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Rolling without slipping & linear acceleration vector

  1. Feb 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Rolling without slipping

    A) Derive the linear acceleration vector equations for points A, B, C, and O in terms of R, ω, α and θ at this instant.

    B) R = 0.5 m, ω=-54 r/s and α = 0. Determine the MPH of the vehicle and the vector accelerations of points A, B, C, and O.

    C) R = 0.5 m, ω=-54 r/s and α = +4.9 rad/sec/sec. Find the magnitude and direction of the acceleration of points O & C.

    D) R = 0.5 m, ω=-54 r/s and α = +4.9 rad/sec/sec. Determine the magnitude and direction of the acceleration of the vehicle in "g" units and how long it would take to stop.

    media-f53-f537a8aa-70c1-4eab-86cd-d6b687139ba8-phpuy2kkq.png

    2. Relevant equations


    3. The attempt at a solution
    A)
    AA = R α j + Rc (-ω2 R) i
    AB = 2 R α i + ω2 R j
    AO = R α i
    AC = ω2 R j

    B)
    V = R ω
    V = 0.5 x 54
    = 27 m/s
    = 60.4 mph

    AA = -ω2 R = -1458 i m/s2
    AB = - 1458 j
    AO = 0
    AC = 1458 j

    C)
    At O AO = R α i ⇒ 0.5 x 4.9 = 2.45 i m/s2
    At C AC = 1458 j

    D)
    A = R α
    = 2.45 m/s2

    A = 0.25 g

    t= ω/α
    t= 11.02 s


    The above was my attempt. Am I on the right track? Or am I doing this wrong?
     
  2. jcsd
  3. Feb 14, 2016 #2

    haruspex

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    I disagree with several of your answers in part A. (What is RC?) Haven't looked beyond that.
    I think the easiest way is to write a vector expression for the location of a point on the circumference in terms of the location of the centre and an angle theta (i.e. theta =0 would be point A, etc.) then differentiate as necessary.
     
  4. Feb 14, 2016 #3
    Thank you.

    So, if i wrote a general velocity vector equation for any point on the periphery of the wheel

    Vp = Vo + ω X Rop

    By differentiating so
    AA = AC + ω × (ω × RAC) + α × RAC
     
  5. Feb 14, 2016 #4

    haruspex

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    Close, but I think you have some sign errors. It's a bit confusing because if omega is positive then v0 is negative, etc.
     
  6. Feb 14, 2016 #5
    Thank

    I am sorry but why would a positive omega means a negative VO ?
     
  7. Feb 14, 2016 #6

    haruspex

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    According to the diagram, theta is measured anticlockwise, so a positive omega would mean rolling to the left. The x axis is positive to the right.
     
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