1. The problem statement, all variables and given/known data You are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be 0.600 m. Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an 8.20-kg block of wood from the free end of the rope. You release the system from rest and find that the block descends 12.0 m in 4.00 s. 2. Relevant equations K1 + U1 = K2 + U2 K = 1/2*m*v2 K = 1/2*I*omega2 v2 = v02+2a(x-x0) v = r*omega 3. The attempt at a solution Variables: d = 0.600 m r = 0.300 m m = 8.20 kg v0 = 0 m/s omega0 = 0 rad/s change in x = 12.0 m t = 4.00s The initial kinetic energy is zero and the final gravitational PE is 0, so we use the equation U1 = K2, which is: mgh = 1/2*m*v2 + 1/2*I*omega2 mgh - 1/2*m*v2 = 1/2*I*omega2 (mgh - 1/2*m*v2)/(1/2*omega2) = I To solve, we need the linear velocity. I used a kinematic equation to calculate the velocity. v2 = v02+2a(x-x0) v2 = 0 + 2(9.81 m/s2)(12.0 m) v2 = 235.44 m2/s2 v = 15.34 m/s Then, v = r*omega omega = v/r omega = (15.34 m/s)/(0.300 m) omega = 51.15 rad/s Now back to the energy equation. I = (mgh - 1/2*m*v2)/(1/2*omega2) I = ((8.20kg)(9.81 m/s2)(12.0 m) - 1/2*(8.20 kg)*(15.34 m/s)2)/(1/2*(51.15 rad/s)2) I = 5.35*10-6 But my homework site says this is wrong.