# Moving ball with rotation

In a numerical simulation, I have a round ball that has velocity of 1 $$\frac{m}{s}$$ in $$x$$-direction on a flat surface. Additionally, it also has a rotational velocity of 1000 RPM around the vertical axis ($$y$$-axis), i.e. if we are looking at the ball from above it is spinning clockwise. Suppose there is no air then the ball curves to the right relatively to its straight path. Why is it and is there a mathematical formula for it? I think it is not the magnus effect as there is no air surrounding the ball. How big of a role do the materials of ball and surface and the friction play? The surface is fixed. Thank you.

• Is the ball sliding on the flat surface or does it have a rotational component about a horizontal axis also? Is this one Earth? Which hemisphere? – Bill N Mar 31 at 19:47
• The ball is only sliding on the flat surface. Everything is happening with standard earth gravitation. The flat surface is fixed, I have added a picture with some more details (different colors represent different velocity). I don't know which hemisphere is the correct one but let's assume it is the north hemisphere. My question is based on a simulation and I would like to know how to correctly calculate the path of the ball, as the simulation fails towards the end (ball gets deformed unnaturally). – sball19 Mar 31 at 21:32

Since the plane is horizontal and the rotation axis is vertical, the point of contact between ball and plane is on the axis of rotation. Supposing an ideal point contact (which is of course not realistic), there would not be a relative velocity between ball and plane in that point. Hence, there would not be any cause for the ball to move sideways. $$\vec v=\vec \omega\times \vec r$$ where $$\vec r$$ is the distance vector from the axis to the point of contact.

Therefore, the assumption that the rotation axis is vertical must be wrong. There has to be a horizontal component of the angular velocity, which is presumably induced by the rolling of the ball on the plane due to friction.

PS: if we are talking about a numerical simulation, the cause of the observed effect can be almost anything, depending on how accurate or inaccurate the simulation is.

• The picture might be confusing, but the surface is horizontal. – sball19 Mar 31 at 22:07
• @sball19: if the plane is horizontal and the rotation axis is strictly vertical, then my first paragraph applies: no lateral motion whatsoever. Of course this is a theoretical example because even the slightest deviation of axis or plane might cause a relative contact velocity, and hence, a lateral motion. But this will not be predictable in the sense of the solution to some equations of motion. – oliver Mar 31 at 22:11
• So the lateral motion does not exist and is caused purely by numerical simulation as the ball is actually divided into tetrahedrons? – sball19 Mar 31 at 22:16
• If it not also rolling in the horizontal direction (which means that the angular velocity vector would not be strictly vertical), then this might be an explanation. You could check the hypothesis by increasing the tesselation. – oliver Mar 31 at 22:21
• By the way, if we are talking about a numerical simulation with contact and friction and all, it gets way more complicated to answer your question, because the answer also depends on the time integrator, the force calculation, the way contacts are modelled and so on. I was initially thinking that we talk about analyitic solutions that are only visualized as shown. – oliver Mar 31 at 22:25

Is the surface rough or smooth?

• If smooth, then friction is not present and plays no role - and I doubt you would see such skewing off from the initial direction. Such a situation would be akin to gravity-free motion through empty space; no rotation could alter the linear direction in such a scenario. (Unless we are dealing with e.g. the Coriolis effect - see discussion on comments.)

• If rough, then a rotation about the z-axis will inevitably be introduced. This z-rotation as well as the y-rotation might be two components that merge into a resulting rotation about a tilted axis - and this new rotation would then, due to friction, introduce a linear velocity in the tilted direction that you see.

I don't think the path should be curved, though, but straight in the x-z plane in a direction slightly angled away from the x-axis.

So, to answer your question, I don't think we require any special math to describe what you observe - rather, the claim that the rotation is solely about the y-axis might be incorrect (it is not clear to see from this gif animation about which axis/axes we are rotating).

• Why would we not see pseudo force effects such as Coriolis? – kbakshi314 Mar 31 at 21:51
• @kb314 Hmm... The Coriolis effect is a result of Earth's rotation, which would mean gradual sideways motion of the flat surface (and would imply a very long piece of surface so that curvature has significant influence). In the shown simulation I'm not sure why I would assume that Earth's rotation has been included as a parameter. Particularly not considering the apparent relative size of ball and surface. Also, the OP states in a comment that the surface is fixed. So the Coriolis effect doesn't seem to be involved here. – Steeven Mar 31 at 21:55
• I agree that the OP states that the 'surface is fixed', but it is tempting to imagine, for instance, that the blue surface is flat with the topmost part located at the earth's south pole so that the initial velocity causes the rightward acceleration due to the Coriolis effect (force) $2 \cdot \text{mass} \cdot \vec{\omega}^{EI} \times \vec{v}$. – kbakshi314 Mar 31 at 21:59
• @kb314 I agree, that could very well be an explanation. This requires more details to what has been modelled. If this is so, then the surface must be assumed smooth as well, though, in order for rotation to be only about the y-axis. – Steeven Mar 31 at 22:01

Although the OP states that the 'surface is fixed', it is tempting to imagine, for instance, that the blue surface is a relatively small flat surface with the topmost part located at the earth's north pole (corrected from the related comment mentioning south pole) so that the initial velocity causes the rightward acceleration due to the Coriolis effect (force) $$2\cdot \text{mass} \cdot \vec{\omega}^{EI} \times \vec{v}$$.

* Thanks to @Steeven for the original answer and related discussion.