Final answer:
After one complete revolution without slipping, the center of mass of a cylindrical can moves a distance of 2πR, the circumference of the can. If slipping occurred, this distance would be smaller because not all rotational motion would contribute to forward movement.
Step-by-step explanation:
The question addresses a common problem found in physics related to rolling motion and friction. When a cylindrical can rolls across a horizontal surface without slipping, the situation can be analyzed using the relationship between linear and rotational motion. The key formulas we need keep in mind are dCM = Rθ and UCM = Rω, where dCM is the distance the center of mass has moved, R is the radius, θ is the angular displacement, and ω is the angular velocity.
For part (a), after one complete revolution (where the angular displacement θ is 2π radians), the distance that the center of mass has moved, dCM, is equal to the circumference of the can, which is 2πR. Therefore, the correct answer is 2πR.
For part (b), if slipping occurred, the can would cover less distance for every revolution because some of the motion that should relate to forward movement is being lost to slipping. Hence, the center of mass would move a smaller distance compared to a non-slip rolling scenario. So, the best answer is (c) 2πR, Smaller.