hey. Moment of Inertia has a square factor on the formula so check your numbers now.
I already have. Have you? I didn't think so.
It is true that rotational inertia of mass at the periphery of wheel is proportional to the square of the radius: I = m x R^2. But you can't compare translational inertia to rotational inertia, but it uses different units. You have to calculate the equivalent inertia, knowing how the translational speed is related to the rotational speed. You can do this with the kinetic energy equation.
The total energy of motion of a spinning wheel is its translational kinetic energy (Kt = 1/2 x m x V^2) plus its rotational kinetic energy (Kr = 1/2 x I x w^2)
Thus, total kinetic energy:
K = Kt + Kr = [1/2 x m x V^2] + [1/2 x I x w^2] = [1/2 x m x V^2] + [1/2 x (m x R^2) x w^2] = 1/2 x m x [V^2 + R^2 x w^2]
But if the wheel rotates without slipping on the ground, the rotational speed w will be equal to the translation velocity V divided by the radius (w = V/R). Substituting this in give:
K = 1/2 x m x [V^2 + R^2 x (V/R)^2] = 1/2 m x [V^2 + R^2 x (V^2)/(R^2)] = 1/2 x m x [ V^2 + V^2] = 1/2 x m x [ 2 x V^2]
Rearranging, this gives:
K = 1/2 x (2 x m) x V^2
In other words, the equivalent inertia of the mass at the periphery of a rotating wheel is twice its mass.
Mass at a smaller radius has an even smaller affect on inertia.