In a playground, there is a small merry-go-round of radius 1.20 m and mass 220 k

Question

In a playground, there is a small merry-go-round of radius 1.20 m and mass 220 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 2.00 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 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)
From I = Mk²,
I = (200 kg)(0.910 m)²
I = 166 kg-m²

(b)
L = r x p
L = r x mv
L = (1.20 m)(44.0 kg)(2.00 m/s) (Radius and momentum are perpendicular for this case.)
L =105.6 kg-m²/s

(c)
L = I
105.6 kg-m²/s = [166 kg-m² + (44.0 kg)(1.20 m)²]() (Total moment of inertia comprises that of the merry-go-round and of the child)
= 0.460 rad/s

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