Yo-yo (maxwell)

1. Dec 18, 2004

Feynmanfan

Yo-yo

Dear friends,

Here's a problem I can't solve. The disk on the picture falls under influence of gravity. The horizontal table creates no friction and the string that binds mass m and the disk is rolled up around the disk. The center of mass of the disk falls vertically.

My question is: what's the condition it has to satisfy, so that this last statement is true?

Do you think it's true that the disk will fall faster if it were attached to the pole?

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Last edited: Dec 19, 2004
2. Dec 18, 2004

dextercioby

Advice:apply Newton's second law for the for the body and for the two disks in both cases (of attachment of the string:either to the side/pole).It actually gives 6 equations.Compare the linear accelerations for the disk in the two cases.The part with the condition i don't understand,though.I guess it's the statement "The center of mass...vertically".The only condition is that the momentum of gravity (of the disk) is null.Besides the tension(which actually tends to rotate the disk) and the gravity i don't see other forces acting on the disk...

Daniel.

3. Dec 18, 2004

Feynmanfan

WHat do you mean that the momentum is null. If it were so ,it wouldn't rotate, would it?
As this is a rigid solid problem, do you think it can be solved using the energy conservation principles?

4. Dec 18, 2004

dextercioby

1.Yep,friend,but the gravity's moment doesn't (normally,at vertical drop/fall of the disk) rotate it (cause it's zero),it's the string's tension moment that rotates the disk.
2.Yes,u can do that.But again i tell u i don't really understand that part with the condition.But,if u do,then it's okay,because u know what to do.

Daniel.

5. Dec 18, 2004

Feynmanfan

I thank you for your help dextercioby; but I'm still a bit confused. I can't see the difference between both problems (the wheel attached to the mass and the regular yo-yo wheel).

I found out that the yo-yo attached to the pole(or just a wall) has a v=(4gh/3)^(1/2)
But I find myself unable to solve the other problem (although it seems a very simple one). How would you calculate both the block's and the disk's movement?
Are the tensions that act on both bodies the same (the lenght of the string changes as it unrolls).

As you can see, I'm quite lost in these rigid body world!

6. Dec 18, 2004

Feynmanfan

Here's what I've thought about the condition so that the mass center moves vertically: the initial velocity must be zero. If this is so the string moves at the velocity of the mass center and there's no slipping.

Do you think it makes sense??

7. Dec 19, 2004

Feynmanfan

I'd be grateful if you could help me

8. Dec 19, 2004

Staff: Mentor

In both cases, the acceleration of the center of mass of the falling disk can be obtained by applying Newton's 2nd law. When the rope is attached to the sliding mass (m): Call the acceleration of m $a_1$ and the acceleration of M $a_2$. The acceleration constraint linking the two masses is $a_2 = a_1 + \alpha R$. Perhaps now you can solve it?

9. Dec 19, 2004

Feynmanfan

Thanks a lot Doc Al for your help. I'm sorry but I didn't quite understand the acceleration constraint. Could you tell me where you got that from (it might be trivial but i can't see it!)

What do you think it's the condition it must satisfy so that the center of mass of M moves vertically.

gracias.

10. Dec 19, 2004

Staff: Mentor

When the disk turns, it accelerates with respect to the rope (that's the $\alpha R$ term), but the rope itself accelerates (that's the $a_1$ term).

I'll have to think about that one!

11. Dec 19, 2004

Feynmanfan

I get the same answer in both cases a2=2g/3

Does that mean that it doesn't matter if the disk is attached to the wall (the pole) or to the block of mass m?

Intuitively it seems to me the disk should fall faster if the string were attached to a fixed point (the pole). AM I right?

THanks again

12. Dec 19, 2004

Staff: Mentor

That's the correct answer when the rope is attached to the pole, but not for when it's attached to the sliding mass.
No. It does matter. (Do it over!)

13. Dec 19, 2004

Feynmanfan

Well, after all these hints you've given me. Let me see if I got it right.

TR=1/2MR^2alpha------>>T=ma1=1/2MRalpha
and Mg-T=Ma2

we get a2=g-Ralpha/2 (if R.alpha=a2 we have the disk attached to the pole)

using a2=a1+alphaR we have

a2=g-(2ma1/M) and also a2=a1+(2ma1)/M

Doing some algebra I get

a2=g(2m+M/3m+M)

If I make m very very massive so that it cannot move, then I get the above result.

I'm anxious to read your opinion. (I don't know if you say that in English)

14. Dec 19, 2004

Feynmanfan

By the way, we have assumed that the center of mass of the disk moves vertically. Under what conditions wouldn't it be so?

We assume that the pole and string are massless. WHat would happen if the disk had an initial velocity?

15. Dec 19, 2004

Staff: Mentor

Right! As you noticed, when m is massive the answer is exactly what you would get if the rope were attached to the pole, which makes sense. Also notice that in the limit as m goes to zero, the acceleration approaches g, which also makes sense.

16. Dec 19, 2004

Feynmanfan

That was good. You acted like Socrates, trying to make me find out the answer. However, I'm still confused about the condition it must satisfy so that the center of mass moves vertically.

WHat do you think of that? I would answer that the initial velocitymust be directed downward, too and that there's no other moment than that of the tension.

Does that make sense?

17. Dec 19, 2004

Staff: Mentor

For a mass to move in a particular direction, the velocity and force normal to that direction must = 0.

So as long as the resultant force acts downward, the CM must move in the direction of that force.