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Planar Motion

  1. May 7, 2010 #1
    1. The collar P slides outward at a constant
    relative speed u along rod AB, which rotates
    counterclockwise with a constant angular
    velocity of 2π/3 rad/s. Knowing that r=10 in
    when θ=0° and that the collar reaches B when
    θ=90°, determine the magnitude of the
    acceleration of the collar P just as it reaches B.



    I think... this is a planar motion problem using polar coordinates. (radial and transverse).

    So my problem, as with most of the subject with physics and dynamics, is I'm not sure how to start this or where to start this.

    What I need right now is a shove in the right direction. If I can get that then I think I can solve the problem. Right now looking for a hint.

    Thanks.



    no attempt at a solution thus far.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 7, 2010 #2

    tiny-tim

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    Hi marshall104! :wink:

    Assuming the rod is frictionless, the force on the collar will be purely tangential, so the radial acceleration will be … ? :smile:
     
  4. May 7, 2010 #3
    radial will be zero? So I am thinking about this correctly?
     
  5. May 7, 2010 #4

    tiny-tim

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    Yup! :biggrin:

    So what's the equation for arad = 0 ? :smile:
     
  6. May 7, 2010 #5
    r double dot minus r*theta dot^2
     
  7. May 7, 2010 #6
    I'm still not totally sure what that means...or how it pertains to this problem...Thanks!
     
  8. May 8, 2010 #7

    tiny-tim

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    Hi marshall104! :smile:

    (just got up :zzz: …)
    That's right :smile:

    so arad = 0 means r'' = r(θ')2,

    and you know θ' = 2π/3,

    so r'' = (2π/3)2r … :wink:
     
  9. May 8, 2010 #8
    So with that being said. I set up my polar coordinates, ur and utheta, and everything in utheta direction (tangentional) equals zero...? Right? Since the velocity u of the bar is constant that is why utheta equlas zero. Then the only thing I have to figure out is the ur direction for acceleration. Which is ar = r'' -( r)(theta')^2. Since ar equals zero that means r'' must equal (r)(theta')^2. And r is the position of the collar AB at theta = 90 And that is it! So the final answer is (20 in *( 2pi/3)^2. Well not exactly because I need to find the magnitude of the acceleration. Is this correct?
     
  10. May 8, 2010 #9

    tiny-tim

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    Hi marshall104! :smile:

    Sorry, but I'm not really understanding any of that.

    What are ur and uθ? And why is uθ = 0?

    And how do you solve r'' = (2π/3)2r ?
     
  11. May 8, 2010 #10
    o.k now I'm more confused than I was before. I was under the impression that I had to use polar coordinates to solve this problem? Where we came up with the fact that r’’ = r(Ɵ’)2…? I attached a photo of the problem. Can we start from the beginning again. There is no tangential acceleration only radial correct? Arad is the acceleration in the radial direction.....Which we said was equal to zero. Correct?

    I really struggle with these problems. Thanks for all of your help.
     

    Attached Files:

    Last edited: May 9, 2010
  12. May 9, 2010 #11

    tiny-tim

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    Hi marshall104! :smile:

    (just got up :zzz: …)
    That's right :smile:

    [STRIKE]atang is given as zero, and arad has to be zero because there is no radial force.

    So you know that r'' = (2π/3)2r, which you now have to solve.[/STRIKE]

    ooops! :redface: … i think i've been answering the wrong question …

    i was wondering, in my first post, why the question didn't say the rod was frictionless (which would be necessary for the tangential acceleration to be zero).

    I've just read the question more carefully, and realised that the collar is forced to move at a constant radial speed r' = u.

    so this isn't a physics question, it's just geometry. :rolleyes:

    we needn't start again, though … our formula for arad (r'' - (2π/3)2r) is still correct, only we don't have arad = 0, instead we have r'' = 0 (because r' = u = constant).

    So you're there … just plug the value of r into the formula for arad. :smile:

    Sorry about the confusion. :redface:
     
  13. May 9, 2010 #12
    Thanks Tiny Tim! Can I ask you another question? I have an answer I would like to have you go over it for me. Thanks :smile:

    Find the equation of motion and the natural
    frequency of the system shown (m and k1, k2, and
    k3 are known). (Hint: The equation of motion will
    have the formx&& +ω2x = 0 , where ω is the natural
    frequency)

    Here is my free body diagram.

    K1 <= particle => k2
    => K3


    My governing equations:
    ∑Fx = -kx1+ kx2 +kx3 = max => kx = mx’’
    ∑Fy = N-mg = may = 0
    Ω2 = k/m = > Ω = sqrt k/m
    Final answer: x’’ – Ω2 = 0

    Is this close? :shy:
     

    Attached Files:

  14. May 9, 2010 #13

    tiny-tim

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    Hi marshall104! :smile:

    (try using the X2 and X2 tags just above the Reply box :wink:)

    Yes, that looks ok (apart from being almost unreadable! :rolleyes:)
     
  15. May 9, 2010 #14
    Sorry about that. Thanks so much for your help Tiny Tim!
     
  16. May 9, 2010 #15
    go Cougs!!
     
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