# What's the motion of this yoyo under external force will be?

A yoyo on a horizontal table is being pulled by a string to the right, the table is not frictionless. If we only know that the object doesn't slip, how do we know if the string is winding up or unwinding? My reasoning is initially the friction is not ever effect so the force pulling the yoyo to the right will act a torque to the object such that the string is unwinding. But someone else said it should be winding up. I don't know why.

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I think that a sketch says more than equations and words

The yoyo will, in an infinitesimal sense, have to move around the red dot (but is kept from doing so by the table). The direction in which the yoyo will move, depends on the angle of the string, and thus the direction of torque around this touching point.

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If the outside radius is $r_O$ and the inside where the string attaches $r_I$ then the no-slip condition is $v + \omega r_O =0$. The string lengthening rate is $\dot{l} = \omega r_I$ so the end of the string motion is $\dot{x}=v+\dot{l} = v + \omega r_I$. Together they make

$$v = \frac{r_O}{r_O-r_I} \dot{x}$$ $$\omega = \frac{1}{r_O-r_I} \dot{x}$$

and their derivatives

$$\dot v = \frac{r_O}{r_O-r_I} \ddot{x}$$ $$\dot \omega = \frac{1}{r_O-r_I} \ddot{x}$$

First if we assume the friction force $F$ acts in the opposite direction as the tension $T$ the equations of motion are

$$T - F = m \dot{v}$$ $$T r_O - F r_I = I \dot{\omega}$$

All of the above are solved for

$$T = \frac{I+m r_O^2}{I+m r_I r_O} F$$ $$\ddot{x} = \frac{(r_O-r_I)^2}{I+m r_I r_O} F$$

This means that if $F$ where to switch signs and work in the same direction as the tesion then the end of the string is going to have a negative acceleration $\ddot{x}<0$ which is impossible because the string would slack.

So the assumption that $F$ acts to the left must be correct.

If we back substitute $\ddot{x}$ into $\dot{v}$ and $\dot{\omega}$ we get an idea which way this thing is going to move

$$\dot{v} = \frac{r_O (r_O-r_I)}{I+ m r_I r_O} F$$ $$\dot{\omega} = \mbox{-} \frac{r_O-r_I}{I + m r_I r_O} F$$

So the center of the yoyo is moving towards the right, and it is spinning clockwise winding the yoyo up.

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How did you include the angle with which the string is pulled? Try with a real yoyo, you can upwind or unwind it, depending on the angle of the string with the horizontal. This is something that you should include in the derivation. – Bernhard Apr 14 '13 at 6:46
I thought OP said horizontal (pulled to the right, not right and up). – ja72 Apr 14 '13 at 19:02
It's not my question, but I think the question is a bit lame when only pulling horizontally. – Bernhard Apr 14 '13 at 19:03