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In the following question, why is the constraint equation used $x^2+(y-R)^2 = R^2$ rather than just $x^2+y^2 = R^2$?

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  • $\begingroup$ so that $(0,0)$ is a point on the trajectory $\endgroup$
    – user256872
    Feb 27 at 18:44
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The coordinate system seems to be defined so that the bottom of the hoop is at $y=0$, as you can see from the potential energy expression. So, the bead is not confined to be on the circle centered at 0 described by $x^2 + y^2 = R^2$, as you assume, but the circle centered on $x=0, y=R$, which is described by $x^2 + (y-R)^2 = R^2$.

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I think it is just due to where they have defined the origin of the coordinate system to be. In this case, at the bottom of the hoop, meaning that the potential energy $V=mgy$.

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