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Coriolis effect, a girl on holding a ball on a carousel

  1. Oct 6, 2009 #1
    A carousel going counter clockwise starts from rest and accelerates at a constant angular acceleration of 0.02 rev/s^2. A girl sitting on a bench on the platform 7 m from the center is holding a 3kg ball. Calculate the magnitude and the direction of the force she must exert to hold the ball 6s after the carousel starts to move. Give the direction with respect to the line from the center of rotation to the girl.

    So the moving carousel simply creates an effective force on the ball opposite its rotation. I think the only force is mw^2r, where m is the mass on the ball w is the angular velocity and r is distance from the center of rotation to the girl. Is this right? the direction of the force the girl puts on the ball is opposite the natural force and tangent the the motion. Im getting .3024 N but this seems very small.

    thanks
     
    Last edited: Oct 6, 2009
  2. jcsd
  3. Oct 6, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi nothingislost! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Why bother with Coriolis force etc?

    Why not just calculate the acceleration of the ball?

    It's being forced to go in a circle of constant radius 7 m, at an angular acceleration of 0.02 rev/s2, so its radial and tangential accelerations are … ? :smile:
     
  4. Oct 6, 2009 #3
    I think you forgot a factor [itex] 2 \pi [/itex] in w

    There's also the tangential acceleration. (And there's no coriolis force if the ball doesn't move with respect to te caroussel)
     
  5. Oct 7, 2009 #4

    tiny-tim

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    Hi nothingislost! :smile:

    Please always reply on the thread, not by PM.
    No

    the force is always parallel to the acceleration, not opposite to it.

    (are you being confused by "centrifugal force"? if so, just remember that centrifugal force is imaginary … only centripetal forces are real)
     
  6. Oct 7, 2009 #5
    ok i see and the total force necessary would just be the tangential and the centripetal forces, basically somewhere between those two vectors right?
     
  7. Oct 7, 2009 #6

    tiny-tim

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    Yes. :smile:
     
  8. Oct 7, 2009 #7
  9. Oct 9, 2009 #8

    Cleonis

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    The most practical approach, I think, is to decompose the vector for the required force in a tangential component and a radial component.

    The tangential component
    The carrousel undergoes uniform angular acceleration. To maintain the same position of the ball with respect to the carrousel a tangential force on the ball is necessary. This force will keep the ball co-rotating with the carrousel, without it the ball would lag behind. Since the carrousel's angular acceleration is constant, and the distance to the central axis of rotation is constant, the required tangential force is constant.

    The radial component
    At each point in time an inward force is required to maintain the same radial distance. The expression for required centripetal force is known to you.

    The total required force to prevent the ball from moving with respect to the carrousel is the vector sum of the tangential and radial component.

    (Not every problem that involves a carrousel and a ball is about Coriolis effect. In this particular case it would be very, very unpractical to try and frame the problem in terms of some Coriolis effect.)

    Cleonis
     
    Last edited: Oct 9, 2009
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