Moment of inertia, I is the resistance of an object to rotational motion. It is the rotational analogue of mass and is given by the following integral, I = [tex]\int[/tex]miri2 . A simple derivation was explained to me as follows. A moment, τ is the product of force, F and lever arm, r. τ = F*r Newton's second law gives force, F as the product of mass, m and linear acceleration, a. F = m*a This can be substituted into the first equation as follows. τ = F*r = (m*a)*r If a particle in a solid body is rotating in a circle (because of an applied moment, τ) then we know it's tangetial acceleration, a_tang is given as below. a_tang = r*α Now, substitute this value into the previous equation to give. τ = (m*a)*r = (m*(r*α))*r = m*r2*α Hence, it is seen that the moment, τ on the particle is proportional to the angular acceleration, α and the constant of proportionality is the term m*r2. For a whole object this constant of proportionality would be the sum of this term over all it's particles, that is, the integral [tex]\int[/tex]miri2 . This constant of proportionality determines the magnitude of the resistance to rotational motion and is termed the moment of inertia, I. Now, I don't understand the following. For a particle rotating in a circle in a rigid solid, it's acceleration, a is a vector which can be decomposed into 2 components. If we use polar co-ordinates then the acceleration can be decomposed into the radial acceleration, a_radial (which I previously called the 'centripetal acceleration') and the tangential acceleration, a_tang. It is well known that for a point rotating in a circle these have the following values; a_tang = r*α and a_radial = r*ω2 = v2/r . So why do we only substitute the tangetial acceleration, a_tang into the term τ = m*a*r ? Why is the radial acceleration, a_radial ignored? I understand that the radial acceleration does no work because there is no displacement in the radial direction. However, I do not understand how this answers my question? Furthermore, if you can answer that question, I'm sure you know the answer to this related problem. Image a cylinder lying on a rough surface that is horizontal. Since the surface is horizontal the normal force, N is equal to the weight of the cylinder, m*g. N = m*g The horizontal surface is rough with an associated coeffecient of friction, μ. Now, a moment, τ is applied to the cylinder and it begins to rotate and move towards the right in pure roll. As such, there is a fricton force, F_fric which acts towards the left. F_fric = μ*N Again, this frictional force does no work, because the point in contact with the ground is always instantanesouly at rest (the centre of the cylinder has a velocity to the right of r*ω and the circumfrence point vertically above the point in contact with the surface has a velocity to the right of 2*r*ω). Now, what balances the frictional force? I can not find any other horizontal force so why doesn't the cylinder slide to the left? Instead, we observe it to roll to the right. Have I misunderstood the concept of the 'standard model of friction'?