Referring to the attached figure; the inner hoop rolls without slipping inside the larger hoop. The large one moves freely on the frictionless table. The mass of the smaller hoop is m while the larger one has a mass of 3m. They start from rest in the position shown and the inner hoop rolls down to the bottom of the large hoop with negligible energy loss for the system as a whole. How fast, relative to the table, is the center of the large hoop moving when the small hoop has its center directly below the center of the large hoop? Is the large hoop rolling clockwise or counterclockwise? If b=radius of the small hoop and 3b is the radius of the large hoop, then the center of mass of the system is -b/2. The moment of inertia (I) of the small hoop I2= m*b^2, and I1=27*m*b^2 for the large one. The moment of inertia of the large hoop about the center of mass of the system is Il=27*m*b^2 + 3*m*(-b/2)^2 ......from the parallel axis theorem. Therefore Il=111*m*b^2/4. The moment of inertia of the small hoop about the center of mass of the system is Is=13*m*b^2/4....parallel axis theorem. The total moment about the center of mass is It = Is + Il = 124*m*b^2/4 = 31*m*b^2 m*g*2*b = .5*4*m*v^2 + .5*It*w^2 => 2*m*g*b = 2*m*v^2 + 2*It*v^2/b^2. Therefore m*g*b = m*v^2 + 31*m*v^2. Therefore g*b = 32*v^2 => v = (g*b/32)^.5. And v for the large hoop is: vl=6*(g*b/32)^.5 My approach to this question may be wrong as the correct answer is vl= 6*(g*b/199)^.5. Any help would be welcome.