Angular Acceleration of a uniform solid sphere in a hemispherical bowl Consider a ball of radius $r$ moving(surfaces are frictionless) inside a hemispherical bowl of radius $R$ and mass $m$. Now if I were to find the angular acceleration ($α$ of the ball I analyse the motion in two ways. I assume the angle is small so the small angle approximation holds. However, The two answers I receive are not consistent
Method 1:
$F=ma$ on Centre of mass in tangential direction
$mgθ=ma$
$gθ=a$
and then I write the acceleration of centre of mass ($a$) as $a=α(R-r)$
therefore angular acceleration of centre of mass $$α=\frac{gθ}{(R-r)}$$
Method 2:
$τ=Iα$ on the sphere  about the centre of the hemisphere $m g θ (R-r) = ( \frac{2}{5} m r^2 +  m (R-r)^2 ) α$
therefore
$$ α = \frac{ g θ (R-r)}{\frac{2}{5} r^2 +  (R-r)^2} $$
I know that the second method is a more fundamental method but where am I wrong in the first one?
 A: The angular acceleration denoted $\alpha$ in method 1) is second derivative of $\theta$. It describes how radius vector of the ball center rotates, but it is not influenced by ball's rotation in any way. 
It can be found using both methods, but only the first one is done correctly in OP.
In the method 2), OP is trying to use the torque equation with torque and inertia moment of the little ball rotating around a bowl center, but this would be legitimate only if the ball was rotating around that point as a rigid body. However, by assumption the little ball does not rotate at all.
Thus the proper moment of inertia to use in the torque equation is that of a mass point, $I=m(R-r)^2$. Then, we get
$$
\tau = I\alpha
$$
$$
mg\theta (R-r) = m(R-r)^2 \alpha
$$
which gives the same result as method 1:
$$
\alpha = \frac{g\theta}{R-r}.
$$
A: In the first method you forget the reaction of the bowl on the sphere which has a non-zero tangential component.
The two situations are not equivalent.
In the second method, you consider the sphere as a pendulum with a link axis without mass up to the axis of rotation. You calculate the moment of inertia around the axis by Huygens' theorem. But in this case the sphere must turn so that the point of contact with the bowl is always the same point of the sphere. It would be the link axis, rigid, that would cause this rotation. The rigid link axis would exert a tangential action on the sphere.
In the first method, as there is no friction, the sphere is in translation (not rectilign): it does not rotate. A diameter always keep the same direction. The point of contact with the bowl is never the same.
