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Suspended compass

  1. Dec 30, 2009 #1
    Hi

    A friend of mine showed this problem he translated from a physics olympiad but I can't solve it:
    "Suspended compass
    Consider a perfectly symmetrical compass. It is constituted by two rigorously equal connecting rods, that join in a vertex the one that we call “pivot”. The opening angle is regulable. Imagine that we suspend the compass for the tip of one of the connecting rods, attaching it to a wire whose other end is attached at the ceiling.
    a) Sketch the position of the bar at various angles of opening.
    b) What is the opening angle of the compass, so that the pivot is the highest possible?
    Hint: The center of mass of the bar, assuming that the rods have densities
    uniform is the point where the bisector of the angle of the bar crosses the line
    passing through the midpoint of both rods. The mass center has to be, for any angle of opening, in the vertical line of the point of suspension of the wire."

    Sorry for the english in the translation. My question is how do you solve this using only knowledge available to a 11th grader??
    Thanks
     
  2. jcsd
  3. Dec 30, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi benf.stokes! Welcome to PF! :wink:

    You can do this with just elementary trigonometry and Pythagoras' theorem.

    Hint: draw the compass as a triangle ABC …

    where is the centre of mass? :smile:
     
  4. Dec 30, 2009 #3
    Hi

    Thanks for the reply, but how would that work out exactly? I couldn't do it.
     
  5. Dec 30, 2009 #4

    tiny-tim

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    Show us how far you get, and where you're stuck, and then we'll know how to help! :smile:
     
  6. Dec 30, 2009 #5
    I draw the triangle and everything but I just can't get to a expression that relates the height of the pivot to the opening angle.
    Thanks
     
  7. Dec 30, 2009 #6

    tiny-tim

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    Start with something easier :wink:

    how far is the centre of mass from the top (for an angle 2θ at the pivot)? :smile:
     
  8. Dec 30, 2009 #7
    I draw the triangle but I see no usable angle :confused:
     
  9. Dec 30, 2009 #8

    tiny-tim

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    Well, it's very simple …

    go to sleep :zzz:, and try again tomorrow. :smile:
     
  10. Dec 30, 2009 #9
    Is it (2lcos(theta))/(cos(45+theta))?? Being 2l the total length of the rod
     
  11. Dec 30, 2009 #10

    tiny-tim

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    (have a theta: θ and a degree: º :wink:)
    Where did 45º come from? And how did 2l get into the answer? :confused:

    Tell us, in words, where the centre of mass is.​
     
  12. Dec 30, 2009 #11
    The centre of mass is in the centre of the triangle. Right?
     
  13. Dec 30, 2009 #12

    tiny-tim

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    What do you mean by the centre of the triangle?
     
  14. Dec 30, 2009 #13
    The point where all the triangle bisectors join; the circumcenter
     
  15. Dec 30, 2009 #14

    tiny-tim

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    Yes, but where exactly is that?

    How far along what line?
     
  16. Dec 30, 2009 #15
    It is l*cos(theta) along the bisector of the opening angle
     
  17. Dec 30, 2009 #16

    tiny-tim

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    (what happened to that θ i gave you? :confused:)

    No, that's the end of the bisector (the centre of AC).

    Call the centre of AC "D" … where is the centre of mass along BD?
     
  18. Dec 31, 2009 #17
    Please tell me how to get there. I never was much of a geometer
     
  19. Dec 31, 2009 #18

    tiny-tim

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    Sorry, on this forum you have to do the work yourself …

    we're only here to help. :smile:
     
  20. Jan 2, 2010 #19
    Hi, I think I got it. Can I use this sketch to get to the answer? Is it arccos(1/3)?
    compasses.jpg
     
  21. Jan 2, 2010 #20

    tiny-tim

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    Sorry, I've no idea what that sketch is supposed to do. :redface:
     
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