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Trying to derive the rotational acceleration.

  1. Jul 26, 2007 #1
    1. The problem statement, all variables and given/known data

    Ok I'm trying to derive [tex]a=v^2/r[/tex]

    2. Relevant equations
    The Boas book gives me two equations to start off with:

    [tex]r^2=\vec{r}*\vec{r}+constant[/tex]

    and

    [tex]v^2=\vec{v}*\vec{v}+constant[/tex]


    3. The attempt at a solution



    Where the hell did those equations come from. I know it gives me the right answer but I don't understand. I can actually do the rest of the derivation easily.

    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    As a grad student I am embarressed I don't know this.
     
  2. jcsd
  3. Jul 26, 2007 #2

    robphy

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    Can you more fully quote the Boas passage?
     
  4. Jul 26, 2007 #3

    berkeman

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    I'm no help on those equations, but can you see how to derive the centripital acceleration just from the nature of the circular motion, and using differentiation?
     
  5. Jul 26, 2007 #4

    VietDao29

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    Are you sure it is [tex]r ^ 2 = \vec{r} . \vec{r} + \mbox{constant}[/tex]?

    It doesn't seem true. =.="

    We have:
    [tex]\vec{r} . \vec{r} = |\vec{r} | \times |\vec{r} | \times \cos ( \vec{r}, \vec{r}) = r ^ 2[/tex].

    Well, so it's either the constant is 0, or there is no need to write constant here.
     
  6. Jul 26, 2007 #5

    George Jones

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    The question says to first expand the rght side of

    [tex]\vec{a} = \vec{\omega} \times \left( \vec{\omega} \times \vec{r} \right),[/tex]

    and then to assume that [itex]\vec{r}[/itex] is perpendicular to [itex]\vec{\omega}.[/itex]

    What do you get when you do the first step?
     
  7. Jul 26, 2007 #6
    Sorry I had a typo in the first one. Here's the complete passage.


    Consider the motion of a particle in a circle at constant speed. We can then write

    [tex]r^2=\vec{r}*\vec{r}=constant[/tex]

    and

    [tex]v^2=\vec{v}*\vec{v}=constant[/tex]


    If we differentiate these two equations using (the rules for differentiating a dot product), we get....


    The context: Chapter 6.4 Differentiation of Vectors.
     
    Last edited: Jul 26, 2007
  8. Jul 26, 2007 #7
    Don't you want to use plane polar coordinates?

    Anyway, with the way you (i.e. Boas) are doing it, the problem is that you don't really understand why the two equations are chosen? The answer is simply that it is a good place to start, and how you know how to start there comes from a experience and thought. There are many ways to derive centripetal acceleration, and you do not have to start with
    [tex]r \cdot r = c_1[/tex]
    and
    [tex]v \cdot v = c_2[/tex].

    Since you are a bit confused, I urge you to derive centripetal acceleration another way. Calc 3 methods can do it very easily, or plane polar.
     
  9. Jul 26, 2007 #8
    I can derive it other ways, but I want to understand where this is coming from.
     
  10. Jul 26, 2007 #9

    robphy

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    Constant speed ( [tex]\vec v\cdot \vec v = const[/tex] , where [tex]const[/tex] is independent of time ) in a circle [of constant radius ( [tex]\vec r\cdot \vec r = const_2[/tex] , where [tex]const_2[/tex] is independent of time ).

    By taking derivatives with respect to t, you obtain [geometrical] relations between the acceleration, [constant speed] velocity, and the radius of the circle.
     
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