In a playground, there is a small merry-go-round of radius 1.20 m and mass 130 k

Question

In a playground, there is a small merry-go-round of radius 1.20 m and mass 130 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 2.50 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 and
kg·m2/s
(c) the angular speed of the merry-go-round and child after the child has jumped on.
rad/s

Explanation / Answer

a) I = Mk^2 = 130kg * (91*10^-2m)^2 = 108 kg*m^2 b) L = r x p = r*mv = 1.2m * 44kg * 2.5m/s = 132 kg*m^2/s c) The total moment of inertia of the merry-go-round and child after jumping on is: I_t = I + mr^2 = 108kg*m^2 + 44kg*(1.2m)^2 = 171 kg*m^2 Since the angular momentum of the system stays the same before and after the child jumped on, we have L = I_t x omega. So the final angular velocity is: omega = L / I_t = 132kg*m^2/s / 171kg*m^2 = 0.77 rad/s

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