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A spring-mass system in circular motion

  1. May 9, 2017 #1
    1. The problem statement, all variables and given/known data
    I uploaded the image of the system to have a visual example. : )
    A point mass ## m ## connected to a spring is bounded to move on a circular guide with radius ## R ## without friction. The guide is fixed to a vertical plane and the other extremity of the spring is attached at point ## B ##, which is the bottom of the guide. The rest lenght of the spring is ## 2 R ##. Initially the mass is in the point ## A ## and at ## t = 0 ## it begins to move with a negligible velocity
    1) Knowing that the mass richeas the point ## D ## with ## v=0 ##, find the spring constant ## k ##;
    2) Find the acceleration ## a_D ## and the reaction ## N_D ## of the guide at point ## D ##;
    3) Let ## \theta ## be the angle between the spring and the vertical. Find the angle at which the velocity is maximum.


    3. The attempt at a solution
    1) I use the energy conservation. First of all I choose the zero of the potential at point ## A ## such that the initial energy is equal to zero. Then I set that equal to the energy at point ## D ##, knowing from the geometry of the system that the length of the spring in this point is ## \sqrt 2 R ##:

    $$ 0 = k R^2 - m g R \quad \Longrightarrow \quad k = \dfrac { m g }{ R } $$

    2) I chose the frame centered in ## B ##. Because the reaction is always perpendicular to the guide, at point ## D ## the reaction force has only a component along the ## x ## axis and then:

    $$ N_D = k R = m g $$

    because the mass is not moving along ## x ## in this position.
    Along the positive direction of ## y ## axis there is an acceleration which is given by:

    $$ m a_D - mg = kR \quad \Longrightarrow \quad a_D = 2 g $$

    where i have substituted the value of ## k ##

    3) Up to now, the only thing that I have in mind is that the velocity is maximum when the acceleration is equal to zero. So I tried to write down that the sum of the forces along both axes is zero but I need another condition. I thought that maybe I have also to impose the minimum potential energy but I am not so sure.

    Can you please give me any hint? Also, do you think that the answers to questions 1 and 2 are correct? Thank you in advance. : )
     

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  2. jcsd
  3. May 9, 2017 #2

    BvU

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    Hello As, :welcome:
    1.
    that is correct. But the potential energy in the spring is then not equal to ##kR^2## !
    2. That would mean the mass stays there ? (the vertical component would also be ##mg##, thus offsetting gravity ?)
     
  4. May 9, 2017 #3
    Hello! I think that the motion should be harmonic between ## A ## and ## D ##. I discussed the problem with a fellow of mine and i'm changing the answers!
     
  5. May 9, 2017 #4

    BvU

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    Please inform us why you think so. What are the criteria for harmonic motion ? Do they apply here ?
    So what derivation for the expression for ##k## do you have now ?
     
  6. May 9, 2017 #5
    I think that the motion is harmonic because at ## A ## and ## D ## the velocity is equal to zero and this means that at these points the acceleration is maximum and the motion changes direction. In fact the spring force is against the direction of motion and is at its maximum value in ## D ##. At least, this is what my intuition tells me. :sorry:
    The new answers are:
    1) I choose the frame of reference centered in ## B ##. The potiential energy in ## A ## is only ## 2 m g R ## because the spring is at rest. The potential energy in D is ## m g R + \dfrac{ 1 }{ 2 } k ( \Delta R )^2 ## and ## \Delta R = 2 R - \sqrt 2 R ## is the elongation of the spring from its rest position. Then I obtain :

    $$ 2 m g R = m g R + \dfrac 1 2 k R^2 ( 2 - \sqrt 2 )^2 \quad \Longrightarrow \quad k = \dfrac { m g }{ ( 3 - 2 \sqrt 2 ) R } $$

    2) In point D I can assume that along the x-axis the acceleration is zero because looking at small movements from ## D ## the particle moves only along the y-axis. Thus, the elastic force along the x-axis is ## k \ \Delta R \ \sin \left( \dfrac \pi 4 \right) = k ( 2 - \sqrt 2 ) R \dfrac {\sqrt 2} {2} = k ( \sqrt 2 - 1 ) R = mg \dfrac { \sqrt 2 - 1 }{ (3 - 2 \sqrt 2) } \ > \ mg ## . The sum of the forces along x eventually gives me:

    $$ N - k \ \Delta R \ \sin \left( \dfrac \pi 4 \right) = 0 \quad \Longrightarrow \quad N = mg \dfrac { \sqrt 2 - 1 }{ (3 - 2 \sqrt 2) } $$

    Knowing that the y-component of the elastic force is equal to x-component, along the y-axis I find instead:

    $$ m a = - m g + mg \dfrac { \sqrt 2 - 1 }{ (3 - 2 \sqrt 2) } \quad \Longrightarrow \quad a = \dfrac { 3 \sqrt 2 - 4 } { ( 3 - 2 \sqrt 2 ) } \ g \ > \ g $$

    which is in accord to what I intuitively think.



    Regarding the third point, I am going to do the calculation as soon as possible. Using the geometry of the system, I expressed the potential in function of ## \theta ##, now I only have to do the derivative and set the result equal to zero in order to find the angle at wich the velocity is maximum. Sorry for the late but I was at University up until now. Also sorry for my english, I am not used to write in this language. : )
    Do you think the reasoning is more or less correct?
     
  7. May 9, 2017 #6

    haruspex

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    I cannot find a definition of "harmonic motion". It is clearly not simple harmonic motion, but it is periodic motion. Is that what you mean?
     
  8. May 9, 2017 #7

    BvU

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    True for ##D##, not true for ##A##. If let go with ##v=0## at D, the mass just reaches ##A## and is at an unstable equilibrium there (you calculated that in part 1)
    Anyway, the restoring force is not proportional to the distance from equilibrium.
    Doesn't really matter where (D or B): potential energy difference counts. Check your math: it's wrong.
     
  9. May 10, 2017 #8
    Haruspec, yeah sorry, I meant periodic motion but it is not like that because the point A is an unstable equilibrium point and the sum of the forces is zero. Thus I think that the mass go back to A and then stays there.
    BvU, is the elastic force proportional to the lenghtening or the compression of the spring, isn't it? I mean, respect to its rest position.
     
  10. May 10, 2017 #9

    haruspex

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    I would not say so...
    Negligible, but not zero. If there really are no losses, what will be its velocity back at A?
     
  11. May 10, 2017 #10
    The same at t=0, yes. This is a good point. So there is always a little bit of kinetic energy that it is negligible in the energy calculation. Anyway, it let the mass move from A. Are you saying that?
     
  12. May 10, 2017 #11

    haruspex

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    Yes, but which way, having returned to A?
     
  13. May 10, 2017 #12
    Counterclockwise?
     
  14. May 10, 2017 #13

    haruspex

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    Right.
     
  15. May 10, 2017 #14

    BvU

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    Do we have a correct answer for parts 1 and 2 already :oops: ? I don't think moving beyond A away from D is within the scope of the exercise !
     
  16. May 10, 2017 #15
    No, because i don't understand what you mean saying that. From what i rember the elastic force is proportional to the compression or the lenghtening of the spring from its rest position. :frown:
     
  17. May 10, 2017 #16

    BvU

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    And that doesn't lead to a factor ##(3-2\sqrt 2)##
     
  18. May 10, 2017 #17
    Mmm, ok! Thanks! I am thinking. :sorry:
     
  19. May 10, 2017 #18

    haruspex

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    I agree with the answers in post #5, but the expressions can be simplified a lot.
    @Aslet, do you know how to avoid having surds in the denominator?
     
  20. May 10, 2017 #19

    BvU

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    My mistake, I read too fast. Apologies. Ignore the first part of #14.
     
  21. May 10, 2017 #20
    Haruspex, yeah sure. Then if this is the only problem I will semplify the equations and as soon as I have the time I will write it here! Thank!
     
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