Vectorial kinematic in cartesian and polar system/notanion

[Jhenrique]

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?

 

[Jhenrique]

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:

But I don't understood how
$$\frac{\mathrm{d} }{\mathrm{d} t}\left (\vec{\omega } \times \vec{r} \right )$$
can be equal to
$$\frac{\mathrm{d} \vec{\omega }}{\mathrm{d} t}\times \vec{r}+2\vec{\omega }\times\left [ \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right ]+ \vec{\omega } \times (\vec{\omega }\times \vec{r})$$
How? Why?

 

[voko]

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:

But I don't understood how
$$\frac{\mathrm{d} }{\mathrm{d} t}\left (\vec{\omega } \times \vec{r} \right )$$
can be equal to
$$\frac{\mathrm{d} \vec{\omega }}{\mathrm{d} t}\times \vec{r}+2\vec{\omega }\times\left [ \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right ]+ \vec{\omega } \times (\vec{\omega }\times \vec{r})$$
How? Why?

$$

\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)

$$

 

[Jhenrique]

$$

\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)

$$

What? Why plus ω×([dr/dt] + ω×r)?

My equation results a different result:
$$\frac{\mathrm{d} }{\mathrm{d} t} \left( \left[ \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right ] + \vec{\omega } \times \vec{r} \right) = \left[ \frac{\mathrm{d}^2 \vec{r}}{\mathrm{d} t^2} \right ] + \frac{\mathrm{d} }{\mathrm{d} t} \left(\vec{\omega } \times \vec{r} \right) = \left[ \frac{\mathrm{d}^2 \vec{r}}{\mathrm{d} t^2} \right ] + \frac{\mathrm{d} \vec{\omega}}{\mathrm{d} t} \times \vec{r}+\vec{\omega} \times \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} = [\vec{a}] + \vec{\alpha }\times \vec{r}+\vec{\omega}\times\vec{v}$$

 

[voko]

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.

 

[Jhenrique]

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.

Now I understood! So, which is correct expression to d²r/dt² ? Is a + α×r + ω×v or a + α×r + 2ω×v + ω×(ω×r) ? My equation in post #4 is incorrect/incomplete?

 

[voko]

The correct formula is given in #3. It is identical to the one in Wikipedia.

 

[Jhenrique]

In Wikipedia I found a variant too:

http://en.wikipedia.org/wiki/Equations_of_motion#Kinematic_equations_for_one_particle

that confirm my equation in post #4. This is more one reason for my doubt...

 

[voko]

You asked for a general derivation using vectors. It was given to you. Is there something you do not understand about the derivation?

 

[Jhenrique]

You asked for a general derivation using vectors. It was given to you. Is there something you do not understand about the derivation?

In actually, yes! Because the 2 derivations below seems be correct...

$$

\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)

$$

$$\frac{\mathrm{d} }{\mathrm{d} t} \left( \left[ \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right ] + \vec{\omega } \times \vec{r} \right) = \left[ \frac{\mathrm{d}^2 \vec{r}}{\mathrm{d} t^2} \right ] + \frac{\mathrm{d} }{\mathrm{d} t} \left(\vec{\omega } \times \vec{r} \right) = \left[ \frac{\mathrm{d}^2 \vec{r}}{\mathrm{d} t^2} \right ] + \frac{\mathrm{d} \vec{\omega}}{\mathrm{d} t} \times \vec{r}+\vec{\omega} \times \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} = [\vec{a}] + \vec{\alpha }\times \vec{r}+\vec{\omega}\times\vec{v}$$

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.

 

[voko]

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.