Q.The oxygen molecule has a mass of $5.30 \times 10^{-26}$ kg and a moment of inertia of $1.94 \times 10^{-46}$ 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.

The problem relates the translational and rotational kinetic energies of an oxygen molecule to find its average angular velocity.

Concept: Relationship between Rotational and Translational Kinetic Energy

1. The translational kinetic energy ($KE_{trans}$) and rotational kinetic energy ($KE_{rot}$) are given by:

2. According to the problem statement, $KE_{rot} = \frac{2}{3} KE_{trans}$. Substituting the formulas:

3. Solve for the angular velocity $\omega$:

4. Substitute the given values: $m = 5.30 \times 10^{-26}$ kg, $v = 500$ m/s, and $I = 1.94 \times 10^{-46}$ kg m$^2$. …

The problem asks us to find the average angular velocity of an oxygen molecule, given its mass, moment of inertia, translational speed, and a specific relationship between its rotational and translational kinetic energies. This requires us to connect the concepts of linear motion (translational kinetic energy) with rotational motion (rotational kinetic energy and angular velocity).

The core idea is to first calculate the translational kinetic energy, which depends on the molecule's mass and linear speed. Once we have this value, we can use the given ratio to determine the rotational kinetic energy. Finally, knowing the rotational kinetic energy and the molecule's moment of inertia, we can directly calculate its angular velocity using the formula for rotational kinetic energy. This step-by-step approach allows us to bridge the information from linear motion to rotational motion.

1. Identify the given quantities.

   We are provided with the following information:

   • Mass of the oxygen molecule, $m = 5.30 \times 10^{-26} \text{ kg}$
   • Moment of inertia about an axis through its centre perpendicular to the line joining the two atoms, $I = 1.94 \times 10^{-46} \text{ kg m}^2$
   • Mean speed of the molecule, $v = 500 \text{ m/s}$
   • Relationship between kinetic energies: $K_r = \frac{2}{3}K_t$, where $K_r$ is rotational kinetic energy and $K_t$ is translational kinetic energy.

2. Calculate the translational kinetic energy ($K_t$).

   The translational kinetic energy of an object is determined by its mass and linear speed. The formula is:

   $K_t = \frac{1}{2}mv^2$

   Substitute the given values for mass ($m$) and speed ($v$):

   $K_t = \frac{1}{2} (5.30 \times 10^{-26} \text{ kg}) (500 \text{ m/s})^2$

   $K_t = \frac{1}{2} (5.30 \times 10^{-26}) (250000)$

   $K_t = (0.5) (5.30 \times 10^{-26}) (2.5 \times 10^5)$

   $K_t = 6.625 \times 10^{-21} \text{ J}$

3. Calculate the rotational kinetic energy ($K_r$).

   The problem states that the kinetic energy of rotation is two thirds of its kinetic energy of translation. We use the value of $K_t$ calculated in the previous step:

   $K_r = \frac{2}{3}K_t$

   $K_r = \frac{2}{3} (6.625 \times 10^{-21} \text{ J})$

   $K_r = 4.41666... \times 10^{-21} \text{ J}$ …

Concept: Relating Translational and Rotational Kinetic Energy

Step 1: Translational KE

Step 2: Rotational KE (given ratio)

Step 3: Solve for $\omega$ …