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183_notes:examples:the_moment_of_inertia_of_a_diatomic_molecule [2014/10/31 14:34] – pwirving | 183_notes:examples:the_moment_of_inertia_of_a_diatomic_molecule [2014/11/05 20:40] (current) – pwirving | ||
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=== Facts === | === Facts === | ||
+ | |||
+ | Mass of nitrogen atom is 2.3 x $10^{-26}$kg | ||
+ | |||
+ | Average distance between nuclei is 1.5 x $10^{-10}$m | ||
=== Assumptions and Approximations === | === Assumptions and Approximations === | ||
+ | |||
+ | The distance between the atoms in the molecule does not change. | ||
+ | |||
+ | The model of the system you are using includes a spring between the atoms but these are not actual springs so the spring has no mass. | ||
=== Lacking === | === Lacking === | ||
+ | |||
+ | The moment of inertia of a diatomic nitrogen molecule $N_{2}$ around its center of mass? | ||
=== Representations === | === Representations === | ||
+ | $I = m_{1}r^{2}_{\perp1}$ | ||
+ | {{course_planning: | ||
=== Solution === | === Solution === | ||
- | For two masses, $I = m_{\perp1}$ | + | For two masses, $I = m_{1}r^{2}_{\perp1}$ |
+ | |||
+ | The distance between the masses is d, so the distance of each object from the center of mass is $r_{\perp1} = r_{\perp2} = d/2$. | ||
+ | |||
+ | Therefore: | ||
+ | |||
+ | $I = M(d/2)^{2} + M(d/2)^{2} = 2M(d/ | ||
+ | |||
+ | Where you substitute in M for $m_{1}$ and $m_{2}$ as it is the same total mass we are talking about. | ||
+ | |||
+ | Substitute in given values for variables. | ||
+ | |||
+ | $I = 2 \cdot (2.3$ x $10^{-26}kg)(0.75$ x $10^{-10}m)^2$ | ||
+ | |||
+ | Compute moment of inertia of diatomic nitrogen molecule | ||
+ | |||
+ | $I = 2.6$ x $10^{-46} kg \cdot m^2$ |