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Dive effect of sliding objects

  1. Apr 7, 2012 #1
    For an object sliding on a horizontal surface, there is a dive effect caused by friction. This means that the front edge of the object is "diving" towards the surface and the back edge has some lift. As a result the friction will be stronger in the front edge of the object, which typically explains curving motion of an object having spin around vertical axis - the front edge pushes the object in one direction stronger that the back end to the other direction.

    For an object such as a cube I think this is easily explained. Friction doesn't work against the center of mass of the cube, but along its bottom side. Hence the friction attempts to accelerate a rotating motion on the cube, but due to the shape of the cube rotation would require either lift of its center of mass or the front edge burying in the surface. While the rotating force is not strong enough to do either, the surface answers the downward push with an equal lift force. Similarly in the back edge the rotating force resists gravitation and reduces the lift required by the surface to keep the object on it. With a crude simplification, the front half of the cube then faces friction force of (½m + x)g and the back half (½m - x)g where m is the total mass of the cube. As sliding continues, x will grow until eventually the back edge will rise from the surface and the cube flips on its other side.

    To this point all is clear (assuming that I'm right). But here's the puzzle.

    After watching some tutorial videos I have learned that this same effect seems to work with a billiard ball, and I can't figure it out. It seems to be a known fact that a ball with a vertical spin tends to curve while its sliding (I don't think its anymore possible in the rolling part of movement where the spin is fixed to the linear motion). Furthermore it is known than this effect can be greatly increased by hitting the ball with the stick in higher vertical angle. A horizontal hit may cause no curve at all, but a hit from say 30° the makes the curve significant. Sounds like a perfect example of creating a push against the surface.

    The problem for me is here that my previous reasoning doesn't seem to work. The first observation is of course that the ball must bury to the surface for a bit to make the contact point circular instead of a point. But even then we still need the force pushing towards the front of the circle and lifting the back of it. While friction causes exactly the same rotational force as with the cube, in this case the rotational acceleration is not resisted by neither gravitation nor the surface. Instead the rotational force works along the perimeter of the ball and doesn't seem to generate any difference on the pressure between the front and back parts of the contact circle.

    So how is this effect explained for a ball?
    Last edited: Apr 7, 2012
  2. jcsd
  3. Apr 7, 2012 #2


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    An idea from me:

    Ok, first we need a coordinate system: x is to the top, z is in the direction of (main) ball movement, and y is orthogonal to both.
    You describe a curved ball movement if there is a rotation both in x- and y-direction, where the y-rotation does not match the usual ball movement (which means that the ball slides over the surface). This can be expressed with a rotational axis somewhere in the xy-plane. Now, friction applys a momentum to the ball, which is in the y-direction only and therefore not aligned with the rotation. It will cause a precession of the rotational axis and give it a z-component. But any rotation around the z-axis gives friction to the left or to the right, which can induce a curved movement.

    I don't know whether this is the explanation or not, but it looks possible to me.
  4. Apr 8, 2012 #3
    While reading your post a very simple explanation popped in to my mind (partly thanks to misreading at first :)

    The simple fact that I had missed is that with a billiard ball there actually is also a z-rotation to begin with. I had started my analysis by looking at a horizontal hit to the ball. In that case you get y-spin (forward/backward) by hitting over or below the center of the mass of the ball. And by hitting left or right of the CoM you get x-spin (possibly no effect). This lead me to think that it's not possible to give z-spin for the ball and that is true with a horizontal hit.

    But I said it myself that the effect is created by elevating the stick and hitting the ball from an angle. First I just thought this creates pressure towards the surface, but the answer is much simpler than that, since by hitting the ball from above you can give z-spin (left/right) with no problem. As a result the ball can have rotation around all axes and can curve anywhere.

    So I guess that the effect I mentioned in the header is not a factor with balls after all. So I can sleep at rest :)
  5. Apr 8, 2012 #4


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    The behavior of the flat-sided cube sliding on a surface that you described in the first part of your post is correctly described using the rules for "sliding friction".

    Now, a billiard ball is goverened by the rules of "rolling friction" (you will need to look up each type)

    Then it gets complicated when you start adding additional spin (english) by striking the ball off center. One more item to consider: a billiard table is covered with "felt", a material which enables the player to apply all manner of english to the ball and expect "follow", "draw", "Masse" and more. So you would need to analyze this additional friction between the spin of the ball and the felt surface.
  6. Apr 8, 2012 #5


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    Flusters new explanation looks good - the combination of an angle between the cue and the board and a non-central hit can induce a spin in the z-direction. And that gives a force towards one side.
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