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The oxygen molecule, O2, has a total mass of 5.30×10-26 kg and a rotational inertia of 1.94×10-46 kg-m2 about an axis perpendicular to the center of the line joining the atoms. Suppose that such a molecule in a gas has a speed of 5.96×102 m/s and that its rotational kinetic energy is two-thirds (2/3) of its translational kinetic energy. What then is the molecule's angular speed aboutthe center of mass?

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Answer:

[tex]w= 8.0433*10^{12}rad/s[/tex]

Explanation:

We need to define the variables.

[tex]m= 5.30*10^{-26}Kg[/tex]

[tex]I= 1.94*10^{-46}Kgm^2[/tex]

[tex]v= 596m/s[/tex]

To use the energy conservative equation we need define [tex]K_r[/tex], that is,

[tex]k_r=\frac{2}{3} k_r[/tex]

So,

[tex]\frac{1}{2}Iw^2 = \frac{2}{3} \frac{1}{2} mv^2[/tex]

[tex]w^2 = \frac{2mv^2}{3*I}[/tex]

[tex]w= \sqrt{\frac{2(5.30*10^{-26})(596)^2}{(3(1.94*10^{-46}))}}[/tex]

[tex]w= 8.0433*10^{12}rad/s[/tex]

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