Galileo's paradox

Isolated system

Galileo’s paradox, also referred to as Galileo’s theorem, asserts that given a vertical circle the time taken for a body to descend to the lowest point of the circle along any chord is the same . The use of Galileo’s paradox as experimental tool to aid lesson is described in Greenslade. The system was also discussed in detail taking into consideration the effect of the environment Aguiar. In both papers solution was obtained using kinematic approach. Yet the experiment can also be used to teach conservation of mechanical energy. To this end we discuss here the approach from energy’s perspective.To avoid ambiguity with other paradoxes attributed to Galileo, we follow the suggestion of the authors in Aguiar in renaming it as Galileo’s kinematical paradox.

To begin with, let us refer to the figure shown below where we assume the chords to be frictionless, a particle traveling to the bottom of the circle along the diameter \(D\), and another particle traveling along either of the chords will arrive at the same time if released, simultaneously, from rest . Generally speaking conservation of total mechanical energy dictates that the sum overall energy must add up to zero \(\begin{aligned} \Delta K+\Delta U =0 \label{eq:one}.\end{aligned}\) Employing the above equation, we arrive at \(v_B=\sqrt{2gD}\). Using which the time taken to travel along the diameter is shown to be \(t_{AB}=\sqrt{\frac{2D}{g}}\). Moreover the time taken for the particle to travel, from \(A\) to \(D\) is readily takes the form \(t_{AD}=\sqrt{\frac{2D}{g}}\). In a similar manner for the motion from \(D\) to \(B\) we note that \(v_B=\sqrt{2gL\sin\theta}\) consequently \(t_{DB}={\sqrt{\frac{2L}{g\sin\theta}}}=\sqrt{\frac{2D}{g}}\).

At this point it is worth recalling that Galileo has also pondered about the travel from top to bottom of the vertical circle through two chords, this means a particle following path \(ADB\) for instance, in such cases we split the total travel into two stages, i.e \(A\rightarrow D\) then \(D\rightarrow B\). Application of the first equation in this page, on both stages separately and collecting what we get yields \(t_{ADB} = \sqrt{\frac{2D}{g}\left(1+\sin^2\theta\right)}\). Therefore even if the distance covered is the same in traveling to the terminal point, either via the diameter (\(AB\)) or through two chords \(ADB\), the time it takes to traverse \(ABD\) is a bit higher.

System with friction

In the previous section we have dealt with an isolated system. But if we are interested in performing the experiment then friction is unavoidable and must be taken into account. To proceed we should make sure that our equation for conservation of total mechanical energy reflects this new information, i.e effect of friction. By this what mean is that, if our particle slides in a chord with friction, the conservation of total mechanical energy is modified to take the form \(\Delta K+\Delta U + \Delta E_{int}=0\). In doing so the speed at the bottom of the circle – for the path across the diameter becomes \(v_B^2=\sqrt{2gD\left(1-\mu_k\right)}\). Though the speed, as expected, is dependent on the coefficient of kinetic friction, what is interesting however is that the result is independent of \(\theta\).

To see if we see similar relation we follow the same footsteps and calculate the time taken to descend across each chords, if for instance the chord in consideration is from point \(A\) to \(D\), we obtain \(\begin{aligned} v_D^2 &= 2g\left[\frac{L\cos^2\theta-\mu_kD\sin^2\theta\cos\theta}{\sin\theta}\right]\\ t_{AD} &= \sqrt{\frac{2D/g}{1-\mu_k\tan\theta}} \end{aligned}\)

Again for the path along the chord \(D\rightarrow B\) the time taken to descend can be shown to be \(t_{DB} = \sqrt{\frac{2D/g}{1-\mu_k\cot\theta}}\). This implies that the inclusion of friction will break the paradox.