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Mechanical engineering mechanics question: Motions of a triangular plate

  1. May 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Shown in picture

    2. Relevant equations
    So far just used w=v/r

    3. The attempt at a solution

    Could only do part a.

    W=v/r
    =0.3/0.2
    =1.5rad/s

    For part b I would assume the velocity is zero? But I don't think that's correct. Any help with part b and c would be great, and is part a correct? Thank you :)
     

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  3. May 4, 2016 #2

    BvU

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    Hi beth, :welcome:
    funny to see 'Shown in picture' for the problem statement, right next to the picture. It doesn't say which half of the picture constitutes the problem :smile:

    You sure about your a answer ? Does it mean ##\omega## is in fact constant ?
     
  4. May 4, 2016 #3
    Could you be more specific in regards to my answer for a? Did I use the right formula? Is there a different formula I should be using?
     
  5. May 4, 2016 #4

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    I asked first ! Where are your v and r and what is your axis of rotation ? Don't you become suspicious if the answer doesn't seem to depend on ##\beta## ?
     
  6. May 4, 2016 #5

    I used the v for the hydraulic cylinder, 0.3, and the radius 0.2, as the edge of the triangle is.

    I think I have found my mistake - I had assumed B was the centre of the circular motion but I now see that B can move and this is wrong. Since there is no fixed point on the triangle, I don't know where to go from here
     
  7. May 4, 2016 #6

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    Oh boy, back to the drawing board :nb).

    There was something else I wanted to bring up: your equation comes from ##\ \ \vec v = \vec \omega \times \vec r\ \ ##, a vector product. You familiar with that ? In other words, if you want to write ##\omega = v/r## you can't just take any ##v##: you need the tangential velocity (the component perpendicular to ##\vec r##).

    With that, do you now see an opening to get going with this exercise ?
     
  8. May 4, 2016 #7
    I see. So when beta=30, the side CB is vertical. However, since CB=r, wouldn't the perpendicular velocity=0, since the machine only operates up and down?
     
  9. May 4, 2016 #8

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    Yes, they made a deceiving drawing. On purpose ?
    Even if CB is vertical, the thing is rotating when A moves upwards: B moves to the right. So ##\omega \ne 0##.

    Can you see a relation between ##\omega## of CD and ##\omega## of BA ? :smile:

    [edit] typo: between ##\omega## of CB and ##\omega## of BA
     
    Last edited: May 4, 2016
  10. May 4, 2016 #9

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    Not because the machine only operates up and down. But because B only moves horizontally. So if you answer b) first you do have an answer for a) -- but you have to take into account that C also moves !
    But perhaps the composer of the exercise wasn't that devious after all and gave you part a) first for a reason....
     
  11. May 4, 2016 #10
    Do you mean of CB and BA? But hmm are they the same?
     
  12. May 4, 2016 #11

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    Very good. Next step: any relation with ##\beta## ?
     
  13. May 4, 2016 #12
    I really can't find a relationship between those two things
     
  14. May 4, 2016 #13

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    What can you say about ##\beta## if the triangle does not rotate (only translates) ?
     
  15. May 4, 2016 #14
    The value stays the same?
     
  16. May 4, 2016 #15

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    Yes. So what is ##{d\over dt} \beta## when it rotates ?
     
  17. May 4, 2016 #16
    Is that the change in β related to time? Where does time come into it?
     
  18. May 4, 2016 #17

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    You want ##\omega## at the moment that ##\beta = {\pi/6}##. A is moving, remember ?
    Do you know the definition of ##\omega## ? It is the time derivative of an angle. What angle in your exercise ?
     
  19. May 4, 2016 #18
    Ah I see, so ω=dθ/dt

    So to calculate the time, use time = distance / speed.
    Not 100% sure on this, but is the distance 0.1m? Since it is half the height of one side of the triangle?
    If that's right, time = 0.1/0.3 = 0.33s

    ω=dθ/dt
    ω=(π/6)/0.33
    ω=π/2 rad/s

    Is this right?
     
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