A jet pilot takes his aircraft in a vertical loop. If the jet is moving at a speed of 1100 at the lowest point of the loop:
A)Determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 5.5 's
B)Calculate the 68 pilot's effective weight (the force with which the seat pushes up on him) at the bottom of the circle.
C)Calculate the 68 pilot's effective weight (the force with which the seat pushes up on him) at the top of the circle (assume the same speed)
Without units on the numbers, it is impossible.
At the bottom of the loop, you have centripetal acceleartion and weight. Centripetal acceleartion = v^2/r
in g's, divide by 9.8m/s^2
A jet pilot takes his aircraft in a vertical loop. If the jet is moving at a speed of 1100km/h at the lowest point of the loop:
A)Determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 5.5g's
B)Calculate the 68kg pilot's effective weight (the force with which the seat pushes up on him) at the bottom of the circle.
C)Calculate the 68kg pilot's effective weight (the force with which the seat pushes up on him) at the top of the circle (assume the same speed)
To solve these problems, we will need to use some basic principles of circular motion and centripetal acceleration.
A) To find the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 5.5 m/s^2, we use the formula for centripetal acceleration:
a = v^2 / r
where a is the centripetal acceleration, v is the velocity, and r is the radius of the circle.
In this case, the speed (v) is given as 1100 m/s, and we need to find the radius (r). Rearranging the formula, we have:
r = v^2 / a
Substituting the given values, we find:
r = 1100^2 / 5.5
r = 2200^2 meters
Therefore, the minimum radius of the circle is 2200 meters.
B) To calculate the pilot's effective weight at the bottom of the circle, we need to consider the net force acting on the pilot. At the bottom of the circle, there are two forces acting on the pilot: the gravitational force (mg) and the normal force (N) from the seat.
The net force is equal to the centripetal force, which is given by:
F = m * a
where F is the net force, m is the mass of the pilot, and a is the centripetal acceleration.
At the bottom of the circle, the net force is the difference between the normal force and the gravitational force:
F = N - mg
Since the normal force provides the necessary centripetal force, we can set N equal to mv^2 / r:
N = mv^2 / r
Substituting this into the equation for the net force, we have:
mv^2 / r - mg = ma
Rearranging the equation and solving for N, we get:
N = m(g + a)
Substituting the given values, we have:
N = m(9.8 + 5.5)
N = 15.3m newtons
Therefore, the pilot's effective weight at the bottom of the circle is 15.3 times the pilot's mass.
C) To calculate the pilot's effective weight at the top of the circle, we can use the same approach as in part B, but with the same speed and opposite centripetal acceleration.
At the top of the circle, the centripetal acceleration is directed towards the center of the circle but opposite in direction to the gravitational force. Therefore, the net force is the sum of the normal force and the gravitational force:
F = N + mg
Using the equation for net force, we have:
N + mg = ma
Rearranging the equation and solving for N, we get:
N = m(g - a)
Substituting the given values, we have:
N = m(9.8 - 5.5)
N = 4.3m newtons
Therefore, the pilot's effective weight at the top of the circle is 4.3 times the pilot's mass.
v =1100 km/h = 305.6 m/s.
a =v^2/R => R = v^2/a = (305.6)^2/5.5•9.8 = 1732 m
R > 1732 m.
The magnitude of the effective weight = the magnitude of the normal force N
bottom: N = m•g – m•a =
= m• (g - 5.5•g) = - 4.5•m•g.
W = - N = 4.5•m•g =4.5•68•9.8 =2998.8 N.
top: N = - m•g – m•a = -m• (g+a) =
= -m• (g+5.5•g),
W = - N = 6.5•m•g =6.5•68•9.8 =4331.6 N.