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Finding frequency of a specific mechanical oscillator -- horizontal rod on pivot

  1. Jan 10, 2017 #1
    1. The problem statement, all variables and given/known data
    domaca_naloga.jpg
    There is a cyllinder with radius 0.5 m fixed on the wall. We put a 6 metres long thin rod with mass 0.3 kg on it, which does not slip. I would like to calculate the oscillating time. It is a part of a clock, so the oscillating time is probably 1 or 2 seconds, but I got around 5 seconds. Please help me find a mistake.

    2. Relevant equations
    equation.jpg

    3. The attempt at a solution
    I got a wrong result
    attempt.jpg
    I am used to write M (torque) and J (moment of inertia)
    attempt1.jpg
     
    Last edited: Jan 10, 2017
  2. jcsd
  3. Jan 10, 2017 #2

    TSny

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    Your work looks correct. As a check, I used a different approach and still got the same answer.
     
  4. Jan 10, 2017 #3
    Thanks for your reply ... I will wait for my schoolmates to solve it in order to compare our results. I am quite sure that everything is correct except the calculation of moment of inertia. Should I treat each part of the rod separately (the longest and the shortest part using a formula for a rod rotating around one end .. (mx^2)/3) and sum moments of inertia of both parts as it does not oscillate accurately around its centre of mass.
    I still think that the answer is not correct, so everyone, please help me solve it.
     
  5. Jan 10, 2017 #4

    TSny

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    You can show that the contribution of the center-of-mass motion to the inertia of the system is negligible for small oscillations. So, you are OK with just taking into account the rotational inertia about the center of mass, ##\frac{1}{12}ML^2##.

    I did a rough experiment and the result for the period was reasonably in agreement with the formula you derived. I used a yard stick on a cylinder of approximate diameter 2.75''. Measured period was about 2.6 s. Compare to the formula prediction of 2.8 s.
     
    Last edited: Jan 10, 2017
  6. Jan 10, 2017 #5
    Thanks for all your help! I really appreciate it! You inspired me to do the experiment. I put a thin tube (L=1.02 metres) on a ball (R=0.107 m). Formula prediction is tequation=1.82 s. I measured it tmeasured=1.92 s ... This is almost 5% accurate.
    As older students write homework for us it is quite possible that the given data are not correct :D . Well, it is everything OK, but I do not know what sense would an oscillator with oscillating time 4,87 s have in relation to a clock. There is still an option that I have made a mistake... Correct me please, if so...
     
  7. Jan 10, 2017 #6

    haruspex

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    I see no difficulty. Pendulum clocks are not expected to be accurate to the second. Within 15 seconds would be considered excellent.
     
  8. Jan 11, 2017 #7
    Do you think my solution is correct or do you just answer to my doubt about the relation to clock?
     
  9. Jan 11, 2017 #8

    haruspex

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    I was just responding to your doubt about suitability as a clock. If TSny says your work is correct, it is.
     
  10. Jan 15, 2017 #9
    The next question is: What happens if we calculate the oscillating time considering that the rod is not thin, but it has width 2b (it is a cyllinder). So b=radius of this cylinder. I think nothing changes except the moment of inertia.
    I am caluculating it in this way:

    I found an equation for a rotating cyllinder around the perpendiculat axis (on the picture marked as xC or yC) in the middle of a cylinder.
    150px-Fizvm_valj000-1.png
    J(moment of inertia) = 1/4 m*b^2 + 1/12 m*L^2
    In our case the cylinder is not rotating around the axis in the middle of it, but around the axis which is a tangent on surface of this cylinder, so I think that the desired moment of inertia for our case considering Steiner's rule is:
    J = m*b^2 + 1/4 m*b^2 + 1/12 m*L^2

    Is my thinking correct?
     
  11. Jan 15, 2017 #10

    TSny

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    Are the moment arms of the forces affected by b?
    OK (for small φ)
     
    Last edited: Jan 15, 2017
  12. Jan 15, 2017 #11
    Yep I forgot to consider that. The moment arms gets longer considering Pitagora's Theorem. arm^2= former arm^2 + b^2, right?
     
  13. Jan 15, 2017 #12

    TSny

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    Does this give the correct moment arm for each side when φ = 0? Does the moment arm on the left side and the moment arm on the right side both get longer due to b?
     
  14. Jan 15, 2017 #13

    haruspex

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    No, that's not a good guess. Draw a diagram.
     
  15. Jan 15, 2017 #14
    image.jpg
    image.jpg
    image.jpg
    I think it gets very very very very complicated if I consider the change of length of the momentum arm
     
  16. Jan 15, 2017 #15
    I do not have any data about b, so I cannot caluclate numbers for alpha or R, so it would get more difficult if I wanted to calculate the omega without numbers.
     
  17. Jan 15, 2017 #16
  18. Jan 15, 2017 #17

    TSny

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    You are interested in how much the moment arm changes as you add the effect of b. Your R does not represent the change in the moment arm. Think about where the force on the left side acts before taking into account b and then after taking into account b.
     
  19. Jan 15, 2017 #18
    The force acts in the center of mass of the left part
     
  20. Jan 15, 2017 #19
    Honestly I do not understand clearly what you want to tell me. Can you help me please with some equations or sketches?
     
  21. Jan 15, 2017 #20

    TSny

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