Answer to Question #301555 in Mechanics | Relativity for sabz

Question

1. A nitrogen molecule can be thought of as two point masses (m of each = 14u, where u = 1.67 x 10-27 kg) separated by a distance of 1.3 x 10-10 m. In air at room temperature the average rotational kinetic energy of such a molecule is about 4.0 x 10-21 J. Find the moment of inertia of such a molecule about its center of mass, and its speed of rotation in revolutions per second. 

Answer 1

$I=\frac{md^2}2=19.76\cdot 10^{-47}~kg\cdot m^2,$

$E=\frac{I\omega^2}2,$

$\omega=\sqrt{\frac{2E}I},$

$\nu=\frac{\omega}{2\pi}=\sqrt{\frac E{2\pi^2 I}}=3.2\cdot 10^{17}~\frac{rev}s.$