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2 months ago

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Physics

1 answer:

Keith_Richards [3.2K]2 months ago

5 0

a) Ver explicación

b) $K_2>K_1$

Justificación:

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.

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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