# Homework Help: Yo-YO being pulled - does my Answer make Sense?

1. Dec 13, 2008

### TFM

1. The problem statement, all variables and given/known data

A yo-yo rests on a level surface, as shown below. A gentle horizontal pull is exerted on the cord so that the yo-yo rolls without slipping. In which direction does it roll and why?

2. Relevant equations

N/A

3. The attempt at a solution

Firstly, the yoyo will go in the opposite direction

My idea is that the string goes around the yoyo, and the weight of the string doesn't effect the bottom of the yoyo, because the weight of the string isn't on the yoyo, but on the top, the weight is. The string at the top is going in the opposite direction, so the force from the string will rotate the yoyo in the opposite direction as the string, and thus the yoyo will move in the opposite direction.

Does this make sense? Are there any formulas that may help prove my point?

Thanks,

TFM

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• ###### YoYo Pull - 2.jpg
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2. Dec 13, 2008

### Q_Goest

Hi TFM. I can't open the attachments yet, they haven't been approved. But consider this. Draw a free body diagram around the yo yo and then take moments around the point where the yo yo touches the ground.

Hope that helps.

3. Dec 13, 2008

### TFM

Okay, so the moment for the string at the bottom is:

$$M = d_1*T$$

At the top, the moement is:

$$M = d_2*T$$

where d2 is d1 plus the diameter of which the string wound round.

This implies that the the moment at the top is greater then the moment at the bottom.

Does this look right?

TFM

#### Attached Files:

• ###### YoYo Pull - 3.jpg
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4. Dec 13, 2008

### Q_Goest

Sorry, but I can't follow your set up as the attachments still haven't been approved. I'll check back later.

5. Dec 13, 2008

### Q_Goest

Ok, looks like your attachments are approved. The first one I'm assuming is your set up (Yo yo pull - 1). There's a single cord or thread pulling towards the right.

The second one I'm assuming is to show the weight of the string and yo yo. You can include the weight of the string if you like, it won't change the answer. But generally in these kinds of problems you can neglect things like cables and strings.

The third one (YoYo pull -3) is just a repeat of the first one showing the tension of the string. Problem here: you show 2 strings coming off the yo yo. Why? There are internal forces at work within the yo yo, but the point of drawing a free body diagram around the yo yo is to identify the forces acting on it, not internal forces. The string as it's wrapped around the yo yo only comes off toward the right, and it's there where you should imagine a break being made in the string and a force acting on it. You shouldn't imagine the string being broken in some other location since every other location only produces forces that are internal to the system in question.

So try your yo yo problem again, but only look at external forces on the yo you. There should only be
1. weight
2. the tension on the string
3. frictional force where the yo yo touches the ground.
Note that the assumption is that these forces produce an acceleration on the yo yo, so if you wanted to determine acceleration and not just direction of motion, we would need to account for the dynamics (ie: inertia) of the yo yo.

Why does taking moments around the point where the yo yo touches the ground make sense? How does this simplify the analysis? Which direction does the yo yo begin to move (right or left)?

6. Dec 14, 2008

### TFM

If we take moments against the ground, it seems that the only moment we need consider i he tension in the string. mg is pulling downwards towards the point its touching, so there is no perpendicular force. the friction will be acting on the ground, so the distance will be 0, This just leaves the tension, which will be multiplied by the distance from the point of touchin t the inner radius of the reel part?

I am not sure if this helps?

Also the reel will move off to the right- which I have proved experimentally with a cotton reel (don't have a yoyo handy)

TFM

7. Dec 14, 2008

### Q_Goest

Good! By taking moments around the point where the yo yo touches the ground, all other moments are eliminated (just as you've found). The result is that there's a moment around this point on the ground equal to the tension (force) in the string times the distance that string acts from the ground which forces the yo yo to roll to the right. Finding the right location where moments can be taken is one of the keys to simplifying a system like this.

8. Dec 14, 2008

### TFM

Okay, but wouldn't this prove the opposite - that the yoyo will roll in the same direction as the tension is pulling though?

TFM

9. Dec 14, 2008

### Q_Goest

Isn't this what you proved experimentally when you said:
The yo yo moves in the direction you pull the string. That's the same direction predicted by summing moments around the point it contacts the ground. Theory should always match the experimental evidence.

10. Dec 14, 2008

### TFM

No, I think I wrote the wrong thing down...When I pull the string, the string rolls out, but the yoyo/cotton reel itself rolls in the opposite direction.

???

TFM

11. Dec 14, 2008

### Q_Goest

I tried it too. Works for me though. Even with the string wrapped around and coming out the bottom as you show in your first attachment, the yo yo rolls in the direction the string is pulled. Try your experiment again, make sure the yo yo doesn't slip on the ground when pulled. It works. Promise.

12. Dec 14, 2008

### TFM

I think I found my problem,I did it on a wooden table, and it wouldn't move forwards, and then rolled back - too much friction, I think. I tried again, on a different, smoother surface, and it does indeed roll forward.

So it moves forwards, as predicted by moments.

TFM

13. Dec 14, 2008

### Q_Goest

Glad you got your experiment to work. Just for fun, I took a video of my own experiment. Unfortunately, the file was too large to attach as a zip file, so I changed the extension to .pdf

If you download the file and change the extension to .mpg you can view it.

#### Attached Files:

• ###### Yoyo.pdf
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14. Dec 14, 2008

### TFM

Will do.

TFM

15. Dec 14, 2008

### Vuldoraq

I'm confused,

Surely the tension causes the string wrapped around the barrel to unwind in an anti-clockwise direction, which would lead to the yo-yo rolling backwards? (If it wasn't slipping.)

I don't see how it can roll forwards?

Please could you explain whats wrong with my interpretation?

16. Dec 14, 2008

### TFM

For a long time, I couldn't understand this either, but I think I've worked it out. We are traking moments of inertia from the point where the reel touches the ground. If you draw a diagram, it has a force acting in the same direction as the pull. This leads to a clockwise moment, making the yo yo roll to the right.

Does this make sense?

TFM

17. Dec 14, 2008

### Staff: Mentor

The string tension is not the only force acting on the yo-yo. (What prevents slipping?)

18. Dec 14, 2008

### Vuldoraq

I think I get it now. I found a reel of wire and did the experiment myself. Most interesting.

I guess it's because the tension always acts at that point of the yo-yo (ignoring changes due to the rope winding), and so it tries to pull it forward, but rather than slide the yo yo rolls (since it can?). If it were a square object in the same situatuation there would be no confusion, but the circular nature of the yo yo adds a (illusiory) complication.

19. Dec 14, 2008

### Vuldoraq

Friction would prevent the slipping. Is it the case that here the friction transforms translational motion into rotational motion?

20. Dec 14, 2008

### Dr.D

There is a key phrase in the original problem statement that is absolutely pregnant with meaning. It says, "A gentle horizontal pull is exerted..." Now, without saying so in so many words, what this is tell the reader is that this is a problem involving rolling friction, not sliding friction. This makes all the difference in the world.

Because the Yo-Yo rolls (that what the problem said, indirectly), the point in contact with the horizontal surface is a point momentarily at rest. This is what makes it permissible to sum moments about this point. If it were a moving point, a much more complicated relation would be required, but it is not; it is a point momentarily at rest and the proper equation of motion can be written by summing moments about this point.

If you pull on the string in an ungentle manner, such as to cause the Yo-Yo to slip on the surface, your best bet for writing the equations of motion is to the the sum of horizontal forces and the sum of moments about the center of mass (presumed to be at the center of the Yo-Yo). If you try to write a moment equation about the contact point, an incorrect result will follow (unless you are very sophisticated about it), because this is an accelerated point when slipping is involved.

When there is no slipping, the Yo-Yo has no option but to move in the direction of the applied force, the direction of the string pull. When there is slipping, then the Yo-Yo can rotate as well as translate, and the net motion can be in either direction, depending on the combined effects of rotation and translation.

This is a somewhat subtle problem, and you have to understand how to read the code in order to work the problem. In particular, you have to understand the implications of "A gentle horizontal pull". This sort of code has been embedded in mechanics textbooks for generations, and everyone who has done these problems for many years knows the code. Those who are new to the code just have to learn it; it is a part of the apprenticeship.