In a playground there is a small merry-go-round of radius 1.25 m and mass 175 kg. Assume the merry-go-round to be a uniform disk. A child of mass 45 kg runs at a speed of 3.0 m/s tangent to the rim of the merry-go-round initially at rest) and jumps on. If we neglect friction, what is the angular speed of the merry-go-round after the child has jumped on and is standing at its outer rim?
A) 0.82 rad/s
B) 2.4 rad/s
C) 0.49 rad/s
D) 1.2 rad/s
E) 0.41 rad/s

Question

Quizplus · Accepted Answer

Explanation:The initial angular momentum of the system is zero. After the child jumps on, the angular momentum of the system is:$$L_i = 0$$$$L_f = I_i \omega_i + I_c \omega_c$$$$0 = (1/2)MR^2(0) + (1/2)mr^2\omega_c$$$$\omega_c = \frac{2Mv}{mr}$$$$\omega_c = \frac{2(175 	ext{ kg})(3.0 	ext{ m/s})}{(45 	ext{ kg})(1.25 	ext{ m})}$$$$\omega_c = 0.82 	ext{ rad/s}$$

In a Playground There Is a Small Merry-Go-Round of Radius