Sphere on a freely spinning turntable

Steve Mould somewhat recently published a video of a sphere rolling on a turntable rotating at constant rate and demonstrated that the path of the ball is, quite surprisingly, a circle. This demonstration was based on a 1978 paper titled Stable circular orbits of freely moving balls on rotating discs, Klaus Weltner. The same problem was also solved in Harvard Physics forum as Ball on turntable.

Its even more amazing that the path of a spherical ball rolling on a freely spinning turntable is a conic section as shown in A ball on a freely spinning turntable, Warren Weckesser [1]. Weckesser's request for a geometric derivation was answered about 20 years later in Comment on "A ball on a freely spinning turntable", Luis Rodriguez. A 2011 paper by Hector Munera titled A ball rolling on a freely spinning turntable: Insights from a solution in polar coordinates [2] further discusses some aspects that were not covered previously.

In this post, we aim to show 'the cone and the plane' of the conic section which has not been noted in any of the quoted papers. The key equations we need are (11) and (5) of [2] which we note here for ease.

$\displaystyle\Omega^2=\frac{\delta+\alpha r_0^2}{\delta+\alpha r^2}$ and $\mathbf{r}'=\alpha\Omega\mathbf{k}\times\mathbf{r}+\mathbf{c}$

For our purposes, we slightly alter these expression by differentiating the second and using some algebraic manipulations in the first.

$\displaystyle\Omega^2\left(\frac{\delta}{\alpha}+r^2\right)=\Omega_0^2\left(\frac{\delta}{\alpha}+r_0^2\right)$ and $\mathbf{r}''=\alpha\mathbf{k}\times(\Omega\mathbf{r})'$

Note that $\Omega_0=1$ by definition and therefore will be ignored at will from now on in this post.

To bring the cone out from hiding, we take the problem to the third dimension. To this end, we define

$\mathbf{p}=\mathbf{r}+\sqrt{\delta/\alpha} \text{ } \mathbf{k}$

Then, our two equations become

$\Omega p=\Omega_0p_0$  and $\mathbf{p}''=\alpha p_0 \mathbf{k}\times\widehat{\mathbf{p}}'$

where $p=|\mathbf{p}|$.

Because $\mathbf{p}$ and $\mathbf{r}$ are just a translation away, they have the same shape. Also, because of our definition of $\mathbf{p}$, we know that the curve defined by $\mathbf{p}$ lies on a plane perpendicular to the z-axis.

We also like to highlight that because of the definition of $\mathbf{p}$, the derivatives of $\mathbf{p}$ and $\mathbf{r}$ coincide and because $\mathbf{r} \cdot \mathbf{k}=0$, the derivatives of $\mathbf{p}$ are also perpendicular to $\mathbf{k}$.

Now we define a new vector $\mathbf{m}$ such that

$\mathbf{m}=\mathbf{p}\times\mathbf{p}'+\sqrt{\delta\alpha}p_0\text{ }\widehat{\mathbf{p}}$

Surprisingly, $\mathbf{m}$ is a constant of motion. This can be proved in multiple ways of which we show one here.

$\begin{align}(\mathbf{p}\times\mathbf{p}')' &= \mathbf{p}\times\mathbf{p}'' \\ &= \mathbf{p}\times (\alpha p_0 \mathbf{k} \times \widehat{\mathbf{p}}') \\ &= \alpha p_0 \bigl[ (\mathbf{p}\cdot\widehat{\mathbf{p}}')\mathbf{k}-(\mathbf{p}\cdot\mathbf{k})\widehat{\mathbf{p}}'\bigr] \\ &= -\sqrt{\delta\alpha} p_0\text{ }\widehat{\mathbf{p}}' \end{align}$

and therefore $\mathbf{m}'=(\mathbf{p}\times\mathbf{p}'+\sqrt{\delta\alpha} p_0\text{ }\widehat{\mathbf{p}})'=\mathbf{0}$.

We have used the fact that $\mathbf{q}\cdot\widehat{\mathbf{q}}'=0$ for any vector $\mathbf{q}$ because

$\displaystyle \widehat{\mathbf{q}}'=\frac{(\mathbf{q}\times\mathbf{q}')\times \mathbf{q}}{|\mathbf{q}|^3}$

Now that we know $\mathbf{m}$ is a constant vector, we can also easily see that $\widehat{\mathbf{p}}\cdot \mathbf{m}=\sqrt{\delta\alpha}p_0$, a constant which shows that the unit vectors of $\mathbf{p}$ have the same angle with $\mathbf{m}$ at all times.

Using these facts, we can finally see that the conic section that results from the path of the ball is the intersection of the $z=\sqrt{\delta/\alpha}$ plane and the cone whose vertex is the origin, $\mathbf{m}$ as its axis and $\mathbf{p} _0$ at its generator.

While the constancy of $\mathbf{m}$ could've been derived directly from (1), (4) and (5) of [2], taking the problem to the third dimension brings out the cone in the conic just beautifully.

Update (28-Mar-25)

The expression for $\mathbf{m}$ can rewritten as

$\mathbf{p}\times\mathbf{p}'=\mathbf{m}-\sqrt{\delta\alpha}p_0\text{ }\widehat{\mathbf{p}}$

As $\mathbf{m}$ is constant and $\widehat{\mathbf{p}}$ traces a circle with its axis along  $\mathbf{m}$, we note that the magnitude of the RHS of the above expression is a constant.

Therefore,

$|\mathbf{p}\times\mathbf{p}'|=|\mathbf{p}_0\times\mathbf{p}'_0|$

Since the cross product of two vectors gives the area of the parallelogram with the two vectors as its side, the curved surface of the cone $dA$ swept in time $dt$ is given by,

$\displaystyle \frac{dA}{dt}=\frac{1}{2}|\mathbf{p}\times\mathbf{p}'|=\frac{1}{2}|\mathbf{p}_0\times\mathbf{p}'_0|$

This shows the 'areal velocity' of $p$ is constant, much like that of the Kepler's laws of planetory motion. In particular, the orbital period $T$ of the sphere rolling on a freely spinning turntable is given by,

$\displaystyle T=\frac{A}{\frac{1}{2}|\mathbf{p}_0\times\mathbf{p}'_0|}$

The curved surface $A$ of the cone can be calculated using the formula given in Finkel's Solution Book as referenced in Problem 336, Mar 1914, American Mathematical Monthly.

To the best of my knowledge, this seems to be first time that an explicit expression for the time period of a sphere rolling on a freely spinning turntable!

Hope you enjoyed the discussion. See ya later.

Until then

Yours Aye

Me

An interesting Probability problem

I recently came across an interesting problem concerning the probability of a random sphere lying inside an unit ball in stackexchange.

The solution methodology was so fascinating that inspired me to come up with the following, almost similar, problem: Given a triangle and three points selected uniformly randomly from the triangle, what is the probability that the circle passing through those points lie completely inside the triangle?

We can characterise the three points the same way as in the SE post. Then,

$\displaystyle\mathbb{P}(\text{circle inside triangle})=\frac{1}{A^3}\iiint\limits_{\sqrt{x^2+y^2}+r \in \triangle}\,dx\,dy\,dr\int\limits_0^{2\pi}\int\limits_0^{2\pi}\int\limits_0^{2\pi}\frac{\partial (x_1,y_1,x_2,y_2,x_3,y_3)}{\partial (x,y,r,\phi_1,\phi_2,\phi_3)}\,d\phi_1\,d\phi_2\,d\phi_3$

where $A$ is the area of the given triangle $\triangle$ and the coordinates $(x,y)$ are w.r.t some arbitrary origin.

The integral involving $\phi$'s can be handled the same way as shown in the SE post which reduces the above expression to,

$\begin{align}\displaystyle\mathbb{P}(\text{circle inside triangle}) &= \frac{1}{A^3}\cdot (2\pi)^3\cdot 2 \cdot \frac{3}{2\pi}\iiint\limits_{\sqrt{x^2+y^2}+r \in \triangle}r^3 \,dx\,dy\,dr \\ &= \frac{24\pi^2}{A^3}\iiint\limits_{\sqrt{x^2+y^2}+r \in \triangle}r^3 \,dx\,dy\,dr \\ \end{align}$

If we now choose the incenter of the triangle to be the origin and $R$ to be inradius of the triangle, then the integral above w.r.t $\,dx\,dy$ is just the area of the set of points that are at the distance of $r$ inward from the triangle. But that is just the same triangle scaled such that its inradius is $R-r$. That is,

$\displaystyle \iint\limits_{\sqrt{x^2+y^2}+r \in \triangle}\,dx\,dy=\left(1-\frac{r}{R}\right)^2\cdot A$

Therefore,

$\begin{align}\displaystyle\mathbb{P}(\text{circle inside triangle}) &= \frac{24\pi^2}{A^3}\int\limits_0^R \left(1-\frac{r}{R}\right)^2\cdot A\,dr \\ \end{align}$

Evaluating this simple integral, we finally get,

$\displaystyle \mathbb{P}(\text{circle inside triangle}) = \frac{2}{5}\left(\frac{\pi R^2}{A}\right)^2$

A similar approach can be used to find the probability that a circle formed by choosing two points uniformly randomly inside a triangle completely lies within the triangle. In this case,

$\displaystyle \mathbb{P}(\text{circle inside triangle}) = \frac{2}{3}\frac{\pi R^2}{A}$

It should be easy to see from here that the above formulas, in fact, works for any tangential polygon. If we take it a step further, the same approach can be made to work for any closed convex curve. The only tricky part is the integral w.r.t '$\,dx\,dy$' which would now be the area enclosed by the 'inward' parallel curve of the given curve, which is not terribly difficult but isn't straightforward either.

An equally interesting but a tad more challenging problem is the following: Choose two points uniformly randomly inside a given triangle. If we now construct the square with the line joining the chosen points as the diagonal, what is the probability that the square lies within the triangle?

We'll see this problem in a separate post. If you figure out the answer, please share it with us in the comments section below.

Until then

Yours Aye

Me