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I Hi, how to find the magnitude of a rotating vector?

  1. Aug 19, 2016 #1
    i get stuck in how to find the magnitude of rotating vector . why say that |dA/dt|=A(dθ/dt) but who we can derive it or interpret this fact
     
  2. jcsd
  3. Aug 20, 2016 #2

    Simon Bridge

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    Start from the definitions.
     
  4. Aug 20, 2016 #3
    the rotating vector have a constant magnitude but it change it's angle by the time
     
  5. Aug 21, 2016 #4

    Simon Bridge

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    OK. So write that out mathematically.
     
  6. Aug 23, 2016 #5
    write the definition mathematically
     
  7. Aug 24, 2016 #6

    Simon Bridge

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    Do you not know how to write down the equation for a rotating vector?
    A vector in rectangular coordinates is written: ##\vec v = v_x\hat\imath + v_y\hat\jmath##
    The magnitude of that vector is ##|\vec v| = v = \sqrt{v_x^2+v_y^2}##

    If the vector makes angle ##\theta## to the x-axis, then:
    ##v_x=v\cos\theta,\; v_y=v\sin\theta##
    If the vector rotates, then ##\theta## is a function of time.

    From there, and the definition of the derivative, you can derive and interpret the relation yourself.
    So here is your task: prove ##\big|\frac{d}{dt} \vec v(t) \big| = v\frac{d}{dt}\theta(t)##
     
    Last edited: Aug 24, 2016
  8. Aug 25, 2016 #7
    let v=r
    dr/dt = lim Δt→0 Δr/Δt
    Δr=2rsin(θ/2)
    Δr≈rθ
    dr/dt=lim Δt→0 Δrθ/Δt
    |dt/dt| = r(dθ/dt)
    is it correct
     
  9. Aug 26, 2016 #8

    Simon Bridge

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    No. dt/dt = 1, your delta-r is a non sequitur... basically, what you wrote is nonsense.
    It looks like you need to revise your notes about writing vectors.
     
  10. Aug 26, 2016 #9
    i saw the explanation from Kleppner and Kolenkow book
     
  11. Aug 29, 2016 #10

    Simon Bridge

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    What explanation?
    Reread post #6. If you will not take advise I cannot help you.
    Do you know how to write a vector?
     
    Last edited: Aug 29, 2016
  12. Aug 29, 2016 #11
    a vector in rectangular coordinates system is written by it's x and y component
    then we should write the vector as vcos(θ)i+vsin(θ) j
     
  13. Aug 30, 2016 #12

    Simon Bridge

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    Well done... so write that out, A= or whatever letter you wsnt to use for a vector.
    Then differentiate both sides with respect to time ... you will need the chain rule because ##\theta## is a function of time.
     
  14. Sep 1, 2016 #13
    dA/dt=-sin(θ)dθ/dt i +vcos(θ)dθ/dt j
     
  15. Sep 1, 2016 #14

    Svein

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    You forgot the v in front of sin(θ). Now figure out what angle there is between A and dA/dt .
     
  16. Sep 2, 2016 #15

    Simon Bridge

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    Don't worry about the angle... it's a shortcut but you don't need to figure that out.
    Your next step is to find the magnatude. Don't forget that |A| = v
    Find the expression for |dA/dt| and simplify.
     
  17. Sep 2, 2016 #16
    now i get it
    |V|= √((-vsin(θ)dθ/dt)^2+(vcos(θ)dθ/dt)^2)
    |V|=vdθ/dt*√(sin^2(θ)+cos^2(θ))
    |V|=vdθ/dt
     
  18. Sep 3, 2016 #17

    Simon Bridge

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    Well done.
    Sometimes you have to just go through the steps without knowing for sure where you will end up.
    Of course I could just have told you ... but then you'd have missed out on that "ahah" moment.
     
  19. Sep 4, 2016 #18
    thank you
     
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