A jet pilot (80 kg) flies a loop-de-loop at a constant speed of 600 m/s.
a) What minimum radius can the loop have so that the centripetal acceleration is no more than 6ag?
b) Suppose that the pilot just happened to be sitting on a spring scale while flying the jet. What will the scale read (in Newtons) when he is b) at the bottom of the loop (and flying right-side-up), and c) at the top of the loop (and flying upside-down)?
Please help ASAP!!
To solve these problems, we can use the following formulas:
a) Centripetal acceleration (ac) is given by the equation ac = v^2 / r, where v is the velocity and r is the radius of the loop. In this case, ac should not exceed 6 times the acceleration due to gravity (6ag).
b) At the bottom of the loop, when the pilot is flying right-side-up, the scale will read the normal force (N) exerted on the pilot. The normal force is equal to the weight (mg) plus the force due to the acceleration (ma). The weight is given by mg, where m is the mass of the pilot and g is the acceleration due to gravity.
c) At the top of the loop, when the pilot is flying upside-down, the scale will also read the normal force. However, since the pilot is upside-down, the normal force is directed in the opposite direction. The normal force is equal to the weight minus the force due to the acceleration (ma).
Now, let's solve each problem step by step:
a) To find the minimum radius, we can equate the centripetal acceleration to 6 times the acceleration due to gravity:
ac = 6ag
Substituting the values, we get:
v^2 / r = 6ag
Rearranging the equation to solve for r:
r = v^2 / (6ag)
Substituting the given values v = 600 m/s and g ≈ 9.8 m/s^2, we can calculate r.
b) At the bottom of the loop, the normal force N is equal to the sum of the weight and the force due to acceleration, given by N = mg + ma.
Substituting the given values, we get:
N = (m * g) + (m * a)
At the bottom of the loop, the acceleration a is equal to the centripetal acceleration (ac) divided by the mass of the pilot (m). Using the value of ac from part a), we can calculate the force exerted at the bottom of the loop.
c) At the top of the loop, the normal force N is equal to the weight minus the force due to acceleration, given by N = mg - ma.
Using the value of ac from part a), we can calculate the force exerted at the top of the loop.
Now, you can use these formulas and the given values to solve parts a), b), and c) of the problem.