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Kinematics of Rigid Body Plane Motion

  1. Jul 18, 2016 #1
    1. The problem statement, all variables and given/known data
    The wheel shown rotates about point O.
    Point A has a velocity=-7j in/sec.
    Point B has a tangential velocity=-4i in/sec^2
    Determine the absolute acceleration of Point C located on the circle.
    radius=6 in
    You will have to look at the figure for more clarification please.
    2. Relevant equations
    rA=rB+rA/B
    vA=vB+vA/B
    aA=aB+aA/B
    vA/B=ω×rA/B
    aA/B=(aA/B)n+(aA/B)t
    (aA/B)n=ω×(ω×rA/B)
    (aA/B)t=α×rA/B
    3. The attempt at a solution
    I tried finding the velocity of C relative to A. I feel like there is an easy way to find the angular acceleration and angular velocity of the wheel but I dont really know how. Thanks for your help.
     

    Attached Files:

  2. jcsd
  3. Jul 18, 2016 #2

    TSny

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    The wheel is considered to be a rigid body rotating about a fixed axis. How does the absolute speed of point C compare to the absolute speed of A? How does the magnitude of the absolute tangential acceleration of C compare to that of B?
     
    Last edited: Jul 18, 2016
  4. Jul 18, 2016 #3
    Since all three points are on the edge of the wheel then they should all have the same absolute acceleration and velocity.
     
  5. Jul 18, 2016 #4

    TSny

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    The magnitudes of the velocity and tangential acceleration will be the same. Directions?
     
  6. Jul 18, 2016 #5
    velocity CW
    acceleration CCW
     
  7. Jul 18, 2016 #6

    TSny

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    OK. But more specifically, can you express the tangential acceleration vector of point C in terms of the unit vectors ##\hat{i}## and ##\hat{j}##?
     
  8. Jul 18, 2016 #7
    No I can't. Isn't that what the problem is asking?
     
  9. Jul 18, 2016 #8

    TSny

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    No, there's more to the problem than just doing that. Does point C have any other component of acceleration besides tangential?
     
  10. Jul 18, 2016 #9
    Yes it has a normal acceleration toward the center of the circle. I guess the tangential acceleration would be atcos60° i and atsin60° j
     
  11. Jul 18, 2016 #10

    TSny

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    You need to draw the tangential and normal acceleration vectors for point C on your diagram. The tangential acceleration of C is not vertically upward (i.e., not +j) and it's not "to the right" (i.e., not i).

    Don't worry about the "hats". You could use bold type instead.
     
  12. Jul 18, 2016 #11
    look at my edited answer please
     
  13. Jul 18, 2016 #12

    TSny

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    That's getting close. But you need to check if 60o is correct here.
     
  14. Jul 18, 2016 #13
    yes it is correct
     
  15. Jul 18, 2016 #14

    TSny

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    Did you actually draw the tangential acceleration vector at C on your diagram? What angle does it make to the horizontal?
     
  16. Jul 18, 2016 #15
    Yes I drew it. I'm not sure how to tell the angle though...
     
  17. Jul 18, 2016 #16

    TSny

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    You'll need to use a little geometry.
     
  18. Jul 18, 2016 #17
    I'm going with 30 degrees
     
  19. Jul 18, 2016 #18

    TSny

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    Did you work it out, or is that a guess?

    Look at the figure below. What is the angle ##\phi##? Form ##\phi## do you see how to get ##\theta##?
    upload_2016-7-18_20-21-15.png
     
  20. Jul 18, 2016 #19
    Yes theta would equal 30 degrees
     
  21. Jul 18, 2016 #20

    TSny

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    Good. Repeat for the normal acceleration.
     
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