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I Quick Question: Rate of Change of a Rotating Vector

  1. Nov 19, 2018 #1
    Is the equation presented (that the time-derivative of a given vector in such a scenario is equal to its angular frequency vector cross the vector itself) true in the case of a vector whose origin is not on the axis of rotation?

    The way I'm visualizing this, if we take such a displaced origin but keep it at rest so that it may still act as an inertial reference frame, we would find that the speed measured from this new reference frame would be the same but that the direction may be in the minus frequency cross vector direction in the case of us displacing our origin far enough so that if we were to construct a cylinder using that given circle of rotation, our origin would be exterior to this cylinder.

    Please tell me where my thinking is flawed.
     

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    Last edited: Nov 19, 2018
  2. jcsd
  3. Nov 19, 2018 #2

    BvU

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    Yes. Do the test with ##\vec A ## and ##\vec B## rotating 'about their origin' (with the same (##\vec \omega##) and look at the time derivative of ##\vec B-\vec A## that in your picture originates at the tip of ##\vec A##.

    You almost (?) lose me here, but the vector product keeps pointing in the same direction (##\vec omega## changes sign if you shoot past the axis of rotation)
     
  4. Nov 19, 2018 #3
    I'm not quite following your description, but if I want to calculate ##\Delta \vec v##, the change in ##\vec v(t)## from time ##t_1## to time ##t_2##, that is over the interval ##\Delta t = t_2 - t_1##, then I would construct a triangle with ##\vec v(t_1)## and ##\vec v(t_2)## tail to tail. The vector ##\Delta \vec v## then goes from the head of ##\vec v(t_1)## to the head of ##\vec v(t_2)##.

    This is independent of where the origin of my coordinate system is.

    And then the rate of change of ##\vec v## is ##\Delta \vec v / \Delta t##, independent of my coordinate system, and the derivative ##d \vec v/dt## is the limit of that as ##\Delta t \rightarrow 0##, again independent of my coordinate system.
     
  5. Nov 19, 2018 #4
    Sorry about my convoluted post. Also, for your post, am I supposed to use A as the displaced vector and B as the vector depicted in my book? I just want to make sure that I understand your post clearly; did you accidentally confuse the alpha angle as the vector depicted in my book (as you reference some vector A directly used in the book).

    I've drawn up something to (hopefully) better depict what my question is; it is attached to my inital post. I use B as it is used in my book and create some arbitrary vector C. When I attempt to directly use the cross-product of omega x C to find the direction of dC/dt, I find it pointing in what I define as the negative theta-hat direction, which shouldn't happen if the tip of the vector is rotating counterclockwise which causes omega to point upward.

    I agree that, given that both origins (the book-defined origin and the one I create) are at rest, the velocity vectors should be have the same speed and magnitude measured from the origins. However, when attempting to use the cross-product definition for the time-derivative (in book), I find that the velocity vector for my newly defined origin should, according to that theorem, point in the minus theta-hat direction, even though it should be pointing in the positive theta-hat direction.

    I have to be misunderstanding something. Am I making a silly cross product error? Am I doing something wrong when I translate my vectors so that they are tail-to-tail before I take the cross product? Is my newly defined vector not classified as a vector "undergoing pure rotation" (as the book would require) because clearly the norm of the position vector from that origin varies as it traces out the particle's path?

    Edit: Ignore the blue, stray line in my drawing on the "cylinder."
     
  6. Nov 19, 2018 #5
    Anyone? Is there something that I'm not being clear about again? I'm mostly wondering why the cross product for my newly defined vector is anti-parallel to the velocity vector, disagreeing with the theorem. (See the new image in the initial post.)
     
  7. Nov 19, 2018 #6
    I don't know what you mean by the origin not being "at the axis of rotation". If that's the case, then the line you're identifying as the "axis" is not in fact the axis it's rotating around.

    In particular, your ##\vec C## appears to be rotating clockwise around an axis which goes through the origin of ##\vec C##, while ##\vec B## is rotating counter-clockwise around its own axis. The directions of rotation are opposite, and that's why the time derivatives point in opposite directions.
     
  8. Nov 19, 2018 #7
    Perhaps this is it. Re-examining the book, the requirement is for our vector to be rotating about some axis with some constant angular frequency. I think the flaw in my thinking is in claiming that my ##\vec C## was rotating about that same axis of rotation. On that topic, it's hard to visualize the motion of ##\vec C##, but I think I agree with you that it is undergoing clockwise motion.

    Would the concept to reinforce then be that we must not only examine the motion of the tip of our vector but the entire direct line segment itself? I think I was implicitly making the argument that ##\vec C## was still making a counterclockwise (from above) circular orbit about that original axis, but this is a conclusion reached when we only examine, once again, the tip of ##\vec C##. I think that I'm a bit confused with this concept because of how some of my previous math textbooks emphasized us viewing only the tip of a vector when examining its motion in space.
     
    Last edited: Nov 19, 2018
  9. Nov 20, 2018 #8
    I in turn started coming around to your way of thinking :-). I'm thinking about what you might be envisioning, and in what sense ##\vec C## could be rotating around ##\hat n##. The question is, what is the tip of ##\vec C## doing? I was thinking you'd taken a snapshot of a clockwise rotating vector in a different circle. But I don't think that's what you had in mind. The tip of ##\vec C## is going around the same circle as the tip of ##\vec B##. So I owe you an apology for completely misunderstanding your diagram.

    So how do we describe that?

    First, we can represent that as ##\vec C = \vec C_0 + \vec c_r## where ##\vec C_0## is constant, pointing to the center of the circle and ##\vec c_r## is the radial component which points from the center to the edge of the circle, and ##\vec B## has the same radial component. Clearly that's the time-varying part and both vectors will have the same time derivative.

    So we should have that ##\frac {d}{dt} \vec C = \frac {d}{dt} \vec c_r = \vec \Omega \times \vec c_r##. But clearly, ##\vec \Omega \times \vec C = \vec \Omega \times \vec C_0 + \vec \Omega \times \vec c_r \neq \vec \Omega \times \vec c_r##. The sign difference you observed is because when the geometry is as in your diagram, those two terms are in opposite directions and ##\vec \Omega \times \vec C_0## is larger. So I conclude that the result does not apply to this vector.

    If you break ##\vec B## into similar components, the vector ##\vec B_0## is parallel to ##\vec \Omega## and the cross product is zero. The problem does not arise.

    Well, my ##\vec c_r## is the motion of the tip, and that's where the time-varying part is. I totally misunderstood your question and your diagram and I apologize for misleading you.

    I'm going to read up a little on rotating coordinate systems and hope to add some words on this theorem and what situations it applies to.
     
  10. Nov 20, 2018 #9
    Oh my gosh! It all makes sense now! Decomposing the vector into those two components, one that gets you to the circle's center and another that tracks the position of that "particle" relative to the center of the circle (let's refer to these as our α and β components respectively), really helps to make things clear. I'm mad that I didn't think of doing that! Thank you!

    Also, it seems that my "cylinder" creation was unnecessary. As long as the origin of our vector is displaced at all from the axis of rotation (not simply translated up/down along the axis of rotation, but taken completely off of it), it will feature an α component that will yield a non-zero cross product when crossed with the Ω vector (in Ω x α), making our equation invalid.

    I typically like to question things and present scenarios to myself while reading the physics textbook to really see if I understand what is going on, but I was very confused initially when I thought of this example but didn't see anything in the textbook explicitly referencing the importance of the origin of our vector. The only disclaimer the book offers is that our vector must be "undergoing pure rotation". So do I then say that any vector in this scenario with some shifted origin (away from that originally defined axis of rotation, so like our ##\vec C## from before and not like the ##\vec B##) is not undergoing pure rotation, or do I just say that the book could have been maybe a bit more clear on this (at least for the students like me who would raise such an unnecessary question, lol)?
     
  11. Nov 21, 2018 #10

    BvU

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    Consider the velocity vector of a person or object off-axis on a carousel.
    Does it qualify as 'shifted'? If not, what does ?
    Which of the expressions in your story is not valid ?
     
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