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How am I supposed to know the direction of this velocity?

  1. Jun 15, 2011 #1

    Femme_physics

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    (((Since my pre-finals test is Friday, this Friday, I decided to skip the energy questions and go over the basic dynamic energy-less stuff since I'll never get enough practice on energy on time. Please excuse me. I'll return to solve my energy-spring topic before that at a later occasion.

    )))

    http://img202.imageshack.us/img202/2360/hhheyn.jpg [Broken]


    I'll quote the question:

    "Controlling the rotation of shaft OA around point O, in the described structure, is performed by analyizing the horizontal motion of hydraulic cylinder shaft piston H, in the described position. The angle, theta, equals 66 degrees. The speed of point B (Vb) is 3 meters per second, and OA is at horizontal position.

    ---

    Now,
    The manual shows Va pointing vertically down (no angle to the Y axis). My question is how am I supposed to know this is the case? That this is really the direction of Va's velocity?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 15, 2011 #2

    I like Serena

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    Hi Fp! :smile:


    Point O is a fixed point, and OA can not bend.
    This means that point A can only make a circular motion around point O.
    In a circular motion the speed is always perpendicular...

    (Btw, what you do and when or if you do anything is always entirely up to you. ;))
     
  4. Jun 15, 2011 #3

    Femme_physics

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    That about clears everything. Thanks :)
     
  5. Jun 15, 2011 #4

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    You're welcome and success with your pre-finals test! :smile:
     
    Last edited: Jun 16, 2011
  6. Jun 16, 2011 #5

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  7. Jun 16, 2011 #6

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    How did you get to the formula [itex]|v_A \sin 30 - v_B \sin 66| = \omega_{AB} \cdot L_{AB}[/itex]?
    It doesn't look right...
     
  8. Jun 16, 2011 #7

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  9. Jun 16, 2011 #8

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    Purrrrrrrr! :)

    Still, how did you get the formula?
     
  10. Jun 16, 2011 #9

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  11. Jun 16, 2011 #10

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    Oh ok. That does seem right. :smile:

    Then could you plug the numbers in and calculate omega again?

    Edit: For reference, what is your solution manual's value?
    I calculated it in a different manner and got omega = 16.7 rad/s

    Edit2: Hold on! Your formula sheet is off!
    As it is drawn, the formula should be: [itex]|v_A \sin 24 + v_B \sin 66| = \omega_{AB} \cdot L_{AB}[/itex]
     
    Last edited: Jun 16, 2011
  12. Jun 16, 2011 #11

    Femme_physics

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  13. Jun 16, 2011 #12

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    Just edited my previous post a couple of times.
    Now I think the solution manual's is wrong....... :wink:

    It should be omega = 18.2 rad/s
     
  14. Jun 16, 2011 #13

    Femme_physics

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    Wow, really? The formulas are wrong? Can you please verify it for sure? I'll send it to the entire class. It's one day before the test! OMG, this is really important!

    Does anyone know the English name of this formula??!?
     
    Last edited: Jun 16, 2011
  15. Jun 16, 2011 #14

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    Yes I'm sure.

    I can even explain.

    Let's start with a stationary stick - it's angular velocity omega is zero.

    Now let one end turn with vA - it will have an angular velocity now.

    Now let the other end turn in the opposite direction with vB - the stick will turn even faster!
    Meaning a higher angular velocity!
     
  16. Jun 16, 2011 #15

    Femme_physics

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  17. Jun 16, 2011 #16

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    I don't think the formula has a name, but I'd call it something like: angular velocity of a moving stick. :)

    Note that the formula is not wrong in itself - it's the choice for beta.
    If beta (and vB) had been drawn on the other side of the stick, the formula would be correct.
     
    Last edited: Jun 16, 2011
  18. Jun 16, 2011 #17

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    Yes. You're right. omega = 16.67 rad/s
    I mixed up the 180 mm with the length of the stick.
    I did it right the first time, but I got excited! ;)

    My point remains that the drawing and the formula on your formula sheet don't match.
     
  19. Jun 16, 2011 #18

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    Btw, if you're interested, I can give you my method to calculate omega.

    It fits in 2 short sentences and 1 simple formula (without trig). :smile:
     
  20. Jun 16, 2011 #19

    Femme_physics

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    Ah, good :)

    I still don't see how the drawing doesn't match the formula. they just used (-sin(beta)) because the vector points in the negative direction. (our x axis is the beam). So it makes perfect sense no?



    I'm good with trig! I'm good at trig! Is your method calculusy? Problem is that I wanna stick with the entire class and don't wanna do the fancy stuff you university grads do! Although I'm curious now, what is it?

    I do have another question for a different exercise that I actually wanna post in this topic because they relate. Or do you wanna give me your method first?
     
  21. Jun 16, 2011 #20

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    What you say makes perfect sense yes! :smile:

    It's just not what's in the drawing.

    How are you supposed to know that you have to add a minus sign?
    It's not something you should gamble about, or do by trial and error. ;)
    There's a reason you did it wrong at first (not you!).




    No, it's not calculousy!
    It's really simple and it fits right in with what you already know and practiced.

    But you'll have to do a bit better than that if you want it! :devil:
     
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