Constraints on a hoop rolling on a cylinder

[LCSphysicist]

Homework Statement

A uniform hoop of mass m and radius r rolls without slipping on a fixed cylinder of radius
R. The only external force is that of gravity. If the smaller cylinder starts rolling from
rest on top of the bigger cylinder, use the method of Lagrange multipliers to find the
point at which the hoop falls off the cylinder.

Relevant Equations

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My question is about the contraint we need to use to solve this problem. The answer to the question use the following constraint:

$$(r+R)\theta = r\phi$$

Where $\theta$ is angle from the radius of the fixed cylinder to, say, the vertical axis. And $\phi$ is the angle that the rolling cylinder has rotated.

My question is, why the contraint is not $$R\theta = r\phi$$ ?

That is, the distance traveled by the point of the rolling cylinder in contact with the fixed cylinder, $R \theta$, equals the length traced by a point on its circunference when it has rotated $\phi$, that is, $r \phi$.

Seems to me that the contraint imposed by the answer says that the ditance traveled by the center of mass of the rolling cylinder is equal to the length it traces while rotate $\phi$, but i can't see why. Shouldn't the condition of "no slipping" be reference to the poins on its circunference?

 

[Steve4Physics]

I think your difficulty is seeing how $\phi$ changes for a given change in $\theta$.
It may help to imagine a complete ‘orbit’ of the hoop around the cylinder (without the hoop falling off).

In this case $\theta = 2\pi$ and the hoop’s centre moves a distance $d = 2 \pi (R+r)$.

If the hoop rolled the same distance $d$ on a flat surface, the hoop (circumference = $2\pi r$) would perform $n$ rotations were $n = \frac {2 \pi (R+r)} {2\pi r} = \frac {R+r}{r}$.

In this situation $\phi = 2\pi n = 2\pi \frac {R+r}{r}$

Look at the ratio $\theta : \phi$

Editted - minor changes.