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# Angular velocity of a yo-yo!?

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

A certain yo-yo can be modeled as a uniform cylindrical

disk with mass M and radius R and a lightweight hub of

radius ½R. A light string is wrapped around the hub.

(a) First, the yo-yo is allowed to fall. Find the angular

velocity of the yo-yo when the string has unwrapped a

distance L.

(b) Now, imagine that that you pull upward on the string such

that the yo-yo remains in the same place. Find the angular

velocity of the yo-yo when you have pulled the string

upward a distance of L.

(c) Explain in words why it makes sense that the answers to

parts (a) and (b) are different.

2. Relevant equations

K (total) = .5 * I (center of mass) *w^2 + .5MR^2

I cm for a uniform cylindrical hub = .5M(R^2 + (.5R)^2)

3. The attempt at a solution

(A)

K (total) = .5 * I (center of mass) *w^2 + .5MR^2 = MgL

W^2 = MgL/ (.5 *I (cm) + .5MR^2)

I cm for a uniform cylindrical hub = .5M(R^2 + (.5R)^2)

So..

W^2 = MgL/(.5 * (.5M(R^2 + (.5R)^2) + .5MR^2)

W^2 = MgL/(1/4Mr^2 + 1/16MR^2 + 1/2MR^2)

W^2 = gL/(13/16R^2)

Does that seem about right? (obviously need to make it the square root but just leaving it squared for now). Actually maybe I should take into account the lightweight hub (but that is just a matter of adding another Inertia :)

(b)

I assume I cannot use conservation of energy, so maybe I could solve this with the Work that is done? I dunno, I'm confused I guess I don't know where to begin

Thanks for any help fellow physics buds!

### 1 Answer

- 1 decade agoFavorite Answer
I won't try to do the whole thing, but I will answer part (c). In the case where we pull the string so the centre of the yo-yo stays put, in building up angular speed, it only has to overcome its angular momentum. From the start of the pull, the force on the rim of the hub from the string is equal to the weight of the yo-yo. However, in the case where we hold the end of the string still and allow the yo-yo to start falling, in building up angular speed, the yo-yo must overcome, not only its angular inertia, but also its inertia. The force at the rim of the hub from the unwinding string changes gradually in this case.

As for your remark that you assume you cannot use conservation of energy, I don't see why you would so assume. Conservation of energy applies.

In the case of pulling the string up, the energy is the weight of the yo-yo times the length L. All this energy will manifest in the rotation.

In the case of allowing the toy to fall, the energy is the same. However, part of it goes into rotation and part into continued falling. The energy of continued falling you can relate to the rotational speed because you know the radius of the hub.