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Velocity of a slider connect to a wheel

  1. Sep 14, 2016 #1
    1. The problem statement, all variables and given/known data
    problem.gif
    A wheel of radius 42 mm is rolling without slipping and its centre O has a velocity 7.2 m/s in the direction shown.

    The slider B is driven by the link AB and A is connected to the wheel.

    If AC= 28.9 mm, AB= 73.9 mm and angle ACB= 73°, what is the velocity of the slider B? Take the right-hand side direction positive.

    2. Relevant equations
    va = vb + va/b
    velocity at edge of circle in line with O = sqrt(2)*v
    Vc = 0
    3. The attempt at a solution

    My initial thoughts are that B first has to move to the left as A is going to pull it rather than push. (So the answer will be negative)
    I first constructed the triangle ABC, but without angle CAB or side CB, I don't know how to complete it.


    Any help would be greatly appreciated, Thanks.
     
  2. jcsd
  3. Sep 15, 2016 #2

    Charles Link

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    Suggestion is to try to determine the velocity vector for the point A. Do you see that the point A is going to be instantaneously moving in a circle centered at point C? Once you figure out how point A moves, go to point B and try relating the coordinates of B (to first order in ## \Delta t ## ) to the coordinates of A.
     
  4. Sep 15, 2016 #3
    Would A be moving straight down as that would be perpendicular to OA?
     
  5. Sep 15, 2016 #4

    Charles Link

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    The pivot point is at C and the radius is CA. The direction of ## v_A ## will be perpendicular to CA.
     
  6. Sep 15, 2016 #5

    Charles Link

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    Just an an additional comment is you could also treat the problem as the center at O translating at constant speed with the wheel rotating about O at a rate that will make it roll without slipping. Considering it as pivoting about point C is a shortcut that should get you the same answer.
     
  7. Sep 16, 2016 #6
    So could I say that the instantaneous centre of OA is at C:
    7.2/0.042 = 171.429 rad/s = Woa
    Va = 171.429*0.0289 = 4.954 m/s
    Then we can construct a right angle trigangle of Va, Vb and Va/b with angle 73 degrees. Then for Vb it would be Va/sin(73) = 5.18m/s => -5.18m/s (right is positive)
    However, this is still wrong any tips on what I did incorrect.
     
  8. Sep 16, 2016 #7

    Charles Link

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    First I recommend you express ## v_A ## as a vector with x and y components. Use C as your origin. Find the location of ## A ## coordinates (x,y) at time t=0. Then at time ## \Delta t ## , ## A'=A_0 +v_A \Delta t ##. Also write location ## B ## (at t=0) in terms of ## A ##, and then ## B' ## (at ## t=\Delta t ## ) in terms of ## A' ##. (Both of these steps are simple=uses Pythagorean theorem.) First get the exact expression for ## B' ## and then expand and simply keep first order term in ## \Delta t ##. Finally, ## B'-B=v_B \Delta t ##. ## \\ ## One helpful hint for the direction of ## v_A ## : CA is perpendicular to ## v_A ## and points in direction (unit vector) ## \hat{t}=\cos(\theta)\ \hat{i} +\sin(\theta) \hat{j} ##. The perpendicular to this is (unit vector) ## \hat{n}=\sin(\theta) \hat{i}-\cos(\theta) \hat{j} ## which will be your direction for ## v_A ##. (Notice that ## \hat{t} \cdot \hat{n}=0 ## ). ... editing... I agree with your result for the amplitude of ## v_A=4.954 \, m/sec ##, but unless the direction of ## v_A ## is horizontal, the computation for ## v_B ## is non-trivial. The method I presented above for finding ## v_B ## should work.... editing... I did get an answer that I think is correct and it (edited=first time I didn't compute it precisely) just slightly greater than +4.954 m/s and in the "+" direction=to the right.
     
    Last edited: Sep 16, 2016
  9. Sep 17, 2016 #8

    Charles Link

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    Additional comment=your answer ## v_B=+5.18 \, m/sec ## is close to the numerical answer I got, but your method of computing it from the ## v_A ## I believe is incorrect.
     
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