This is the problem I solved sometime ago. Same problem as you show. The angle $\theta$ used is measured from x-axis, positive anti-clockwise, as it was simpler to do so, but it does not affect the solution ofcourse. Since there is no Mathematica stuff in this, I can add a Manipulate later on if needed to show the ball falling of?
There are two coordinates $r,\theta$ (polar) which is the position vector of the ball, and one constraint
\begin{equation}
f\left( r,\theta\right) =r-R=0 \tag{1}
\end{equation}
$R$ above is the radius of the hemisphere.
Now we set up the equations of motion for $m$
\begin{align*}
T & =\frac{1}{2}m\left( \dot{r}^{2}+r^{2}\dot{\theta}^{2}\right) \\
U & =mgr\sin\theta\\
L & =T-U\\
& =\frac{1}{2}m\left( \dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)
-mgr\sin\theta
\end{align*}
Hence the Euler-Lagrangian equations are (we need to add contraint)
\begin{align}
\frac{d}{dt}\frac{\partial L}{\partial\dot{r}}-\frac{\partial L}{\partial
r}+\lambda\frac{\partial f}{\partial r} & =0\tag{2}\\
\frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}}-\frac{\partial L}
{\partial\theta}+\lambda\frac{\partial f}{\partial\theta} & =0 \tag{3}
\end{align}
But
\begin{align*}
\frac{d}{dt}\frac{\partial L}{\partial\dot{r}} & =m\ddot{r}\\
\frac{\partial L}{\partial\dot{\theta}} & =mr^{2}\dot{\theta}\\
\frac{d}{dt}\left( \frac{\partial L}{\partial\dot{\theta}}\right) &
=m\left( 2r\dot{r}\dot{\theta}+r^{2}\ddot{\theta}\right) \\
\frac{\partial L}{\partial r} & =mr\dot{\theta}^{2}-mg\sin\theta\\
\frac{\partial L}{\partial\theta} & =-mgr\cos\theta\\
\frac{\partial f}{\partial r} & =1\\
\frac{\partial f}{\partial\theta} & =0
\end{align*}
Hence (2) becomes
\begin{equation}
m\ddot{r}-mr\dot{\theta}^{2}+mg\sin\theta+\lambda=0 \tag{4}
\end{equation}
And (3) becomes
\begin{align}
m\left( 2r\dot{r}\dot{\theta}+r^{2}\ddot{\theta}\right) +mgr\cos\theta &
=0\nonumber\\
r\ddot{\theta}+2\dot{r}\dot{\theta}+g\cos\theta & =0 \tag{5}
\end{align}
We now need to solve (1,4,5) for $\lambda$. Now we have to apply the constrain
that $r=R$ in the above to be able to solve (4,5) equations. Therefore, (4,5)
becomes
\begin{align}
-mR\dot{\theta}^{2}+mg\cos\theta+\lambda & =0\tag{4A}\\
R\ddot{\theta}+g\cos\theta & =0 \tag{5A}
\end{align}
Where (4A,5A) were obtained from (4,5) by replacing $r=R$ and
$\dot{r}=0$ and $\ddot{r}=0$ since we are using that $r=R$ which is constant
(the radius).
From (5A) we see that this can be integrated giving
\begin{equation}
R\dot{\theta}^{2}+2g\sin\theta+c=0 \tag{6}
\end{equation}
Where $c$ is constant. Since if we differentiate the above with time, we
obtain
\begin{align*}
2R\dot{\theta}\ddot{\theta}+2g\dot{\theta}\cos\theta & =0\\
R\ddot{\theta}+g\cos\theta & =0
\end{align*}
Which is the same as (5A). Therefore from (6) we find $\dot{\theta}^{2}$ to
use in (4A). Hence from (6)
$$
\dot{\theta}^{2}=-2\frac{g}{R}\sin\theta+c
$$
To find $c$ we use initial conditions. At $t=0$, $\theta=90^{0}$ and
$\dot{\theta}\left( 0\right) =0$ hence
$$
c=2\frac{g}{R}
$$
Therefore
\begin{align*}
\dot{\theta}^{2} & =-2\frac{g}{R}\sin\theta+2\frac{g}{R}\\
& =2\frac{g}{R}\left( 1-\sin\theta\right)
\end{align*}
Plugging the above into (4A) in order to find $\lambda$ gives
\begin{align*}
-mR\left( 2\frac{g}{R}\left( 1-\sin\theta\right) \right) +mg\sin
\theta+\lambda & =0\\
\lambda & =m\left( 2g\left( 1-\sin\theta\right) \right) -mg\sin\theta\\
\lambda & =2mg-2mg\sin\theta-mg\sin\theta\\
& =mg\left( 2-3\sin\theta\right)
\end{align*}
Now that we found $\lambda\,,$we can find the constraint force in the radial
direction
\begin{align*}
N & =\lambda\frac{\partial f}{\partial r}\\
& =mg\left( 2-3\sin\theta\right)
\end{align*}
The particle will leave when $N=0$ which will happen when
\begin{align*}
2-3\sin\theta & =0\\
\theta & =\sin^{-1}\left( \frac{2}{3}\right) \\
& =41.8^{0}
\end{align*}
Therefore, the angle from the vertical is
$$
90-41.8=48.2^{0}
$$
r*θ''[t] == g*Sin[θ[t]], the solution will be $\theta(t)=0$, so the output of the first code sample is correct. $\endgroup$θ[t0]==0, why doesn'tL = 1/2 m (R θ'[t0])^2+m*g*R (1 - Cos[θ[t]]) - m*g*R*Cos[θ[t]]work?", as someone never learned about Lagrangian mechanics, I don't know the answer, either. Once again, I believe this should be asked in physics.SE. $\endgroup$