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Show that velocity is perpendicular to rxv

  1. Aug 24, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle moves so that its pos. vector is gn by r = cos wt i+sin wt j. show that the velocity v of the particle is perpendicular to and that rxv is a constant vector.

    2. Relevant equations



    3. The attempt at a solution

    I've tried working this out assuming [B]v[/B] = xi + y j + zk and proceeding with the vector product, but i'm not able to solve for sin wt = 1. (rxv / mod r mod v)
    Actually this problem is under the chapter differential vector calculus, but i dont know if perpendicularity can be found usin DVC. Please help.
     
  2. jcsd
  3. Aug 24, 2011 #2

    tiny-tim

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    welcome to pf!

    hi msslowlearner! welcome to pf! :wink:

    oooh, you're missing the obvious …

    v = dr/dt …

    so what is dr/dt ? :smile:
     
  4. Aug 26, 2011 #3
    ah ahhh .. yes, i get it ..the tangent , i.e., the velocity ... how cud i hv missed it ?? anyways, tats the first part .. what about the second ? r x v is a constant vector??
     
  5. Aug 26, 2011 #4
    how do you say a vector product is constant? i understand that the vector product is the area of the parallelogram formed by the 2 vectors r and v. but how is this constancy represented mathematically ?
     
  6. Aug 26, 2011 #5

    tiny-tim

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    hi msslowlearner! :smile:

    (just got up :zzz: …)
    hmm … you're understanding a visualisation of the https://www.physicsforums.com/library.php?do=view_item&itemid=85"

    visualisations are there to help you, they are not what the mathematics "really is"

    a vector product is mathematics, and to prove things about it, you usually need to use the mathematical definitions, and write out the mathematical equations :wink:
    you just write out r x v mathematically …

    from that, it should be obvious that it's constant

    (if something is constant, it usually is obvious! :wink:) …

    so what do you get? :smile:
     
    Last edited by a moderator: Apr 26, 2017
  7. Aug 26, 2011 #6

    HallsofIvy

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    I think when tiny time says "write out r x v mathematically" he means "in terms of the components of r and v".
     
  8. Aug 26, 2011 #7
    so, here we go..
    r x v = (z sin wt)i + (-z cos wt)j+ (y cos wt-x sin wt)k

    since rxv is a vector perpendcular to both r and v, v is perpendicular to rxv. right ?? tis one's from the definition itself ...

    now, rxv is a fn(t). d/dt (rxv) = zw()r) +yw(-2sin wt)k ... hope i'm right so far... now i assumed that as change in t i.e., del t tends to zero, the k component becomes zero , and
    d/dt(rxv=zw (r)

    this is what i've arrived at.. how do i say rxv is constant ???
     
  9. Aug 26, 2011 #8

    tiny-tim

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    hi msslowlearner! :smile:
    where does this come from?? :confused:

    (and what is "z" supposed to be?)

    start again …

    1) write out r

    2) write out v

    3) then multiply them …

    what do you get for 1) 2) and 3) ? :smile:
     
  10. Aug 26, 2011 #9
    1) r= cos wt i+sin wtj

    2) i assumed v=xi+yj+zk

    3) i used the matrix form to calculate r x v = (z sin wt)i + (-z cos wt)j+ (y cos wt-x sin wt)k
     
  11. Aug 26, 2011 #10
    i mean am i getting the cross product atleast correct?? vectors do stump me.
     
  12. Aug 26, 2011 #11

    tiny-tim

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    no, use v = dr/dt = … ? :smile:
     
  13. Aug 27, 2011 #12
    oh yes ... v=dr/dt ... it escaped my mind .. i got rxv = w, which is a constant ... so simple .. thanks tiny-tim :)
     
  14. Aug 27, 2011 #13

    HallsofIvy

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    Good! What constant did you get? What [itex]is[/itex] r x v?
     
  15. Aug 27, 2011 #14

    tiny-tim

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    he said w ! :wink:

    (but he left out the direction)
     
  16. Aug 27, 2011 #15


    r = cos wt i+sin wt j
    dr/dt=v=w (-sin(wt)) i + w (cos(wt)) j

    r dot product V = 0

    rXv = i j k
    cwt swt 0
    -wswt -wcwt 0
    i j k

    cwt swt 0


    w cwt cwt k + w swt swt k= w k

    rxv = w


    and is a constant
     
  17. Aug 27, 2011 #16

    tiny-tim

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    hi stallionx! :smile:

    no harm done (since msslowlearner got the answer in post #12),

    but pleeease don't give out full answers on this forum! :redface:
     
  18. Aug 27, 2011 #17
    It's probably important to mention that [itex]\vec{r}\times\vec{v} = w\hat{k}[/itex]. That is, it is a vector in the z direction... not just a number.
     
  19. Aug 27, 2011 #18
    Hiii, thanks.

    I am terribly sorry...
     
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