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Mechanics behind laminar motion of rigid body

  1. Apr 5, 2014 #1
    A ridig sphere is rolling from left to right down an incline plane with an angle elevation of β

    In the y -direction, FN = mg cos β.
    In the x-direction, mg sin β - f

    (where f is the frictional force: μsFN)

    The laminar motion is then given by Icmdω/dt = Rf (where R is the radius of the sphere)

    What is the significance of the laminar motion and what is the reasoning behind it being derived?(how is it derived?)
  2. jcsd
  3. Apr 5, 2014 #2


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    Gold Member

    The sphere has translational and rotational motion.

    Laminar motion would be the rotational motion, in which case all points on the sphere move parallel to one another, with the velocity of the point a function only of its distance from the centre of rotation. This is similar to laminar flow of a fluid where the fluid moves in what can be described as sheets parallel to one another, except for the sphere the points are moving in a circle. That's the best I can explain it.

    Can you see now that this is just the torque (Rf) producing an acceleration ( dw/dt) about the centre of rotation, from the perspective of an observer moving with the centre of rotation.

    Of course, as the sphere moves down the ramp, to obtain the movement of a point from the perspective of an observer on the ramp, one adds together the translational and rotational movements.
  4. Apr 5, 2014 #3
    I just realized laminar motion is another fancy name for torque or Newton's version(second law) for rotation.

    Torque is the first time derivative of angular momentum.
    τ = I(dω/dt) = r.F

    How does the above then connects to:

    dx/dt = rω → d2t/dt = r(dω/dt) = r2f/I ?

    where f = friction.
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