A jet pilot takes his aircraft in a vertical loop (Fig. 5-43).
(a) If the jet is moving at a speed of 1900 km/h at the lowest point of the loop, determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.0 g's. _____m
(b) Calculate also the 70 kg pilot's effective weight (the force with which the seat pushes up on him) at the bottom of the circle. _____N
(c) Calculate the pilot's effective weight at the top of the circle. (Assume the same speed.) _____N
The centripetal acceleration at the bottom will be v^2/r + g. Be certain to change velocity to m/s
at the top, the force will be
mv^2/r - mg
(a) 19+6.0007858787577=287767566354526
a) You don't want the centripetal accelartaion force to exceed 6g so you set it to 6g
6g = v`2/r
then you can simply solve for r
r = v`2/6g
To find the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.0 g's, you can use the formula:
centripetal acceleration = v^2/r
where v is the velocity in m/s and r is the radius of the circle.
First, convert the speed from km/h to m/s:
1900 km/h * (1 hr/3600 s) = 527.78 m/s
Next, substitute the values into the formula and solve for r:
6g = (527.78 m/s)^2 / r
Since g is the acceleration due to gravity and is approximately 9.8 m/s^2, we can rewrite the equation as:
6 * 9.8 m/s^2 = (527.78 m/s)^2 / r
Now, solve for r:
r = (527.78 m/s)^2 / (6 * 9.8 m/s^2)
r ≈ 1438.44 meters
Therefore, the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.0 g's is approximately 1438.44 meters.