http://www.aapt.org/physicsteam/2010/upload/2009_F-maSolutions.pdf 1. The problem statement, all variables and given/known data Two discs are mounted on thin, lightweight rods oriented through their centers and normal to the discs.These axles are constrained to be vertical at all times, and the discs can pivot frictionlessly on the rods. The discs have identical thickness and are made of the same material, but have diﬀering radii r1 and r2. The discs are given angular velocities of magnitudes ω1 and ω2, respectively, and brought into contact at their edges. After the discs interact via friction it is found that both discs come exactly to a halt. Which of the following must hold? Ignore eﬀects associated with the vertical rods. 2. Relevant equations Conservation of rotational energy Conservation of angular momentum (although I don't think I need this. Momentum is not conserved because there are forces acting). Torque = Iα = rF 3. The attempt at a solution Tried it two ways. Once with conservation of rotational energy, once with torque. The answer is ω1r13 = ωr23 1) If they slowed each other to a halt, their energies must've been = I1 ω12 /2 = I2 ω22 /2 So mass times radius squared times angular velocity squared is equal. Densities are equal, heights are equal, pi cancels out, so (using area of circle) r12*r12ω12 = r22*r22ω22 Which simplifies to one of the answer choices, C: ω1r12 = ω2r22 2) Here the reasoning is that if they halted each other, the torques must've been equal: T1 = T2 Torque equals moment of inertia times angular acceleration, which equals moment of inertia times angular velocity over time I1ω1 / t = I2ω2 / t The times are equal; they cancel out. Using the same technique as last time, I get M1 = (r12 / r22) M2 ((r12 / r22) M2)r12ω1 = M2r22ω22 Which turns into ω1r14 = ω2r24 Answer choice E. But the correct answer is D: ω1r13 = ω2r23 How? How did both methods fail? Gah! Thanks in advance.