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Vector product question in cylindrical coordinates

  1. Apr 18, 2012 #1
    I am trying to work the following problem;
    A rigid body is rotating about a fixed axis with a constant angular velocity ω. Take ω to lie entirely on th z-axis. Express r in cylindrical coordinates, and calculate;
    a) v=ω × r
    b)∇ × v
    The answer to (a) is v=ψωρ and (b) is ∇ × v = 2ω

    Firstly, I am not sure if we need r=[ρcosψ, psinψ, z] or simply [p, ψ, z]. Secondly, are they saying that ω is simply [0, 0, ω]? if so then I don't get the result so I am doing something wrong. Any help? I am sure that tackling this problem will deepen my understanding of curvilinear coordinates.
     
  2. jcsd
  3. Apr 18, 2012 #2
    the first expression ist good ... r=[ρcosψ, psinψ, z] ...keep going with that ;)
    next describe omega as a vector ... and calc. velocity with cross product :>
     
  4. Apr 18, 2012 #3

    I like Serena

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    I think the answer to (a) should be: ##\boldsymbol{\vec v}={\boldsymbol{\hat \psi}}\omega\rho##, where ##\boldsymbol{\hat \psi}## is the local unit vector in the ψ direction.

    Similarly the answer to (b) should be: ##\nabla \times \boldsymbol{\vec v}=2 \boldsymbol{\vec \omega}##, where ##\boldsymbol{\vec \omega}## is the vector in the z direction with length ##\omega##.

    I expect that they want you to express r as [ρcosψ, ρsinψ, z], so you can calculate the curl in cartesian coordinates.
    (Alternatively, you could simply use r=[ρ, ψ, z] and use the relevant formula from: http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates)

    And yes, they are saying that ##\boldsymbol{\vec \omega}## is simply [0, 0, ω].

    You say you didn't get the same result they do, so what did you try?
     
  5. Apr 19, 2012 #4
    I get the answer you describe for part (a) (that v=[0,ρω,0] which seems the obvious answer) if I use r=[ρ,0,0] not with r=[pcosψ,ρsinψ,0] is ω=[0,0,ω] unless I am really dumb and can't comput vector products (which is likely the problem b.t.w!) I am guessing that if I use the cartesian representation for r then ω has to be re-expressed in some way to resolve this.
    The answer that v=[0,ρω,0] seems to give a curl of 2ω in one of the components I can't remember which (making use of the cylindrical coordinate version of curl).
    Regarding the problem for (a) I guess there is a typo in the textbook?
     
  6. Apr 19, 2012 #5

    I like Serena

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    Not a typo, but I'm missing vector notations for things that are vectors and, in particularly, unit vectors.


    The challenge would be in the curl in cylindrical coordinates.
    Can you express v in cartesian coordinates (x,y,z)?
    Then take the curl, followed by a conversion to cylindrical coordinates?


    The curl of 2ω is in the z-direction, which is the same direction as ω points.
    That is: ##2ω\boldsymbol{\hat z}=2\boldsymbol{\vec \omega}##, where ##\boldsymbol{\hat z}## is the unit vector in the z-direction.
     
    Last edited: Apr 19, 2012
  7. Sep 12, 2012 #6
    Can be deleted
     
    Last edited: Sep 12, 2012
  8. Sep 12, 2012 #7
    So I know this problem is old, but it is cut & paste one of my HW problems with which I am having a lot of trouble. If I need to start a new thread I can, my apologies if I've broken a rule, I looked and didn't see one.

    Anyways,

    a) I have been letting:

    [itex]\vec{r}=[\rho*cos(\psi),\rho*sin(\psi),0] [/itex]

    [itex]\vec{\omega}=[0,0,\omega][/itex]

    Then I get [itex]\vec{v}=\vec{\omega}\times\vec{r} [/itex] [itex]=[-\rho\omega*sin(\psi),\rho\omega*cos(\psi),0] [/itex]

    If I have performed this correctly, then I don't see how I get to the asnswer: [itex][0,\omega\rho,0][/itex]?

    If [itex]\psi=0[/itex] then I get the answer, but I think that is just a coincidence. Does anyone see anything wrong with my logic.
     
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