In a playground, there is a small merry-go-round of radius 1.20 m and mass 190 k

Question

In a playground, there is a small merry-go-round of radius 1.20 m and mass 190 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
(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and
(c) the angular speed of the merry-go-round and child after the child has jumped on.

Explanation / Answer

values are different but method is same The radius of gyration (R) is defined from the moment of inertia (I) and mass (M) as: R = sqrt(I/M) Part a asks for the mass moment of inertia, alsp called the rotational inertia here. From above, this is: I = M R^2 = 397.1 kg-m^2 Angular momentum is defined as: H = m * (r x V) Where x is the cross product of the vectors. Since you have a tangential velocity, the scalars will give the correct answer: H = m r v = 39 kg * 2.5 m * 3.4 m/s = 331.5 kg m^2 /s This momentum is conserved when the kid jumps on. The angular momentum (H) relates to the moment of inertia (I) and angular velocity (w) by: H = I w The moment of inertia must include the merry go round and the kid so now: I = 397.1 kg-m^2 + 39 kg * (2.5 m)^2 = 640.85 kg-m^2 The final angular velocity then is: w = H/I = (331.5 kg m^2 /s)/(640.85 kg-m^2) = 0.517 rad/sec

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