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Consider the diatomic molecule oxygen O2 which is rotating in the xy plane about the z axis passing through its center, perpendicular to its length. The mass of each oxygen atom is 2.66 × 10−26 kg, and at room temperature, the average separation distance between the two oxygen atoms is 1.9 × 10−10 m (treat the atoms as point masses). If the angular speed of the molecule about the z axis is 5.75 × 1012 rad/s, what is its rotational kinetic energy? Answer in units of J.

Respuesta :

Answer:

  K = 8.35 10⁻¹¹ J

Explanation:

The rotational kinetic energy is

           K = ½ I w²

Where I is the moment of inertia and w the angular velocity

The moment of inertia of the two particles is

           I = m₁ r₁² + m₂ r₂²

The distance from the center to each particle is

          r₁ = r₂ = r₀ = 1.9 10⁻¹⁰ / 2

          r₀ = 0.95 10⁻¹⁰ m

calculate          

          I = 2 m₀ r₀²

          I = 2 2.66 10⁻²⁶  0.95 10⁻¹⁰

          I = 5,054 10⁻³⁶ kg m²

 

Let's calculate the kinetic energy of rotation

          K = ½  5,054 10⁻³⁶ (5.75 10¹²)²

          K = 8.35 10⁻¹¹ J

         .

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