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 2.2 × 10−10 m (treat the atoms as point masses). If the angular speed of the molecule about the z axis is 5.49 × 1012 rad/s, what is its rotational kinetic energy? Answer in units of J.

9.7*10^-21 J

Explanation:

Given

Mass of oxygen atom, m = 2.66*10^-26 kg

Separation distance between the two atoms, r = 2.2*10^-10 m

Angular speed, w = 5.49*10^12 rad/s

First, we start by finding our moment of Inertia, I, using the relation

I = 1/2mr²

I = 1/2 * 2.66*10^-26 * (2.2*10^-10)²

I = 1/2 * 1.2874*10^-45

I = 6.437*10^-46 kgm²

The Rotational Kinetic Energy is given by

KE = 1/2Iw²

Now, using this value of I, we plug it in the energy equation.

KE = 1/2 * 6.437*10^-46 * (5.49*10^12)²

KE = 1/2 * 1.94*10^-20

KE = 9.7*10^-21 J

Thus, the rotational kinetic energy is 9.7*10^-21 J

increasing transnational kinetic energy. because the transnational kinetic energy formula consists of only two variables, mass and velocity, increasing one of those properties is the only way to increase an object's transnational kinetic energy. increases to mass and velocity, however, do not have the same impact.