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Conical Pendulum with lift and tension

  1. Dec 2, 2015 #1
    1. The problem statement, all variables and given/known data
    A model airplane has a small gas powered motor in it to allow it to fly. It is tethered its controller by a long
    cord. The plane will fly in a circle at the end of this cord around the controller who uses the tether to control
    the altitude of flight of the plane. Currently the plane is flying in level flight while the cord makes an angle
    above the horizontal. The plane’s speed is 13.0 m/s and its mass is 1.25 kg. The cord is 10.0 meters long.

    A) What is the tension on the cord?
    B) What is the angle the cord makes with the horizontal?
    C) What is the amount of lift the wings are providing to the plane?

    2. Relevant equations
    Fx= Tcosθ + Lsinθ= m(v^2)/r
    Fy = Tsinθ + Lcosθ - mg = 0
    r = 10cosθ
    3. The attempt at a solution
    I've been working on this problem for a while. I am starting to become convinced that there are too many unknowns in the problem and that it is flawed. You could solve for T in each equation and then set the equations equal to each other to get ((mv^2/r) - Lsinθ)/cosθ = (-Lcosθ +mg)/sinθ . You eliminate 1 unknown and now you have 3 unknowns. Then you can eliminate r by substituting 10cosθ and then you would have 2 unkowns. If I was only looking for the angle, I"m pretty sure I could set something up including a quadratic equation. The problem is defining L. Maybe I set the problem up wrong in the beginning. I'm new to this forum but any help would be appreciated. I ended up here so far

    (mv^2sinθ)/10(1-sin^2θ)-mg = -L(cos^2θ-sin^2θ)
  2. jcsd
  3. Dec 3, 2015 #2
    The main job is to pull the aircraft inward and L lifting the aircraft. Without the string the aircraft has to bank so that there are components forces.
    Here the centripetal force is provided by the string with added force to the weight.
    Last edited: Dec 3, 2015
  4. Dec 3, 2015 #3
    Ok. Well I was assuming that the aircraft was banking. I've thought about what you are saying. In essence, you are saying that the centripetal force is only provided by the horizontal component of the tension such that Fx= Tcosθ = m(v^2)/r ( and since there is no banking, lift only comes into play in the y dimension such that Fy = Tsinθ + Lcosθ - mg = 0 ?
  5. Dec 3, 2015 #4
    (Mv^2sintheta/10cos^2theta) -mg = L according to your set up but still a dead end. I would love to solve for theta but we dont know r. Circular logic...
  6. Dec 3, 2015 #5
    TCos(x)=mv^2/r , T=?
    Cox(x)=r/10, Sinx=?, Tanx=?
    Since c and T are both in same direction

    (Mg +TSin(x))/(mv^2/r)=Tan(x)

    L=mg +TSin(x)
  7. Dec 3, 2015 #6
    I'm looking at your reply and I've done something similar with the trig but it gets me nowhere. I also don't see L in the equation. What you did is tan(theta)= Ty/Tx but Ty is not equal to mg + Tsin(theta).
  8. Dec 3, 2015 #7
    Just substitute T in term of r and only one unknown "r" in the equation. I use computer to solve it.
  9. Dec 3, 2015 #8
    I don't think your equation is correct. You are equating L with Ty. Also, to substitute T in terms of r, you bring an unknown angle back into it getting you nowhere since r = 10costheta.
  10. Dec 7, 2015 #9
    So does anybody out there have a way to solve this? I just noticed 153 views and I am coming to the conclusion that there is not enough info to solve after several days of rearranging and rearranging.
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