Final answer:
The rotational kinetic energy of a uniform cylinder rolling along level ground with constant velocity and without sliding is half its translational kinetic energy. A. half its translational kinetic energy
Step-by-step explanation:
When a uniform cylinder rolls with constant velocity and without sliding along level ground, the condition that arises is known as pure rolling motion. In this state, the rotational kinetic energy (associated with the cylinder's rotation around its axis) is directly related to its translational kinetic energy (associated with its movement across the ground). By evaluating the energies for a rolling solid cylinder, it emerges that the rotational kinetic energy is half the translational kinetic energy.
The derivation of this relation is due to the fact that there is a distribution of mass within the cylinder that contributes to both forms of kinetic energy. The translational kinetic energy of the cylinder is given by Kt = (1/2)mv2 and the rotational kinetic energy is Kr = (1/2)Iω2, with 'I' being the moment of inertia for the cylinder and 'ω' being its angular velocity. For a solid cylinder, I = (1/2)mR2 and ω = v/R, with 'm' being the mass, 'R' being the radius, and 'v' the linear velocity. Substituting these relationships into the equations for kinetic energy leads to the realization that Kr = (1/2)Kt.