Q. 203.9( 8 Votes )

# The oxygen molecule has a mass of 5.30 × 10^{-26} kg and a moment of inertia of 1.94×10^{-46} kg m^{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 500 m/s 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 :

We need to find average angular velocity, ω of oxygen molecule.

We know that,

V = r ω

Given

Mass of oxygen molecule, M = 5.30 × 10^{-26} kg

Moment of inertia of oxygen molecule, I = 1.94 × 10^{-46} kg m^{2}

Mean speed of molecule, V = 500 m s^{-1}

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 = mr^{2} + mr^{2}

= 2mr^{2}

But m is mass of an oxygen atom,

∴ m = M/2

Thus, I = 2(M/2) r^{2}

= Mr^{2}

i.e. r = (I/M)^{2}

= [ (1.94 × 10^{-46})/5.30 × 10^{2}] 1/2

= .606 × 10^{-10}

KE_{rot} = 2KE_{tran}/3

1/2 I ω^{2} = 2/3 × (m v^{2}/2)

1/2 m r^{2} ω = 2/3 × mv^{2}/2

ω = (2/3)^{1/2}× v/r

we know,

v = 500 ms^{-1}

r = .606 × 10^{-10} m

Thus, ω = (2/3)^{1/2} × 500/ (.606 × 10^{-10})

= 6.68 × 10^{12} rad s^{-1}

Average angular velocity of molecule is 6.68 ×10^{12} rad s^{-1}

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A solid sphere is set into motion on a rough horizontal surface with a linear speed u in the forward direction and an angular speed u/R in the anticlockwise direction as shown in figure (10-E 16). Find the linear speed of the sphere (a) when it stops rotating and (b) when slipping finally ceases and pure rolling starts.

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