A solid cylinder merry-go-round of mass 250kg and radius of 1.9m spinning at 1 revolution every 5 seconds has a 40kg child sitting at 1.1m from the axis. A 50kg child, running tangentially at 3m/s, jumps on the merry go round at the outer edge. What is the new rotation rate in rev/s?
The Attempt at a Solution
(1/2MR^2 +m1r^2)ω1 + (m2r^2)(v/r)=(1/2MR^2 +m1r^2 +m2r^2)ω2
I plugged in all the variables and solved for ω2 and got 0.214 rev/s. Did I set up the equation correctly and come about the right answer?