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Door and energy efficiency

  1. Jul 16, 2012 #1
    To understand, in a specific case, which motion spends less energy - angular or linear - I want to ask an unusual question.

    Which kind of door - hinged or sliding door - needs less energy to change its state from closed to fully opened, in a specific interval of time "t"?

    If I want to close, fully open, close, fully open... a door in some process should I choose a hinged door or a sliding door, in order to spend the least amount of energy?

    Thanks for the answers :)

    With hinged door, I mean the usual rotating door in our houses. Sliding door has just linear motion.
    (Friction ignored)
     
  2. jcsd
  3. Jul 16, 2012 #2

    berkeman

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    Why do you want to ignore friction?
     
  4. Jul 16, 2012 #3
    I just would like to understand if linear motion is more energy efficient than angular motion in this situation or not. Friction would just complicate the answer, I think. And you can develop doors with very small friction.

    And, by the way, if hinged doors are worse, why are them so spread around?
     
  5. Jul 16, 2012 #4

    Integral

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    So you find that it requires to much effort to swing a door open? Why else would you worry about this?

    Actually the friction involved in opening a pocket door (one that slides into the framework) can vary a lot over time. If the tracks get rusty or dirty it can have a significant effect on the effort required.

    Since the the hinges on a regular swinging door are easily accessible, maintenance is much easier.
     
  6. Jul 16, 2012 #5
    Of course. I know it doesn't require much effort xD. But, knowing this simple situation I can extrapolate to more complex situations, that's why I made up this simple example.

    Anyway, without friction, I believe hinged doors would spend more energy since the translational component of energy is lost in the hinges (wasted energy). Although the useful energy is less in the hinged doors compared with the sliding door. Am I right?
     
  7. Jul 16, 2012 #6

    phyzguy

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    Without friction, I can make the energy expenditure of opening and closing either type of door arbitrarily small by just doing it arbitrarily slowly.
     
  8. Jul 16, 2012 #7
    No, no. I meant in a specific interval of time "t". I want it opened after "t" seconds.

    In both ends there would be springs to make an idealized perpetual system. Which kind of door would need the smaller energy input to start this repeated system with "2t" periods?
     
  9. Jul 16, 2012 #8
    Ignore friction??
    If you ignore friction, you can spend as little energy as you want, you just have to open/close it very slowly. So your question has no answer: both doors can be pushed and set in motion with just a puff of air, as light as you want.
    You might compare which door is easier to open and close N times in a given time for example. In this case, provided you don't have a way to reuse the kinetic energy of the moving door, the hinged door requires just a bit more energy, [itex]\frac{\pi^{2}}{8}[/itex] times the sliding door. It's not worth the fuss of building and maintaining a sliding door.
    Hey, but do you find doors so hard to push usually? :uhh: :biggrin:
     
  10. Jul 16, 2012 #9
    Oh, I saw that this is what you actually meant.
    My answer is based on the assumption that the sliding door opens completely (it moves as much as it's wide) and the rotating door rotates 90 degrees.
     
  11. Jul 16, 2012 #10
    Thanks! How did you get (pi^2)/8?

    The problem is not with simple doors xD (some are quite hard to push actually :P), I'm trying to work on possible misconceptions I have about energy
     
  12. Jul 16, 2012 #11

    jbriggs444

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    I have serious doubts that the metric you are after is a useful one.


    Let's try the "distance moved" metric...

    With a hinged door you are moving the center of gravity of the door over a trajectory of total length pi * w / 4 where w is the door width.

    With a pocket door you are moving the center of gravity of the door over a trajectory of total length w where w is the door width.

    So the hinged door wins on distance covered by the center of mass. [It also wins on net displacement of the center of mass, by a somewhat greater margin]


    Let's try the "energy required" metric...

    Assume that the hinged door is accelerated by a uniform torque up until it is half open and then decellerated by a uniform torque up until it comes to a stop, fully open. The net energy required to accomplish this is precisely zero.

    Assume that the pocket door is accelerated by a uniform force up until it is half open and then decellerated by a uniform force up until it comes to a stop, fully open. The net energy required to accomplish this is precisely zero.

    So it is a wash on energy required.


    Let's try the "peak energy" metric...

    Impose a time limit for the opening operation and assess energy required for the process of opening the door to the halfway point only. This is equivalent to determining the kinetic energy of the door at the halfway point.

    Let us suppose that the time limit is t, the mass of the door is m and its width is w.

    Take the hinged door. If we take the axis of rotation to be the hinge then the kinetic energy will be entirely rotational kinetic energy. The moment of inertia of the door will be 1/3 m w^2. The peak rotation rate will be given by pi/t.

    So the peak kinetic energy of the hinged door is 1/2 ( 1/3 m w^2 ) omega^2
    = 1/6 m pi^2 w^2 / t^2
    = (pi^2 / 6) m w^2 / t^2
    ~= 1.645 m w^2 / t^2

    Take the pocket door. The peak velocity will be given by 2w/t.

    So the peak kinetic energy of the pocket door is 1/2 m v^2
    = 1/2 m 4 w^2 / t^2
    = 2 m w^2 / t^2

    So the hinged door wins on peak kinetic energy.


    You are welcome to work the "force required" metric yourseof.
     
  13. Jul 16, 2012 #12
    Thanks a lot! I get it. But didn't you forgot the translational energy lost in the hinges? If so, in the end, hinged door may lose. Or is all energy input used in rotational motion?
     
  14. Jul 16, 2012 #13
    Working backwards, I assumed that both doors are starting with same initial kinetic energy and calculated which would be closed first.
    I assumed that the sliding door need to move a distance equal to its base dimension, and the hinged door needs to rotate 180 degrees about its hinge.

    Result: hinged door takes almost twice as long to close given same initial KE. (see attachment).
     

    Attached Files:

    Last edited: Jul 16, 2012
  15. Jul 16, 2012 #14
    jbriggs444,
    Looks like we were working on this in parallel, hope our answers are close.
     
  16. Jul 16, 2012 #15
    So if you had assumed that the hinged door needs to rotate 90 degrees, it would take a little less time with same initial KE.
     
  17. Jul 16, 2012 #16
    Is "force required" still favoring hinged door? That's the answer I ultimately seek. I don't know if all energy from the force is used in rotational motion or if some is lost since the hinges don't let the door move forward (so some wasted energy here)?
     
  18. Jul 16, 2012 #17
    This seems like a variation on the Brachistochrone problem. Which force profile minimizes the closing time and peak energy in each case?

    Edit...on second thought it is not so heady. Minimum peak energy given a time target would be to simply give it all of the energy in the form of impulse at t=0.
     
    Last edited: Jul 16, 2012
  19. Jul 16, 2012 #18

    By mistake... I dropped a 2 exponent along the way. It's actually (pi^2)/12, so the hinged door is better.
    Calculate the kinetic energy of the two doors. The one for the rotating door is [itex]\frac{1}{2}Iω^{2}[/itex] where [itex]I=\int^{l}_{0}ρx^{2}dx[/itex] with ρ linear density and [itex]l[/itex] width of the door.
    I was writing a lengthier answer when I hit something on the page and lost it all so you'll forgive me if I let you do the details :)
     
  20. Jul 17, 2012 #19

    haruspex

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    OTOH, if you don't need to open the door all the way, sliding may be better. E.g. if the upright on the catch side of the door is set against an orthogonal wall and you need clearance w < l between wall and door then the break even is (1-w/l) = cos(w√3/l), i.e. w is about 78% of l.
     
  21. Jul 17, 2012 #20

    sophiecentaur

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    They don't use swing doors on any space ship I have ever seen in a film. Could that be relevant?:wink:
    (And they always go Hisssss)
     
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