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The Tangential Coriolis Force

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Hello all,
    I was hoping to get some help on this difficult question I encountered. Thanks in advance.
    Cheers.

    "Imagine you are taking a ride on a horizontally spinning carousel. Suppose that at a certain moment, you are moving with linear speed [itex]\vec{v}[/itex] and

    [itex]\frac{d}{dt}\vec{v}=\vec{b}+.5\vec{v}[/itex] where [itex]\vec{b}⋅\vec{v}=0[/itex].

    Let [itex]\vec{C_t}(t)[/itex] denote the tangential component of the Coriolis force exerted on your body at time t and let its magnitude equate 1. Calculate

    [itex]\frac{d}{dt}|\vec{C_t}|^2(t) [/itex]
    "

    The normal component of angular velocity is constant.


    3. The attempt at a solution

    [itex]\vec{C_t}(t)=-2\vec{\Omega}\times\vec{v}[/itex]

    I really don't know where to even begin. I can't seem to get any questions regarding the Coriolis effect.
     
  2. jcsd
  3. Sep 20, 2011 #2
    What are the relevant equations for the coriolis force?
     
  4. Sep 20, 2011 #3
    [itex]\vec{C}=-2\vec{\Omega}_T\times\vec{v}-2\vec{\Omega}_N\times\vec{v}[/itex], however, we are only interested in the tangential component, no?

    We know [itex]|-2\vec{\Omega}_T\times\vec{v}|=1[/itex] and [itex]\vec{\Omega}_T⋅\vec{\Omega_N}=0[/itex]
     
  5. Sep 20, 2011 #4
    Sorry---I didn't see the equation you had under 'attempt at a solution' previously.
    Why do you have two angular velocities?

    If you're initially with a linear velocity [itex]\vec{v}[/itex], what direction are [itex]\vec{v}[/itex] and [itex]\vec{b}[/itex] in?

    If you find the total coriolis force, how can you then reduce it to only the tangential component?
     
  6. Sep 20, 2011 #5
    There are two angular velocities, one tangential, denoted T, and one normal, denoted N. This is the equation that my textbook provides. The velocity vector is tangent to the circular motion of the carousel, and the b vector either is directed to the center of the carousel or away from the center.

    I'm really not getting much further with this.
     
  7. Sep 20, 2011 #6
    There's only one component to the carousel's angular velocity. You're right about the direction of v and b. So if you have the total coriolis force, what vector operations can you use (along with the vectors v and b) to find only the tangential component. it might help to draw a diagram.
     
  8. Sep 20, 2011 #7
    Are we supposed to assume that vector b is the position vector?

    [itex]|-2\vec{\Omega}\times\vec{v}|=1[/itex]
     
    Last edited: Sep 20, 2011
  9. Sep 20, 2011 #8
    No... What does the cross product do? What about the dot product?
     
  10. Sep 20, 2011 #9
    Cross product produces a vector perpendicular to the two vectors. In my textbook that is how the write the formula for the Coriolis force.
     
  11. Sep 20, 2011 #10
    Let me make this more clear: you can use the dot-product, and or cross-product to find the component of the coriolis force that you're looking for.
     
  12. Sep 20, 2011 #11
    I'm not getting this question at all.. Which direction is the force vector going to be pointing? Will it be tangent to the carousel, parallel with the velocity vector?
     
  13. Sep 20, 2011 #12
    [tex]C \propto \Omega \times v[/tex]
    [itex]\Omega[/itex] points perpendicularly upwards from the carousel.
    So what direction is C in, relative to [itex]\Omega[/itex] and v?
     
  14. Sep 20, 2011 #13
    C will be directed radially inward? If that's the case, then the dot product of angular velocity and the C force will equal zero?
     
  15. Sep 20, 2011 #14
    Scratch that, the force will be directed out
     
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