Answer to Question #241365 in Mechanics | Relativity for Anas

Question

A jet traveling at a speed of 1.20 X 10° m/s executes a vertical loop with a radius of 5.00 x 10° m. Find the magnitude of the force of the seat on a 70.0-kg pilot at (a) the top and (b) the bottom of the loop.

Answer 1

gravitational acceleration = g = 9.81 m/s²

Constant speed of the stealth fighter $= V = 1.2 \times 10^2\; m/s = 120 \;m/s$

Radius of the vertical loop $= R = 5.00 \times 10^2 \; m = 500 \; m$

Mass of the pilot = M = 70 kg

Weight of the pilot = W (downwards)

W = Mg

(a) Force exerted by the seat on the pilot at the top of the loop = N (downwards)

Centripetal force on the pilot = F_b

$F_b = \frac{MV^2}{R}$

At the top of the loop the centripetal force will be directed downwards.

The centripetal force on the pilot is the net force on the pilot from the force the seat exerts on him and his own weight.

$F_b = N+W \\ \frac{MV^2}{R} = N+Mg \\ \frac{70 \times 120^2}{500}= N +70 \times 9.81 \\ N= 1.33 \times 10^3 \;N$

(b)

$\sum F_y = F_n -Mg = F_c \\ F_N = F_c +Mg \\ F_N = \frac{MV^2}{R} + Mg \\ F_N = 70(\frac{120^2}{500} + 9.81) \\ F_N = 270.2 \;N$