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Homework Help: Circular loop problem: finding the radius of a small circle

  1. Sep 15, 2008 #1
    1. The problem statement, all variables and given/known data

    In an air show, a pilot is to execute a circular loop at the speed of sound(340 m/s) . The pilot may black out if his acceleration exceeds 8g. Find the radius of the smallest circle he can use. [Take g=10 m s^-2

    2. Relevant equations

    possible equations: u^2 >= 2*M*G/a
    3. The attempt at a solution

    u=340 m/s
    a=dv/dt=>=80 m s^-2
    M=6.00e24 kg

    acceleration=dv/dt=> dv/dt=-M*G/r^2

    option one
    r=sqrt(M*G/(dv/dt)=7.07e12 meters
    option 2
    a=2*M*G/u^2, a being minimal radius and not the acceleration

    a= 28284271.25 meters

    actual answer: r>=1445 meters

    what did I do wrong? should I have used polar coordinates since the problem states that the plane is going around a loop?

    Perhaps I should write:

    m*dv/dt=F(z) , F(z) representing the sum of all forces. the only two forces acting on the object is the gravitational force and the centripetal force ; so my equation looks like:


    dv/dt=dv/dr*dr/dt=dv/dr*v. Now I can integrate in terms of the velocity vector and radius vector.


    I get something that looks like this:

    dv/v=dr/r-dr/g ==> ln v = In r -r/g
    Last edited: Sep 15, 2008
  2. jcsd
  3. Sep 16, 2008 #2


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    What equations do you know of that deal with circular motion?

    FYI, the "r" in m*M*G/r^2 is not necessarily a radius, it's the distance between two masses.
  4. Sep 16, 2008 #3

    the equations that deal with circular motion are : F(centripetal)=m*v^2/r and the equation for a particle in general motion:
    Well, since the airplane is obviously flying around a circle on earth, then doesn't the radius of the earth , plus the height the airplane is some distance above the surface of the earth become relevant in finding the total R=R(earth)+height?
  5. Sep 16, 2008 #4


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    Hi Benzoate! :smile:

    I'm really confused :confused:

    the question gives you the acceleration (8g) … which you can assume is constant (they're telling you g = 10, which is about 2% out anyway :wink:)

    you don't need G or M!

    Try again (and don't forget you'll have to add or subtract up to 1g for the gravity) :smile:
  6. Sep 16, 2008 #5
    could I just write:

    a=v^2/r and since I know v and a I can easily calculate r correct?
  7. Sep 16, 2008 #6


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    :biggrin: Woohoo! :biggrin:
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