Question

The oxygen molecule has a mass of 5.30 × 10⁻²⁶ kg and a moment of inertia of 1.94 × 10⁻⁴⁶ kg m² 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 500 m/s and that it's autumn 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 1

We have to find average angular velocity, ω of oxygen molecule.
We know that,v= r ω

It is given that, Mass of oxygen molecule, M = 5.30 × 10^-26 kg
Moment of inertia of oxygen molecule, I = 1.94 × 10^-46 kg/m²
Mean speed of molecule, v = 500 m/s

To find ω we need value of r (distance between axis of rotation and oxygen atom in the molecule)
The equation for moment of inertia of an oxygen molecule (system consisting of 2 equal masses (m) separated at a distance of r from axis),
I = mr² + mr²
= 2mr²
But m is mass of an oxygen atom,
∴ m = M/2
Thus, I = 2(M/2) r²
I = Mr² ........(1)

And r = [ (1.94 × 10^-46)/5.30 × 10^-26]½
= .606 × 10^-10 m

We know, 3 × rotational kinetic energy = 2 × translational kinetic energy
[ Diatomic molecule has 5 degree of freedom, 2 translational and 3 rotational]
so, 3 × 1/2 Iω² = 2 × 1/2 Mv²

Or, 3 × Iω² = 2mv²

Or, 3 × mr²ω²= 2mv²

Or, ω² = 2v²/3r²

Thus, ω = (2/3)1/2 × 500/ (.606 × 10^-10)
= 6.68 × 10¹² rad/s
Average angular velocity of molecule is 6.68 ×10¹² rad/s

Answer 2

Answer:

We have , L'=\Sigma r'_i\times P'_iL

′

=Σr

i

′

×P

i

′

Differentiating both sides with respect to time,

\begin{gathered}\frac{dL'}{dt}=\frac{d}{dt}\left(\Sigma r'_i\times p'_i\right)\\\\=p'_i\times\Sigma\frac{dr'_i}{dt}+\Sigma r'_i\times\frac{dp'_i}{dt}\\\\=\Sigma m_ir'_i\times v'_i+\Sigma r'_i\times\frac{dp'_i}{dt}\end{gathered}

dt

dL

′

=

dt

d

(Σr

i

′

×p

i

′

)

=p

i

′

×Σ

dt

dr

i

′

+Σr

i

′

×

dt

dp

i

′

=Σm

i

r

i

′

×v

i

′

+Σr

i

′

×

dt

dp

i

′

Where r'_ir

i

′

is the position vector with respect to centre of mass of system of particles.

But from definition of centre of mass,

\Sigma m_ir'_i=0Σm

i

r

i

′

=0

so, \frac{dL'}{dt}=\Sigma r'_i\times\frac{dp'_i}{dt}

dt

dL

′

=Σr

i

′

×

dt

dp

i

′

[ hence, proved]

We know, \bf{\tau=r\times\frac{dv}{dt}}τ=r×

dt

dv

So, \tau_{ext}=\Sigma r'_i\times\frac{dp'_i}{dt}τ

ext

=Σr

i

′

×

dt

dp

i

′

And hence, \frac{dL'}{dt}=\tau_{ext}

dt

dL

′

=τ

ext