Centripetal Force: 80kg Pilot in Jet at 305m/s

In summary, the question is asking for the apparent weight that an 80kg pilot feels while making a loop the loop in a jet traveling at a constant speed of 305 m/s with a radius of 1.900km. The relevant equations are Acp=w^2/r, w=(v/r), and F=ma or F=mg. After solving for omega and Acp, the correct calculation for the force the pilot feels at the bottom of the loop is mg+Acp.
  • #1
electricheart
6
0

Homework Statement



An 80 kg pilot, makes a loop the loop in a jet, the jet maintains a constant speed of 305 m/s and the radius of the loop is 1.900km. What is the apparent weiht that the pilot feels( Force) at the bottom of the loop the loop.

Homework Equations


Acp=w^2/r
w=(v/r)
F= ma or F= mg


The Attempt at a Solution


First I solved for omega w=0.1605, then I solved for Acp=0.0000136
the Force the pilot feels at the bottom of the loop would be
mg+ Acp or is this wrong?
 
Physics news on Phys.org
  • #2
I figured out what I was doing wrong, I calculated the Acp wrong.
 
  • #3


I would like to clarify a few things in this scenario. Firstly, it is important to note that the pilot's weight is not changing throughout the loop. The force of gravity, mg, remains constant regardless of the pilot's motion. Therefore, the pilot's apparent weight at the bottom of the loop should be equal to their actual weight, 80 kg.

Now, let's look at the forces acting on the pilot at the bottom of the loop. The pilot is experiencing two forces: the force of gravity, mg, and the centripetal force, Fc. The centripetal force is the force that keeps the pilot moving in a circular path. It is given by Fc = m(v^2/r), where m is the mass of the pilot, v is the velocity, and r is the radius of the loop.

In this case, we have all the necessary values to calculate the centripetal force. Plugging in the values, we get Fc = 80(305^2/1900) = 3,088 N. This is the force that the pilot is feeling at the bottom of the loop.

Now, if we want to calculate the apparent weight of the pilot, we can simply add the force of gravity and the centripetal force. The apparent weight, Fapp, is given by Fapp = mg + Fc. Plugging in the values, we get Fapp = 80(9.8) + 3,088 = 3,868 N. This is the apparent weight that the pilot feels at the bottom of the loop.

In conclusion, the pilot's apparent weight at the bottom of the loop is 3,868 N, which is the sum of their actual weight and the centripetal force acting on them. It is important to note that the pilot's actual weight remains constant throughout the loop, but the apparent weight changes due to the additional centripetal force.
 

1. What is centripetal force?

Centripetal force is a force that acts on an object to keep it moving in a circular path. It is directed towards the center of the circle and is necessary to maintain the object's velocity and prevent it from flying off in a straight line.

2. How is centripetal force calculated?

The formula for calculating centripetal force is F = mv^2/r, where F is the force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.

3. What is the centripetal force acting on an 80kg pilot in a jet traveling at 305m/s?

Using the formula from the previous question, we can calculate the centripetal force acting on the pilot to be approximately 19,806,250 N.

4. Can centripetal force be greater than the weight of the object?

Yes, centripetal force can be greater than the weight of the object. This is because centripetal force is dependent on the mass and velocity of the object, not just its weight.

5. How does centripetal force affect the motion of an object?

Centripetal force is what keeps an object moving in a circular path. Without it, the object would continue in a straight line. The magnitude and direction of the centripetal force determine the speed and direction of the object's motion.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
1K
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
12
Views
201
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top