Question

Chapter 2, Problem 2/125 At the bottom of a loop in the vertical (r, θ) plane at an altitude of 495 m, the airplane P has a horizontal velocity of 625 km/h and no horizontal acceleration. The radius of curvature of the loop is 1045 m. For the radar tracking at O, determine the recorded values of r and θ for this instant. 625 km/h 495 m 1040 m Answers: m/s2 rad/s2

Answer 1

At this point, the plane has only one acceleration, radial.

a = v^2 / R

where v = speed = 625 kmph = 625 x 5/18 = 173.61 m/s

R = 1045 m

a = 28.84 m/s^2... first answer. This acts towards center. That means upwards.

to find angular acceleration, we use the relation a = r$\alpha$

r = $\sqrt{495^2 + 1040^2 }$ = 1151.8 m

$\alpha$ = a/r = 28.84 / 1151.8 = 0.025 rad/s^2 ... second answer