Calculating normal force in a loop-the-loop situation

In summary: Now that you have the equations for the two points in the loop, plug them in for the total force on the pilot and solve for the magnitude.For the top of the loop, the equation is:Ft = mg*\omega*\Delta\thetaFor the bottom, the equation is:Ft = mg*\omega*\Delta\theta-\omega*\Delta\thetaFor the total force on the pilot, the equation is:Ft = mg*\omega*\Delta\thetaNow solve for the magnitude of the force.Ft = (2.93*mg)*\omega*\Delta\theta = 9.29*mg*\ome
  • #1
jeanjacket
1
0

Homework Statement



Here's the situation:

A jet pilot puts an aircraft with a constant speed into a vertical circular loop. If the speed of the aircraft is 700 km/h and the radius of the circle is 2.0 km, calculate the normal forces exerted on the seat by the pilot at the bottom and top of the loop. Express your answer in terms of the pilot's weight mg.

a) at the bottom
b) at the top

I actually have the answers to the problem, but of course it doesn't help me understand the problem nor the process in getting there. Not sure which angular motion/centripetal force equations to use. But if it helps you verify results, here are the answers:

At bottom: 2.93*mg
At the top: .929*mg

I would really appreciate any help, even if it's just pointing me in the right direction as far as which equation to use, I'm not necessarily asking anyone to work it out. Thanks in advance!

Homework Equations



Not necessarily sure which to use, here's some conversions just for quick reference:
700 km/h = 194.444 m/s
2.0 km = 2000 m

Possible eqns:

Angular Velocity = [tex]\omega[/tex] = [tex]\Delta[/tex][tex]\theta[/tex]/[tex]\Delta[/tex]t
Angular Acceleration = [tex]\alpha[/tex] = [tex]\Delta[/tex][tex]\omega[/tex]/[tex]\Delta[/tex]t
[tex]\textbf{F}[/tex][tex]_{centripetal}[/tex] = mass x accel(centrip) = m[tex]v^{2}/r[/tex]
[tex]\textbf{F}[/tex][tex]_{gravity}[/tex] = mg

The Attempt at a Solution



Can't quite figure out where to start...
 
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  • #2
Start with 2 free-body diagrams, 1 for the forces on the pilot at the top of the loop and one for the bottom. Keep in mind what force(s) provide the centripetal acceleration for the pilot at those points.
 
  • #3
At all times the centrifugal acceleration pushes the pilot onto the seat. The difference is the direction of the weight of the pilot, which is always the directed towards the centre of the earth. So at the bottom force = centrifugal + weight, at top its centrifugal - weight.
 

What is normal force in a loop-the-loop situation?

Normal force is the force that a surface exerts on an object that is in contact with it. In the context of a loop-the-loop situation, it refers to the force that the track exerts on the object as it travels through the loop.

How do you calculate normal force in a loop-the-loop situation?

To calculate normal force in a loop-the-loop situation, you can use the formula N = mg + mv^2/r, where N is the normal force, m is the mass of the object, g is the acceleration due to gravity, v is the velocity of the object, and r is the radius of the loop.

What factors affect the normal force in a loop-the-loop situation?

The normal force in a loop-the-loop situation is affected by several factors, including the mass of the object, the velocity of the object, the radius of the loop, and the acceleration due to gravity. Additionally, the angle of the loop and the friction between the object and the track can also affect the normal force.

Why is it important to calculate the normal force in a loop-the-loop situation?

Calculating the normal force in a loop-the-loop situation is important because it helps us understand the forces at play and determine if the object will successfully complete the loop or not. It also allows us to identify potential safety concerns and make adjustments to ensure the object's safety.

Are there any limitations to using the normal force formula in a loop-the-loop situation?

Yes, there are limitations to using the normal force formula in a loop-the-loop situation. The formula assumes that the track is a perfect circle and that there is no friction between the object and the track. In reality, these conditions may not be met and can affect the accuracy of the calculation.

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