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What are the ##\vec{u}_r## and ##\vec{u}_θ## vectors?

  1. Jun 16, 2015 #1
    < Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

    In textbook I am troubling understanding the question.Actually I dont know the terms ##\vec{u}_r## and ##\vec{u}_θ##
    Here the question.
    A particle moves according to the polar equation ##r=1+cosθ##, ##θ=e^t## where t is in seconds are r in feet.What are the ##\vec{u}_r## and ##\vec{u}_θ## components of acceleration of this particle at the instant ##t=ln(π/2)##.
    I dont want you to answer the question.I write the question to clearify my question.
    I know write accelaration vector in terms of ##\vec{T}## and ##\vec{N}## not ##\vec{u}_θ## and ##\vec{u}_r##.If you can show me how to write it,I can do the question(Thats why I didnt open this thread in Homework section)
    Thanks
     
    Last edited by a moderator: Jun 16, 2015
  2. jcsd
  3. Jun 16, 2015 #2

    ur
    is the vector along the radial velocity direction of the particle and u## \theta ## is the vector along the tangential velocity direction.

    when you will differentiate the radius vector (as given in polar coordinates); you will get two components of velocity; one along the radius vector itself, responsible for changing the magnitude of the vector and one perpendicular to the radius vector, responsible for changing the direction of radius vector. On further differentiation of the velocity vector, you will again get an expression for acceleration; a vector which will have components in the above-mentioned two directions.

    P.S. For the cylindrical coordinate system; the two vectors you are looking for in this case are the radius vector along the radial direction and the tangential vector along the azimuthal direction.
     
  4. Jun 16, 2015 #3
    This is not quite right. The tangential velocity is the component of the velocity vector tangent to the trajectory of the particle. This is not generally in the direction of u## \theta ##. u## \theta ## is the unit vector in the θ direction.

    Chet
     
  5. Jun 16, 2015 #4
    So If I differantiate it respect to ##t## I will find speed vector.##d\vec{R}/dt=(d\vec{R}/dθ).(dθ/dt)=-sinθe^t## and I will differantiate respet to ##t## again to find acceleration vector thats is .##d\vec{V}/dt=(d\vec{V}/dθ).(dθ/dt)=(-sinθ+-cosθθ).e^t##
    Is it true so far ?
     
  6. Jun 17, 2015 #5

    Nathanael

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    Homework Helper

    The time derivative of a vector should be another vector, right? I'm guessing you meant to say [itex]d\vec{R}/dt=(-sinθe^t)\vec u_r[/itex] right?

    Anyway this is not correct because you're assuming [itex]\vec u_r[/itex] is a constant with respect to time. This is an important difference between polar and cartesian coordinates: cartesian unit vectors point in a constant direction whereas polar unit vectors point in a direction that depends on θ. (And so if θ changes with time, the unit vectors will change with time.)

    You have to use the product rule: [itex]\frac{d}{dt}(r\vec u_r)=\frac{dr}{dt}\vec u_r +r\frac{d\vec u_r}{dt}[/itex]

    I will leave it to you to try and figure out what [itex]\frac{d\vec u_r}{dt}[/itex] is equal to.
     
  7. Jun 17, 2015 #6
    ##\vec{R}=r\vec{u}_r## then as you said ##v=(dr/dt)\vec{u}_r+r(d\vec{u}_r/dt)##
    so its ##\vec{v}= (-sinθe^t)\vec{u}_r+(1+cosθ)e^t\vec{u}_θ.##
    so acceleration vector.##\vec{a}=(d^2r/dt^2)\vec{u}_r+dr/dt(d\vec{u}_r/dt)+dr/dt(d\vec{u}_r/dt)+r(d^2\vec{u}_r/dt^2)##
    .##\vec{a}=((-sinθ+-cosθθ).e^t)\vec{u}_r+2(-sinθe^t).(\vec{u}_θ.e^t)+(1+cosθ).(-\vec{u}_re^t+\vec{u}_θ)e^t##

    Is this true ?
     
  8. Jun 17, 2015 #7

    Nathanael

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    Homework Helper

    Yeah it is correct, good job.
     
  9. Jun 17, 2015 #8
    Thanks
     
  10. Jun 17, 2015 #9
    I must have used the word azimuthal. Thanks for pointing out.
     
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