Suppose in the figure attached View attachment Doc1.doc that the two hoops are rigidly fastened together. The system is released from the position shown and there is negligible friction between the large hoop and the table. How fast relative to the table is the center of the large hoop moving when the centers of the two hoops lie on a vertical line? If the radius of the small hoop is b and its mass is m and the radius of the large hoop =3b and its mass is 3m. My attempt: If x=0 is the present position of the hoops then their center of mass is x_cm = -b/2. I=I1+I2=3*m*(3b)^2 + m*b^2 = 28*m*b^2. Ug = Ktrans +Krot 4*m*g*b/2 = .5*4*m*v^2 + .5*I*v^2/(-b/2)^2 2*m*g*b = 2*m*v^2 + 2*28*b^2*v^2/b^2 = 2*m*v^2 + 56*m*v^2. Therefore 29*v^2=g*b. Therefore v = (g*b/29)^.5; Is this the correct answer, if not where did I go wrong? Thanks.