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Homework Help: Equations of motion two springs pendulum system

  1. Jul 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Give governing equations for the system about its static equilibrium, assuming small vibrations

    System consists of two springs located under 45 degrees to the vertical (both have same k-value) in undisturbed situation. Lower ends of the springs are attached to each other and to massless rod with length D, upper ends to immovable supports. The massless rod has point mass (m) at the lower end. System can move vertically and horizontally and is exposed to gravity.

    2. Relevant equations
    Displacement method: mx"+cx'+kx = F(t)

    3. The attempt at a solution
    System has 3 degrees of freedom (horizontal (x2) , vertical (x1) and rotation of pendulum around supension (theta))

    I suggested:
    mx"1 = -1/2*(k*sqrt(2)*x1) - 1/2*(k*sqrt(2)*x1) + mg
    mx"2 = -1/2*(k*sqrt(2)*x2) + 1/2*(k*sqrt(2)*x2)
    J(theta)" = -m*g*D*(theta)

    The last equation under the assumption of small angles of the massless rod.

    However, in this case it results in a decoupled dynamic system which I suppose is incorrect?

    Thanks in advance for all support!
     
  2. jcsd
  3. Jul 3, 2015 #2
    Can you draw it?
    I can't see what system you are looking at, I have some ideas but guessing is not efficient.

    Are you using the Newton equations or the Euler-Lagrange equations? (I know which I would use)
     
  4. Jul 3, 2015 #3
    It is mentioned to use the Newton equations. The schematization is like this:

    upload_2015-7-3_17-32-41.png
     
  5. Jul 4, 2015 #4

    haruspex

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    Are your x1 and x2 displacements of the mass or displacements of the Y junction?
     
  6. Jul 7, 2015 #5
    I assumed displacement of the Y junction as the mass is connected (fixed connection) to the rod.
     
  7. Jul 7, 2015 #6

    haruspex

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    Your first two equations treat the mass as being at the Y junction. You need to allow that the rod is not always vertical.
     
  8. Jul 9, 2015 #7
    That was the reason I used the third equation for small angles of the rod to it's static (vertical) equilibrium. How do you suppose I should link the horizontal and vertical displacement of the whole system with the mass on certain distance, expressed in the two first equations?
     
  9. Jul 9, 2015 #8

    haruspex

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    Adding an equation doesn't correct a wrong equation. Let the tension in the rod be T and consider the FBDs for the junction and the mass separately.
     
  10. Aug 12, 2015 #9
    upload_2015-8-12_19-7-36.png

    As suggested, you would consider the body diagram as schematized in the figure above as starting point for this issue, Haruspex? For the upper part, there will be two equations for displacement in horizontal direction and vertical direction due to Force T and for the lower part a reaction Force R and gravity Force=m*g?
    Appreciate your help!
     
  11. Aug 12, 2015 #10

    haruspex

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    Ok, but you should show the angle of the rod as a variable too.
     
  12. Dec 9, 2015 #11
    Dear Haruspex,

    As you proposed, I had the same idea as you can see in the attachment. The angle between the rod and the vertical is indicated as theta. In this case, should you get two governing equations, one for the upper part including theta1 and one for the lower part including theta2 where theta1=theta2?
     

    Attached Files:

  13. Dec 9, 2015 #12

    haruspex

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    Yes.
     
  14. Jan 4, 2016 #13
    Hi Haruspex,

    When I make the FBD of the upper part (1) and lower part (2), I get the following equations (where u=displacement in vertical direction):

    1) mu" = -2*(k*sqrt2*u)/2 + T; T = m*g*sin(theta)
    2) mu" = -R + T; R = reaction force, T = m*g*sin(theta)

    However, the displacements in horizontal direction are not incorporated in the equations, which is not correct right?
    Thank you for your help!
     
  15. Jan 4, 2016 #14

    haruspex

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    Your vertical equations are incorrect.
    T is not equal to m g sin(theta), in general. The FBD for for the junction doesn't 'know' about the mass. It just feels the tension in the rod. So the left part of (1) is right, but throw away the second part.
    The rod is massless, so. What is the relationship between T and R?
    Likewise, in (2), you don't want both T and R. All the mass knows about are gravity, the pull from the rod, and its own acceleration.
     
  16. Jan 6, 2016 #15
    I reconsidered the equations as you suggested. The relationship between T and R is that they have to make a force balance; in other words T = -R. With this input, I get the following for the upper part (1) and lower part (2):

    1) mu" = -2*(k*sqrt2*u)/2 + T
    2) mu" = m*g*sin(theta) - R

    Is this what you meant, where you can balance the forces of T and R?
     
  17. Jan 6, 2016 #16

    haruspex

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    I'm sorry, I made a mistake in my previous post. The left hand side of your (1) was not right either.

    There is no mass at the junction, so forces there must completely balance.
    We must consider an x displacement and a y displacement of the junction. What is the sum of forces that results in each (x and y) direction? Each of these sums will be zero.

    You can express the displacement of the mass in terms of the above x and y and the angle theta. What are the vertical and horizontal sums of forces there?
     
  18. Jan 7, 2016 #17
    Okay so when I just consider the junction, you get the following:

    For displacement y=vertical displacement
    my" = -2*k* 0,5*y*sqrt(2)

    For displacement x=horizontal displacement
    mx" = -k* 0,5*x*sqrt(2)

    In this consideration, the change in angle of 45 degrees of the springs to their support is assumed as small. Is this reasoning correct?
     
  19. Jan 7, 2016 #18

    haruspex

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    There is no mass at the junction. Instead, there is a force from the rod, which you have now omitted. You had that previously.

    Other than that, the y equation is right - it's not exact, but it is a valid approximation for small displacements. The exact form would use both x and y displacements to find the spring extension, then find the vertical component of the resulting force. The approximation is ok because the x displacement increases the vertical component of one tension and reduces the vertical component of the orher in roughly equal amounts.
    The same arguments apply in the horizontal direction, but this time you did not get quite the right expression for the force. There are two springs.
     
  20. Jan 8, 2016 #19
    So let's call it the left spring and right spring with corresponding k_left and k_right.
    For example, applying a displacement of the junction to the right (x = horizontal small displacements) gives than:

    mx" = -0,5*k_left*x*sqrt(2) + 0,5*k_right*x*sqrt(2)
    This means that the left spring extends and the right spring shortens, right?

    I am sorry, indeed I omitted T which is the force from the rod. So the y=vertical displacement gives the following equation:
    my" = -2*k*0,5*y*sqrt(2) + T

    These equations are for the upper part but how do I combine these with the lower part (where the mass is located)? In y=vertical direction, the only vertical forces are m*g and reaction force R as far I know. In x=horizontal direction, due to the angle theta that results from the displacement of the junction, the horizontal forces of the mass are R*cos(theta) right?

    Thanks a lot.
     
  21. Jan 8, 2016 #20

    haruspex

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    You don't seem to be grasping the significance of the fact that there is no mass at the junction. It means there is no mass to be associated with the acceleration. The applied forces there, two springs and rod, must exactly balance. ##\Sigma F=0##.
    Also, you have left out the rod's tension in the X equation.
    I don't know you introduced kleft and kright for horizontal but use k for both in the verical. Each spring only has one constant, and the springs are the same,
     
  22. Jan 8, 2016 #21
    Hi Haruspex,

    In the previous post, you mentioned that "There is no mass at the junction. Instead, there is a force from the rod, which you have now omitted. You had that previously." For the vertical displacement (which is y here) I considered this "force from the rod" as T, in combination with the reaction forces in the springs. Is that incorrect? In other words; as there is no mass associated with the acceleration, there is no force from the rod?

    Another question, the exercise mentions that the lower ends of the springs are connected to each other and to a rigid massless rod. In which way can I incorporate the rod's tension in horizontal direction (which is x here)? I assumed that the rod (with length L but massless) is just a tool to get inertia as it is at certain distance from the junction.
     
  23. Jan 8, 2016 #22

    haruspex

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    Focus on the junction. There are three forces acting on it, the tensions in the two springs and the tension in the rod. The mass itself is not in this FBD. Since there is no mass, these three forces must be in balance, and it is unnecessary to consider the acceleration of the junction.
    Since we are only considering small displacements, we can take the spring tensions as always acting at 45 degrees. The rod tension acts at varying angle theta.
    You have correctly determined the horizontal and vertical components of the spring tensions. If the tension in the rod is T, you can easily write down the vertical and horizontal components of that. So write down two equations, one for the vertical balance of forces at the junction, the other for the horizontal balance there.
     
  24. Jan 10, 2016 #23
    As you said in the previous post, the horizontal and vertical components of the spring tension are correctly determined so focussing on the tension T in the rod gives for:

    Vertical component => y = T*cos(theta)
    Horizontal component => x = T*sin(theta)

    When this is combined with the horizontal and vertical displacement of the springs, you will get the following:

    Vertical balance of forces at junction => y = -2*k*0,5*y*sqrt(2) + T*cos(theta)
    Horizontal balance of forces at junction => x = -0,5*k*x*sqrt(2) + 0,5*k*x*sqrt(2) + T*sin(theta)

    Based on the principle of SigmaF=0 in the junction
     
  25. Jan 10, 2016 #24

    haruspex

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    Nearly right, but you have the x components of the spring tensions cancelling. That cannot be right, can it? Bear in mind that both springs will be under tension because of the vertical pull, but in the horizontal direction they will cancel when x=0. What we need to find is the change in the horizontal components of the tensions when x is small nonzero. One will increase, while the other decreases.
     
  26. Jan 11, 2016 #25
    Hi Haruspex,

    That is the question I had as well. So if you consider a small displacement "delta" under the assumption that the angle between the springs and their supports stays the same (45 degrees), you'll get the following equation for the horizontal balance of forces at the junction:

    x = -0,5*k*sqrt(2)*(x+delta) + 0,5*k*sqrt(2)*(x-delta)

    In this case, the two components will not cancel out each other. However, you will get another variable delta. Do you suppose another way to express this or would you apply the same line of reasoning?
     
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