Question

ltem 11 A 43-kg child stands on the edge of a 450-kg playground merry-go-round that is turning at the rate of 1 rev every 2.2 s. He then walks to the center of the platform. Part A If the radius of the platform is 1.5 m, what is its rotational speed once he arrives at the center? Assume that the platform is a uniform disk, and the child is "point-like", that is, his dimensions are so small compared to the platform size, that he can be considered to be a point-mass. are cnisdered to be a point-mass.his Hints wf = Submit My Answers Give Up Continue

Answer 1

Mass of the merry-go-round = M = 450 kg

Radius of the merry-go-round = R = 1.5 m

Moment of inertia of the merry-go-round = I

I = MR²/2

I = (450)(1.5)²/2

I = 506.25 kg.m²

Mass of the child = m = 43 kg

The platform is rotating at a rate of 1 revolution every 2.2 sec.

Initial angular speed of the merry-go-round = $\omega$₁ = 1/2.2 rev/s = 0.4545 rev/s = 0.4545 x (2$\pi$) rad/s = 2.856 rad/s

Final angular speed of the merry-go-round = $\omega$₂

Initially the child is at the edge of the platform therefore the child is at a distance equal to the radius of the platform from the center of the platform.

By conservation of angular momentum,

(I + mR²)$\omega$₁ = I$\omega$₂

(506.25 + (43)(1.5)²)(2.856) = (506.25)$\omega$₂

$\omega$₂ = 3.4 rad/s

A) Rotational speed of the platform after the child arrives at the center of the platform = 3.4 rad/s