Physics, 06.04.2021 05:20 robert7248

A child of mass 51.9 kg sits on the edge of a merry-go-round with radius 2.4 m and moment of inertia 215.24 kg m2 . The merry go-round rotates with an angular velocity of 2.1 rad/s. The child then walks towards the center of the merry-go-round and stops at a distance 0.864 m from the center. Now what is the angular velocity of the merry-go-round

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Answer from: tasnimsas3

4.25 rad/s

Explanation:

Given that.

Mass, m = 51.9 kg

Radius, r1 = 2.4 m

Moment of inertia, I = 215.24 kgm^2

Angular velocity, ω = 2.1 rad/s

Radius, r2 = 0.864 m

To start with, we are going to use the Conservation of angular momentum to solve the question, which is

l(initial) = l(final)

[I₁ + I₂](initial)*ω(initial) = [I₁ + I₂](final)*ω(final)

Making ω(final) the subject of formula, we have

ω(final) = [I₁ + I₂](initial)*ω(initial) / [I₁ + I₂](final)

ω(final) = [215.24 + (51.9)(2.4)²](2.1) / [215.24 + (51.9)(0.864)²]

ω(final) = [215.24 + 298.944]2.1 / [215.24 + 38.74]

ω(final) = 514.184 * 2.1 / 253.98

ω(final) = 1079.786 / 253.98

ω(final) = 4.25 rad/s

= 5.273 rad/s

Answer from: Quest

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Answer from: Quest

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A child of mass 51.9 kg sits on the edge of a merry-go-round with radius 2.4 m and moment of inertia...

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