A Hoop's Transitional and Rotational Kinetic Energy Calculation
What are the translational and rotational kinetic energies of a 2.25 kg hoop rolling across a level surface at 4.50 m/s with a radius of 0.220 m?
Translational kinetic energy (Kt) is calculated using the formula 0.5 * m * v^2, where m is the mass and v is the speed of the object. For the hoop in this scenario: Kt = 0.5 * 2.25 kg * (4.50 m/s)^2 = 22.9 J Rotational kinetic energy (Kr) is determined by the formula 0.5 * I * w^2, where I is the moment of inertia and w is the angular speed of the object. For a hoop, the moment of inertia is expressed as m * r^2 and the angular speed as v/r: Kr = 0.5 * 2.25 kg * (0.220 m)^2 * (4.50 m/s / 0.220 m)^2 = 11.45 J Therefore, the hoop's translational kinetic energy is 22.9 J, and its rotational kinetic energy is 11.45 J.Translational Kinetic Energy Calculation
Rotational Kinetic Energy Calculation