Question 12.SP.11: NASA flies a reduced-gravity aircraft (affectionately known ......
NASA flies a reduced-gravity aircraft (affectionately known as the Vomit Comet) in an elliptic flight to train astronauts in a microgravity environment. The plane is being tracked by radar located at O. When the plane is near the bottom of its trajectory, as shown, values from the radar tracking station are \dot{r} = 120~ m/s~ ,~ \dot{θ} = − 0.090~rad/s~,~\ddot{r} = 34.8~m/s²,~and~\ddot{θ} = 0.0156~rad/s². At the instant shown, determine the force exerted on the 80-kg pilot by his seat.
STRATEGY: You want to find the force the pilot experiences at this instant and you can calculate the accelerations, so you should use Newton’s second law. Because you know that the radial distance and the angle are changing with time, use radial and transverse components.
MODELING: Choosing the pilot as the system, draw the free-body and kinetic diagrams as shown in Fig. 1. You could choose to put the forces and the pilot in the r and θ direction or the x and y direction (we chose F_{x}~and~F_{y} to represent the forces from the seat back and bottom, respectively).
ANALYSIS: Before you apply Newton’s second law, determine r and θ from the geometry.
r={\sqrt{800^{2}+600^{2}}}=1000\,{\mathrm{m}}\qquad\theta=\tan^{-1}(600/800)=36.87^{\circ}
Kinematics. Determine the components of the accelerations as
a_{r}=\ddot{r}-r\dot{\theta^{2}}=34.8\,\mathrm{m}/s^{2}-(1000\,\mathrm{m})(-0.090\,\mathrm{rad}/{\bf s})^{2}=26.7\,\mathrm{m}/s^{2}
a_{\theta}=r{\ddot{\theta}}+2{\dot{r}}{\dot{\theta}}=(1000\ m)(0.0156\,\mathrm{rad}/s^{2})+2(120\,\mathrm{m/s})(-0.090\,\mathrm{rad/s})
= − 6.00 m/s²
Kinetics. Obtain scalar equations by applying Newton’s second law in the horizontal and vertical directions. Thus,
\underrightarrow{+}\Sigma F_{x}=m a_{x}\qquad\qquad\qquad F_{x}=m a_{r}\cos\theta-m a_{\theta}\sin\theta (1)
+↑\Sigma F_{y}=m a_{y}\qquad\qquad F_{y}-m g=m a_{r}\sin\theta+m a_{\theta}\mathrm{cos}\ \theta (2)
You have two equations, (1) and (2), and two unknowns, F_{x}~and~F_{y}. Substituting the known values into Eqs. (1) and (2) gives
F_{x}=(80~k{\mathrm{g}})(26.7~\mathrm{m/s^{2}})\cos36.87^{\circ}-(80~{\mathrm{kg}})(-6.00~\mathrm{m/s^{2}})\sin36.87^{\circ}
{F}_{{x}} = 1997 N → ◂
F_{\mathrm{y}} = (80 kg)(9.81 m/s²) + (80 kg)(26.7 m/s²) sin 36.87° + (80 kg)(−6.00 m/s²) cos 36.87°
F_{\mathrm{y}} = 1682 N ↑ ◂
REFLECT and THINK: These forces correspond to a forward acceleration of 2.54 g and a vertical acceleration of 2.14 g. Although this is a bit high for a passenger aircraft, it is within the flight characteristics for the Vomit Comet. If you had been asked to determine whether the plane was speeding up or slowing down, you would need to find the component of the acceleration in the tangential direction, which is defined by the direction of the velocity vector.
