Final answer:
The rotational kinetic energy of a uniform cylinder rolling without slipping and with constant velocity on level ground is given by K = 1⁄2 Iω². This energy, along with translational kinetic energy, is conserved in the system assuming no nonconservative forces.
Step-by-step explanation:
When a uniform cylinder rolls with constant velocity and without sliding along level ground, its rotational kinetic energy can be understood in the context of energy conservation. In this case, where there are no nonconservative forces at work, the total energy of the rolling object is conserved.
The rotational kinetic energy is expressed using the formula K = 1⁄2 Iω², analogous to the translational kinetic energy formula K = 1⁄2 mv². Here, 'I' represents the moment of inertia, 'ω' is the angular velocity, and 'v' is the linear velocity of the object.
The moment of inertia for a cylinder depends on whether it is hollow or solid.
For a hollow cylinder rotating around its central axis, the moment of inertia is I = mR²,
whereas for a solid cylinder, it is I = 1⁄2 mR², where 'm' is mass and 'R' is radius.
Since the question suggests the cylinder rolls without slipping, the linear velocity 'v' is directly related to the angular velocity ω through the relation v = ωR.
This means that the translational kinetic energy and rotational kinetic energy are equally shared for a cylinder rolling without slipping.
Using these relationships, one can calculate the rotational kinetic energy of the cylinder, which is essential in various applications such as the use of flywheels to store energy in vehicles. It also has implications for objects rolling down inclines, as both translational and rotational energies must be accounted for to determine the final velocity of the object.