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Relative Motion and acceleration

  1. May 30, 2005 #1
    I tried doing this, but I can't get the exact same answer as the one given


    I use relative acceleration
    [tex]a_a = a_b + \dot \omega \cross r + \omega \cross (\omega \cross r) + 2 \omega \cross V_{rel} + a_{rel}[/tex]
    and as

    [tex]V_{rel} = 0 [/tex]
    [tex]\dot \omega = 0[/tex]
    [tex]a_{rel} = 0[/tex]
    [tex]a_b = (\doubledot r - r\omega^2) \widehat e_r + (2\dot r+r\dot\omega) \widehat e_\theta[/tex]
    so [tex]a_b = b\omega^2 \widehat e_r = b\omega^2 \widehat i[/tex]

    and what I ended up for the answer [tex]a_a[/tex] is [tex]b \omega^2 \widehat i + \omega pr \widehat j - p^2r \widehat k[/tex]
    fits funny, cause I got the right answer for [tex] a_b[/tex] using the same formula.... :confused:
  2. jcsd
  3. May 31, 2005 #2
    oh no the 2nd order version of the BKE. I dont like that form of the equation. This is what I suggest you do... You need to define 3 sets of unit vectors here for each problem. Your first is inertial, your second is in the frame of the turn table, and your thirds are attached to the rotating discs. If you are good at taking cross products use the third as ur working frame, if you are better at taking derivatives use the frame fixed in the turntable as your working frame. Now just apply the BKE
    dR/dt(inertial) = dR/dt(working frame) + w(intertial->working) X R
    that is
    Va = Vb + w(ab) X R
    Aa = Ab + w(ab) X Va

    Ab = dVb/dt in the frame of b
    Vb = dR/dt in the frame of b
    b is the working frame
    a is the interial frame

    (I tried to kinda us ur notation with the a and b stuff)

    X is the cross product operator

    notice if you substitue Va into my equation for Aa, and take the correct derivatives you will get

    Aa = Ab + w(ab) X (w(ab) X R) + 2w(ab) X Vb + dw(ab)/dt X R

    which is your equation.
  4. May 31, 2005 #3
    well..ok...did you get the answer written on the question with that equation ?..I think that's not the problem though, as I'll be using the same formula anyway
  5. May 31, 2005 #4


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    It's been a long time since I last looked at these multiple rotating frames things, and I'm having trouble following the notation in abercrombiems02's post. In particular I'm unsure how to interpret

    Which frame is a and which frame is b and what is w(ab) in this problem?

    What if you write the velocity in the inertial frame at the moment in time of interest, and the two angular velocity vectors related to the two rotations. Is it valid to add the angular velocity vectors and add the velocity components to find a net velocity vector and find the linear acceleration as the cross product of angular velocity and velocity? I did that and got the same answers as you

    [tex] \overrightarrow {v_A } = pr\widehat i - b\omega \widehat j [/tex]

    [tex] \overrightarrow {\omega _A } = p\widehat j + \omega \widehat k [/tex]

    [tex] \overrightarrow {\omega _A } \times \overrightarrow {v_A } = \left( {p\widehat j + \omega \widehat k} \right) \times \left( {pr\widehat i - b\omega \widehat j} \right) = b\omega ^2 \widehat i + rp\omega \widehat j - rp^2 \widehat k [/tex]

    [tex] \overrightarrow {v_B } = \left( {r - b} \right)\omega \widehat j - pr\widehat k [/tex]

    [tex] \overrightarrow {\omega _B } = p\widehat j + \omega \widehat k [/tex]

    [tex] \overrightarrow {\omega _B } \times \overrightarrow {v_B } = \left( {p\widehat j + \omega \widehat k} \right) \times \left( {\left( {r - b} \right)\omega \widehat j - pr\widehat k} \right) = \left[ {\left( {b - r} \right)\omega ^2 - rp^2 } \right]\widehat i [/tex]
    Last edited: May 31, 2005
  6. May 31, 2005 #5
    OMG !!! :surprised : you make it look so easy !! :bugeye:

    ok..how do you get this equation [tex] \overrightarrow {v_A } = pr\widehat i - b\omega \widehat j [/tex] ? is it simply angular speed cross radius ? but why is it [tex]b\omega \widehat j [/tex] ? is b the radius ?..looks funny, bcause b doesn't even touch the end of the circle.. :uhh:
    and how is acceleration = [tex]\omega \times v[/tex] ??
  7. May 31, 2005 #6


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    Don't ignore my caveats. I am not certain that what I have done is valid in the general case. The velocity vector is fairly simple. Point A is a distance b from the vertical axis in the figure. Since the rotation about that axis is with angular speed [itex]\omega [/itex], the linear velocity associatied with that rotation is [itex]b \omega [/itex], directed in the negative y direction. In addition, the spinning of the small wheel of radius r with angular velocity p gives a linear velocity component of rp in the positive x direction.

    As for the rest of it, I don't have a good feeling about it. [tex]\overrightarrow{\omega} \times \overrightarrow{v}[/tex] gives you the centripetal acceleration for circular motion at constant speed, and you can see that it is a contributing term in other equations, but it may not be the whole story. In fact it does not hold for central force motion in a plane in general, like orbital motion, when [itex]\omega [/itex] is not constant. If the velocity is not tangential, the direction of the cross product will not be central. I thought maybe the constant angular velocity magnitudes in this case would make the approach valid. I guess I'm going to have to pull out the old text books and see what I can see. I was hoping someone else might jump in here and straigten us out.
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