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Circular motion with a spring

  1. Dec 13, 2016 #1
    1. The problem statement, all variables and given/known data
    image.png
    http://[URL=http://www.siz.co.il/][PLAIN]http://up416.siz.co.il/up3/w2jymjm5nqxy.png [Broken]

    A body with a mass that equals to m is inside a smooth tube and is attached to two identical springs with constants that equal to k. Their length when they are limp is L0. When the body is in the middle of the tube both springs are limp.
    The tube is attached to a pole that spins around with an angular velocity that equals to ω (The motion is only horizontal)
    1. In what distance the body needs to be from the center of the tube so he will spin around along with the tube without moving in relative to him?
    Answer: mω2L0/(2k-mω2)
    2. Now we attach the tube in an angle of 37 degrees to the pole (α=37 degrees), what the angular velocity ω0 needs to be so both springs will remain limp?
    2b. The angular velocity in increased to 2ω0, in what distance the body needs to be from the center of the tube so he will spin around along with the tube without moving in relative to him?

    Honestly I tried a lot but I can't seem to solve the first question, can anyone help please?
    Sorry for my English, I know it's very poor.
    Equations:
    mar=F
    ar2r
    F=KΔL
    ma=F

    How I tried to solve this:

    I tried to say that the force that the springs enables on the body is 2KΔL, but it didn't work.
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Dec 13, 2016 #2

    PeroK

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    Can you start by calculating the centripetal force required for circular motion of the mass ##m##?
     
  4. Dec 13, 2016 #3
    Honestly I have know idea how to do it, we have just started to learn this topic.
    Can you post a solution to the first question and I will try to understand from it?
     
  5. Dec 13, 2016 #4

    PeroK

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    The purpose of this forum is to help you do problems that you perhaps can't quite do yourself. We assume you have some familiarity with the material.

    The best thing to do is to find your notes on circular motion and centripetal force. There is also lots of material online about this subject. For example:

    http://www.physicsclassroom.com/class/circles

    You can always ask specific questions here if there is something in particular you don't understand.
     
  6. Dec 13, 2016 #5
    centripetal force is ##mr\omega^2## where r - radius, m - mass, ##\omega## - angular velocity.
     
  7. Dec 13, 2016 #6
    Sorry, I didn't explain myself quite well. We did work on this subject for a bit, but this question is much harder than what we did so far. I know the basics but this question is really hard for me.
     
  8. Dec 13, 2016 #7

    PeroK

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    You appear to be stuck on the basics. @Buffu has given you the answer to my first question above.
     
  9. Dec 13, 2016 #8
    You just need to find the displacement of the object inside the tube due to its rotation.
    Forget everything else written in the question especially "the tube without moving in relative to him".
     
  10. Dec 13, 2016 #9
    You should consider the object when all the forces acting on it will be at equilibrium.
    So first write out all the forces acting on the body. :smile::smile:
     
  11. Dec 13, 2016 #10
    This I realized, I tried to do it but I didn't get the correct answer. Can you explain the forces in the question please, obviously gravity and normal, but to which direction the springs enable the force on the body, and what is the size of those forces?
     
  12. Dec 13, 2016 #11
    Also, when I try this I don't get the angular velocity in any of the forces, and it appears in the final answer twice.
     
  13. Dec 13, 2016 #12
    What answer did you get ?
    Show the gist of what you did to get it.
     
  14. Dec 13, 2016 #13
    I said the left string enables a force that equals to kΔl=k(l0+x) (x is the distance from the center), then I said the right string enables a force that equals to kΔl2=k(l0-x), I also think both are in the same direction, but then they won't be equal to zero, if they are not and I want to check when they are equal I get an answer that is not even close. It looks like I am really missing something here, please, can you help?
     
  15. Dec 13, 2016 #14
    First problem is that both of them wont be in same direction. Secondly you did not mention centrifugal force on the object.
    The horizontal force on the object are :-
    1) force by left spring that is toward the center of axis of rotation.
    2) force by left spring that is away the center of axis of rotation.
    3) centrifugal force (Find its direction)

    Now can you equate them and get the answer ...
     
  16. Dec 13, 2016 #15
    Obviously it will be toward the center, but I still don't get the right answer. The equation isn't mω2r+kΔl1-kΔl2=0 and Δl1=-x and Δl2=x?
     
  17. Dec 13, 2016 #16
    Anyone?
     
  18. Dec 13, 2016 #17

    haruspex

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    yes, but what do you have for r there?
    Please do not just say "I did not get the right answer". Post all your working so that we can see exactly what you did. We are not mind readers.
     
  19. Dec 13, 2016 #18
    Obviously you are wrong, It is not towards the center, It is away from center.
     
  20. Dec 14, 2016 #19
    Okay I understand that a centrifugal force will be directed away from the center, but when the force is centifugal and when it's centripetal? I was taught that the body is able to hae a circual motion thanks to the centripetal force, how there is a circual motion without it and how to know if the force is centrifugal or centripetal? We didn't discuss about this in class at and that's the last thing I don't understand here. Thanks in advance, you helped me a lot.
     
  21. Dec 14, 2016 #20

    PeroK

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    If the mass is displaced outwards, then both springs will apply a force on it inwards, creating the necessary centripetal force.
     
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