Which direction will the yoyo move?

This question has been around the net for a while, and I haven't seen a good explanation for it:

A yo-yo is initially at rest on a horizontal surface. A string is pulled in the direction shown in the figure. In what direction will the yo-yo rotate and move?

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This problem is often posed in classes (both introductory and upper division) because there is always a wide diversity of opinion about the expected behavior and the correct way to analyze the problem. – dmckee Dec 2 '12 at 5:03
Some of the answers below have assumed that there will be no slipping. Is that an intended assumption from your point of view? – LarsH Mar 8 at 12:18

Imagine the string runs over a pulley and is connected to a weight, like this:

The weight must fall, which means the red dot must move to the right. To make that happen, the yo-yo must roll to the right.

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Is this a power point sketch? Just curious. – ja72 Dec 2 '12 at 5:24
geogebra geogebra.org/cms – Mark Eichenlaub Dec 2 '12 at 6:57
I don't see why the yo-yo must roll to the right. I would have expected it could slip to the right while rolling to the left, with the net effect on the centre of the yo-yo still unclear. – Henry Dec 2 '12 at 8:52
Thinking about the torque about the centre of the yo-yo, I am afraid I still cannot see that it can roll to the right, i.e. clockwise. – Henry Dec 2 '12 at 22:57
@Henry Of course it rolls to the right for the reason I gave. If you want to think about the torque, it is far easier to consider the torque about the point of contact with the floor. If you absolutely must calculate the torque about the center of mass, it rolls to the right because the torque from friction exceeds the torque exerted by the string, even though the force of friction is smaller. – Mark Eichenlaub Dec 3 '12 at 1:05

There is a similar version with a bicycle, that is easier to figure out experimentally: in what direction will a bicycle move if you pull back on a pedal that is at the bottom of the stroke? You are pulling back, but in doing so, are creating a torque on the back wheel that should propel it forward...

The answer is simpler than it seems. In both cases, if there is any movement, it will be in the direction of the external force. In both cases there is a frictional force that opposses movement, and that will prevent any from happenning until a threshold is reached, and then you will start having sliding, not rolling, and the movement will happen in the direction of the outside force.

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"Sliding, not rolling" - How do we know that when the threshold is reached, we will not also get rolling in addition to sliding? – LarsH Mar 8 at 12:07

Let's say the bigger radius is $r_1$ and the smaller one $r_2$. The question really has to do with the direction of friction force. I assign the friction coefficient $\mu$, make the direction the force is acting as positive motion and write the sum of the forces in the horizontal direction, and sum of moments about the center:

$$F-\mu\, m g\, {\rm sign}(v+r_1 \omega) = m \dot{v} \\ F r_2 - \mu\, m g\, r_1\, {\rm sign}(v+r_1 \omega) = I \dot{\omega}$$

where ${\rm sign}()$ returns -1,0 or 1 depending on the sign of the argument, and $v$ is linear velocity of center, $\omega$ angular velocity of center, $m$ the total mass, and $I$ the mass moment of inertia at the center.

There are three posibilities:

1. Yo-yo is rolling with the contact point velocity zero, $v+\omega r_1=0$ and motion $$v = \frac{F}{m} t \\ \omega = \frac{F r_2}{I} t$$
2. Yo-yo is slipping with the contact point moving to the right, $v+\omega r_1>0$ and motion $$v = \frac{F-\mu\,m g}{m} t \\ \omega = \frac{F r_2-\mu\,m g\,r_1}{I} t$$
3. Yo-yo has backspin with the contact point moving to the left, while the center moves to the right, $v+\omega r_1<0$ with motion $$v = \frac{F+\mu\,m g}{m} t \\ \omega = \frac{F r_2+\mu\,m g\,r_1}{I} t$$

Now I can check the conditions where is possibility can exist.

First if the friction is zero or less than $\mu = \frac{F (I+m r_1 r_2)}{m g (I+m r_1^2)}$ then we have the slipping condition since $$v+\omega r_1 = \frac{F-\mu\,m g}{m} t + r_1 \frac{F r_2-\mu\,m g\,r_1}{I} t > 0$$

Otherwise if the friction is higher than $\mu = \frac{F (I+m r_1 r_2)}{m g (I+m r_1^2)}$ then we have an interesting situation. The problem is that $\mu$ is not constant, as the yo-yo goes through a slip-stick scenario.

The best I can do is treat friction as linear to slip velocity by some arbitrary coefficient (damping) $d$ such that $\mu = d v_c = d (v + \omega\,r_1)$. I can find the acceleration of the contact point by $\dot{v}_c = \dot{v} + \dot{\omega}\,r_1$ which is used to find the equation of motion of the contact point:

$$\dot{v}_c = \left(\frac{F-d\,m g\,v_c}{m}\right) + \left( \frac{F r_2 - d\, m g\,r_1\,v_c}{I}\right) r_1 \\ \dot{v}_c = \left(\frac{1}{m}+\frac{r_1 r_2}{I}\right) F - d\,g\,\left(1+\frac{m r_1^2}{I}\right) v_c \\ \dot{v}_c = a_0 - \beta v_c$$ where $a_0 = \left(\frac{1}{m}+\frac{r_1 r_2}{I}\right) F$ and $\beta = d\,g\,\left(1+\frac{m r_1^2}{I}\right)$. This differential equation has solution $$v_c = \frac{a_0}{\beta} \left(1-\boldsymbol{e}^{-\beta t}\right) \\ \dot{v}_c = a_0 \boldsymbol{e}^{-\beta t}$$

With that and the equations of motion the center of the yo-yo moves by

$$\dot{v} = a_0 \left(\frac{I}{I+m r_1^2}\boldsymbol{e}^{-\beta t} + \frac{(r_1-r_2) I m r_1} {(I+m r_1 r_2)(I+m r_1^2)}\right)$$

This is just a decreasing function. More interesting is the angular acceleration and velocity which both cross the zero line

$$\dot{\omega} = a_0 \left(\frac{m r_1}{I+m r_1^2}\boldsymbol{e}^{-\beta t} -\frac{m I (r_1-r_2)}{(I+m r_1 r_2)(I+m r_1^2)}\right)$$ and $\omega = \int \dot{\omega}\,{\rm d} t$

$$\omega = a_0 \left( \frac{m r_1}{\beta (I+m r_1^2)} \left(1-\boldsymbol{e}^{-\beta t}\right) - \frac{m I (r_1-r_2)}{(I+m r_1 r_2)(I+m r_1^2)} t \right)$$

Lastly the coefficient of friction varies by

$$\mu = \frac{I a_0 \left(1-\boldsymbol{e}^{-\beta t}\right) }{g (I+m r_1^2)}$$

which starts from zero and stabilizes quickly at

$$\mu = \frac{I+m r_1 r_2}{m g (I+m r_1^2)} F$$

The above can also be expressed in terms of the angular velocity as $\mu = \frac{r_2}{r_1} \frac{F}{m g} - \frac{I \dot{\omega}}{m g\,r_1}$.

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If you assume that there is no slippage, then it becomes a purely geometrical problem.

• $r =$ string rolling radius
• $R =$ ground rolling radius
• $s_{rs} =$ distance reel travels relative to string
• $s_{rg} =$ distance reel travels relative to ground
• $s_{sg} =$ distance string travels relative to ground

From the two equations:

• $r/R = s_{rs}/s_{rg}$
• $s_{rg} = s_{sg} + s_{rs}$

You get:

$$s_{rg} / s_{sg} = \frac{1}{1 - (r/R)}$$

The right side is positive if $r < R$ so $s_{rg}$ & $s_{sg}$ have the same sign (reel moves in the same direction as the string).

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