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I am reading the solution to a problem that involves the figure

enter image description here

During the solution the author makes the statement that $ OAP=\frac{r_{0}\,\theta}{{r}} $, where $r_{0}$ is the radius of the big circle and and $r$ is the radius of the small circle. Why is the angle $ OAP=\frac{r_{0}\,\theta}{{r}} $?

gbd
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  • How is the point P determined? – Bernard Dec 25 '16 at 22:00
  • Depending on what is $;P;$ : is it true that arc of (the big) circle determined by $;\theta;$ equals the arc of the small circle determined by $;\angle OAP;$ ? – DonAntonio Dec 25 '16 at 22:04
  • The point P is the intersection of the small circle with some curve. How is that curve defined? – quasi Dec 25 '16 at 22:04
  • @Bernard, Just updated my question. – gbd Dec 25 '16 at 22:05
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    Ok, then it's easy -- follow DonAntonio's comment.

    Use the formula for arc-length of a sector in terms of the radius and the central angle.

     – quasi Dec 25 '16 at 22:15
  • @quasi, I am not sure what you mean? – gbd Dec 25 '16 at 22:17
  • Let H = (r0,0), and let Q be the variable point of contact of the two circles. The physical interpretation of "rolling" implies that the arc length of the sector HOQ from the big circle is the same as the arc length of the sector PAQ from the small circle. Hence you get the equation (r0)(theta) = (r)(angle OAP). Solve for angle OAP. Voila! – quasi Dec 25 '16 at 22:25

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That's quite simple: let $T$ be the point of contact of the two circles whent the centre of the small circle is in position $A$, and $\Omega$ the initial position of $P$.

Observe the lengths of arc $\Omega T$ on the bigcircle and arc $TP$ on the small circle are equal: $$\theta r=\widehat{TAP}\, r_0, \text{ whence}\quad\widehat{TAP}=\frac{\theta r}{r_0}.$$

Bernard
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