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Rodmechanism relative movements problem

  1. Jul 13, 2013 #1
    1. The problem statement, all variables and given/known data
    g73rnWh.png
    The rods are hengedly connected. O is fixed, C is free to move in the x-direction and AB is parallel to OC. OA turns around O with an angular velocity of ##\omega##.

    further given: $$\theta = 30^{\circ}$$ $$|OA|=|BC|=l_1=0.30m$$ $$|AB|=l_2=0.40m$$ $$\omega=15rad/s$$ $$v_c = 6 m/s$$
    Question: Find ##v_{B}##.

    2. Relevant equations
    $$v_{B,\text{abs}} = v_{B,\text{sleep}} + v_{B,\text{rel}}$$ $$v_{B,\text{sleep}} = v_{0'} + \omega '\cdot |OB| = v_A = \omega\cdot l_1$$

    3. The attempt at a solution

    I define a non-rotating coordinate system o'x'y' that is fixed in point A.
    k5eCsg7.png

    $$v_{B,\text{abs,x}} = -v_{B,\text{tan}}\cos{\beta} + v_c = v_{B,\text{sleep}}\cos{\beta}$$ $$v_{B,\text{abs,y}} = v_{B,\text{tan}}\sin{\beta} = -v_{B,\text{rel}} - v_{B,\text{sleep}}\sin{\beta}$$ $$v_{B,\text{sleep}}=\frac{-\omega l_1\cos{\beta}+v_c}{\cos{\beta}}=7.5m/s$$ I think this is incorrect. $$v_{B,\text{rel}}=-v_{B,\text{tan}}\sin{\beta}-v_{B,\text{sleep}}\sin{\beta}$$ This can't be correct, because this is a negative number and I know from the solution that ##\omega_{BC} = \omega_{OA}## so, I believe, the velocity of B relative to o'x'y' should work out to zero.

    What am I doing wrong?

    (extra) written version: http://i.imgur.com/WAeCf8L.jpg
     
    Last edited: Jul 13, 2013
  2. jcsd
  3. Jul 13, 2013 #2

    TSny

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    Hello, wouterbeke. Welcome to the forum!

    It appears to me that your ##\vec{v}_{B,tan}## is the velocity of B relative to C. So, ##\vec{v}_{B,tan}## is perpendicular to rod BC (rather than "tangent").

    Shouldn't the y-component of ##\vec{v}_{B,tan}## be negative? See your drawing of the vectors at B. You have written the y-component as positive.
     
  4. Jul 14, 2013 #3

    haruspex

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    I must be misunderstanding something - it seems to be overspecified (and contradictory).
    If AB is held parallel to OC then the length of AB is irrelevant for most purposes. A moves at speed 15*0.3 = 4.5m/s. Its horizontal component should be half that, and C's speed should be double that, so also 4.5m/s. Should it say OA=BC=0.40m?
     
  5. Jul 14, 2013 #4

    TSny

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    I agree that the horizontal component of velocity of A is 2.25 m/s, but I don't see how to get that C's speed should be 4.5 m/s. The horizontal rod AB is rotating.

    I get an answer that I believe is consistent with the given data, but I don't get that rods OA and BC have the same angular speed as stated by the OP.
     
    Last edited: Jul 14, 2013
  6. Jul 14, 2013 #5

    haruspex

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    OK, I see - it's only transiently parallel to OC. Thought I must have had something wrong.
     
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