Question

(4) A playground ride, known as merry-go-round, consists of a disk with a mass of 40.0 kg and a radius of 1.80 m mounted on a frictionless axle. The frictionless axle is a vertical cylinder of radius kg, and has the same rotational axis as the disk. A child with a mass of 3.50 m/s on a line tangential to the outer rim of the disk and jumps onto the outer edge of the ride while it is at rest causing the ride to rotate. (15 points) (a) What is the moment of inertia of the ride, before the child jumps on? (5 points) (b) What is the angular speed of the ride after the child jumps onto it? Treat the child as a point particle. (5 points) (c) While the ride is spinning, the child walks from the outer edge to a distance 0.600 m from the center, what is the new angular speed of the ride? (5 points)

Answer 1

② - moment of inertia initially I = I disk & fcye = x 40x1.8² + + x 10 x 0.152 .: I = 64.9125 kg ml 6 After he jumps, angular momentum remains constant Li=bg → 35 x 3.5 X 1.8 = 664.9125 + 35x1.8%) af Hence, as = 1.2366 radys Again by the conservation of angular momentum. 35x3.5 x 1.8 = (64.9125 + 35*0.63) af Tag = 2.845 rad/s) ©