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Relating radius and angular freq of an rotating object

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data
    xa2s8.png

    an ball of mass m is connected by a string with spring constant k, to a rotating shaft.

    Find a relation between the radius of the circle, and the angular frequency.


    2. Relevant equations



    3. The attempt at a solution

    Let:
    Natural length of spring = x0
    Extension of spring = x
    Radius of circle = r = x + x0

    Centripetal force = Force due to spring
    mrω2 = kx = k(r-x0)
    Make r the subject,
    mrω2 = kr-kx0
    kx0=kr-mrω2
    kx0=r(k-mω2)
    r=[itex]\frac{kx_{0}}{k-m\omega^{2}}[/itex]
    --------------
    The expression indicates that there is an asymptote when [itex]\omega=\sqrt{\frac{k}{m}}[/itex]

    Does this means the rotation can never hit this angular velocity?
     
  2. jcsd
  3. Apr 23, 2012 #2

    ehild

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    Yes. You can also express ω in terms of r:

    [itex]\omega=\sqrt{\frac{k}{m}(1-\frac{x_0}{r})}[/itex]
    ω would be equal to the "natural frequency" at infinite radius.

    ehild
     
  4. Apr 23, 2012 #3
    but expressing ω in terms of radius, i get an asymptote.
    what is the physical meaning of this mathematical constraint?
     
  5. Apr 23, 2012 #4

    ehild

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    If the object performs SHM, the angular frequency is √(k/m) neither higher nor less. Now the object moves along a circle. r is its distance from the origin. It can not be negative. So ω can not exceed √(k/m)


    ehild
     
  6. Apr 23, 2012 #5
    sorry, I don't understand your explanation.
    You provided a mathematical description, but I am looking for a physical explanation.

    I know that r cannot be negative and hence ω can not exceed √(k/m). I can see that mathematically.

    So what happens if I attach a motor to the rotating shaft and drive the shaft at an angular freq higher than √(k/m)?
     
  7. Apr 23, 2012 #6
    The spring would break or deform. Hooke's law isn't an exact law, its just an experimental approximation for stresses below a certain threshold. You would cross that threshold by pumping omega, since the spring can only stretch out so far. Springs stop being springs (perhaps permanently) when you stretch them beyond their limit.

    If you had an infinitely long wire and wound it into an infinitely long spring and could somehow drive it up to that omega with the mass stretching out that far I don't know what would happen since that it isn't really physical.
     
  8. Apr 24, 2012 #7

    ehild

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    The relation between r and ω holds for the stationary solution. In real life, everything is in rest initially, before switching the motor on.
    The motor exerts some torque τ, and the torque accelerates rotation: dω/dt=τ/I. The moment of inertia I=mr2 increases with r as ω increases, and the applied torque becomes less and less effective to increase ω.There is a limit for ω that can not be exceeded by a given torque.

    ehild
     
    Last edited: Apr 24, 2012
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