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--- 183_notes:examples:the_moment_of_inertia_of_a_diatomic_molecule [2014/10/31 14:43] – pwirving
+++ 183_notes:examples:the_moment_of_inertia_of_a_diatomic_molecule [2014/11/05 20:40] (current) – pwirving
@@ Line 5: / Line 5: @@
 === 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 ===
+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 ===
+The moment of inertia of a diatomic nitrogen molecule $N_{2}$ around its center of mass?
 === Representations ===
+$I = m_{1}r^{2}_{\perp1}$
+{{course_planning:course_notes:mi3e_09-018.jpg|?300}}
 === Solution ===
-For two masses, $I = m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2}$. The distance between masses is d, so the distance of each object from the center of mass is $r_{\perp1} = r_{\perp2} = d/2$. Therefore:
+For two masses, $I = m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2}$.
+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:
 $I = M(d/2)^{2} + M(d/2)^{2} = 2M(d/2)^{2}$
+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 $10^{-46} kg \cdot m^2$