1. The problem statement, all variables and given/known data So rigid body (ring, cylinder, sphere), who's radius of gyration is known, rolls down an inclined plane without slipping. Find velocity as a function of height. m - mass of the body. y0 - initial height, g - gravitational acceleration, v - speed. I - moment of inertia, K - radius of gyration. T - torque, a - linear acceleration, alfa - angular acceleration, w - angular speed. theta - inclined plane angle with horizontal. 2. Relevant equations conservation of mechanical energy: 1/2mv2+1/2Iw2+mgy = mgy0 I = mK2 v = wR -> w = v/R T = I*alfa 3. The attempt at a solution using conservation of mechanical energy i get: 1/2mv2+1/2mK2v2/R2+mgy = mgy0 from this velocity is: v2 = 2g(y0-y)/(1+(K2/R2)) which is correct answer as my textbook says. However I get in trouble trying do this problem in different way - using forces. As i think, if body rolls without slipping that means that force of friction is equal to the m*g*sin(theta), only friction makes this body roll. All torques about center of mass except that of friction equal to zero, and I get: T = R*m*g*sin(theta) I*alfa = R*m*g*sin(theta) K2alfa = R*g*sin(theta) and if i use relation a = alfa*R, and multiply both sides of last equation by R, I get: a = R2/K2*g*sin(theta) to get velocity as a function of height, i rewrite a = dv/dt = dx/dt*(dv/dx) = v(dv/dx): vdv = R2/K2*g*sin(theta)dx if initial speed is zero and initial coordinate along the plane also zero, i get: v2/2 = R2/K2*g*sin(theta)*x and sin(theta)*x is just y0 - y so my final solution is bit different from true one: v2 = 2g(y0-y)*R2/K2 So, where did I mistake?