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Rod/Stick's Velocity

  1. Dec 23, 2013 #1

    Nugso

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    1. The problem statement, all variables and given/known data
    http://i.imgur.com/toGAIGP.png?1
    BC is vertical at the given position

    Calculate the velocity and acceleration of the point C.


    2. Relevant equations

    V = ω*r




    3. The attempt at a solution

    I guess, AB, BC and CD rods have the same angular speed ω = 12 rad/s. And now I'm sort of up the creek and have no idea what to do. I, however, think that I can write the following equations:

    Va/(30√3) = Vb/30


    Vc = VC/B + VC/D + VB + VD
     
  2. jcsd
  3. Dec 23, 2013 #2

    tiny-tim

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    Hi Nugso! :smile:

    For a start, what are the lengths of AB BC and CD ?
     
  4. Dec 23, 2013 #3

    Nugso

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    Hi tiny-tim and merry Christmas!

    AB = 20[itex]\sqrt{}3[/itex] cm

    BC = 30 cm

    CD = 16[itex]\sqrt{}3[/itex] cm
     
  5. Dec 23, 2013 #4

    tiny-tim

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    hmm … am i missing something?

    the diagram doesn't seem to say how high up D is :confused:
     
  6. Dec 23, 2013 #5

    Nugso

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    Ummm, well, how about CD2 = (30-10[itex]\sqrt{}3[/itex])2 + 242

    I think the equation is correct only when A and D have the same height.
     
  7. Dec 23, 2013 #6

    tiny-tim

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    yes, but i don't think they do have the same height :confused:

    anyway, we can make a start …

    we know completely the velocity of B

    and we can find at least one of the components of the velocity of C relative to B
     
  8. Dec 23, 2013 #7

    Nugso

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    Can we calculate the length of CD without assuming they have the same length or AB and CD are parallel to each other?

    How do we know the velocity? From the equation V = ω*r which is then 12*20[itex]\sqrt{}3[/itex]? Or do you mean the angular velocity of B which is 12 rad/s?
     
  9. Dec 23, 2013 #8

    tiny-tim

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    yes, V = ω*r gives you the speed, and the direction is … ? :smile:
     
  10. Dec 23, 2013 #9

    Nugso

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    The direction is upwards I think but when want to do it with i, j, k I can't find it.

    w= -12k

    r = (30i - 10√ 3j)

    Hence when we multiply, we get 2 directions one is - j and the other one is +i , right?
     
  11. Dec 23, 2013 #10

    tiny-tim

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    you mean vB = 12(-30j + 10√3i) ?

    actually, no :redface:

    but wouldn't it be simpler just to say that it's obviously perpendicular to AB? :wink:
     
  12. Dec 23, 2013 #11

    Nugso

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    Well when I think of it logically I guess you are right, but why can't I write it with I j k versions? Also shouldn't we put k after w since its direction is in k?
     
  13. Dec 23, 2013 #12

    tiny-tim

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    you can, but you're much more likely to make a mistake (with all those minuses) if you do!
    using your method, yes :smile:
     
  14. Dec 23, 2013 #13

    Nugso

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    I don't seem to get the right direction with my method :frown:
     
  15. Dec 23, 2013 #14

    tiny-tim

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    you got a minus wrong!
     
  16. Dec 23, 2013 #15

    Nugso

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    All right, I'd not bother with minuses from now on especially when there's much easier way! So we know the velocity of B and now how do we calculate the C's velocity? Vc = 12*30? so VC/B = 12*30 - 12*20√3
     
  17. Dec 23, 2013 #16

    tiny-tim

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    try calculating the velocity of C relative to B …

    what can you say about it?
     
  18. Dec 23, 2013 #17

    Nugso

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    Vc/b= 12*20√3 - 12*30?
     
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