How Does Angular Velocity Change in Different Yo-Yo Maneuvers?

[Physics122]

Homework Statement

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.

Homework Equations

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

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

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)

(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!

 

[Physics122]

Just realized that I probably should have taken into account the Inertia of the lighweight middle area but that will just change the general format of the first problem. It's the second problem that remains confusing