A cylinder rolling down an inclined plane A few questions popped into my mind while studying rotational motion.
Take a cylinder to the top of an inclined plane. Suppose there is friction. Let go of the cylinder. If it is rolling without slipping, is its acceleration constant over the time interval it is rolling down? If so, why? Why does the acceleration depend on the rotational inertia of the body in this case? And the final and most important question that had me struggling: why can't we simply apply $F = ma$ on these objects and get the same result on every object, regardless of their rotational inertia, since all the forces acting on the object in this system are proportional to the mass?
 A: Newton's second law is stated for point objects where acceleration has no ambiguity. When you are studying a system like the cylinder which is composed of many points, the problem is to choose which point are you going to choose to calculate the acceleration. The centre of mass theorem states that the point to consider if you are considering only the external forces is the centre of mass of the system.
In the case of your cylinder, you can certainly apply the centre of mass theorem. You have three forces acting on the cylinder: weight, normal reaction and friction. The normal reaction is only there to cancel out the normal component of weight. You therefore have a tangential acceleration of the center of mass of the cylinder (which is on its axis if you're assuming rotational symmetry). The subtlety is that the determination of the friction force is not evident as it originates from the non slip condition. This is the part that will have a non intuitive dependence on the moment of inertia.
This is why a moment method / energy is more appropriate as it easily handles the non-slip condition. It does not need to calculate this tricky friction force, but rather focuses directly on the relevant angular velocity. In fact, once you've solved the equations of motion using these methods, you can revert back to the center of mass theorem to figure out the friction force.
Hope this helps.
A: The acceleration of the center of mass (CM) is the net force divided by the mass; the net force is the component of gravity down the incline minus the force of friction up the incline.  You do just apply $\vec F = m \vec a_{CM}$ to determine the acceleration of the CM, $\vec a_{CM}$; however, you need to consider the rotational motion to evaluate the force of friction, which is not constant for rolling without slipping. Friction provides a torque that causes rotational motion, so the force of friction depends on the moment of inertia. The force of friction is not constant and equals $ma_{CM}/2$ for a cylinder of mass $m$. The acceleration of the CM is constant, equal to ${2 \over 3} g sin(\theta)$ for a cylinder, where $\theta$ is the angle of the incline.  For rolling without slipping the force of friction does no work and the potential energy at the top of the incline is converted to kinetic energy of the CM plus rotational energy around the CM.   This problem is evaluated in numerous physics textbooks, such as one of the many textbooks by Halliday and Resnick.
See Consistent Approach for Calculating Work By Friction for Rigid Body in Planar Motion for more details.
