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what is the magnitude of its angular velocity, in radians per second, after a 17 kg child gets onto it by grabbing its outer edge? the child is initially at rest.

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Final answer:

The angular velocity of the merry-go-round after the child gets onto it is approximately 0.404 rev/s.

Step-by-step explanation:

The angular velocity of the merry-go-round after the child gets onto it can be found using the principle of conservation of angular momentum.

The initial angular momentum of the merry-go-round can be calculated using the formula L = I * ω, where L is the angular momentum, I is the moment of inertia, and ω is the initial angular velocity.

The final angular velocity can be found using the equation L = I' * ω', where I' is the new moment of inertia and ω' is the final angular velocity.

In this case, the initial angular momentum of the merry-go-round is the product of its moment of inertia and initial angular velocity, L_initial = I * ω = (120 kg * (1.80 m)^2) * (0.500 rev/s).

When the child gets onto the merry-go-round, the new moment of inertia is equal to the sum of the moment of inertia of the merry-go-round and the child, I' = I + M * R^2 = (120 kg * (1.80 m)^2) + (17 kg * (1.80 m)^2).

To find the final angular velocity, we equate the initial and final angular momentum equations and solve for ω', L_initial = L_final:

(120 kg * (1.80 m)^2) * (0.500 rev/s) = [(120 kg * (1.80 m)^2) + (17 kg * (1.80 m)^2)] * ω'.

Simplifying, we can divide both sides of the equation by [(120 kg * (1.80 m)^2) + (17 kg * (1.80 m)^2)]:

0.500 rev/s = ω' * [(120 kg * (1.80 m)^2) / [(120 kg * (1.80 m)^2) + (17 kg * (1.80 m)^2)]].

Calculating the value of ω', we find:

ω' ≈ 0.404 rev/s.

User Eric Falsken
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