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A gymnast is in a tucked position to complete her somersaults. While tucked her moment of inertia about an axis through the center of her body is 16.0 kg-m² and she rotates at 2.5 rev/s. When she kicks out of her tuck into a straight position, her moment of inertia becomes 19.5 kg-m². What is her rate of rotation after she straightens out?

Respuesta :

Answer:

[tex]\omega_s=12.8886\ rad.s^{-1}[/tex]

Explanation:

Given that:

  • moment of inertia of tucked body, [tex]I_t=16\ kg.m^2[/tex]
  • rotational speed of the body, [tex]N_t=2.5\ rev.s^{-1}[/tex]
  • i.e. [tex]\omega_t=2\pi\times 2.5=15.708\ rad.s^{-1}[/tex]
  • moment of inertia of  the straightened body, [tex]I_s=19.5\ kg.m^2[/tex]

Now using the law of conservation of angular momentum:

angular momentum of tucked body=angular momentum of straight body

[tex]I_t.\omega_t=I_s.\omega_s[/tex]

[tex]16\times 15.708=19.5\times \omega_s[/tex]

[tex]\omega_s=12.8886\ rad.s^{-1}[/tex]

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