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Decomposition of the acceleration vector

  1. May 2, 2013 #1
    A basis for any 3-dimensional vector space must have 3 vectors in it.

    So the acceleration of any object in [itex]ℝ^{3}[/itex] can be decomposed into the standard basis vectors for [itex]ℝ^{3}[/itex].

    However, I have seen another decomposition, namely, into the tangential and normal (centripetal) acceleration vectors. This set contains only two vectors. This seems to contradict the fact that at least three vectors are necessary to span [itex]ℝ^{3}[/itex].

    Either I am missing a vector, or something is terribly wrong. All insight is appreciated.

    EDIT: The decomposition I am referring to is [tex] a = \frac{dv}{dt}T + κ||v||^{2}N[/tex]

    where [itex] κ [/itex] is the curvature of the path.

    Thanks!

    BiP
     
    Last edited: May 2, 2013
  2. jcsd
  3. May 2, 2013 #2
    Tangent and normal to what?

    Two components are not enough to specify a 3D vector, so you're right, but it's hard to give any more information than that without reading your source material.
     
  4. May 2, 2013 #3

    jtbell

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    Staff: Mentor

    Draw a circle on a sheet of paper. Mark one point on the perimeter of the circle. At it, draw the radial unit vector that points away from the center of the circle. Draw the tangential unit vector.

    Now, what is the third, "missing", unit vector? (Hint: you can't draw it. :wink:)
     
  5. May 2, 2013 #4
    The decomposition in tangential and one normal works for plane motion. In this case the acceleration is always in the plane of the trajectory and two components are sufficient.
    For a general, 3D trajectory, there are two "normal" components. One example of this will be the helical motion.

    See "osculating" and "rectifying" planes.
     
  6. May 2, 2013 #5
    Is it the binormal vector?

    BiP
     
  7. May 2, 2013 #6

    WannabeNewton

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    Hi Bip! When dealing with ##\mathbb{R}^{3}## we can also use the Frenet-Serret frame which is what you speak of. Let ##\gamma:J\rightarrow \mathbb{R}^{3}## be a regular curve parametrized by arc-length. The unit tangent ##\dot{\gamma}##, the unit normal ##n = \frac{\ddot{\gamma}}{\left \| \ddot{\gamma} \right \|}##, and the unit binormal ##b = \dot{\gamma}\times n## form an orthonormal basis field for ##\mathbb{R}^{3}##; such a basis field is also called a frame field.
     
  8. May 2, 2013 #7

    AlephZero

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    At any instant in time, the object has an arbitrary velocity vector and an arbitrary acceleration vector.

    Those two vectors always lie in a plane (and in the general case the two vectors define a unique plane), so in that sense the motion at that any instant in time can be entirely described in that plane.

    But that plane has nothing to do with basis vectors for space, and in general the plane is different at every different instant in time (imagine a point travelling aling a helix, for example).
     
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