Galileo's kinematical paradox and the expanding circle of simultaneity

Consider a vertical circle and the diameter BA and representative chords BC, BD, which start from point B at the top of the circle, and also the chord EA and the line segment GF (see figure 1). If we simultaneously release two or more beads from B in such a way that one of the beads falls along diam...

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Bibliographic Details
Published in:Physics education 2013-11, Vol.48 (6), p.702-704, Article 702
Main Authors: Francisquini, M, Soares, V, Tort, A C
Format: Article
Language:English
Subjects:
Acceleration
Beads
Circumferences
Education
Horizontal
Paradoxes
Segments
Theorems
Triangles
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Summary:Consider a vertical circle and the diameter BA and representative chords BC, BD, which start from point B at the top of the circle, and also the chord EA and the line segment GF (see figure 1). If we simultaneously release two or more beads from B in such a way that one of the beads falls along diameter BA and the others slide along chords BC and BD or EA, then the beads will hit the circumference of the circle at the same instant, even though the chords have different lengths [1, 2]. This apparent paradox can easily be explained if we relate the length of diameter BA, let us call it A, to the length of a chord, say EA, let us call it [scriptl]. Then, by making use of Thales' theorem, which states that any triangle inscribed in a semicircle is a right triangle, we can write [scriptl]/h = |EA|/|BA| = sin [straighttheta], where [straighttheta] is the measure of the angle between chord EA and horizontal line GF. It also follows that the acceleration along the diameter and the chord are related in the same way a/g = sin [straighttheta].
ISSN:0031-9120
1361-6552
DOI:10.1088/0031-9120/48/6/F01