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 (see Problem 85 of Chapter 10) is 91.0 cm. A child of mass 44.0 kg runs at a speed of 3.00 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.

(a) Calculate the rotational inertia of the merry-go-round about its axis of rotation.
kg·m2

(b) Calculate the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round.
kg·m2/s

(c) Calculate the angular speed of the merry-go-round and child after the child has jumped on.
rad/s

Explanation / Answer

a) moment of inertia = m ( radius of gyration)^2

I = 220 x 0.91^2 =182.18 kg m^2

b)

for child I = mr^2 and w = (v/r)

angular momentum = Iw = mr^2 (v/r) = mvr

= 33 x 1.20 x 3 =118.8 kg m^2 /s

c) Using angular momentum (Iw) conservation,

initial = final angular momentum

0 = (182.18 w) - (33 x 1.20^2 x (3/1.2))

w = 0.65 rad/s

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