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2012 F = ma Contest, Problem 24

  1. Jan 6, 2013 #1
    1. The problem statement, all variables and given/known data

    24. Three point masses m are attached together by identical springs. When placed at rest on a horizontal surface the
    masses form a triangle with side length l. When the assembly is rotated about its center at angular velocity ω, the
    masses form a triangle with side length 2l. What is the spring constant k of the springs?
    http://aapt.org/physicsteam/2013/upload/exam1-2012-unlocked-solutions.pdf

    2. Relevant equations
    No clear idea


    3. The attempt at a solution
    No clear idea
     
  2. jcsd
  3. Jan 6, 2013 #2

    Delphi51

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    Start with a diagram of the thing with the springs extended to length 2L.
    Write expressions for the three forces on one mass - the centrifugal force and the other two springs pulling on it. Put these on a diagram for one mass. The three forces must sum to zero since it is all in equilibrium from the point of view of this mass (circular motion frame of reference). I would consider the components in the direction of the centrifugal force to reduce it to one dimension. To get the relationship between L and Radius, split the 120 degree angle in half to make a right triangle and note that sin(60) = root 3 over 2.
     
  4. Jan 6, 2013 #3
    Thanks Delphi51. Now I understand it completely.
    I didn't picture it correctly before: for each spring, one end is tied to the
    center of the triangle, and a point mass m is tied to the other end.
     
  5. Jan 6, 2013 #4

    Delphi51

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    Now that is fascinating - I pictured it as nothing tied to the center; springs connected between masses. And I got the key answer. Amazing that it would be the same answer for both situations.
     
  6. Jan 6, 2013 #5
    Are you getting the same answer by keeping one end tied to the center of the triangle.
    If one end is tied , then k*l = m*2l*(ω^2) which means k = 2m(ω^2)
     
  7. Jan 6, 2013 #6
    No, you don't get the same answer both ways. Only the way you are describing gives the correct answer. I think the OP just hadn't worked through the problem yet after they misunderstood your explanation.
     
  8. Jan 6, 2013 #7

    Delphi51

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    Okay, thanks. It was quite a tricky problem. I had considerable trouble with #1 on that test, getting (10 minus the correct answer) for one thing. Most cleverly composed questions!
     
  9. Jan 6, 2013 #8
    Thanks everyone for the quick responses.:-)

    Originally I thought that one end of each of the three spings tied together
    at the center, and I got the answer (A): 2mω^2, whcih was not (C):2/3mω^2.
    I had been checking what mistake I made on principle/formula/calculation,
    but couldn't find any.
    After having read Delphi51's first post, I realized that I had pictured the problem
    incorrectly, and was able to get the correct answer (C) right away.

    Obviously, Delphi51's picture is more meaningful.
    However, will the springs be curved in this picture? Are the springs mass free?
     
  10. Jan 6, 2013 #9
    Not in this problem, but real springs would be slightly curved by having their centre points deflected outwards by the rotational motion. However, the effect would be very small because of the springs not being very massive. The centripetal force required to keep the center portion of the spring moving in a circle is proportional to the mass of that piece. This force would be provided by the tension in the spring causing the neighbouring segments to pull the centre of the spring back towards the centre of the triangle, and the tension force is independent of mass. I.e. if the spring curves outward then a component of the tension force acting at the centre will be directed to the triangle's centre, providing the necessary centripetal force. As the spring is taken to be lighter and lighter, the spring needs to curve less and less to balance all the forces properly. In real life, the effect would be quite small though possibly noticeable. In physics problems such as this, we use ideal springs which are taken to be massless. In this limit, the springs stay straight. Which answers your second question too.
     
  11. Jan 7, 2013 #10
    Thanks LastOneStanding!
     
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