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Vertical Circular Motion

  1. Nov 28, 2013 #1
    1. The problem statement, all variables and given/known data

    A 76-kg pilot at an air show performs a loop de loop with his plane. At the bottom of the 52-m radius loop, the plane is moving at 48 m/s. Determine the normal force acting upon the pilot.


    2. Relevant equations

    (ƩF)R = maR = m(v2/r)

    3. The attempt at a solution

    I drew a diagram and I know I have to figure it out considering the position of the plane at the bottom of the plane. At the bottom of the plane FR and FN are pointing towards the center of the circle while Fg is pointing the opposite direction. Since the pilot has no movement in the y-direction we know:

    FR + FN = Fg

    After that I tried putting in the values I have but I got Fg's value was smaller than the total of FR + FN which does not make sense meaning I've made a mistake. I'm confused on how to tackle the problem after this part.
     
  2. jcsd
  3. Nov 28, 2013 #2

    Doc Al

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    Staff: Mentor

    Only two forces act on the pilot: The upward normal force and the downward gravitational force. Their sum equals the net force, which in this case equals what you call FR.

    Show what you got for FR, FN, and Fg.
     
  4. Nov 28, 2013 #3

    BruceW

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    Homework Helper

    FR is the resultant force, right? And I'm guessing you're taking FR, FN and Fg to be the absolute values. So the equation FR + FN = Fg is not quite right. Remember that the resultant force is not an extra force on its own. The resultant force is the force due to the sum of all other forces.

    edit: whoops, Doc Al got there first.
     
  5. Nov 28, 2013 #4
    Oh, ok I sorta see what you're getting at.

    So by doing Fnet = FN - Fg

    Then we use algebra to make it:

    Fnet + Fg = FN

    Which is then:

    m(v2/r) + mg = FN

    I then sub in the values I got from the question arriving to the answer of:

    FN = 4112.18 N

    Am I on the right track?
     
  6. Nov 28, 2013 #5

    Doc Al

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    Staff: Mentor

    Exactly right. :thumbs:

    (Now compare that to what the normal force would be at the top of the loop. What's the key difference?)
     
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