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Alex17521
2 months ago
13

Platform

Physics
1 answer:
Keith_Richards [3.2K]2 months ago
5 0

a) Ver explicación

b) K_2>K_1

Justificación:

a)

La velocidad angular de un objeto en rotación se refiere a la variación en su desplazamiento angular:

\omega=\frac{\Delta \theta}{\Delta t}

donde

\Delta \theta representa el desplazamiento angular

\Delta t indica el tiempo transcurrido

El momento angular de un objeto en rotación se define como

L=I\omega

donde I es el momento de inercia del objeto.

A medida que se avanza de la figura 1 a la figura 2, el momento de inercia del atleta disminuye al acercar sus extremidades hacia su cuerpo. Dado que el atleta representa un sistema aislado, el momento angular L debe mantenerse constante, por lo tanto, a medida que I desciende, \omega (velocidad angular) tiene que ascender.

En contraste, al pasar de la figura 2 a la figura 3, el momento de inercia del atleta incrementa nuevamente y, por lo tanto, dado que L debe mantenerse constante, la velocidad angular disminuirá.

b)

La energía cinética rotacional de un objeto en movimiento rotacional se establece como

K=\frac{1}{2}I\omega^2

donde

I es el momento de inercia

\omega representa la velocidad angular

Utilizando

L=I\omega

podemos reescribir la energía cinética rotacional como:

K=\frac{1}{2}L\omega

En la parte a), afirmamos que el momento angular L permanece invariable; sin embargo, la velocidad angular \omega aumenta mientras pasamos de la figura 1 a la figura 2. Dado que la energía cinética rotacional es proporcional tanto al momento angular como a la velocidad angular, y considerando que el momento angular se mantiene constante, significa que la energía cinética rotacional también aumenta al transitar de la figura 1 a la figura 2.

Por lo tanto, la respuesta es

K_2>K_1

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