1. a)Two children, each with mass m = 21.5 kg, sit on opposite ends of a narrow board with length L = 5.6 m, width W = 0.15 m, and mass M = 5.9 kg. The board is pivoted at its center and is free to rotate in a horizontal circle without friction. What is the rotational inertia of the board plus the children about a vertical axis through the center of the board? b) What is the magnitude of the angular momentum of the system if it is rotating with an angular speed of 1.79 rad/s? c) The system is rotating, the children pull themselves toward the center of the board until they are half as far from the center as before. What is the resulting angular speed? d) What is the change in kinetic energy of the system as a result of the children changing their positions? The attempt at a solution I know the answers for the first two parts. Answer for part a is 352.5kg*m^2 and the answer for part b is 632 J*s. I found part A by 1/12m(a^2+b^2)+2(m(L/2)^2) and found part b by multiplying the inertia found with the angular velocity. However, I am stuck on part C, I realize it has to do with conservation of momentum and know that I needed to the changing inertia which i calculated to be 99.71 with the same equation as before except it became L/4 instead of L/2. so the equation i have for this is KE= .5I(final)(angular speed ^2) = 5I(initial)(angular speed(initial) ^2). I am stuck on how to find the angular veolcity.