Systems of Particles and Rotational Motion

Question : 20

Total: 33

The oxygen molecule has a mass of 5.30×10–26 kg and a moment of inertia of 1.94×10–46kgm2 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.

Solution:

Here,m=5.30×10–26kg,
I=1.94×10–46kg m2,v=500ms–1

is mass of each atom of oxygen and 2r is distance between the two atoms as shown in figure, then
I=

r2+

r2=mr2
r=√

=√

1.94×10−46

5.3×10−26

=0.61×10−10 m
As K.E. of rotation =

K.E. of translation
∴12Iω2=

mv2
⇒12mr2ω2=

mv2
ω=√

=√

500

0.61×10−10

=6.7×1012‌rad‌s−1

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