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Physical Interpretation of a Vector Quantity

  1. Jul 3, 2014 #1
    1. The problem statement, all variables and given/known data
    If [itex]\mathbf{r}, \mathbf{v},\mathbf{a}[/itex] denote the position, velocity, and acceleration of a particle, prove that

    [itex]\frac{d}{dt} [\mathbf{a} \cdot (\mathbf{v} \times \mathbf{r})] = \dot{ \mathbf{a}} \cdot (\mathbf{v} \times \mathbf{r})[/itex]


    2. Relevant equations



    3. The attempt at a solution

    I have already proven the result, but am now wondering the what the physical significance of this vector quantity is. I believe I have some idea:

    The vectors [itex]\mathbf{v}[/itex] and [itex]\mathbf{r}[/itex] create their own two dimensional subspace (a plane) in [itex]\mathbb{R}^3[/itex], with [itex]\mathbf{v} \times \mathbf{r}[/itex] being normal to this plane. The dot product of this with the acceleration vector [itex]\mathbf{a}[/itex] gives the projection of the acceleration vector onto the normal vector, it is a measure of how parallel they are. If the dot product is zero, then this implies [itex]\mathbf{a}[/itex] is orthogonal to the vector [itex]\mathbf{v} \times \mathbf{r}[/itex], and that it lies in the plane [itex]\mathbf{v}[/itex] and [itex]\mathbf{r}[/itex] create. If it is not zero, then the this implies that the acceleration vector has a component lying in the plane, and a component parallel to the normal of the plane. This causes the particle to spiral away from the plane.

    Does this seem correct?
     
    Last edited: Jul 3, 2014
  2. jcsd
  3. Jul 3, 2014 #2

    AlephZero

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    Can you define your notation better?

    The question says ##\mathbf{a}## denotes the position. In your solution you say it is the acceleration vector.

    What do ##\mathbf{b}## and ##\mathbf{c}## have to do with the question?

    And what are ##\mathbf{v}## and ##\mathbf{r}##?

    :confused::confused::confused:
     
  4. Jul 3, 2014 #3
    Whoops, I am terribly sorry. I just edited my original post.
     
  5. Jul 3, 2014 #4

    AlephZero

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