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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?

User Mouthpiec
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1 Answer

5 votes

Answer:

w = w₀ M / (M + 2m)

Step-by-step 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)

User Ziauz
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