# Homework Help: Rotational motion of a yo-yo

1. Nov 24, 2008

### chronorec

1. The problem statement, all variables and given/known data
A yo-yo is made from two uniform disks of raduis R with a combined mass of m. A short massless shaft of radius r connects the disks. A long thin string is wrapped around the shaft several times by a yo-yo player, who releases it with zero speed. Assuming that the string is vertical at all times, find the tension in the string during the descent and subsequent ascent of the yo-yo.

2. Relevant equations
moment of inertia of the yoyo is mR2/2

3. The attempt at a solution
Another one of my exam questions.
Let tension in string be T.
Tr=(mR2/2)(a/r)
mg-T=ma
Solving the 2 equations, I get T=(mR2g/(R2+2r2)
For ascent, T-mg=ma
Solving the equations again, I get T=(mR2g/(R2-2r2)
Would like to know if this is the correct way to do this

2. Nov 24, 2008

### naresh

Looks good to me.

(Why is the angular acceleration a/r?)

3. Nov 24, 2008

### chronorec

I'm assuming it's pure rotational motion with no slipping, hence a=r(alpha)
however I have doubts on the ascending part, is the acceleration of the yoyo really upwards or downwards? I'm taken it to be upwards in my calculations

4. Nov 24, 2008

### naresh

Yes, that's right (kind of). Why does it roll without slipping here? Can it slip? Do you see the geometrical constraints?

5. Nov 25, 2008

### chronorec

I don't really get what you mean by geometrical constraints.
My understanding of the question is such that the string is wound tightly around the shaft so that slipping will not occur.
If the end of the string is not connected to the shaft itself, then I think it may be possible that slipping can occur. However, I'm not sure how to apply this to the equation of motions for the yoyo

6. Nov 25, 2008

### naresh

Maybe the attached picture (ignore the fact that it is called pulley and not yoyo) will help a bit to explain what I meant by geometrical considerations.

Let us choose the direction of the velocity arbitrarily - I have chosen different directions for the descent and ascent. This doesn't matter, you just need to be consistent.

On the descent,
mg-T = ma: the force mg - T increases v, therefore a is positive.
Tr = I $$\alpha$$: The torque increases the angular velocity.
v = r$$\omega$$: If $$\omega$$ is in the direction shown, so is v.

On the ascent,
(maybe you can fill this in?)

Notice that in the expression you have for the ascent, T = mR2g/(R2-2r2), if I set R = r$$\sqrt{2}$$ (which is a perfectly legal choice), you end up with a superstring. (Sort of like Superman, not physical)

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