Question

Consider an O2 rotating in x,y plane about the z-axis the rotation passess through the centre of molecule perpendicular to its length the mass of each O2 is 2.26 X10 ​-26Kg and at room temperature the average separation between the 2 atom
a.find the moment of inertia of the molecule about the z-axis
b.if the angular Speed of the molecule about the z-axis is 4.6piX10^12 rad/s.what is its rotational kinetic energy

Answer 1

Answer on Question #63868, Physics / Mechanics | Relativity

Consider an $O_2$ rotating in $x, y$ plane about the $z$-axis the rotation passes through the center of molecule perpendicular to its length the mass of each $O_2$ is $2.26 \times 10^{-26} \, \mathrm{kg}$ and at room temperature the average separation between the 2 atoms.

a) Find the moment of inertia of the molecule about the z-axis.

b) If the angular Speed of the molecule about the z-axis is $4.6\pi \times 10^{12} \, \mathrm{rad/s}$. What is its rotational kinetic energy?

Solution:

a) The moment of inertia of the molecule:

$$J = m \left(\frac {d}{2}\right) ^ {2} + m \left(\frac {d}{2}\right) ^ {2} = \frac {1}{2} m d ^ {2}$$

Radius of an atom of oxygen $r = 48 \times 10^{-12} \, \mathrm{m}$

$$d = 4 \times 48 \times 10^{-12} \, \mathrm{m} = 1.92 \times 10^{-10} \, \mathrm{m}$$

Then

$$J = \frac {1}{2} \times 2.26 \cdot 10^{-26} \times (1.92 \cdot 10^{-10}) ^ {2} = 4.17 \cdot 10^{-46} \, \mathrm{kg/m^2}$$

b) The kinetic energy of rotational motion of molecules:

$$KE = \frac {1}{2} I \omega^ {2}$$

Then

$$KE = \frac {1}{2} \cdot 4.17 \cdot 10^{-46} \times (4.6\pi \cdot 10^{12}) ^ {2} = 4.32 \cdot 10^{-20} J$$

Answer: a) $4.17 \cdot 10^{-46} \, \mathrm{kg/m^2}$; b) $4.32 \cdot 10^{-20} J$

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