Oscillation of a rolling sphere in a bowl This is a homework task. I already came to a result but I am very unsure.
The task: In a bowl with the shape of a semi-circle ($R$ = 0.5m) a sphere (there is no specification for the size of the sphere) is rolling without friction. Calculate the period of the oscillation. Use approximation for a small amplitude.
Original problem description (German):
In einer Wanne mit halbkreisförmigem Querschnitt (Radius $R = 0.5m$) rollt ein kleines
Kügelchen reibungsfrei hin- und her.


Berechnen Sie die Schwingungsfrequenz dieser Bewegung für kleine Auslenkungen.

My solution: Using Newton 2 I make up the equality of torque: The first term is the inertia of the sphere according to the linear movement, the second term is the torque caused by gravity and the third term is the inertia according to the rotation of the sphere, $r$ is the radius of the sphere.
$$mR^2*\frac{d^2\phi}{dt^2} + mgR*\phi + \frac{d\phi}{dt^2}*\frac{R}{r}*\frac25mr^2 = 0$$
$$\frac{d^2\phi}{dt^2}*(mR^2+\frac25mr^2\frac{R}{r}) = - mgR*\phi$$
$$\frac{d^2\phi}{dt^2} = - \frac{mgR}{(mR^2+\frac25mr^2\frac{R}{r})}*\phi$$
$$\frac{d^2\phi}{dt^2} = - \frac{g}{R+\frac25r}*\phi$$
This is a standard differential equation for oscillation and solving it yields:
$$T = 2\pi\sqrt{\frac{R+\frac25r}{g}}$$
for the period $T$. It makes sense that $T$ is higher the bigger the radius of the sphere is but I can't calculate it with $r$ being unspecified.
Is there an error in my solution or is the task unsolvable?
 A: Here's a different approach, giving a different answer.
I will choose a coordinate system so that the angle $\phi=0$ is downward, and normalize the center of the bowl to have 0 gravitational potential. I will also assume $r\ll R$.
The gravitational potential energy of the bead is
$$
-mgR\cos\phi.
$$
The kinetic energy of a rolling (solid) ball of constant density is $\frac{7}{10}mv^2$ (where $v$ is the velocity of the center of the ball), so the kinetic energy of the marble is
$$
\frac{7}{10}m\left(R\frac{d\phi}{dt}\right)^2.
$$
Setting the derivative of the total energy equal to 0 gives
\begin{eqnarray*}
mgR\sin\phi\frac{d\phi}{dt}+\frac{7}{5}mR^2\frac{d\phi}{dt}\frac{d^2\phi}{dt^2}&=&0,\\
\sin\phi+\frac{7}{5}\frac{R}{g}\frac{d^2\phi}{dt^2}&=&0.
\end{eqnarray*}
Using the approximation $\sin\phi\sim\phi$ for $|\phi|$ small:
$$
\phi+\frac{7R}{5g}\frac{d^2\phi}{dt^2}=0,
$$
so the period of oscillation is
$$
T=\frac{2\pi\sqrt{7R}}{\sqrt{5g}}.
$$
I believe the reason we have arrived at two different answers is the following: the problem statement seems to imply that (i) the marble is rolling, and (ii) there is no friction. Statments (i) and (ii) are inconsistent, so we must abandon one of them. In my opinion, the question is more interesting if we we keep (i) and drop (ii). If we assume the marble is rolling instead of sliding, then there necessarily must be friction between the bowl and the marble. The friction causes a torque, which is not included in the OP's computation. If one calculates the frictional force (using the fact that the marble rolls) and includes it in the OP's computation, I believe this gives the same answer I gave.
A: A 1-D analysis will suffice. We have $U(x)\approx\frac{mg}{2R}x^2$, and thus we solve the relevant DEQ $mx''(t)=-U'(x(t))$ to obtain $$\tau=2\pi\sqrt{\frac{R}{g}}=1.418\mbox{ sec}.$$
Side note: The reasoning for the reduction to the 1-D case follows since the 2D potential is $$U(\mathbf{r})=\frac{1}{2}mg \left(\begin{array}{cc}
\frac{1}{R} & 0\\
0 & \frac{1}{R}
\end{array}\right):\mathbf{r}\otimes\mathbf{r}+O(\mathbf{r}^3)$$ which means the system is a 2D isotropic oscillator for small oscillation.
EDIT: This assumes the small ball is sliding around the bowl (due to the frictionless surface not being able to impart a torque on it as it moves), and not rolling around. See Julian Rosen's answer for the rolling interpretation.
