If you pour water into a glass half full and put a tennis ball there it would aim to its sides. But if you pour water completely full a ball would aim to center.

It's definitely related to a surface tension: when a glass is not full water surface is concave, and when a glass is full it's convex. So a ball aims to highest regions. Why does he do it?

Here's the video illustrating it: https://www.youtube.com/watch?v=M4ZATwHfheQ


2 Answers 2


It's definitely related to a surface tension

That is correct.

Surface tension means that the total surface area wants to be minimised.

Ball in glass.

Although the water surfaces aren't very well drawn, the surface is slightly concave in the half filled glass, slightly convex in the brim filled glass. The presence of the ball reduces the total surface area.

As a result, in the half filled glass total surface area is minimised when the ball is near the edges but it the brim filled glass surface area is minimised when the ball is dead centre.

  • 1
    $\begingroup$ Great! Let's try to understand it "more locally". We know that pressure under a convex/concave surface of water differs from pressure under it according to a Laplace's formula. So as we can see from your picture, in a concave case this difference of pressures sucks a ball to a side. But why it pushes it to a center in a convex case? $\endgroup$
    – vanger
    Commented Jun 9, 2016 at 15:53
  • $\begingroup$ Because (hydrostatic) pressure is a little higher at the top of the 'hill'. $\endgroup$
    – Gert
    Commented Jun 9, 2016 at 16:24
  • $\begingroup$ Maybe lower? It seems reasonable. But consider an energy argument. Water surface area increases when a ball deviates from a center. So a surface energy is the least when a ball in on the top of the hill. So there should be a force pushing it there. This is actually your answer. But this reasoning doesn't use hydrostatic pressure. $\endgroup$
    – vanger
    Commented Jun 9, 2016 at 16:37
  • $\begingroup$ Because my answer doesn't involve an equation of motion. I only state why the end-state is what it is, not how you get from ball-in-the-centre to ball-at-the-edge in time. The vid clearly shows there's acceleration, so that means there's force. But don't ask me to calculate that force because I don't know how to. :-) $\endgroup$
    – Gert
    Commented Jun 9, 2016 at 16:41
  • $\begingroup$ Hydrostatic pressure is definitely slightly higher just below the surface of the top of the 'hill': $p=p_0+mgh$ (Pascal's law). $\endgroup$
    – Gert
    Commented Jun 9, 2016 at 16:46

Area of a base of a spherical cap of a height $h$ is $S = \pi (2Rh - h^2)$. This is the value water surface decreases as the ball dives in. This means that according to the principle of virtual work the force water sucks the ball in is $$ F_{suck} = \frac{d}{dh} \sigma S = 2 \pi \sigma (R - h), $$ where $R = 2$ cm is a ball's radius, $h \approx 1$ cm -- immersion depth, $\sigma \approx 0.072$ N/m -- surface tension of water against air. It's almost vertical, but directed towards the center of curvature (towards the edge when the surface is concave and towards the center when the surface is convex).

Let us assume that the surface has spherical form with the radius of curvature $R_{curv}$ of order of a meter. Let $\alpha$ be a (small) angle between the line from the center of curvature and a ball center and the vertical. $F_{suck}$ is constant in the first order by $\alpha$. Then projection of Newton's second law to the horizontal is $$ m R_{curv} \ddot \alpha = - F_{suck} \alpha, $$ where $m = 2.7$ g is the ball mass. This is the equation of a harmonic oscillator with period $$ T = 2 \pi \sqrt{\frac{m R_{curv}}{F_{suck}}} \approx 4.8 \text{ s}. $$ Which is reasonable. The experiment shows that the time ball moves from the edge to the center (quarter of the period) is of the order of a second. Then damping becomes important.

  • $\begingroup$ +1 You mean centimeters when you say $R=2sm$, (i.e. it should be $R=2{\rm cm}$) right? You might also want to say that you're using the method of virtual work in the first equation. There will be a deformation of the water's convex surface too, and you haven't taken it's change in energy into account, but it's very interesting that you get the right period, so this tells us that the deformation of the water's "main" surface is not the dominant change in total energy (this is meant to be a comment, not a criticism: I don't know without doing a great deal of thinking how one would account .... $\endgroup$ Commented Jun 9, 2016 at 23:50
  • $\begingroup$ .... for all the surfaces.) $\endgroup$ Commented Jun 9, 2016 at 23:51
  • $\begingroup$ Yeah. Thanks for your remarks. Surely it's approximate reasoning. It's straightforward but very tedious to take into account the deformation of the "main" surface if we consider it as spherical, but it seems to be the excess of precision. Actually, this is the sphericity of the surface and it's radius of curvature that are main approximations. And it's hard to estimate how big mistake it gives. $\endgroup$
    – vanger
    Commented Jun 10, 2016 at 0:30
  • $\begingroup$ @WetSavannaAnimalakaRodVance: "but it's very interesting that you get the right period". I disagree. The numbers for $h$, $R_{curve}$ and $m$ are all plucked out of thin air. Also, watch the video and see if you can see any oscillation, please. This is dismissed as "Then damping becomes important" but water isn't very viscous at all. I've done some experiments and can't for the life of me see any oscillation going on. Will run some more experiments with lower surface tension tomorrow. $\endgroup$
    – Gert
    Commented Jun 10, 2016 at 1:14
  • $\begingroup$ @Gert. $m$ and $R$ are mass and radius of a standard table tennis ball: en.wikipedia.org/wiki/Table_tennis#Ball $h$ can be easily measured precisely. But we don't need such precision and it can be seen by an eye that it's about $0.5$ cm. (It also can be calculated from equating Archimedes' force with gravity force.) $R_{curv}$ is really a bit of plucked out of air. As the assumption of sphericity itself -- the surface is not spherical. It's rough appraisal and it'd be great if it would be correct right up to the order (as it is). $\endgroup$
    – vanger
    Commented Jun 10, 2016 at 1:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.