physics
posted by Sasha .
A yoyo is made of two solid cylindrical disks, each of mass 0.053 kg and diameter 0.073 m, joined by a (concentric) thin solid cylindrical hub of mass 0.0052 kg and diameter 0.012 m. Use conservation of energy to calculate the linear speed of the yoyo when it reaches the end of its 1.1 m long string, if it is released from rest.

The decrease in potential energy equals the increase in kinetic energy, most of which is rotational KE. First compute the moment of inertia of the yoyo, I.
If it falls a distance L (which is 1.1 m in this case) then
M g L = (1/2) I w^2 + (1/2) M V^2
The angular rotation rate w can be replaced by V/r, where r is the radius of the cylindrical hub. Then solve for V.
V^2[1 + I/(Mr^2)] = 2*g*L
Take it from there. 
I still don't get it.Moment of inertia of the yoyo is 1/2 mr^2: 1/2(.053kg)(.0365^2) + 1/2(.0052kg)(.006^2)
which equals to 3.54*10^5
so i plugged that into = square root of (2*9.8*1.1)/ (1 + (3.54*10^5/(mr^2))
and i got something like 4.6 which is not a correct answer. Can you help me clarify this question please?
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