a merry-go-round of radius 2 m and a moment of inertia 651 kgmm is stationary. a 78 kg woman runs at 1.6 m/s and jumps on tangentially to the rim of the merry-go-round. what is the final angular velocity of the merry-go-round??

Question

0.383 rad/s? 17.672 rad/s? 0.554 rad/s? 0.259 rad/s? 0.309 rad/s?

Answer 1

final angular momentum about axis of rotation = initial angular momentum

initial ang mom = 78*1.6 * 2
= 250 kg m^2/s

final ang mom = 78 (2)^2 omega + 651 omega

omega ( 963) = 250
omega = .26 radians/second