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Doing something wrong Newtons law proof

  1. Sep 10, 2007 #1
    if r, v, a denote the position, velocity, and acceleration of a particle, prove that

    d/dt [a . (v x r)] = d/dt(a) . (v x r ) (the periods between a and (vxr) denotes a dot product, and the actual question shows a dot over the a instead of d/dt(a)



    I know the point is that you get v x v which = 0 ,but this is what i get:

    d/dt(a . (vxr) = a [ (dv/dt)(r) + (dr/dt)(v) ] + (da/dt)(vxr)

    = a(ar + vv) + (da/dt)(vxr)
    = a^2r + (da/dt)(vxr) and i shouldn't have the first part...

    thanks alot!
     
  2. jcsd
  3. Sep 10, 2007 #2
    Your first error is
    with my correction added in bold-style. Later on, you forgot that you are dealing with vectors. Effectively, you could have seen that your result cannot be correct from seeing that your two addends are different objects (the 1st one is a vector, the 2nd is a real).

    EDIT: hehe, misread your orignal equation :D.
     
    Last edited: Sep 10, 2007
  4. Sep 10, 2007 #3
    i still get stuck...

    a [ (dv/dt) (rxv) + (dr/dt) (rxv) ] + (da/dt) (vxr)
    =a [a (rxv) + v(rxv)] + (da/dt)(vxr)
    = a^2(rxv) + 0 + (da/dt)(vxr)

    is there some reason why that first part should be eliminated?
     
  5. Sep 10, 2007 #4

    learningphysics

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    Timo has pointed out the main error. Keep the dots and crosses in your equations... also remember that:

    [tex]\vec{a} \times \vec{b} \ne \vec{b} \times \vec{a}[/tex]

    [tex]\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})[/tex]


    This is important for your derivative definition:

    [tex]\frac{d}{dt}{(\vec{a} \times \vec{b})} = (\frac{d}{dt}\vec{a}) \times \vec{b} + \vec{a} \times (\frac{d}{dt}\vec{b})[/tex]

    so you need to have: [tex]\vec{a}\times(\frac{d}{dt}\vec{b})[/tex] in your derivative as the second term, not:
    [tex](\frac{d}{dt}\vec{b})\times\vec{a}[/tex]
     
    Last edited: Sep 10, 2007
  6. Sep 10, 2007 #5

    learningphysics

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    See Timo's updated post. He fixed the equation. Also, don't mix up dot product and cross product...
     
  7. Sep 10, 2007 #6
    There is, but you got it wrong - which might be because my original correction was wrong.
    1) Don't just take my correction for granted. Figure out if and why it is correct.
    2) Scalar products are not associative, i.e. in general [tex] \vec a \cdot (\vec b \cdot \vec c) \neq (\vec a \cdot \vec b) \cdot \vec c[/tex]. You will not encounter a similar situation in the proof, but it seems to be somthing you are having trouble with and surely is something that one should be aware of.
    3) There will be two reasons why the first term will =0. You have already mentioned the 1st reason in your original post (namely that v x v = 0). You seem to already have used the other reason (which cancels another term) in your 2nd post. So the proof should be pretty straightforward for you if you get the 1st step correct and overcome your lazyness and coherently write out the symbols for the scalar products and the vectors.

    EDIT: @learningphysics: The symbol "\times" imho looks much nicer than a capital X.
     
    Last edited: Sep 10, 2007
  8. Sep 10, 2007 #7

    learningphysics

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    Just thought I should point out: [tex] \vec a \cdot (\vec b \cdot \vec c) [/tex] is undefined. You can't do a dot product of a vector with a scalar.
     
  9. Sep 10, 2007 #8

    learningphysics

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    Thanks!
     
  10. Sep 10, 2007 #9
    ok, i definitely have been mixing them. My bad.

    So, I get to (in the first part) a [ a X r + v X v ]
    the v X v cancels out, which I understand to give: a [ a X r ], now I can't mix these and they cancel out how?

    ( I probably sound like an idiot, sorry. My weekend was horrible and I'm not thinking straight yet).
     
  11. Sep 10, 2007 #10

    learningphysics

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    That term is 0, becuase [tex]\vec{a}\times\vec{r}[/tex] is perpendicular to [tex]\vec{a}[/tex] (that's one of the properties of the cross product... it's perpendicular to each of the vectors that are being crossed)

    So when you take the dot product of two perpendicular vectors, their angle is 90... cos(90) = 0.

    No, you don't sound like an idiot at all. It takes a lot of time before one can absorb this stuff.
     
    Last edited: Sep 10, 2007
  12. Sep 10, 2007 #11
    thanks a ton learningphysics and everyone.

    I didn't know that the cross product of a X r is perpendicular to the dot product of a and a X r ... can you think that through or do i need to memorize that fact?
     
    Last edited: Sep 10, 2007
  13. Sep 10, 2007 #12
    That's a well-defined and well-understood term :biggrin:. I've deliberately not quoted the rest of your post because it would prove your point correct and show my mistake :wink:. No, honestly: Thx for the correction.
     
  14. Sep 10, 2007 #13
    - You should memorize it; it's a nice property.
    - Never trust anyone on the internet (perhaps this is the point where I should claim that the error in the original version of my 1st post was on purpose ...). Prove it for yourself. Just write down the term in component form (meaning [tex] \vec a \cdot (\vec a \times \vec x) = a_1 (a_2 x_3 - a_3 x_2) + \dots [/tex] for any vector x - you can use the symbol r as well, of course) and check that for each resulting addend there is another addend exactly countering it => ... = 0.
     
    Last edited: Sep 10, 2007
  15. Sep 10, 2007 #14

    learningphysics

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    did you change your post... I think what you had written before was correct, but now incorrect...

    a X r is a vector... it is perpendicular to a. It is also perpendicular to r.

    a.(a X r) is a scalar (dot products are scalars... just ordinary numbers)... so you can't have a vector perpendicular to a scalar... ie a X R can't be perpendicular to a.(a X R)

    But a is perpendicular to a X R, hence a.(a X R) = 0

    I think you just made a typo.
     
    Last edited: Sep 10, 2007
  16. Sep 10, 2007 #15
    I'm remembering that now, (yeah, I did change it in an attempt to make it more clear).

    So, I'm correct in saying that the result of all cross/vector products are perpendicular to both vectors being multiplied?
     
  17. Sep 10, 2007 #16

    learningphysics

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    yup. exactly.
     
  18. Sep 10, 2007 #17

    learningphysics

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    lol. no prob.
     
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