Acceleration of ball rolling down incline Suppose you have some object (which can roll like a ball,cylinder,wheel,etc) rolling down an incline without slipping (moment of intertia $I=kmr^2$. I want to find the accleration of the ball as it rolls, and I calculate this in two different ways, and I want to know if these ways are equivalent.
First I consider torques about a pivot point at the point of contact between the ball and incline. The only torque is due to gravity so $\tau=I\alpha=Ia/r=mgr\sin(\theta)$ which simplifies to $a=g\sin(\theta)/k$.
Alternatively I consider torques about a pivot point at the center of the object. The only torque is due to friction so $\tau=fr=Ia/r$. I sense that this alternative method is flawed for some reason. If it is not, it can be used to calculate $f$, using the $a$ we derived via the first method, right?
 A: Rather, I think your first method is flawed. Because your $\alpha$ is always about the pivot you select. Since you select the contact point as pivot, then $\alpha$ should be about the contact point. So $\alpha$ is not $a/r$ of course.

Here is my approach for this problem.
First, I select the CM (center of mass) as the pivot. Let $f$ be the friction at the contact point. Every moment the total torque about the pivot is 
$$\tau= fR= I \alpha$$
Here is another matter. How to determine $I$ ? My opinion is that since the shell is completely filled with frictionless fluid, then the fluid will not rotate with the shell at all since there is literally no force to push it to. So we ought to count only the shell's part. (I'm not definitely sure here, though) And thus $I=\frac{2}{3}MR^2$. Hence
$$f=\frac{I\alpha}{R}=\frac{2}{3}M R\alpha=\frac{2}{3} Ma$$
Then, consider the famous theorem of kinetic energy (which doesn't require conservation
!). Let $l$ be the distance the shell (or the CM, to be exact) has moved from the point where it is released, $v$ be the speed of the CM  and $\omega$ be the angular speed. Note that the gravity and the friction are the only net forces acting on the shell together with the fluid contained, we have
$$(2Mg\sin\theta-f)l=\frac{1}{2}(2M)v^2+\frac{1}{2} I \omega^2 $$
Note that for a rotation without sliding, $v=R\omega$ always holds, thus the above equation simplifies to
$$(2Mg\sin\theta-f)l=\frac{4}{3}Mv^2$$
Differentiate both sides by $dt$
$$(2Mg\sin\theta -f)v=\frac{8}{3}Mv \frac{dv}{dt}=\frac{8}{3} Mva$$
Put $f=\frac{2}{3}Ma$ into it and note that $v$ cancels out, thus
$$a=\frac{3}{5}g\sin\theta$$
A: Crap I am stupid. This problem is not hard at all. 
Consider pivot point at the center of mass. By torques, $fR=I\alpha=I(a/R)$. By linear acceleration, $mgsin(\theta)-f=ma$. Now solve for $a$ and $f$. 
