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Please help settle this dispute

  1. Dec 16, 2009 #1


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    Homework Helper

    1. The problem statement, all variables and given/known data
    I'm involved in a dispute about conical pendulums, and my friend, for whatever reason, doesn't want to believe me until somebody backs me up. Assume t is the angle that the string makes with the horizontal.

    2. Relevant equations

    3. The attempt at a solution

    Tsint = mg
    Tcost = mw^2*r = mw^2*Lcos(t)
    T=mw^2*L, where L is the length of the string

    Dividing gives:

    sin t = g/(w^2*L)

    This implies that if w^2*L is smaller than the acceleration of gravity, a conical pendulum cannot maintain a constant angle t with the horizontal.

    P.S. this is actually not a homework question. I'm almost certain my calculations are right, but my friend insists that there's no minimum rotation speed for a conical pendulum.
  2. jcsd
  3. Dec 16, 2009 #2
    I'm not iterested in weighing in one way or another, but the math seems ok by my reckoning--I get tan (x) = g/(w^2*L) but assume that since both approximate to t, you are using that. But if you use that than the angle must be small which implies that sin (t)= tan (t) is small. May be I'm not understanding the last line of the proof.
  4. Dec 16, 2009 #3


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    Science Advisor

    You could also say that w^2 cannot be less than g/L. When the angle with the horizontal is close to 90 degrees (that is, when the bob is moving slowly and hanging close to straight down, you have sin t being close to 1 and the angular velocity w of the pendulum approaches the minimum of
    [tex]w = \sqrt{\frac{g}{L}}[/tex]​

    This is why pendulums (including conical pendulums) make good time pieces.

    Cheers -- sylas

    PS. You could also show your friend the wikipedia article on Conical pendulum. It uses the opposite angle convention, measuring the angle from the vertical rather than from the horizontal, but the equation at the end of the article for the period of the pendulum is
    [tex]\textrm{period} = 2 \pi \sqrt{\frac{L \cos \theta}{g}}[/tex]​
    which has the same implication; there is a maximum possible period for the conical pendulum.
    Last edited: Dec 16, 2009
  5. Dec 16, 2009 #4


    Staff: Mentor

    Here is how I did it:
    Given a bob of mass m at the end of a string of length L making an angle [itex]\theta[/itex] with the horizontal. The mass is travelling in uniform circular motion in the horizontal plane, calculate the frequency of rotation.

    Newton's 2nd law in the vertical direction
    [tex]T \, sin(\theta) - mg = 0[/tex]

    Newton's 2nd law in the horizontal direction
    [tex] T \, cos(\theta) = ma [/tex]

    Centripetal acceleration
    [tex]a = r\omega^2[/tex]

    Radius of circle
    [tex]r=L \, cos(\theta)[/tex]

    So we have four equations which we can solve for [itex]\omega[/itex] eliminating a, T, and r. This gives
    [tex]\omega = \sqrt{\frac{g}{L \, sin(\theta)}}[/tex]

    So, I agree with your conclusion. To me it was surprising to find that, as the bob hangs almost vertical the frequency does not go to zero. However, I was OK once I realized that the velocity does go to zero.
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