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lukranit [14]
3 years ago
13

A merry-go-round can be modeled as a solid disk of radius R and mass M. Initially, it is rotating on a sturdy frictionless axle

with angular velocity ω0 . A rock of mass m is dropped vertically onto the merry-go-round and it sticks to the surface at a radius r from the center. What is the final angular velocity ω of the merry-goround + rock system?
Physics
1 answer:
Crazy boy [7]3 years ago
3 0

Answer:

w = w₀ M / (M + 2m)

Explanation:

This exercise can be solved using the concepts of conservation of angular momentum

          L = I w

Let's write in angular momentum in two points

Initial. Before impact

        L₀ = I w₀

Final. After the rock has stuck

        L_{f} = I w + (m r²) w

The system is formed by the disk and the rock, so that the forces and moments during the crash are internal and the angular momentum is preserved

        L₀ =  L_{f}

        I w₀ = (I + m r²) w

       w = w₀ I / (I + m r²)

The roundabout is a disk so its moment of inertia is

         I = ½ M r²

        w = w₀ ½ Mr² / (½ M r² + mr²)

        w = w₀ ½ M / (½ M + m)

        w = w₀ ½ M2 / (M + 2m)

        w = w₀ M / (M + 2m)

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