Which way does the force of friction point on a turning cylinder?

[Ryker]

I'm just doing dynamics of rotational motion and was wondering if anyone could help me with the direction the friction points when you have a rope looped around a cylinder and it starts unwinding. The book says the force of friction does no work and is also responsible for the cylinder to actually turn. My first guess was that friction would point towards the centre of the cylinder, since it's undergoing circular motion and the acceleration would therefore have to be pointed there, but on the other hand it seems more intuitive that friction would point tangential to the circle. Or is it actually somewhere in between, because the cylinder's linear speed (of points at its rim) is increasing and isn't constant? Perhaps so that the normal force of the rope points towards the center of the cylinder and the friction tangentially to it, the resulting net force being somewhere in between?

 

[rcgldr]

The direction issue was recently discussed in this thread:

https://www.physicsforums.com/showthread.php?t=441881

The issue of static friction doing no work was discussed in this thread:

https://www.physicsforums.com/showthread.php?t=435212

This article explains the concept of rolling static friction and work if forces and accelerations are involved. The net work done by friction is zero, because the linear work component is equally opposed by the angular work component. In the case of an internal energy source, like a car on a level road, the friction force could be considered to do negative work against the engine, equally opposed by the positve work in accelerating the car.

http://cnx.org/content/m14391/latest

 

[Ryker]

Thanks, I'll be sure to look at the threads and the article you linked. I guess I should've done a search first, since this stuff (apart from the article) was already available at PF.

 

[Ryker]

Hmm, well, after reading this I have to say I am unfortunately still unable to clearly see the direction friction is acting in.

I also have to say the example I envisioned differs a bit from the examples discussed in that I would have a cylinder suspended in air (say on a beam through its axis) and then I would have a rope looped around it. I would pull this rope on one side and this pull would result in the rope and the cylinder moving without slipping. You can imagine a pulley with enough friction between it and the rope to make the wheel of the pulley turn to see what I mean. I guess I should have been more clear in my original post as to what situation I wanted to apply this.

Now I'm not sure if that changes anything or not. From what I can see, there's no linear acceleration of the cylinder in relation to an observer's frame of reference, but there is in relation to that of the rope.

I've tried breaking down forces that act on the cylinder. If there was no friction between it and the rope, then the rope would slip and would move without causing the cylinder to rotate, right? But if it does rotate then that means there must be a net force that exerts torque, and that force would seem to be in the same direction as the force of tension. But the force of tension acts on the rope, not on the cylinder, doesn't it? So since the direction the surfaces of rope and cylinder want to move is in the direction of tension, friction of the cylinder acting on the rope would be in the direction opposite to that (that is, opposite to tension). But this causes a reaction force of the rope on the cylinder in the direction of the force of tension (but is not equal to tension). Is this correct?

What also seems to confuse me is the fact that when you have a circular motion, the net force needs to act towards the centre of that circle in order for the object to travel in a circular path. But here I guess the net force doesn't act towards the centre, right? Is it because it all comes down to torque, the direction of whose vector is obtained by using the right hand rule, so the same logic cannot be applied?

 

[rcgldr]

I was assuming you meant a rope going around a cylinder while it also rested on a horizontal plane, with two points of static friction.

For a cylinder in space, free from any external forces other than the rope, then the friction area is the point of application of a Newton third law pair of forces. The rope exerts a tension force on the cylinder, and the cylinder's reaction force to linear, circular (pendulum like motion), and angular acceleration exerts an equal and opposing force on the rope.

 

[Ryker]

I've tried breaking down forces that act on the cylinder. If there was no friction between it and the rope, then the rope would slip and would move without causing the cylinder to rotate, right? But if it does rotate then that means there must be a net force that exerts torque, and that force would seem to be in the same direction as the force of tension. But the force of tension acts on the rope, not on the cylinder, doesn't it? So since the direction the surfaces of rope and cylinder want to move is in the direction of tension, friction of the cylinder acting on the rope would be in the direction opposite to that (that is, opposite to tension). But this causes a reaction force of the rope on the cylinder in the direction of the force of tension (but is not equal to tension). Is this correct?

So like this then?

 

[rcgldr]

The friction force on the cylinder is in the direction of tension. The friction force on the rope is in the opposite direction.

 

[Ryker]

Alright, thanks for clearing that up and this is in fact what that quoted of paragraph was trying to say