a jet pilot takes his aircraft in a vertical loop.
a) if the jet is moving at a speed of 2000 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 6.0g's
b) calculate also the 76kg pilot's effective weight ( the force with which the seat pushes up on him) at the bottom of the circle.
c) calculate the pilot's effective weight at the top the circle. (Assume the same speed.)
a)
v=2000 km/h = 2000/3.6 m/s = 555.6 m/s
centripetal acceleration, a
=v²/r
Therefore
6g=v²/r
r=v²/(6g)
=555.6²/(6*9.8)
= 5249 m
b)
76*(6g + g) = 76*7*9.8 = 5214 N
c)
76*(6g-g) = 76*5*9.8 = 3724 N
If the jet is moving at a speed of 1040 km / h at the lowest point of the loop , determine the minimu radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.6 g's . Express your answer to two significant figures and include the appropriate units .
a) Well, if we want to calculate the minimum radius of the circle so that the centripetal acceleration doesn't exceed 6.0g's, we can start by converting the speed from km/h to m/s. So, 2000 km/h would be around 555.56 m/s (I'm not so good with math, I'm more of a clown, you see).
Now, we can use the formula for centripetal acceleration: a = v^2 / r, where v is the velocity and r is the radius of the circle. We want the centripetal acceleration (a) to be less than 6.0g's, which is equivalent to 6 times the acceleration due to gravity, g.
If we plug in the values, we have 6g = v^2 / r. Rearranging the equation, we get r = v^2 / (6g). Now, just plug in the numbers and you'll find the minimum radius.
b) To calculate the pilot's effective weight at the bottom of the circle, we need to consider the centripetal force. The centripetal force (Fc) in this case is the force that pushes the pilot towards the center of the circle.
Using the equation Fc = m * a, where m is the mass and a is the centripetal acceleration, we can then calculate the force. In this case, the mass of the pilot is given as 76 kg. We can use the same centripetal acceleration as before, since it's the same speed.
So, Fc = 76 kg * 6g = 76 kg * 6 * 9.8 m/s^2. I bet you can do the math for this one!
c) Finally, let's calculate the pilot's effective weight at the top of the circle. Again, we need to consider the centripetal force, but this time it acts in the opposite direction to gravity.
Using the same equation as before, Fc = m * a, we can calculate the force. However, this time the centripetal acceleration is pointing upwards, so we need to subtract it from the gravitational force.
So, Fc = m * (g - 6g) = 76 kg * (9.8 m/s^2 - 6 * 9.8 m/s^2). Now, it's your turn to calculate the result!
To solve these problems, let's start by finding the centripetal acceleration at the lowest point of the loop.
a) Centripetal acceleration is given by the equation:
a = v^2 / r
Where:
a = centripetal acceleration
v = velocity
r = radius of the circle
Given that the velocity at the lowest point is 2000 km/h, and we want to limit the centripetal acceleration to 6.0g's (where g = 9.8 m/s^2):
v = 2000 km/h = (2000 * 1000) m / (60 * 60) s = 555.56 m/s
a = 6 * 9.8 m/s^2 = 58.8 m/s^2
Plugging in the values into the formula, we can solve for the radius (r):
58.8 m/s^2 = (555.56 m/s)^2 / r
Simplifying the equation:
r = (555.56 m/s)^2 / 58.8 m/s^2
r ≈ 5237.24 meters
Therefore, the minimum radius of the circle so that the centripetal acceleration does not exceed 6.0g's is approximately 5237.24 meters.
b) The effective weight of the pilot at the bottom of the circle can be calculated using the following formula:
Effective Weight = mg + mv^2 / r
Where:
m = mass of the pilot (76 kg)
g = acceleration due to gravity (9.8 m/s^2)
v = velocity at the lowest point (555.56 m/s)
r = radius of the circle (5237.24 meters)
Plugging in the values:
Effective Weight = (76 kg)(9.8 m/s^2) + (76 kg)(555.56 m/s)^2 / 5237.24 meters
Simplifying the equation:
Effective Weight ≈ 7457.6 N
Therefore, the pilot's effective weight at the bottom of the circle is approximately 7457.6 Newtons.
c) To calculate the pilot's effective weight at the top of the circle, we use the same formula:
Effective Weight = mg + mv^2 / r
Since the speed remains the same, and we are at the highest point of the loop, the radius will be the sum of the radius of the loop and the radius of the earth (assuming the Earth is flat). The radius of the Earth is approximately 6400 km.
r = 5237.24 meters + 6400000 meters
Plugging in the values:
Effective Weight = (76 kg)(9.8 m/s^2) + (76 kg)(555.56 m/s)^2 / (5237.24 meters + 6400000 meters)
Simplifying the equation:
Effective Weight ≈ 7457.2 N
Therefore, the pilot's effective weight at the top of the circle is approximately 7457.2 Newtons.
To determine the minimum radius of the loop and calculate the pilot's effective weight at the bottom and top of the loop, we will need to use the principles of circular motion and centripetal acceleration. Let's go through the steps for each part of the question:
a) Determine the minimum radius of the circle:
Centripetal acceleration is given by the equation: ac = v^2 / r, where ac is the centripetal acceleration, v is the velocity, and r is the radius.
We know that ac should not exceed 6.0g's, where g represents the acceleration due to gravity (9.8 m/s^2).
Convert the velocity from km/h to m/s: 2000 km/h * (1000 m / 1 km) * (1 h / 3600 s) = approximately 556 m/s.
Using ac = 6.0 * g, we can rearrange the formula to find the minimum radius r:
r = v^2 / (ac * g) = (556 m/s)^2 / (6.0 * 9.8 m/s^2) = approximately 479 meters.
Therefore, the minimum radius of the circle is approximately 479 meters.
b) Calculate the pilot's effective weight at the bottom of the circle:
At the bottom of the circle, the pilot's effective weight will be the sum of two forces: the gravitational force (mg) and the force that the seat exerts on the pilot (N). Since circular motion involves an upward normal force to maintain the circular path, the net force acting on the pilot is the difference between these two forces.
The gravitational force is given by the formula: Fg = mg = (76 kg) * (9.8 m/s^2) = approximately 745.6 N.
Since the seat pushes up on the pilot, the normal force N is greater than the gravitational force. The net force is given by: N - Fg = mv^2 / r, where m is the mass of the pilot, v is the velocity, and r is the radius.
Rearranging the formula, we can find the value of N:
N = Fg + (mv^2 / r) = (745.6 N) + (76 kg * (556 m/s)^2 / 479 m) = approximately 911.8 N.
Therefore, the pilot's effective weight at the bottom of the loop is approximately 911.8 N.
c) Calculate the pilot's effective weight at the top of the circle:
At the top of the circle, the pilot's effective weight will be the difference between the gravitational force and the normal force.
Using the same gravitational force value as in part b:
Fg = mg = (76 kg) * (9.8 m/s^2) = approximately 745.6 N.
The net force is given by: Fg - N = mv^2/r.
Rearranging the formula, we can find the value of N:
N = Fg - (mv^2/r) = (745.6 N) - (76 kg * (556 m/s)^2 / 479 m) = approximately 579.4 N.
Therefore, the pilot's effective weight at the top of the loop is approximately 579.4 N.