An oxygen molecule has a mass of 5.30xx10^(-26) kg and a moment of inertia of 1.94xx10^(-46)kgm^2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500ms^(-1) and that its kinetic energy of rotation is two thirds of its kinetic energy of translation Find the average angular velocity of the molecule.

An oxygen molecule has a mass of 5.30xx10^(-26) kg and a moment of ine

1.	An oxygen molecule has a mass of 5.30xx10^(-26) kg and a moment of inertia of 1.94xx10^(-46)kgm^2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500ms^(-1) and that its kinetic energy of rotation is two thirds of its kinetic energy of translation Find the average angular velocity of the molecule.
Answer» Solution :Inertia of the SYSTEM of ATOMS, `I=(m/2)r^2+(m/2)t^2` `I=mr^2` `:.r=SQRT(I/m)=sqrt(1.94xx10^(-46)/(5.30xx10^(-26))=0.60xx10^(-10)m` given `1/2(mr^2)(omega^2)=1/3mv^2` i.e `1/2(mr^2)(omega^2)=1/3mv^2""omega^2=2/3(v^2)/(r^2)` i.e `omega=v/rsqrt(2/3)`i.e. `omega=(0.8165xx500)/(0.6xx10^(-10))=680.4xx10^(10)` i.e `omega=6.804xx10^(12)rads^(-1)`