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Coriolis Acceleration of a Mechanism

  1. Nov 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Rod AB rotates with the angular velocity and acceleration CCW as shown. Points A and D are pin connected. The collar C is pin connected to the link CD and slides over the link AB. At the instant shown the link CD is vertical and the link AB has an angular velocity of 2 rad/s, and an angular acceleration of 4 rad/s2

    determine,

    1) The angular velocity of link CD
    2) The relative velocity of point C with reference to A
    3) The velocity of point C
    4) The angular acceleration of link CD
    5) The relative acceleration of point C with reference to A
    6) The acceleration of point C

    http://imgur.com/6lTfnYV --> Here is a picture of the question so you can see what the mechanism looks like.


    2. Relevant equations
    v = ωr (1)
    an = ω2r (2)
    at = αr (3)
    vb = va + vb/a (4)
    ab = aa + ab/a (5)


    Coriolis Accel. eqn

    acorn = -2ω(vb/a t) (6)
    acort = 2ω(vb/a n) (7)


    3. The attempt at a solution

    Since this solution involves drawing various vectors I will include a picture of my work to help make things a bit easier. All of my work is in this picture as well so you can either look at my work their or at what I have written below. http://imgur.com/PAjUJoh,4PSWKHt

    VELOCITY ANALYSIS

    First off I started with the velocity analysis and drew where I thought the velocity vectors of each point would be. From my vectors vc, vc', and vc/c' I found,

    vc' = vc cos 30 = 1.386 m/s
    vc/c' = vc sin 30 = 0.8 m/s

    next I found my angular velocity of CD,

    ωCD = vc' / rCD = 4.62 rad/s

    ACCELERATION ANALYSIS

    I started off by writing down all the relevant equations

    acn = rAC ωAC2
    act = rAC αAC

    ac'n = rCD ωCD2
    ac't = rCD αCD

    ac/c'n = ?
    ac/c't = ?

    acorn = -2ωCD(vc/c'n)
    acort = 2ωCD(vc/c't)

    What I am stuck on is that I don't know where my vectors are supposed to go. I'm not sure what I've done wrong and don't really know where to go from here.
     
    Last edited: Nov 2, 2014
  2. jcsd
  3. Nov 2, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    The ω^2-related accelerations are always towards the center of rotation, the others orthogonal to it. And you know the orientations of those vectors with your 30°-sketch, this is similar to the velocities.
     
  4. Nov 2, 2014 #3
    So since I drew my vc/c' directly down my ac/c'n would be down in the same direction and I would have no tangential ac/c't?
     
  5. Nov 2, 2014 #4

    mfb

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    2016 Award

    Staff: Mentor

    I'm not sure if I understand your notation (why don't you use coordinates like x and y?). You have both horizontal and vertical acceleration at point C.
     
  6. Nov 2, 2014 #5
    It's ok I think I have found what I did wrong!
     
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