Particle in a cylinder with a spring, sign convention in potential energy (Lagrangian multipliers)

Question

I'm trying to get the force of constraint. The problem I have is when defying the sign of the potential energy using cylindrical coordinates $(\rho,\phi,z)$, what I have is: $$ V=mgy-\frac{1}{2}k\left(\rho^2+R\rho \sin\phi+\left(\frac{R}{2}\right)^2+z^2\right)= $$ $$ =mg\rho\sin{\phi}-\frac{1}{2}k\left(\rho^2+R\rho \sin\phi+\left(\frac{R}{2}\right)^2+z^2\right). $$ But the solution in theory is: $$ V=-mg\rho\sin{\phi}-\frac{1}{2}k\left(\rho^2+R\rho \sin\phi+\left(\frac{R}{2}\right)^2+z^2\right). $$ I don't get why $mg\sin{\phi}$ is negative, considering the axes in the frame of reference from the diagram below, shouldn't be positive if the $y$-axis has the same direction as the gravity.

Answer 1

Since the $y-$axis is pointing downwards, the height will increase when the $y$ coordinate decreases, so the correct gravitational potential energy is $V_g=-mgy$, as it is greater at higher height.

You can check it this way: The azimuthal angle $\phi$ is usually defined from the $x-$axis, so $V_g=-mgy=-mg\rho\sin\phi$ will be minimum when $\phi=\pi/2$ (the situation of the diagram) and maximum when $\phi=3\pi/2.$