Correct operation for coordinates of hoop particle system?

[FallenApple]

Homework Statement

All in the pic below. Part of the solution presented. Didn't present the whole thing as that would clutter the page.

I just want to know how they set up the x coordinate for the particle.

Homework Equations

This problem is just about using the lagragian. So my issue is with the first line in the solution. x=R*theta+R*sin(phi) where x is presumably the position of the particle relative to the lab frame. This is just using gallian relativity.

The Attempt at a Solution

Shouldn't it be x=R*theta--R*sin(phi) instead? We have to minus the x position of the particle relative to the hoop frame from the x position of the hoop itself since we know that they are in opposite directions. As the particle slides down from the right moving to the left, the CM of the hoop must move right.

 

[TSny]

Part of the setup of a problem is to define the positive direction for the kinematic variables. In the problem statement and in the solution, the variables $x, y, \theta$ and $\phi$ are not defined. In particular, there is no indication whether $\theta$ increases in the counterclockwise direction or the clockwise direction. Likewise, for $\phi$. Also, there is no indication if $x$ is positive to the right or to the left. These definitions will determine the signs in $x = \pm R\theta \pm R \sin \phi$. The fact that the CM of the hoop moves opposite to the particle's motion is not relevant here. This fact will follow from solving the equations of motion with the appropriate initial conditions.

 

[FallenApple]

Part of the setup of a problem is to define the positive direction for the kinematic variables. In the problem statement and in the solution, the variables $x, y, \theta$ and $\phi$ are not defined. In particular, there is no indication whether $\theta$ increases in the counterclockwise direction or the clockwise direction. Likewise, for $\phi$. Also, there is no indication if $x$ is positive to the right or to the left. These definitions will determine the signs in $x = \pm R\theta \pm R \sin \phi$. The fact that the CM of the hoop moves opposite to the particle's motion is not relevant here. This fact will follow from solving the equations of motion with the appropriate initial conditions.

Thanks, that makes sense. Everything is relative to the coordinate set up.