homework and exercises - Write a specific Lagrangian

I'm pretty new to Lagrangian mechanics, and after some thoery I'm facing the first exercises, as the following:

Consider a p.particle $P$ of mass $m$ which is constrained to a semi-circle of radius $R$ of equation $x^2+z^2=R^2$, where $z<0$, under the action of the gravity.

Write then the Lagrangian of the system using $x$ as Lagrangan coordinate.

My solution

Since the text asks for use $x$ as Lagrangian coordinate, I think to express the $z$-coordinate as a function of $x$, and from the equation of the (semi) circle, I get $z=-\sqrt{R^2-x^2}$

Then, I write the Kinetic energy $T(x,\dot x)=\frac{1}{2}m \dot z^2=\frac{1}{2}m \frac{\dot x^2 x^2}{R^2-x^2}$

while the potential energy $V(x)$ I assume to be simply $V(x)=-m g z(x)=+mg\sqrt{R^2-x^2}$

So, the Lagrangian would be $L(x,\dot x)=\frac{1}{2}m \frac{\dot x^2 x^2}{R^2-x^2}+mg\sqrt{R^2-x^2}$

Now, writing the Lagrange equations is just a matter of computations. I'd like to know if I moved correctly, or if I am completely wrong

Answer

You forgot a term in the kinetic energy. You should have

$$T=\frac{1}{2}mv^{2}=\frac{1}{2}m\left(\dot{x}^{2}+\dot{z}^{2}\right)$$

since $\boldsymbol{v}=\dot{x}\hat{x}+\dot{z}\hat{z}$.

As for you potential term, the gravitational energy increases with height

$$V=mgz$$

Finally, note that $\mathcal{L}=T-V$ (you used + instead).