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Hooke's law and circular motion.

  1. May 24, 2014 #1
    1. The problem statement, all variables and given/known data

    In a zero gravity experiment a spring has a rest length of 1.0 metre. It is attached at
    each end to 1.0 kg masses. The combination is then set rotating about its centre of
    mass at 1.0 revolution per second. Each mass undergoes uniform circular motion with
    a radius of 70 cm. Neglecting the mass of the spring and assuming Hooke’s Law, the
    spring constant must be approximately:

    3. The attempt at a solution

    Fc = FH

    (1kg)(2pi(f))/(1m) = k(1m)
    I tried equating them and solving for k but it's wrong.
     
    Last edited: May 24, 2014
  2. jcsd
  3. May 24, 2014 #2

    Doc Al

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    Not sure what formulas you are using.

    How much is the spring stretched? What force does it exert? What's the centripetal acceleration?
     
  4. May 24, 2014 #3
    I was equating the centripetal force to hook's law.

    I do no know how much is the spring being stretched but I know that there is a centripetal force pulling the spring to the center.
    The centripetal acceleration is v2/r.
     
  5. May 24, 2014 #4

    Doc Al

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    That's what you should have been doing, but your posted equation does not seem to match that.

    You are told the radius of each mass's revolution. (Draw a picture for yourself.) By comparing that to the unstretched length, you should be able to figure out how much the spring is stretched.

    The spring exerts a force on the masses. That force is the centripetal force.

    That works. But you might find the alternate expression, in terms of ω, more suited for this problem.
     
  6. May 24, 2014 #5
    That's the problem! I cannot even begin to figure out the physical interpretation.
    Well, ω = 2πf
     
  7. May 24, 2014 #6

    Doc Al

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    Imagine the two masses swinging around in a circle, the center of which is at the middle of the spring.

    OK, you'll need that.
     
  8. May 24, 2014 #7
    You can just plug 1m for things you don't know yet. You have to find theirs values and then plug it in.
     
  9. May 24, 2014 #8

    HallsofIvy

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    You are told that the spring has a rest length of 1 m. You are told that the two masses are rotating in a circle with a radius of 0.7 m. What is the distance between the masses? How much has the spring been stretched?

    (I think dauto meant to say "you cannot just plug in 1m for thing you don't know yet.")
     
  10. May 24, 2014 #9
    If the spring has a rest length of 1m and during the rotation the radius is 0.7m then could it be that there is a change in length of 0.3m?
    By this, FH = k(r-x) = k(r-0.3m) where r is the original radius (rest length of the spring and x is the change in radius)
     
  11. May 24, 2014 #10

    Doc Al

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    If the radius is 0.7 m, what is the full length of the stretched spring?
     
  12. May 24, 2014 #11
    Not quite. Draw a picture following the description in the problem.
     
  13. May 25, 2014 #12
    I really have no idea. The next best is to guess but I never like blundering my way through.
     
  14. May 25, 2014 #13

    CAF123

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    Consider a spring of relaxed length 1m with two masses at either end. The CoM is in the middle of the spring, a distance 0.5m from each mass. We are told that each mass undergoes uniform circular motion with a radius of 0.7m (i.e each mass is now at a distance of 0.7m from the middle of the spring). How much did the spring stretch?

    I think this is the only possible interpretation, since in your analysis the force from the spring on the masses would not be pointing radially inward, and hence there would be no possibility for circular motion.
     
  15. May 25, 2014 #14

    Doc Al

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    Did you draw yourself a picture?
     
  16. May 25, 2014 #15
    I'm starting to develop a physical intepretation. Tje spring stretch 0.2m.
     
  17. May 25, 2014 #16

    I did but it doesn't capture fully the demand of the question. I'm refining the physical intepretation as I go.
     
  18. May 25, 2014 #17

    Doc Al

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    How did you determine this?
     
  19. May 25, 2014 #18

    Doc Al

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    What does your diagram show?
     
  20. May 25, 2014 #19

    Initially it the mass is positioned at 0.5m from the origin. The hint was the COM. During the spin, it is displaced to a position 0.7m from the origin which is also the COM.
    Tell me if I am wrong
     
  21. May 25, 2014 #20

    Doc Al

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    Excellent. So what's the total length of the stretched spring? Then, how much has it stretched from its original length?
     
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