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Humber I beg you to get a kid's yo-yo and try this:
I will find a child and steal his or her yo-yo.
Put the yo-yo on the table with the unwound string going off to your right. The yo-yo represents the cogwheel, the table is the rail and the string is the chain. OK?
Now gently pull the string to the right and watch which way the yo-yo moves and also how quickly it moves relative to your hand which is pulling the string.
Yes, I know about that JBTH. I have really lived.
Try it with the string wound under the yo-yo spindle and also flip it over and try with the string wound over the spindle.
With the string emerging from under the spindle (like the chain in (1) and (2) above) the yo-yo will move to the right, faster than your hand.
With the string running over the spindle (Like (3) and (4)) the yo-yo will still move to the right, but will move slower than the hand that pulls it.
Please try it.
Yes, I am aware of what you are saying, but also that the entire process is not what you may think it is.
To convince yourself that
nothing special is going on, take the yo-yo and remove the string, but wind one turn around the axle. Maintain a little tension on the thread and pull it left and right. Like this.
Don't let the thread or the reel slide or allow any lifting. It helps if the reel is weighted, by a bolt through the central hole, perhaps.
ETA (Edit)
Now watch what happens to the hand that pulls the threads. One moves away from, and one towards the reel. Now flip it over to get the "over the top" version
The paper and reel video is misleading. I will finish that video, but in the mean time...The thread represents the chain.
A reel can be rolled towards the hand which is applying F, but may also turn counter-clockwise (CCW), and roll away from the hand.
Assuming that the force is applied parallel to the floor ( no lifting of the reel), which way will it go?
a = inner radius
b = outer radius
Fd = Friction
F = Applied force
m= reel mass
I = reel moment of inertia
v= linear velocity
w = angular velocity (radians/sec)
Linear component:
Net force to the right = F - Fd
Acceleration dv/dt = F - Fd/m
Angular Component:
Net clockwise (CW) moment = b*Fd - a*F
Angular Acceleration dw/dt = (b*Fd - a*F/)I
When a wheel rolls with out sliding (pure rolling) the axle of the reel moves a distance equal to the circumference. So the linear velocity (the speed the axle moves) and the angular velocity ( the speed it turns) are related by
v = b*w
Meaning it travels 2pi*r with each turn.
The linear and angular accelerations are related as:
dv/dt = b*dw/dt
From a = f/m
(F - Fd)/m = b(b*Fd - a*F)/I
Rearranging to relate Fd in terms of F
Fd( I + mb^2) = F( I + mab)
Fd = F(I + mab)/(I + mb^2)
If the reel moves to the right, then both the linear and angular forces will be positive (to the right). That depends on the friction to the surface and how the force is applied. If pulled slowly, the reel will tend to roll with the hand, but a jerk will send it away.
It's a balance of the inertia of rotation and the friction to the surface, where the angular component is either positive or negative.
There is a third case, where there reel does not roll, but slides, so the angular component is zero. The reel is then just a simple mass, and dragged along as if it were a block. That may happen when the friction is very low. Wheels cannot roll without adequate friction.
The geared version is going to be bad. The most likely outcome is that the gear will turn CCW, and move to the left. To move it to the right, would mean it would be dragged by the chain at the speed of the chain, and horribly, if at all possible.
