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Grr springs

  1. Jul 23, 2004 #1
    hello there
    consider a spring that is fixed to a point (i have tried to draw a picture, so to make it clearer as to what i am talking about) that rotates anti clockwise in a horizontal fashion. there is a ball on the end of the spring with mass m, the spring has spring constant k and unstretched length L and the ball moves with angular velocity omega
    i am trying to find an expression for theta, the angle that the spring makes with the vertical, as a function of L, w, k, m and g.
    me being a maths student and not a physics student just treated this as simaltaneous equations and plugging some stuff together using

    W=F*s *cos(theta)
    F=m*v^2/L where v is the velocity perpendicular to the centripetal force
    F=-k*theta

    after a bit of algebra we get that
    cos(theta)=-k/(L^2*m*omega^2)
    of course one can arrange for theta but i am not sure if this expression is correct, more than likely it is wrong. so if someone could shed some lighton the matter it would be greatly appreciated.

    thanks
     

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    Last edited: Jul 23, 2004
  2. jcsd
  3. Jul 23, 2004 #2

    Doc Al

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    Staff: Mentor

    Here's how I would approach this problem. Apply Newton's 2nd law to the vertical and horizontal forces on the mass:
    [itex]T sin\theta = mg[/itex], where T is the tension in the spring
    [itex]T cos\theta = m\omega^2 R[/itex], where R is the radius of circle
    Combine this with what you know about the spring: its length is L + T/k, thus [itex]R = (L + T/k)sin\theta[/itex].

    You should be able to solve this for for [itex]\theta[/itex], but it looks messy.
     
  4. Jul 24, 2004 #3
    thanks for the helpful advice
     
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