Paradoxic cylinder rolling in opposite direction

I just solved the following exercise from my textbook and I don't really see the intuition behind the result:

In which direction will the cylinder roll (without slipping) if a constant force is applied on the rope? Inner radius, lower 'r'; outter radius, upper 'R'; moment of inertia, 'I'; and mass, 'm'.

The force is applied below the rotation axis and thus, if there weren't any friction one would expect it to rotate counter-clockwise. It turns out that with friction it does not.  I tried to solve it using Newton's Laws and Momentum:

F: $ma\vec{i} = F_{\text{applied on the rope}} \vec{i} + F_{\text{friction}} (-\vec{i})$

M: $I\alpha(-\vec{k}) = r(-\vec{j}) \times F_{\text{applied on the rope}} \vec{i} + R(-\vec{j})F_{\text{friction}} (-\vec{i})$

Rolling without slipping: $\vec{\alpha} = \frac{\vec{a}}{R}$

However, solving these equations I get that $$a = \frac{F_{rope}R(R-r)}{mR^2+I}$$ (and $a\gt 0$ since $R\gt r$). Thus it will roll to the right.

However I would expect it not to roll at all, but to stay still. This is what you get if $r = R$ but just in that case. (Or maybe rotate to the left because of the torque generated by the rope?) Is it there any intuitive way of understanting what is going on here? It seems that the friction force is taking over control and rotating the cylinder the way it wants.

• What do you understand by intuition? Concentrate on the equations, they and they only tell the story! – Gert Nov 23 '17 at 20:42
• @Gert I don't agree. A great teacher once told me 'don't make any calculations before you know what you'll get'. – J. Doe Nov 23 '17 at 20:48
• For example, if the equations told us that the cylinder would rotate to the left instead of the right I would be surprised because 'a priori' I would expect the cylinder to roll (or move) the same way (or just not the opposite) the force is applied. – J. Doe Nov 23 '17 at 20:49
• Sticking to the equations is good sometimes but, as I see it, in physics it is as important as keeping a qualitative model in mind (which can later help you to do approximations or studying edge cases). I chose to solve the problem this way (instead of the Lagrangian for instance) because I thought this way I could see better what it is all about. – J. Doe Nov 23 '17 at 20:54
• As it happens, my 'intuition' told me it would roll to the left, so what does that make me? ;-) Agree to disagree, I say. :-) – Gert Nov 23 '17 at 21:51

We can show this without resorting to numbers at all!

First off, you have been told that the coefficient of friction is large enough that the wheel does not slip.

Now assume it rolls to the left, and you can show by contradiction that it does not. Think about what would have to happen for it to move to the left. Since $F=ma$, the net force would have to be to the left, which means the frictional force must be larger than $F_{\rm rope}$. But then the torque due to friction is clearly larger than torque due to the rope, since it has both a greater force and a greater lever arm, and so it rolls clockwise. If it is rolling clockwise but moving left, it must be slipping, so we have broken our assumptions.

We can show the same way that it can't remain still. If the two forces are equal, the torques are not equal (since they have a different lever arm), and so it will begin to roll.

Thus, the only way it can roll without slipping is to the right.

• @sammygerbil In the limit as friction goes to infinity, it's obvious that it will roll without slipping if it rolls at all, isn't it? And it can roll to the left or the right if it is allowed to slip, so I don't understand your objection. – Chris Nov 24 '17 at 2:12
• Sorry, I thought it must slip. My objection was that I could not see how the rope could wind itself up. But seeing is believing : youtube.com/watch?v=Bwf3msm7rqM ... I have retracted my down-vote but I had to make a minor edit to your question in order to do so. – sammy gerbil Nov 24 '17 at 2:30
• @Chris Sweet! I just realised of that; the key is that they are applied at diferent distances from the axis. – J. Doe Nov 24 '17 at 2:48
• @sammy gerbil cool video ;) about the torque thing, I think I see your point: let's call it momentum then. – J. Doe Nov 24 '17 at 2:53
• @sammygerbil There are two separate forces, both of which are applied at different distances from the center of the wheel and so give different torques via $\vec\tau=\vec r\times\vec F$. You certainly can have torque with just one force as long as it's not applied through the center of mass. – Chris Nov 24 '17 at 3:10