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Momentum vector always parallel to velocity

  1. Mar 30, 2014 #1
    For an obejcting in a state of rotation, the velocity is always tangential to the distance acceleration to the axis of rotation.
    From what I read, the momentum vector, p = mv, is always parallel.
    Would it then be right to state that the momentum vector, p, is nothing more than a scalar product of m and v?
    In other words, the momentum vecotr is "superimposed" onto the velocity vector and therefore parallel.

    In looking at this from the cross product, v x vm (where both v are vector), the product = 0 because the angle between them is zero. Sin(0°) = 0.

    Am I looking at this from the right angle or is there a better way to look at it?
  2. jcsd
  3. Mar 30, 2014 #2

    Simon Bridge

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    Only in uniform circular motion - the object could be increasing or decreasing it's distance from the center of rotation, then it will have a radial component to it's velocity as well as a tangential component.

    If it's angular velocity increases, then it's total acceleration will no longer point towards the center either.

    For objects with mass ##\vec p = m\vec v## yes.
    For light ##\vec p = \hbar \vec k## ... where ##\vec k## is the wave-vector - it points in the direction the light travels and has magnitude ##k=2\pi/\lambda##.

    I don't know what you mean by "nothing more", this is quite sufficient.

    It is also valid to say that the velocity is nothing more than the momentum multiplied by a scalar too. However, momentum is important because it is a conserved quantity.

    It points in the same direction as the instantaneous direction of travel yes.

    ... this is for linear momentum.
    There is also angular momentum.

    in general $$\vec\omega = \frac{\vec v \times \vec r}{r}$$

    You seem to be overthinking the concepts here.
  4. Mar 30, 2014 #3
    From what I gathered from the lecture.

    An object in a state of rotation can be decomposed into torque and angular momentum.

    Let L→ be angular momentum

    L→ = r→ x p→ = r→ sin Θ x mv→

    Taking the derivative of L:

    dL/dt = dr/dt x p + r x dp/dt

    v x mv + r x F

    I'm a little confused as to why you said vx mv is true only in linear momentum. I did worked out v x mv = 0 from angular momentum. I might be missing something here.
  5. Mar 30, 2014 #4

    Simon Bridge

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    ... Oh I think I got confused by your notation.
    Last edited: Mar 30, 2014
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