Final answer:
After one complete revolution without slipping, the center of mass of a cylindrical can of radius R moves a distance equal to the circumference of the can, which is 2πR.
Step-by-step explanation:
The question pertains to the concept of rolling motion in Physics, specifically analyzing the movement of a cylindrical can as it rolls across a horizontal surface without slipping. After one complete revolution of the can, the center of mass of the can will have moved a distance equal to the circumference of the can's base. When a cylindrical can of radius R rolls without slipping for one complete revolution, each point on the circumference of the can make contact with the ground once, mapping a circular path onto the surface. The distance moved by the center of mass is equal to the circumference of the can, which is calculated using the formula C = 2πR, where R is the radius.
Therefore, the center of mass travels a distance dCM = 2πR after one complete revolution. If slipping occurred, this distance would be different, as slipping would imply that the distance traveled by the can's center of mass would not match the arc length traced by a point on its circumference. The concept of rolling motion dictates that dCM = Rθ, where θ is the angular displacement in radians, which, for one complete revolution (θ1 revolution = 2π radians), reinforces the relationship dCM = 2πR.