Question:

# Rolling a cylinder using a rope having one end fixed and another end mobile

Eli: 3 days ago

A cylinder is being rolled(without slipping) with the help of a massless rope towards the left direction. An attractive force is effective on the cylinder towards the right direction. One end of the rope is kept fixed at a certain position. so part of the rope below the center of mass of the cylinder will get smaller and smaller with the passage of time. That means the fragments of the rope which are contact with the lower surface of the cylinder will go upwards eventually. so there should be a difference in velocity/force on between the to parts of the rope(the lower and the upper part). I have considered that the difference of tension between the two ropes is extremely small($$dx$$). if the cylinder is rolling without slipping with a negligible velocity, how much energy am I applying to the system? I am pulling the upper part of the rope. If I am applying $$(T+dx)$$ force on the rope then a component of it must push the body towards left. Again according to my perspective total of a component of my applied force and a component of tension of lower part of the rope must equal $$F$$ so that the body can move towards left. But if am applying only a part of the force to keep it moving that means in terms of energy contribution I am partially contributing, then is the tension of the lower part of the rope also expending energy? if it is expending energy, from where is it getting the energy at first?

I am quite puzzled regarding this topic. Please point out where I am wrong. Are my assumptions wrong here?

Thank you

Christian: 3 days ago

The force in the rope (tension) $$F_R$$ is half the attractive force on the cylinder $$F_C$$.

$$F_R=F_C/2$$

But the upper left end of the rope is moving twice as far as the cylinder.

$$X_R=2X_C$$

So the work being done by the hand pulling the rope ($$W_R$$) is equal to the change in potential energy of the cylinder ($$E_C$$).

$$W_R = X_R F_R = (2X_C) (F_C/2) = X_C F_C = E_C$$

Now all the work done and change in energy of the system is accounted for. The wall holding the other end of the rope is not doing any work.