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Math Help - Circular motion?

  1. #1
    Super Member fardeen_gen's Avatar
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    Circular motion?

    A circular table with smooth horizontal surface is rotating at an angular speed ω about its axis. A groove is made on the surface along a radius and a small particle is gently placed inside the groove at a distance l from the centre. Find the speed of the particle with respect to the table as its distance from the centre becomes L.
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  2. #2
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    Hmm I don't think this is a very well defined problem. Could you be more specific or elaborate? Is there a diagram you could upload, perhaps?
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  3. #3
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    Quote Originally Posted by Mush View Post
    Hmm I don't think this is a very well defined problem. Could you be more specific or elaborate? Is there a diagram you could upload, perhaps?
    The table is rotating.

    In the co-rotating reference frame of the table top the particle is constrained to move on a fixed radius.

    So taking for the co-rotating reference frame a coordinate system with origin at the centre of rotation and x axis oriented along the radius the particle moves under a psuedo-force:

    f=m x \omega^2

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  4. #4
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    Unless im misinterpreting this, this is a relative velocity problem.

    Let \vec{v}_{TE} be the velocity of the table relative to Earth. Let \vec{v}_{PE} be the velocity of the particle relative to Earth. And let \vec{v}_{PT}=\vec{v}_{PE}-\vec{v}_{TE} be the velocity of the particle with repsect to the table.

    The tangential speed of the table is given by: \vec{v}_{TE} = r\omega where r is the radius of the table.

    Since every point on a rotating object has the same angular speed, the tangential speed of the particle at radius L is given by: \vec{v}_{PE} = L\omega.

    Hence the speed of the particle with repsect to the table is:
    \vec{v}_{PT}=\vec{v}_{PE}-\vec{v}_{TE} = L\omega - r\omega.
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  5. #5
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    Quote Originally Posted by Pn0yS0ld13r View Post
    Unless im misinterpreting this, this is a relative velocity problem.
    It is not a simple relative velocity problem as there are real forces acting on the particle to constrain it to remain in the groove, it is a problem on kinematics/dynamics in an non-intertial reference frame.

    .
    Last edited by Constatine11; December 28th 2008 at 09:44 PM.
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  6. #6
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    Quote Originally Posted by Pn0yS0ld13r View Post
    Unless im misinterpreting this, this is a relative velocity problem.

    Let \vec{v}_{TE} be the velocity of the table relative to Earth. Let \vec{v}_{PE} be the velocity of the particle relative to Earth. And let \vec{v}_{PT}=\vec{v}_{PE}-\vec{v}_{TE} be the velocity of the particle with repsect to the table.

    The tangential speed of the table is given by: \vec{v}_{TE} = r\omega where r is the radius of the table.

    Since every point on a rotating object has the same angular speed, the tangential speed of the particle at radius L is given by: \vec{v}_{PE} = L\omega.

    Hence the speed of the particle with repsect to the table is:
    \vec{v}_{PT}=\vec{v}_{PE}-\vec{v}_{TE} = L\omega - r\omega.
    I would check your notation, there appears to be some confusion about what \vec{v}_{TE} denotes, is this the velocity of the table centre of mass with respect to the Earth, or the table surface at the point currently occupied by the particle, or some point on the rim of the table.

    (and some confusion between vectors and scalars, or at least what kind of vector product is implied by juxtaposition)

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