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A playground merry-go-round with a radius of 2.0 m and a rotational inertia of 100 kg m2 is rotating at 3.0 rad/s. A child with a mass of 22 kg jumps onto the edge of the merry-go-round, traveling radially inward. What is the new angular speed of the merry-go-round

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okpalawalter8

Answer:

The new angular speed of the merry-go-round = [tex]1.6rad/sec[/tex]

Explanation:

From angular momentum conservation

[tex]Iw_1 = (I + mr^2)w_2\\\\3*100 = (100 + 22*2^2)w_2\\\\300 = (100 + 88)w_2\\\\w_2 = \frac{300}{188}\\\\w_2 = 1.6rad/sec[/tex]

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https://brainly.com/subject/physics

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