Analyzing the motion of a ball rolling without slipping inside a hemispherical bowl Consider a solid ball of radius $r$ and mass $m$ rolling without slipping in a hemispherical bowl of radius $R$ (simple back and forth motion). Now, I assume the oscillations are small and so the small angle approximation holds. I wish to find the period of oscillation and I analyze the motion in two ways, first using conservation of energy and secondly using dynamics. However, I receive two inconsistent answers. One or both of the solutions must be wrong, but I cannot figure out which one and more importantly, I cannot figure out why. 
Method 1: We write the energy conservation equation for the ball
$mgh + \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = Constant$
from the center of mass, we take the height as $h = R-(R-r)cos\theta$ where $\theta$ is the angle from the vertical. Applying the no slip condition $v = r\omega$ and taking the moment of inertia for a solid sphere $I = \frac{2}{5}mr^2$ we can write the energy equation as
$mg(R-(R-r)cos\theta) + \frac{7}{10}mr^2\omega^2 = Constant$
Differentiating with respect to time:
$mg(R-r)sin\theta\cdot\omega + \frac{7}{5}mr^2\omega\cdot\alpha = 0$
taking the small angle approximation $sin\theta = \theta$ we get
$g(R-r)\theta + \frac{7}{5}r^2\alpha=0$
$-\frac{5g(R-r)}{7r^2}\theta = \alpha$
from which we can get $T = 2\pi\sqrt{\frac{7r^2}{5g(R-r)}}$
Method 2: The only torque acting on the ball at any point in its motion is the friction force $f$. So we can write
$\tau = I\alpha = fr$
again using the rolling condition $a = r\alpha$ and the moment of inertia for a solid sphere,
$\frac{2}{5}ma = f$
The net force acting on the system is the tangential component of gravity and the force of friction, so
$F = ma = mgsin\theta - f$
$\frac{7}{5}a = gsin\theta$
taking the small angle approximation and converting $a$ to $\alpha$ we get
$\alpha = \frac{5g}{7r}\theta$
and a corresponding period of $T = 2\pi\sqrt{\frac{7r}{5g}}$
Now the solutions are very different and I would appreciate it if someone would point out where I went wrong.
 A: The answer you got in first method was wrong.
The answer in second is nearly reached, we have to take $R-r$ instead of $r$ because you have to take $r$ from centre of the sphere. 
The correct answer was $2\pi\sqrt{7(R-r)/5g}$
A: Your first derivation, using energy, uses two different meanings for the same symbol $\omega$.  In one place, you interpret it as
$$\omega = \dot{\theta}$$
the time derivative of the angle of the line from the center of the ball to the center of the bowl with the vertical.
In another place, you interpret $\omega$ as the time derivative of the unnamed angle through which the ball itself has rotated.
These two angles are related to each other by the $r/(R-r)$ factor by which you are off.
A: A problem I see in your first solution is that $v$ is the velocity of the center of mass. You assume $v=r\omega$ but that is the velocity of the bottom of the ball. The center of mass is moving parallel to this in a circular motion, the radius of which is $R-r$, $v=((R-r)/R)r\omega$.
A: Both answers are incorrect. The correct answer taking into account both angles as mentioned by others, has R-r in the numerator. The limiting case, in which the ball gets really small, produces a period that cannot go to zero, and this consideration alone can be used to eliminate both of your answers. 
