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I Caculate the reaction force of a radially moving object

  1. Dec 6, 2017 at 7:51 AM #1
    Hello,

    An object moves outward in shaft radially due to centrifugal force.
    lest say it weights 1kg . in a timestep=0.0001 it travels 0.0017 meters
    this is actually the difference of two radius,
    let say first radius r1 = 0.25 meters
    second r2 = 0.25 + 0.0017 meters
    rpm of the rotating system is: 500.0

    velocity difference by the distance it travelled from r1 to r2 vd = ((500 / 60.0 * r1*2.0*pi) - (500 / 60.0 * r2*2.0*pi))
    now this velocity have to be delivered to the system by decelerating it.
    What would be the torque?
    v=a*t
    a=v/t = vd/timestep
    f = m*a
    T = f * r2 (its ok to use the higher one not much difference)
    the problem is that it makes pretty big anti torque, it seems that it is impossible even for the timesteps..
    in a system 0.5kg moving outwards with the same velocity would create like 100+ NM anti torque which seems too much work for such a small object.

    My question would be then that is it valid to calc this way or any other valid ways?

    Thanks.
     
  2. jcsd
  3. Dec 6, 2017 at 9:25 AM #2

    A.T.

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    Torque is not work, even though they have the same units.

    Use the rotating frame where there is only radial motion, compute radial velocity based on centrifugal force, then Coriolis force to get the tangential contact force required to counter it.
     
  4. Dec 6, 2017 at 9:50 AM #3

    A.T.

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    Coriolis force is perpendicular to the frame rotation axis and the object's velocity in that frame. In the frame with radial motion only, Coriolis is tangential and gives the required force by the shaft wall.
     
    Last edited by a moderator: Dec 6, 2017 at 12:00 PM
  5. Dec 6, 2017 at 10:32 AM #4
    maybe you misunderstood.
    Velocity dir = centrifugal force dir, object moving outward or inwards would cause the system decelerate or accelerate but the rotating frame remains the same...

    In physics, the Coriolis force is an inertial force[1] that acts on objects that are in motion relative to a rotating reference frame. In a reference frame with clockwise rotation, the force acts to the left of the motion of the object. In one with anticlockwise rotation, the force acts to the right.

    So it is not valid right, maybe in one scenario. This should be the same vice-versa.
    I calculate and see anyways thanks.

    Well in some motors this force as the weight moves outwards and inward periodically creating a summa zero anti torque-torque . not feeling like Coriolis force

    also if I use cross the force will be out of plane upwards or downwards.

    well I calculated after and seen the using the d = 0.5 *a*t^2 , f=m*a gives the same as the Coriolis. So this two actually gives different result which seems more valid then the velocity based.
    So its okay thanks.

    ((rpm/60.0)*(r1*2*pi))-((rpm/60.0)*(r2*2*pi))
    d = 0.00785446*0.0001 as double
    7.85446d-007
    timestep = 0.0001 as double
    0.0001d0
    d/timestep^2/0.5
    157.089d0 m/s2
    power = (1570.89 * 0.5) * 52.3599 // torque * angvel
    41125.8 watts much less than a megawatts for this simple things. (plus some more because there are more weights in the system.)
     
    Last edited: Dec 6, 2017 at 11:17 AM
  6. Dec 6, 2017 at 1:44 PM #5

    jbriggs444

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    An object which is moving radially in the rotating reference frame is moving in the rotating reference frame and is subject to the Coriolis force.
     
  7. Dec 6, 2017 at 2:15 PM #6

    A.T.

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    What plane are you talking about? The Coriolis force has no component perpendicular to the plane of rotation.
     
  8. Dec 7, 2017 at 10:48 AM #7
    Okay, so vector force = [0,1,0] vector angvel = [-50,0,0]
    In math.
    cross [0,1,0] [-50,0,0] = [0,0,50] that is out of the plane.
    Fc = [0,v,0] × [angvel,0,0] * 2.0 * m
    (min or plus depends on rot dir

    okay so how is it for you that it is in plane?
    So this is an official calculus
    Freactionforce would be (lets not call it coriolis) = (angvel*v) * 2.0 * m
     
  9. Dec 7, 2017 at 11:34 AM #8

    jbriggs444

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    That angular velocity. You specified 500 rpm. That's not -50 radians per second and it should not be in the plane. Angular velocity is perpendicular to the plane of rotation.
     
    Last edited: Dec 7, 2017 at 11:43 AM
  10. Dec 7, 2017 at 1:36 PM #9
    angular velocity is not perpendicular to the rotation it is the rotation itself.

    https://www.symbolab.com/solver/vec...trix}\times\begin{pmatrix}-5&0&0\end{pmatrix}

    x/y plane the velocity and the angular velocity the cross is out of plane.
    Wrong description.
     
  11. Dec 7, 2017 at 1:39 PM #10
    It doesnt matter that is an other example and its ok, the point is that by cross calculation you cant get it right in terms of direction but only values.
    Its really getting annoying and stupid that every second thing have wrong units then we say it doesn't have units or direction that it is not the direction but the values ok. So it is this.
     
  12. Dec 7, 2017 at 2:01 PM #11

    jbriggs444

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    The angular rotation vector is perpendicular to the plane of rotation. The cross product of the angular rotation vector with any vector in the plane of rotation will produce a result that is in the plane of rotation.

    Not much I can do about that.
     
  13. Dec 7, 2017 at 2:02 PM #12

    A.T.

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