In a playground, there is a small merry-go-round of radius 1.20 m and mass 170 k

Question

In a playground, there is a small merry-go-round of radius 1.20 m and mass 170 kg. Its radius of gyration is 91.0 cm. (Radius of gyration k is defined by the expression I=Mk2.) A child of mass 44.0 kg runs at a speed of 4.00 m/s along a path that is tangent to the rim of the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round.
Calculate
(a) the rotational inertia of the merry-go-round about its axis of rotation (kg•m2)
(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round (kg•m2/s) and
(c) the angular speed of the merry-go-round and child after the child has jumped on. (rad/s)

Explanation / Answer

a) I = Mk2 = 170kg*(0.91m)^2 = 140.8kg*m^2 b) L = r x p = mvr = 44kg*4m/s*1.2m = 211.2kg*m^2/s c) L = Iw + mr^2*w, w = L/(I+mr^2) = 211.2/(140.8+44*1.2^2) = 1.03 rad/s.

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