Hi, When a circular section object (cylinder, sphere of radius r) rolls down an inclined plane (at angle theta) it experiences both linear and angular acceleration due to gravity. I have read multiple times that: mg sin(theta) - friction = m a where friction = Torque/r From there on, everything is clear: moment of inertia * angular acceleration = friction * radius of ball = Torque a= angular acceleration * r and finally: a = (mg sin(theta)) / (m + moment of inertia * r^-2); friction = moment of inertia * a * r^-2 But, going back to the first equation there is a question that has always profoundly bothered me: mg sin(theta) - friction = m a Why does friction appears here? Isn't it a force applied at distance r in the contact point with the surface and directed down, parallel to it? I understand that friction is needed to avoid slipping so it should exist. Moreover, that the weight must be reduced by some force to represent the final rotational energy. I mean, if we had just mg sin(theta) = m a and presume rotation then we would have the same final kinetic energy as a pure traslation with no friction (slipping) but with an impossible extra rotational energy. Summarizing, I recognize that: At centre of mass: mg sin(theta) - Unknown force = m a At contact point: Torque = moment of inertia * angular acceleration = friction * radius of ball But how do we get Unknown force = friction? Thanks in advance!