Calculating Total Kinetic Energy of a Rolling Hoop

What is the total kinetic energy of a rolling hoop?

Given data: A 2.0 kg hoop rolls without slipping on a horizontal surface with a constant linear speed of 6.0 m/s. The radius of the hoop is 0.5 m.

Answer:

The total kinetic energy of the rolling hoop is 72 Joules.

To calculate the total kinetic energy of the rolling hoop, we need to consider both its rotational and translational kinetic energy. The rotational kinetic energy is given by the formula:

Rotational Kinetic Energy = 0.5 * I * ω^2

where I is the moment of inertia (I = m*r^2) and ω is the angular velocity. In this case, the angular velocity can be expressed as: ω = v/r, where v is the linear velocity and r is the radius of the hoop.

Substituting the expressions for I and ω into the formula for rotational kinetic energy, we get:

Rotational Kinetic Energy = 0.5 * m * r^2 * (v/r)^2 = 0.5 * m * v^2 = 0.5 * m * v^2

This is the same as the formula for translational kinetic energy, which is given by 0.5 * m * v^2. Therefore, the total kinetic energy of the rolling hoop is the sum of its rotational and translational kinetic energy:

Total Kinetic Energy = Rotational Kinetic Energy + Translational Kinetic Energy = m * v^2 = 2 * 6^2 = 72 J

Hence, the total kinetic energy of the rolling hoop is 72 Joules.