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Homework Help: Conical pendulum problem

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data

    A particle is suspended from a fixed point by a light inextensible string of length a. Investigate 'conical motions' of this pendulum in which a string maintains a constant angle [tex]\theta[/tex]with the downward vertical. Show that , for any acute angle theta
    ex], a conical motion exists and that the particle speed u is given by u^2=a*g*sin([tex]\theta[/tex])*tan([tex]\theta[/tex])

    2. Relevant equations

    dv/dt= (r''-r([tex]\theta[/tex])^2)r-hat + (r([tex]\theta[/tex])''+2r'([tex]\theta[/tex])')[tex]\theta[/tex]-hat

    3. The attempt at a solution

    since the motion of the particle goes around in a circle, I know that r''=r'=0; therefore :

    dv/dt='-r([tex]\theta[/tex])^2)r-hat + (r([tex]\theta[/tex]'[tex]\theta[/tex]-hat

    let \rho be the radius of the cricle; therefore \rho = L*sin([tex]\theta[/tex], L is the length of the string. the pendulum makes a circle in they xy plane. In the z direction, from the circle that lies on the xy plane to the top of the pendulum , k=L*cos(\theta)

    I think there are three forces that act on the string: centripetal force, the gravitational force, the force of the inextensible string, and I think air resistance is small , so I can neglect air resistance.

    -m*L*([tex]\theta[/tex])''=mg*cos([tex]\theta[/tex]) - T+ mv^2/(L*cos([tex]\theta[/tex])
    -m*L*([tex]\theta[/tex])''=mg*sin([tex]\theta[/tex])

    I not sure if I need to integrate dv/dt to calculate the equation of motion and then square v after I have integrated dv/dt
     
  2. jcsd
  3. Sep 29, 2008 #2

    tiny-tim

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    Hi Benzoate! :smile:

    I'm not really following your equations :confused:

    There are two forces … tension, and gravitation … they have to match the acceleration

    and the acceleration is purely radial (zero tangential) …

    and you need to find T anyway …

    try again! :smile:
     
  4. Sep 29, 2008 #3
    I would used the equation of acceleration in polar for right where:

    a=(-r*([tex]\theta[/tex])^2')r-hat+(r*[tex]\theta''[/tex])theta-hat

    Would the sum of the gravitational and tension forces be the sum of the centripetal force? If so then may=T*cos([tex]\theta[/tex])-mg=0==> T=mg/cos([tex]\theta[/tex]), and max=T*sin([tex]\theta[/tex])==>ax= mg/cos([tex]\theta[/tex])*sin([tex]\theta[/tex])=mg*tan([tex]\theta[/tex]) right?

    r=L*sin([tex]\theta[/tex])
     
  5. Sep 29, 2008 #4

    tiny-tim

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    Hi Benzoate! :smile:

    (have a theta: θ and a squared: ² :smile:)
    Wow! That's perfect! :smile:

    Now (since θ'' = 0 and rθ' = v), just use v²/r instead of rθ'² and you have the answer!
     
  6. Sep 29, 2008 #5
    Don't quite understand why [tex]\theta[/tex]'' is zero. Here are my calculations :


    mv2/L*sin([tex]\theta[/tex])=mg*tan([tex]\theta[/tex]) , m cancel out and therefore I am left with: v2=g*tan([tex]\theta[/tex])*L*sin([tex]\theta[/tex])
     
  7. Sep 29, 2008 #6

    tiny-tim

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    Hi Benzoate! :smile:

    (what happened to that θ I gave you?)
    That's right! That's the given answer (with L instead of a). :smile:

    (and θ'' = 0 because the string is always radial, and there's never any tangential acceleration :wink:)
     
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