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A uniform circular disk of moment of inertia 8.0 kg.m² is rotating at 4.0 rad/s. A small lump of mass 1.0 kg is dropped on the disk and sticks to it at a distance of 1.0 m from the axis of rotation.
What is the new rotational speed of the combined system?

Respuesta :

Answer:

[tex]\omega'=32\ rad.s^{-1}[/tex]

Explanation:

Given:

  • moment of inertial of a uniform circular disk, [tex]I=8\ kg.m^2[/tex]
  • angular speed of rotation, [tex]\omega=4\ rad.s^{-1}[/tex]
  • Mass of lump, [tex]m'=1\ kg[/tex]
  • position of the lump from the center of the disk, [tex]r'=1\ m[/tex]

Using the conservation of the angular momentum:

[tex]I.\omega=I'.\omega'[/tex]

[tex]8\times 4=m'.r'^2\times \omega'[/tex]

[tex]32=1\times 1\times\omega'[/tex]

[tex]\omega'=32\ rad.s^{-1}[/tex]

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