Angular momentum of particle rolling around inside of sphere I have a hemispherical bowl in which I roll a small particle around the edge, starting from the top at point A with a velocity $v_o$. It travels halfway around the sphere and reaches point B, which is a vertical distance h below A, with a velocity $v_f$. Point A is a radial distance of $r_o$ from the vertical centerline and point B is a radial distance of $r$ from the vertical centerline. There is no friction. The goal is to solve for the angle, $\theta$, between the horizontal and the velocity $v_f$.
Here is a diagram of the problem scenario:

My solution relies on the assumption that angular momentum only relies on the velocities in the plane perpendicular to the vertical centerline. Is that a safe assumption? Also, when dealing with energies, is rotational KE and linear KE the same? Should I be taking RKE into account?

$$
L_o=L_f
$$
$$
mr_ov_o=mrv_f\cos \theta
$$
$$
\theta = \arccos(\dfrac {mr_ov_o}{mrv_f}) = \arccos(\dfrac {r_ov_o}{rv_f})
$$

$$
KE_o + PE_o = KE_f
$$
$$
\frac 12 mv_o^2 + mgh = \frac 12 mv_f^2
$$
$$
v_o^2 + 2gh = v_f^2
$$
$$
\sqrt {v_o^2 + 2gh} = v_f
$$

$$
\theta = \arccos(\dfrac {r_ov_o}{rv_f}) = \arccos(\dfrac {r_ov_o}{r\sqrt {v_o^2 + 2gh}})
$$
 A: Your solution looks fine to me.
Yes: the angular momentum is preserved in the horizontal plane (the weight is vertical and the reaction of the sphere surface is a central force) so your first relation is fine, just remember that $\theta$ is not the vertical angle, but lies on the plane tangent to the sphere at point B.
There are two kinds of rotational energy: the one of the particle spinning on itself which would require the particle's mass distribution to be computed, which is not given so I assume it should be neglected. What may be confusing is the particle's rotational energy around the centre of the sphere:
$$E=\frac{1}{2}I\omega^2$$
For a point like mass the moment of inertia is:
$$I = mr^2$$
Remembering the relation between angular and tangential velocity:
$$\omega= \frac{v}{r}$$
we get exactely:
$$E=\frac{1}{2}mv^2$$
In fact for a point like mass it is exactly the same to consider the kinetic energy related to the tangent velocity, or the rotational kinetic energy related to the angular velocity, for this problem I approve your choice for the first one.
A: I do not have a solution, just some steps to get there.
I have parametrized the problem with spherical coordinates, $\varphi$ is the azimuthal angle (around the hoop), $\psi$ is the nutation angle (drop from horizontal plane) for a position vector
$$ \vec{r} = \begin{pmatrix} r \cos \varphi \cos \psi \\ -r \sin \psi \\ -r \sin \varphi \cos \psi \end{pmatrix} $$
The derivate gives us velocity
$$ \vec{v} = \begin{pmatrix}
-r \dot\psi \cos\varphi \sin\psi - r \dot\varphi \sin\varphi \cos\psi \\
-r \dot\psi \cos \psi \\
r \dot\psi \sin\varphi \sin\psi - r \dot\varphi \cos\varphi \cos\psi
\end{pmatrix} $$
The energies are
$$ \begin{aligned} PE & = -m g r \sin \psi \\
KE & = \frac{1}{2} m r^2 \left( {\dot\varphi}^2 \cos^2 \psi + {\dot\psi}^2 \right) 
\end{aligned} $$
At point A we know that $\varphi_0=0$, $\psi_0=0$, $\dot\varphi_0 =\frac{ v_0}{r_0}$ and $\dot\psi=0$.
At point B we know that $\varphi_1=\pi$, $\psi_1=\sin^{-1} \left(\frac{h}{r}\right)$, and that $r=\sqrt{r_0^2-h^2}$ in order for the object to remain on the surface.
Now the angle $\theta$ is kinda tricky to derive and my best guess is
$$ \begin{aligned} \tan \theta & = \frac{-\vec{v}_y}{\sqrt{\vec{v}_x^2+\vec{v}_z^2}} \\
 & = \frac{\dot\varphi \cos\psi}{\sqrt{\left({\dot\varphi}^2-{\dot\psi}^2\right)\cos^2\psi + {\dot\psi}^2}} \end{aligned} $$
Now when you have $PE_0 + KE_0 = PE_1 + KE_1$ there are to variables that you need for $\theta$ ($\dot\varphi$ and $\dot\psi$) and only 1 equation. The 2nd equation must come from the angular momentum conservation, so I think you are on the right path.
A: First, you are right: in this problem angular momentum around the vertical axis is conserved. This is because all forces acting on the particle have no azimuthal component.
Note, that even if we assume that the particle is rolling on the bowl surface without slippage, the angular momentum of its own rotation would be of negligible compared with the angular momentum of its movement around the bowl axis.
Second, for neglecting the rotational kinetic energy, there are ambiguities in the formulation of the problem. You wrote: "I roll a small particle". To my mind its meaning is  that the particle would be rolling in a bowl, without slipping. The phrase "There is no friction" thus should mean there is no rolling friction, that is, mechanical energy is conserved. However, one may also prioritize the phrase "There is no friction" and assume that the "I roll" part to be imprecise way of saying that the particle will be sliding, without rotating. The first interpretation means that we have to make an assumption about the structure of a particle (or at least its tensor of inertia). Assuming that the particle is a solid uniform ball, and that its radius is small compared with the radius of curvature of the bowl $r_0$, the rotational energy due to the no-slipping rolling will be proportional to linear kinetic energy: $RKE = \frac 25 \frac {m v^2}{2}$. Then, your energy balance equation would be written:
$$
 KE_0+RKE_0 + PE_0 =  KE_f+RKE_f.
$$
Substituting the expression for rotational energy:
$$
\frac 75 \frac{m v_0^2}{2} + m g h = \frac 75 \frac {m v_f^2} 2
$$
Note, that this equation could be made looking like your original energy conservation equation (without rolling) if we redefine parameters $m$ and $g$: 
$$
m' = \frac75 m, \qquad g' = \frac57 g, \qquad m g = m'g',
$$
will gives us energy equation same as without rotation. Since mass can be eliminated from all equations, and energy equation is the only one containing $g$, such 'renormalization' is consistent with the rest of the equations.
Third point, is that you made a  mistake. To understand that, take a limit $v_0 = 0$, $h\ne 0$, which correspond to particle sliding down the bowl without any angular momentum. Your answer will always give $\theta=\pi/2$, while the correct answer should be $\theta = \arcsin (h/r_0)$.
This mistake appeared in writing angular momentum conservation equation. You wrote:
$$
mr_ov_o=mrv_f\cos \theta.
$$
However, here $\theta$ is not the angle between the velocity and horizontal plane. The velocity of the particle has two components in the spherical coordinate system:
$$
\mathbf{v}=\mathbf{e}_{\psi}\, v_\psi + \mathbf{e}_{\phi}\, v_\phi
$$
where $\mathbf{e}_{\psi,\phi}$ are unit vectors  in the directions of increasing nutation  and azimuthal angles. $v_\phi$ enters the (correct) equation for angular momentum conservation:
$$
m v_0 r_0  = m (v_f)_\phi r,
$$
however, because the $\mathbf{e}_\psi$ vector is not vertical,  $(v_f)_\phi \ne v_f \cos \theta $. Instead, since $\mathbf{e}_\psi$ forms an angle $\psi$ with vertical axis:
$$
 \sin \theta = \frac{(v_f)_\psi\, \cos \psi }{v_f} = \frac{(v_f)_\psi\, r }{v_f\, r_0},
$$
where we used $\cos \psi = r / r_0$. The unknown component $(v_f)_\psi$ could be easily found, since from the momentum conservation equation we know $(v_f)_\phi$, and from energy conservation (either your original or my for rolling ball) we can find $v_f$.
So after substitution we get
$$
\sin \theta = \pm \sqrt{\frac{r^2}{r_0^2}-\frac{v_0^2}{v_f^2}},
$$
where $v_f$ is found using the energy equation: $v_f^2 = v_0^2 + 2 g \sqrt{r_0^2-r^2}$, (substitute $g$ with $5g/7$ to account for no-slip rolling). The $\pm$ sign reflects the fact that the velocity could have either positive or negative vertical component.
