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Spring and centrifugal force

  1. Jun 6, 2012 #1
    hi

    i found the following equation that a spring(with constant k) , a mass m rotating with angular velocity ω, will have a constant angle [itex]\alpha[/itex] to the vertical axis that is given by
    [itex]cos(\alpha)=\frac{g}{\omega^2 l}(1-\frac{m \omega^2}{k})[/itex]
    this looks similar to something like [itex]cos(\alpha)=F_{gravitation}/F_{centripetal}-F_{gravitation}/F_{spring}[/itex] but i do not see the geometrical idea behind this.

    i am not sure whether this equation is correct at all, I found it in my physics schoolbook
     

    Attached Files:

  2. jcsd
  3. Jun 6, 2012 #2
    The equation seems to be OK.
    In order to find it you could apply Newton's second law for the vertical and horizontal directions. Use the equation for elastic force and centripetal acceleration, too.
     
  4. Jun 6, 2012 #3
    thank you for this hint. now, i found it by myself.
     
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