291.5

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
Fr (lb)	-356.9	-408.8	-441.7	-457.0	-464.0	-478.9	-520.7	-604.9	-738.2

Plot,

Table of $F_{\theta }$, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
F? (lb)	26.1	24.4	18.0	10.4	5.8	7.7	17.4	32.8	49.2	60.4	61.4	49.4

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
F? (lb)	26.0	-3.0	-29.4	-44.3	-41.5	-19.5	17.5	60.2	96.5

Plot,

Conclusion:

Plots are mentioned the explanation part.