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Potential energy in a mouse trap

  1. May 11, 2017 #1
    1. The problem statement, all variables and given/known data
    So I have been building a mouse trap car at my school, and I need to hand in a report on it tomorrow. :biggrin: I don't have all of the measurments at the moment, but the only thing I want to know is how to calculate this. I want to see how much energy is lost to friction/warmth/sound by dividing the total kinetic energy(I have figured that out!) by the potential energy.

    The mouse trap is 180° or π rad from the closed position when it's set.
    So [​IMG]=π rad
    I only have the average speed.

    2. Relevant equations
    Potential energy in a torsion spring: [​IMG]
    Hooke's law: [​IMG]
    Work: W=F*s
    Torque: [​IMG]
    [​IMG]
    3. The attempt at a solution
    So at first I was thinking that that k must be equal to T / [​IMG], and that T must be equal to the arm length times the average force used to set the mouse trap. k should therefore be k=T/π=(r*π*F)/π. Using the torsion spring equation, I then get U=(r*F*π)/2.

    I then thought that the work in order to set the mouse trap should be equal to the potential energy. I used the circumference of a semicircle as the stretch. I then got W=F*π*r, but this is twice as much as I got with my other ""solution"".

    I then thought that I might be able to just use that the X in Hooke's law is equal to the circumference of the semicircle. k should therefore be F/(π*r). Using the first equation, I then get U=(1/2) * (F/(π*r)) * π^2= (F*π)/(r*2)

    I'm also not sure about how F should be measured. Should I take the force used when you're barely lifting the mouse trap + the force used when it's fully set and divide that by 2?

    Sorry if anything was unclear!:confused: What do you reckon I should do?
     
  2. jcsd
  3. May 11, 2017 #2

    scottdave

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    The (1/2)*k*theta2 comes from integrating k*[​IMG]*d[​IMG]. Note that the force necessary to hold the spring at [​IMG] is k*[​IMG]. Since you probably don't know what k is, replace k[​IMG] with (force), and integrate (force)d[​IMG]. Approximate this integral with a summation. If you have some fish scales (the kind which you hang the fish), then hook that onto the arm of the trap, and take some force measurements at equal angles. Try at least 8 to get decent accuracy - more if you can. You'll need a protractor. Use the midpoint or trapezoidal rule of sums. And you should pull at right angles to the moment arm (mousetrap arm). If not, then you are going to need to do some trig. Hopefully that should be pretty close to the amount of energy put into the spring.
    As far as taking the average of the two extremes, that would be what I would do if no other options are available.
     
  4. May 11, 2017 #3

    JBA

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    You should measure the force, as you described, at the point where you barely lift it and then where it is fully cocked, then subtract the first force from the second one and divide that value by the number of degrees between those two points, that will give you the correct units for the spring factor κ in lbs/degree; and, multiply that by the square of number of degrees you measured between the two load points to get the potential energy U of the trap in its cocked position.
     
  5. May 11, 2017 #4

    JBA

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    Please note that I edited my initial post to read: "then subtract the first force from the second one" if you opened that post before I made my correction.
     
  6. May 11, 2017 #5
    Thank you! This was extremely helpful! However, won't pounds not being an SI unit affect the results? Souldn't I measure it in Newtons (or kilograms)? Same question for using degrees instead of radians.

    I should also probably mention that I don't need this to be extremely precise by the way! :) We haven't really learned a lot about energy in school yet, so sorry if I seem extremely stupid haha.
     
  7. May 11, 2017 #6

    JBA

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    Sorry about the units confusion, I am one of those USA people who has used imperial units all my life and still do. You should use the equivalent SI unit for force.
     
  8. May 11, 2017 #7

    JBA

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    And for degrees as well
     
  9. May 11, 2017 #8

    scottdave

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    Good point about the units. And looking back at your formulas, the spring constant (small kappa) for rotational spring has the dimension of Energy / angle, rather than force/angle as I originally thought.
     
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