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

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

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.

• Thought of everything but didn't notice that minor part🙂
– MSKB
Sep 22, 2021 at 18:28
• It's always easy to forget a detail in these sorts of problems. You probably already know this, but your problem is a classic example of a pulley. You can read more on wikipedia if you are interested. en.wikipedia.org/wiki/Pulley Sep 22, 2021 at 18:39
• Well a little bit enquiry....why does the curve formed by pully does not influence the tension of the rope? I mean why do the act as if the shape of the rope is "V" instead of a circular bump?
– MSKB
Sep 22, 2021 at 18:42
• @MSKB: Here are a couple of questions that might help. physics.stackexchange.com/questions/510771/… and physics.stackexchange.com/questions/426027/… Sep 22, 2021 at 19:12

If the velocity is constant, then 2T – F = 0. The positive work being done by the upper segment of rope combined with the negative work being done by (F) gives no change in the kinetic energy. If the cylinder is accelerating, and you assume no friction between the rope and the cylinder, then you need a static friction force from the supporting surface to provide the angular acceleration.