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Homework Help: Understanding forces

  1. Feb 26, 2012 #1
    Here is the situation:

    An airplane flies in a loop (a circular path in a vertical plane) of radius 150 . The pilot's head always points toward the center of the loop. The speed of the airplane is not constant; the airplane goes slowest at the top of the loop and fastest at the bottom.

    Here is the question:

    At the top of the loop, the pilot feels weightless. What is the speed of the airplane at this point?

    There is a similar question already asked on this forum but i had a few additional questions about this.

    I have a free body diagram with the plane at the top of the loop and the force that is acting on it is its weight (gravity). The similar post does not include the normal force that would oppose weight. Is the normal force excluded because the plane is in the air and therefore there is no normal force?

    Thank you for helping me understand this concept.
  2. jcsd
  3. Feb 26, 2012 #2
    Without seeing the other post I cannot comment on it. Do you know what centrifugal force is?
  4. Feb 26, 2012 #3
    All of the information that you see is what i was given.

    part two of the question states:

    At the bottom of the loop, the speed of the airplane is 300 . What is the apparent weight of the pilot at this point? His true weight is 620 .

    I am completely lost on how to start this.

    Can you just explain what forces I need to use and how to figure out how to choose them.

    So far I have in the Y direction (with axis pointing up)

    Normal (airplane) - Weight (pilot) = Mv^2/R

    Normal - 620 = Mass ((620/9.8)* 300^2)/ 150

    is this set up correct?

    also here is a link to the post I am referring to

  5. Feb 26, 2012 #4
    For part 1 of the problem the forces you need to use are his weight and centrifugal. When they balance, the pilot is weightless. The airplane weight is not an issue. Write an equation that balances the above forces and solve for the speed.

    For part 2, the situation is similar except the direction of centrifugal force is down, not up. So his apparent weight is computed how?
  6. Feb 26, 2012 #5
    Thank you by the way for all of you help

    The first part I figured out.

    For the second part I am confused about the centrifugal force, the centrifugal force is the force directed towards the center of the circle correct?

    If so I would have the normal force pointing up, the weight pointing down, and the centrifugal force pointing up correct?
  7. Feb 26, 2012 #6
    The equation is keep coming up with is this:

    N - mg = m(v^2/R)

    with N being normal force from plane to pilot, mg being weight of pilot, and the other side being mass of pilot times his velocity squared divided by the radius. The axis is pointing up towards the center of the circle.
  8. Feb 26, 2012 #7
    It's the force pointed away from the center due to changing direction of velocity. I believe the word centrufugal means to flee the center. Anyway, semantics aside, if you had a rock on a string and swung it in a vertical plane, the tension in the string is greater at the bottom because you have centrifugal force added to the weight of the rock. String tension is least at the top because weight is opposite centrifugal force. This is what occurs with the airplane.

    Getting back to the airplane and your last statement, you have only two forces. One is weight and the other is centrifugal. Both point down and their sum is the apparent weight.
  9. Feb 26, 2012 #8
    Ok I understand that

    Then both of these forces are equal to the mass times acceleration in the Y direction then?

    Fc + W = M*V^2/R

    Fc + 620 = (620/9.8) * (300^2)/150

    I still am getting a very large number what am I missing? I am not looking for an answer by the way just the concept that i am not catching.
  10. Feb 26, 2012 #9
    You never mentioned the units of the speed of the airplane. Is it meters/second or kilometers/hour?

    The apparent weight is the sum of normal weight and centrifugal force.
  11. Feb 26, 2012 #10
    300 km/h the units must not be text and didnt copy when i copied the original problem
  12. Feb 26, 2012 #11
    I understand that the apparent weight is the weight plus the centrifugal force but I keep getting a very large number for my centrifugal force.

    is this equation correct: Fc + W = M*V^2/R? with Fc being my centrifugal force
  13. Feb 26, 2012 #12
    Fc + W = M*V^2/R

    Above equation has the W on the wrong side assuming Fc is the apparent weight. The force due to circular motion should add to the weight when the airplane is at its lowest point.

    Since it's km/hour, the answer is reasonable. You can get that kind of centrifugal acceleration in some race cars.

    So now complete the problem.
  14. Feb 26, 2012 #13
    Ok thanks for all of your help so far but I am still completely confused.

    According to what you said

    Fc = M*V^2/R + W

    Plugging number in i get

    Fc = (620/9.8) * (300^2/150) + 620

    Fc = 38579 N ?
  15. Feb 26, 2012 #14
    You may prefer to use centripetal force
    There are two relevant forces on the pilot:

    1.The normal force which is the apparent weight(N).The question implies that this acts towards the centre of the circle
    2.The pilots weight (mg).This acts vertically downwards.

    At the top N +mg both act towards the centre and the resultant force(N+mg) provides the centripetal force(F).
    The pilot experiences weightlessness at a speed such that N becomes zero.He is in a momentary state of free fall.

    At the bottom the forces point in opposite direction and the resultant force (N-mg) provides the centripetal force.
    From this N can be found
  16. Feb 26, 2012 #15
    You implied the speed is 300 km/hr. You need to change the units to m/second before plugging into the equation.
  17. Feb 26, 2012 #16

    I was so frustrated that I didnt even bother looking at the units which is exactly why my number was so large. Thank you Lawrence for sticking with this post and showing me my rookie mistake.

    also thank you Dadface for helping me clear up my understanding of the forces that are applied.
  18. Feb 26, 2012 #17
  19. Mar 30, 2014 #18
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