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Relative Velocity coordinate system

  1. Nov 27, 2007 #1
    (a) A point is observed to have velocity [tex] v_A [/tex] relative to coordinate system [tex] A [/tex]. What is its velocity to coordinate system [tex] B [/tex] which is displaced from system [tex] A [/tex] by distance [tex] R [/tex]? ([tex] R [/tex] can change in time)

    I think its [tex] v_B = v_A - \frac{dR}{dt} [/tex]. But I am not completely sure why this is the case.

    (b) Particles [tex] a [/tex] and [tex] b [/tex] move in opposite directions around a circle with angular speed [tex] \omega [/tex], as shown. At [tex] t = 0 [/tex] they are both at the point [tex] r = l \bold{j} [/tex], where [tex] l [/tex] is the radius of the circle. Find the velocity of [tex] a [/tex] relative to [tex] b [/tex].

    So [tex] v_B = v_A - \frac{dR}{dt} [/tex]

    [tex] = (\sin t \bold{i }+ \cos t \bold{j)} \omega - (\cos t \bold{i} - \sin t \bold{j}) [/tex].

    Is this correct?
  2. jcsd
  3. Nov 27, 2007 #2

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    Ra = R +Rb, by vector addition, where Ra is the posn vector of the pt wrt frame A, Rb is the posn vector of the pt wrt frame B and R is the posn vector of the origin of frame B wrt A. So, differentiating,

    Va = dR/dt + Vb, which is what you've got.

    An easier way to remember is that V_a/b = Va –Vb, where the latter velos are wrt the same frame, and V_a/b represents velo of point a wrt point b. This is a vector eqn.

    Ra = l*[sin(wt) i + cos(wt) j], if a is moving clockwise.
    Rb = l*[-sin(wt) i + cos(wt) j], if b is moving counter-clockwise.

    You can now take the time derivatives, apply the formula, and see if the result tallies with your answer.
  4. Nov 27, 2007 #3
    So [tex] v_B = -l \omega \cos \omega t \bold{i} - \omega \sin \omega t \bold{j} [/tex].

    Is this correct?
  5. Nov 28, 2007 #4

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    'l' is missing in the 2nd term, otherwise it's correct.
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