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Circular motion

  1. Jan 1, 2007 #1
    1. The problem statement, all variables and given/known data

    A figure is attached.

    The symmetric system consists of 4 massless rodes with length 'L',the purple thing is a spring(the spring constant is k),when the system is at rest theta=45 deg.
    the system starts to rotate with angular speed 'w',the rodes are pushing on the spring and theta is getting smaller.

    I need to express theta using:L,m,k,w.

    3. The attempt at a solution

    the length of the spring when the system is at rest is: sqrt(2)*L
    so the force is: k(2*sin(theta)*L - sqrt(2)*L) = 'F'

    I tried to apply newton's second law:
    F = ma = m*w^2*R = 2m*w^2*Lcos(theta)
    where Lcos(theta) is the radius.

    but the answer is wrong..
     

    Attached Files:

  2. jcsd
  3. Jan 1, 2007 #2

    andrevdh

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    Where do the rods pivot?

    How are they pushing on the spring?

    How is the spring mounted with respect to the rods?
     
  4. Jan 1, 2007 #3
    the rods are connected to the balls on one side and to the spring on the other side.

    because the system is rotating(imagine it's spining like a top) the balls are getting farther from each other and the rods are pushing on the spring.
     
  5. Jan 1, 2007 #4

    andrevdh

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    Are there no pivots involved?

    How do the ends of the rods push on the spring?

    Is the spring vertical and the balls swing out and up from a central axis of rotation?
     
  6. Jan 1, 2007 #5
    I think you can call the central axis a pivot,and imagine that it is nailed to floor,and the system rotates about this central axis.

    I guess the rods are pushing on the string because the system rotates(it makes sense)..there's a remark that says that this "device" can explain why Earth is flattened near the poles,because it's rotating about it's axis.

    this is exectly the same situation here,the system rotates about the central axis.and the balls want to "get away" from each other...and that's what's causing the rods to push on the string.

    and yes the spring is vertical,the central axis goes right through it.
     
    Last edited: Jan 1, 2007
  7. Jan 2, 2007 #6

    andrevdh

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    I have not been able to solve it yet, but here is a few remarks in the mean time.

    The fact that the problem states that the angle is 45 degrees when the system is at rest means that the bottom slider on the axis (where the bottom two rods come onto the rotation axis) must rest on some support. This means that the top slider comes down as the system rotates (balls rotate further away from the rotation axis as their speed increases). This need not cause too much concern since the forces at the two ends of the spring will be the same (one only need to consider the sine component of the tensions in the top two rods as far as the compression of the spring is concerned).

    Also note that the top rod will be under tension while the bottom rod will be under compression under all circumstances (rotating or not).
     
  8. Jan 3, 2007 #7

    andrevdh

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    Some further thoughts on the problem:

    When the system starts rotating the tension in the top rod will increase while the compression in the bottom rod will decrease (the ball is pulling more on the top rod and resting less on the bottom rod). Eventually at some speed the compression in the bottom rod will become zero. Increasing the speed even more will result in a tension in the bottom rod. The tension in the two rods are not the same as can be seen from an analysis of the vertical force components:

    [tex]T_{top}\sin(\theta) = W_{ball} + T_{bottom}\sin(\theta)[/tex]

    (this analysis is at a point where the system rotates at such speed that both rods are in tension).

    Confusion arises when one realizes that the spring will experience two forces of different magnitudes due to the tensions in the rods - the top component of the tension is larger than the bottom. Since the spring is in equilibrium vertically it means that the forces at the top of the spring needs to be the same as at the bottom. This means that an additional upwards force is necessary at the bottom to keep the spring in equilibrium. This forces again need to be supplied by the support at the bottom slider (or if the bottom slider is fixed to the axis).

    So lets assume that the top slider is free to move on the rotation axis, but the bottom one is fixed to it.
     
    Last edited: Jan 3, 2007
  9. Jan 3, 2007 #8

    andrevdh

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    Looking at the horizontal force components one gets that

    [tex]T_t + T_b = m\omega ^2 L[/tex]

    To bring the spring into play consider the vertical tension component of the top rod acting on the spring. At a particular rotating speed this component is matched by the restoring force of the spring. This fact together with the known lenght of the spring when the system is at rest enables you to construct the required relationship.
     
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