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Find Normal and Tangential components of accelaration

  1. Jun 15, 2015 #1
    1. The problem statement, all variables and given/known data
    The Particle moves in the plane in a such a way that its polar equation of motion is ## \vec{R}=t\vec{i}+(t^2+1)\vec{j}##.
    a)What are its normal and tangential components of accelaration any time t ?
    b)Whats the curvature of path any time t ?
    2. Relevant equations
    ##\vec{a}=d^2s/dt^2\vec{T}+k\vec{N}d^2s/dt^2##
    ##k=dθ/ds##

    3. The attempt at a solution
    I couldnt find ##ds/dt## I tried but I stucked.
    I have no idea to find k but I couldnt find s
     
    Last edited: Jun 15, 2015
  2. jcsd
  3. Jun 15, 2015 #2

    ChrisVer

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    One note to avoid confusion. The [itex]\frac{ds}{dt}[/itex] you wrote is the magnitude of the velocity, whereas the [itex]\frac{d\vec{R}}{dt}[/itex] is its vector.

    Where did you get that equation for [itex]k[/itex]?
     
  4. Jun 15, 2015 #3
    Yeah I remember equations false I checked again it was ##d\vec{R}/ds##
    I remembered that false too it will be ##k=dθ/ds##
     
  5. Jun 15, 2015 #4
    And Is this equation true every time ? ##d\vec{T}/dθ=\vec{N}## (N is not unit vector).I now its a stupid qustion but the derivation of it come to me not generally true
     
    Last edited: Jun 15, 2015
  6. Jun 15, 2015 #5

    ChrisVer

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    Let me put it else... what does [itex]\theta[/itex] stand for?
     
    Last edited: Jun 15, 2015
  7. Jun 15, 2015 #6

    you can look the graph
     
  8. Jun 15, 2015 #7
    I got the idea in 21:49 look the magnitude of ##d\vec{T}/ds=dθ/ds=k## Is that mean the magnitude of ##\vec{N}## equal ##k##.I am confused one time he says ##d\vec{T}/dθ=\vec{N}##(In 10:49) now he says ##d\vec{T}/ds=\vec{N}## Is it a special case which I mentioned ?
     
  9. Jun 15, 2015 #8

    ChrisVer

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    The magnitude of [itex]\vec{N}[/itex] is unity (it's a unit vector). [itex]k= \Big| \frac{d\vec{T}}{ds} \Big|[/itex] is some number and so:
    [itex] \vec{N} = \frac{d \vec{T}}{ds} \Big/ \Big| \frac{d\vec{T}}{ds} \Big|[/itex] has magnitude equal to 1.
     
  10. Jun 15, 2015 #9

    ChrisVer

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    First of all, did you find the tangent and normal vector to the acceleration?(question a)
     
  11. Jun 15, 2015 #10

    ChrisVer

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    It is just because the normal vector is:

    [itex] \vec{N} \equiv \frac{d\vec{T}}{dt} \Big/ \Big|\frac{d \vec{T}}{dt} \Big|[/itex]
     
    Last edited: Jun 15, 2015
  12. Jun 15, 2015 #11
    Tangent vector is [itex] \vec{T} = \frac{d \vec{R}}{dt} \Big/ \Big| \frac{d\vec{R}}{dt} \Big|[/itex] so Its ##\vec{T}=1/√5\vec{i}+2/√5t\vec{j}##.Normal vector will be then ##\vec{N}=\vec{j}##
     
    Last edited: Jun 15, 2015
  13. Jun 15, 2015 #12
    beg me but I didnt understand first ds (in your #8 post) now we can use dt. Whatever we want we can use that equation I guess.

    In previous post I did something wrong cause dot product of T and N must be zero and here its not
     
    Last edited: Jun 15, 2015
  14. Jun 15, 2015 #13

    ChrisVer

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    Your T looks wrong. What's dR/dt ? and what's |dR/dt| ?
     
  15. Jun 15, 2015 #14

    ChrisVer

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    I am using the parameter of your curve.
     
  16. Jun 15, 2015 #15
    dR/dt=i+4j If I divide the lenght which its √17 I find dT/dt=i/√17+2/√17j I made a huge mistake I must be an idiot soort
     
  17. Jun 15, 2015 #16

    ChrisVer

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    Either you use the parameter [itex]t[/itex]/[itex]\lambda[/itex],[itex]\theta[/itex] or whatever you call it, or the arc-length [itex]s[/itex] it doesn't matter....
    [itex] \frac{d \vec{T}}{ds}[/itex] will point to the same direction as [itex] \frac{d \vec{T}}{dt}[/itex]. That can be seen by rewritting:
    [itex] \frac{d \vec{T}}{ds} = \frac{d \vec{T}}{dt} \Big/ \frac{ds}{dt} [/itex] so you are only getting a multiplicative factor of [itex] (ds/dt)^{-1}[/itex] between the vector [itex] \frac{d \vec{T}}{ds}[/itex] and [itex]\frac{d \vec{T}}{dt}[/itex]. You don't care about that factor as long as you normalize them.
     
  18. Jun 15, 2015 #17

    ChrisVer

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    this is wrong, what's the derivative [itex] \frac{d}{dt} (t^2+1)[/itex]?
     
  19. Jun 15, 2015 #18
    I made a huge mistake I must be an idiot so so soory
     
  20. Jun 15, 2015 #19

    ChrisVer

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    so?
     
  21. Jun 15, 2015 #20
    dR/dt=i+4j If I divide the lenght which its √17 I find dT/dt=i/√17+2/√17j
     
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