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Vectorial kinematic in cartesian and polar system/notanion

  1. Dec 28, 2013 #1
    I was trying to study vectorial kinematics in all its fullness, without decorating formulas, only deducting all vectorially through mathematical definitions. I felt much difficulty, because it's a puzzle of many pieces and I not found a embracing explanation in any book. Someone could explain to me how the concepts of position vector, displacement vector, velocity vector, angular velocity vector, radial velocity vector, acceleration vector, angular acceleration vector and radial acceleration vector are connected geometrically and algebraically? Or, at least, could indicate me to some book that discusses these questions with clarity?
     
  2. jcsd
  3. Dec 29, 2013 #2
    I found 2 excelents explanations!!
    http://en.wikipedia.org/wiki/Equations_of_motion#General_planar_motion
    http://en.wikipedia.org/wiki/Centripetal_force#General_planar_motion
    http://en.wikipedia.org/wiki/Vector...ates#Second_time_derivative_of_a_vector_field

    However, still remained one doubt, in this topic (http://en.wikipedia.org/wiki/Centrifugal_force_(fictitious)#Acceleration) there is the follow deduction:

    fd210a5722c5cfbd937aa2451527531d.png


    But I don't understood how
    [tex]\frac{\mathrm{d} }{\mathrm{d} t}\left (\vec{\omega } \times \vec{r} \right )[/tex]
    can be equal to
    [tex]\frac{\mathrm{d} \vec{\omega }}{\mathrm{d} t}\times \vec{r}+2\vec{\omega }\times\left [ \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right ]+ \vec{\omega } \times (\vec{\omega }\times \vec{r})[/tex]
    How? Why?
     
  4. Dec 29, 2013 #3
    $$

    \frac{\mathrm{d} }{\mathrm{d} t} \left( \left[ \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right ] + \vec{\omega } \times \vec{r} \right)

    = \left[ \frac{\mathrm{d} }{\mathrm{d} t} \left( \left[ \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right ] + \vec{\omega } \times \vec{r} \right) \right] + \vec{\omega } \times \left( \left[ \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right ] + \vec{\omega } \times \vec{r} \right)

    \\

    = \left[ \frac{\mathrm{d} }{\mathrm{d} t} \left[ \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right ] \right]

    + \left[ \frac{\mathrm{d} }{\mathrm{d} t} \left( \vec{\omega } \times \vec{r} \right) \right]

    + \vec{\omega } \times \left[ \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right ]

    + \vec{\omega } \times \left( \vec{\omega } \times \vec{r} \right)

    \\

    = \left[ \frac{\mathrm{d} ^2 \vec r }{\mathrm{d}^2 t} \right]

    + \left[ \frac{\mathrm{d} \vec \omega }{\mathrm{d} t} \right] \times \vec{r}

    + \vec \omega \times \left[ \frac{\mathrm{d} \vec r}{\mathrm{d} t}\right]

    + \vec{\omega } \times \left[ \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right ]

    + \vec{\omega } \times \left( \vec{\omega } \times \vec{r} \right)

    \\

    = \left[ \frac{\mathrm{d} ^2 \vec r }{\mathrm{d}^2 t} \right]

    + \frac{\mathrm{d} \vec \omega }{\mathrm{d} t} \times \vec{r}

    + 2 \vec \omega \times \left[ \frac{\mathrm{d} \vec r}{\mathrm{d} t}\right]

    + \vec{\omega } \times \left( \vec{\omega } \times \vec{r} \right)

    $$
     
  5. Dec 29, 2013 #4
    What? Why plus ω×([dr/dt] + ω×r)???

    My equation results a different result:
    [tex]\frac{\mathrm{d} }{\mathrm{d} t} \left( \left[ \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right ] + \vec{\omega } \times \vec{r} \right) = \left[ \frac{\mathrm{d}^2 \vec{r}}{\mathrm{d} t^2} \right ] + \frac{\mathrm{d} }{\mathrm{d} t} \left(\vec{\omega } \times \vec{r} \right) = \left[ \frac{\mathrm{d}^2 \vec{r}}{\mathrm{d} t^2} \right ] + \frac{\mathrm{d} \vec{\omega}}{\mathrm{d} t} \times \vec{r}+\vec{\omega} \times \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} = [\vec{a}] + \vec{\alpha }\times \vec{r}+\vec{\omega}\times\vec{v}[/tex]
     
  6. Dec 29, 2013 #5
    Because for any vector ##\vec x##, $$ {\mathrm{d} \vec x \over \mathrm{d} t} = \left[ {\mathrm{d} \vec x \over \mathrm{d} t} \right] + \vec \omega \times \vec x , $$ where the square brackets denote differentiation in the co-rotating frame.

    In this case we have $$ \vec x= \vec v = {\mathrm{d} \vec r \over \mathrm{d} t} = \left[ {\mathrm{d} \vec r \over \mathrm{d} t} \right] + \vec \omega \times \vec r , $$ so that formula is applied twice.
     
  7. Dec 29, 2013 #6
    Now I understood! So, which is correct expression to r/dt² ? Is a + α×r + ω×v or a + α×r + 2ω×v + ω×(ω×r) ? My equation in post #4 is incorrect/incomplete?
     
  8. Dec 29, 2013 #7
    The correct formula is given in #3. It is identical to the one in Wikipedia.
     
  9. Dec 29, 2013 #8
  10. Dec 29, 2013 #9
    You asked for a general derivation using vectors. It was given to you. Is there something you do not understand about the derivation?
     
  11. Dec 29, 2013 #10
    In actually, yes! Because the 2 derivations below seems be correct...
    I know that your answer is correct, but I don't know why mine isn't. I followed the sum rule and the product rule correctly, not should have two different results for same equation.
     
  12. Dec 30, 2013 #11
    In your derivation, you assumed that $$ \mathrm d \over \mathrm d t $$ and $$ \left[ \mathrm d \over \mathrm d t \right] $$ are the same thing. They are not, as shown in #5.
     
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