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I Tangential velocity in polar coordinates

  1. Nov 2, 2018 #1

    marksyncm

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    Hello,

    I am in need of some clarification on tangential velocity in polar coordinates. As far as I know, the tangential velocity vector is ##\vec{v} = v\vec{e_t}##, where ##\vec{e_t} = \frac{\vec{v}}{v}##. This gives us the ##\vec{e_r}## and ##\vec{e_\varphi}## coordinates of the tangential velocity vector.

    For example, the velocity for the Archimedes spiral is ##\vec{v} = s\omega \vec{e_p} + s\omega^2 t\vec{e_\varphi}##. But the tangential velocity is ##\frac{1}{\sqrt{1+\omega^2 t^2}}\vec{e_p} + \frac{\omega t}{\sqrt{1+\omega^2 t^2}}\vec{e_\varphi}##

    What I do not understand is why are the two different? I was under the impression that the velocity (ie. the derivative of the location) is already and by definition tangential to the curve/trajectory. What is the difference between the two in this case?

    Thank you.
     
  2. jcsd
  3. Nov 2, 2018 #2
    The second equation you gave is not the tangential velocity. It is the unit vector in the tangential direction.
     
  4. Nov 2, 2018 #3

    marksyncm

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    So we would obtain the tangential velocity by multiplying the velocity by the tangential vector? If so, why do need to do that, ie. why isn’t the ‘original’ velocity already tangential?

    Thank you.
     
  5. Nov 2, 2018 #4
    The tangential velocity is the "magnitude of the tangential velocity" multiplied by the "unit vector in the tangential direction." In the present case, the magnitude of the tangential velocity is ##\omega s##. Sometimes we like to express the tangential velocity this way. For example, in the present case, it looks like the magnitude of the tangential velocity is constant, and only its direction is changing. That might be of interest to know.
     
  6. Nov 2, 2018 #5

    haruspex

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    The velocity along a given curve depends on the relationship between the path distance (s in your notation?) and time, which is completely independent of the curve. It seems to me your two formulations give the same answer if the relationship is s2ω2(1+ω2t2)=1.
     
  7. Nov 3, 2018 #6

    marksyncm

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    Thank you. Doesn't the ##\vec{e_\varphi}## vector point in the tangential direction as well?

    EDIT:

    I am not sure I understand where we obtain this from. The velocity for motion along the Archimedes spiral is (as far as I know) given by ##s \omega \vec{e_p} + s \omega^2 t \vec{e_\varphi}##. The magnitude of this vector seems to be ##\omega s \sqrt{1+\omega^2 t^2}##, as opposed to ωs. Do I have my terminology wrong? I know you said "magnitude of the tangential velocity", so if what I gave is not the magnitude of the tangential velocity, then I'm still confused - isn't velocity tangential by definition?
     
    Last edited: Nov 3, 2018
  8. Nov 3, 2018 #7

    haruspex

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    Nice of you to like my post #5 but I get the impression you have not understood it.
    For you to have written down an expression for the tangential velocity along a given curve you must have made an assumption about the function s=s(t). If the assumption matches the relationship I quoted then there is no mystery: the two expressions are really the same.
    What assumption did you make?
     
  9. Nov 3, 2018 #8

    marksyncm

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    Sorry, but I'm not sure I understand the question. I think my understanding of the theory behind polar coordinates is lacking at the moment, so I will revisit that first before attempting to answer the question above. Thank you!
     
  10. Nov 3, 2018 #9
    No. For that to be the case, the particle would have to be traveling in a circle about the origin.
    EDIT:


    Oops. My mistake. You are correct that this is the magnitude of the velocity. However, what I said about the unit vector in the tangential direction is correct.
     
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