1. The problem statement, all variables and given/known data A person grabs on to an already spinning merry-go-round. The person is initially at rest and has a mass of 29.5 kg. They grab and cling to a bar that is 1.70 m from the center of the merry-go-round, causing the angular velocity of the merry-go-round to abruptly drop from 45.0 rpm to 19.0 rpm. What is the moment of inertia of the merry-go-round with respect to its central axis? 2. Relevant equations Angular Momentum = I * ω Momentum is conserved, so L(initial) = L(final), and so, I(initial) * ω(initial) = I(final) * ω(final) Inertia = MR^2 3. The attempt at a solution Inertia of the system changes when the person grabs on. So I(final) = I(merry) + I(person) The inertia of the person is given by MR^2 = 29.5 kg * 1.7^2 = 85.255. Using the conservation of momentum, here is what I put together: L(final) = L(person, final) + L(mgr, final) = I(merry) * ω(final) + I(person) * ω(final) L(final) = I(merry) * 19.0rpm + (82.255) * (19.0) Solved for L to get L = 60.1094. If L is conserved then that should be the inertia of the merry-go-round. So I solved 60.1094 = I * 45.0rpm for I, which came out to be I = 1.33576. kgm^2. I am not sure where I could have gone wrong with this.