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A A x ds = v x dv or a • ds = v • dv?

  1. Jul 30, 2018 #1
    I have this question for 3D curvilinear motion of particles. I know that the relation ads = vdv is commonly used in rectilinear physics, but what about in 3D curvilinear motion?

    Would it be a x ds = v x dv or a • ds = v • dv for describing curvilinear motion?
     
  2. jcsd
  3. Jul 30, 2018 #2

    andrewkirk

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    It will be the latter - a dot product. Consider uniform circular motion, in which the acceleration is always perpendicular to the velocity. ds will be zero because the speed is constant, but dv will not, as the velocity is changing direction. So the simple multiplicative equality will not hold because the LHS is zero but the RHS is not. But the dot product equality will hold, as both sides will be zero (because dv is perpendicular to v).
     
  4. Jul 30, 2018 #3
    Hi sir, I was wondering why ds would be zero? Wouldn't a particle in circular motion have some sort of change in displacement over time? (e.g at 90° or so)
     
  5. Jul 30, 2018 #4

    andrewkirk

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    Ah, I was interpreting your s as meaning 'speed' (=|v|) but based on your response, I see you mean it to refer to displacement.

    In that case the equation has to be a dot product, not a multiplication, as one cannot multiply vectors. We have an equation for each of the three dimensions:

    \begin{align}
    a_1\,ds_1 &= v_1\,dv_1\\
    a_2\,ds_2 &= v_2\,dv_2\\
    a_3\,ds_3 &= v_3\,dv_3
    \end{align}

    Adding them together, we can represent this as:
    $$\vec a\cdot \vec {ds} = \vec v\cdot \vec {dv}$$
     
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