I know the answer to this question, I\'ve put it below, I\'m looking for why. Th
ID: 2207430 • Letter: I
Question
I know the answer to this question, I've put it below, I'm looking for why. Thanks. A solid cylinder (I=(1/2)MR2) is rolling along a flat, horizontal plane. The center of mass of the cylinder is moving toward the south at constant velocity. Which one of the following statements concerning the translational and rotational kinetic energies of the cylinder is true? The answer is: a) The translational kinetic energy is greater than the rotational kinetic energy. I would please like an explanation as to whyExplanation / Answer
The energy at the lower end of the incline is rotational plus kinetic, this is Total Energy = 1/2 m V^2 + 1/2 I w^2 = mg h where V is the velocity of the center of mass, I is the moment of inertia, and h is the initial height at top of incline. The moment of inertia for a sphere is I = 2m R^2 /5 substituting in the energy equation we get 1/2 m V^2 + 1/2 (2 m R^2/5) w^2 = mg h 1/2 m V^2 + 1/5 m R^2 w^2 = m g h now, for rolling motion, the point of contact of the sphere with the incline has zero speed, thus the angular velocity and the velocity of the center of mass are related as V = w R thus, the expression for energy is just 1/2 m V^2 + 1/5 m V^2 = m g h this can be rewritten as 1/2 m V^2 + 2/5 (1/2 m V^2) = m g h or KE + 2/5 KE = PE which means that the rotational energy is 2/5 of the kinetic energy. This is essentially the result you found. As you can see, this result does not depend on the radius of the sphere.