Sphere on a freely spinning turntable (again)..

After my initial post on this topic, I still felt that there is some more to the problem than what we've explored. I kept returning to this problem hoping to find something more and am glad I did.

We start with a following 'force' equation that we saw earlier.

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

Taking a dot product on both sides with $\mathbf{p}'$, we have

$\mathbf{p}'\cdot\mathbf{p}''=\alpha p_0 \mathbf{p}'\cdot(\mathbf{k}\times\widehat{\mathbf{p}}')$

Permuting the RHS and integrating the above, we see that

$\displaystyle \frac{1}{2}\left\vert{\mathbf{p}'}\right\vert^2+\alpha p_0 \int\mathbf{k}\cdot(\mathbf{p}'\times\widehat{\mathbf{p}}')\,dt=\text{const.}$

For any vector $\mathbf{q}$, with straightforward differentiation, we can show that

$\displaystyle \mathbf{q}'\times\widehat{\mathbf{q}}'=\frac{\mathbf{q}\cdot\mathbf{q}'}{\left\vert\mathbf{q}\right\vert^3}(\mathbf{q}\times\mathbf{q}')$

Using this result, the constant vector $\mathbf{m}$ from our earlier post and the fact that $\mathbf{p}\cdot\mathbf{k}=\sqrt{\delta/\alpha}$, we see that

$\displaystyle \mathbf{k}\cdot(\mathbf{p}'\times\widehat{\mathbf{p}}')=\frac{\mathbf{p}\cdot\mathbf{p}'}{p^3}\left(\mathbf{k}\cdot\mathbf{m}-\frac{\delta p_0}{p}\right)$

where $p=\left\vert\mathbf{p}\right\vert$.

Let $u=1/p$. It is easy to see that $u'=-(\mathbf{p}\cdot\mathbf{p}')/p^3$. This makes it almost trivial to integrate the above expression.

With the above, our 'energy' equation becomes

$\displaystyle E=\frac{1}{2}\left\vert{\mathbf{p}'}\right\vert^2-\alpha (\mathbf{k}\cdot\mathbf{m}) \frac{p_0}{p}+\frac{\delta\alpha}{2}\frac{p_0^2}{p^2}$

Now comes the interesting part. If we interpret the first term of the RHS as a measure of kinetic energy, the remaining terms gives us the 'effective potential' energy of the system. That is,

$\displaystyle V_{\text{eff}}(p)=-\alpha (\mathbf{k}\cdot\mathbf{m}) \frac{p_0}{p}+\frac{\delta\alpha}{2}\frac{p_0^2}{p^2}$

But that is exactly in the same form as that of the effective potential of the radial equation of an inverse square central force (Central forces). Therefore, the problem of a sphere on a freely spinning turntable is a Kepler problem in disguise which helps us in utilizing many known results.

For example, it known that the orbital period of a kepler problem can be written in terms of the 'energy' and the coefficient of the '$1/r$' term in the effective potential. Using the same, we can see that the orbital period $T$ for our problem is given by

$$\displaystyle T=\frac{2\pi\text{ }\alpha\text{ }p_0\text{ }(\mathbf{k}\cdot\mathbf{m})}{\left\vert2E\right\vert^{3/2}}$$

UPDATE (4-Aug-2026):

From (13) of Wessecker's paper, we can easily infer that

$L=\alpha p_0 p + \mathbf{k}\cdot(\mathbf{p}\times\mathbf{c})$

where $L$ is the reduced angular momentum about the $z$-axis which is shown to be a constant and $\mathbf{c}$ is the constant of integration in (10).

Now we define a vectorized form of $L$ such that,

$\mathbf{L}=\alpha p_0 p \mathbf{k}+\mathbf{p}\times\mathbf{c}$

We now show that $\mathbf{L}$ is a constant of motion by showing its components perpendicular and parallel to $\mathbf{k}$ are constants.

It's easy to see that $\mathbf{L}\cdot\mathbf{k}=L$ is a constant. Now,

$\mathbf{L}_{\perp}=\mathbf{k}\times(\mathbf{L}\times\mathbf{k})$

Using the fact that $\mathbf{c}\cdot\mathbf{k}=0$, we have

$\mathbf{k}\times(\mathbf{p}\times\mathbf{c})=(\mathbf{p}\cdot\mathbf{k})\mathbf{c}$

which is clearly a constant. Thus we see that $\mathbf{L}_{\perp}$, and hence, $\mathbf{L}$ is a constant.

We now start with the velocity equation which is (10) of Wessecker's paper expressed in terms of $\mathbf{p}$.

$\displaystyle\mathbf{p}'=\alpha p_0 \text{ }\mathbf{k}\times\hat{\mathbf{p}}+\mathbf{c}=\frac{\alpha p_0}{p} \text{ }\mathbf{k}\times\mathbf{p}+\mathbf{c}$.

Crossing the velocity equation with $\mathbf{p}$, with simple manipulations its easy to see in fact that $\mathbf{L}=\mathbf{m}$.

Dotting the velocity equation with itself and simplifying, we have,

$\displaystyle\vert \mathbf{p}'\vert^2=\vert\mathbf{c}\vert^2-\alpha^2 p_0^2+\frac{2\alpha p_0 L}{p}-\alpha\delta\frac{p_0^2}{p^2}$

Comparing with our energy equation, we see that

$\displaystyle E=\frac{1}{2}(\vert\mathbf{c}\vert^2-\alpha^2 p_0^2)=\frac{1}{2}\vert \mathbf{p}'\vert^2-\frac{\alpha p_0 L}{p}+\frac{\alpha\delta p_0^2}{2p^2}$

Hope you enjoyed the discussion. See ya later.

Until then

Yours Aye

Me