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December 22, 2014

December 22, 2014

Posted by **Sasha** on Monday, November 29, 2010 at 2:05am.

- physics -
**drwls**, Monday, November 29, 2010 at 3:02amThe decrease in potential energy equals the increase in kinetic energy, most of which is rotational KE. First compute the moment of inertia of the yo-yo, 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.

- physics -
**Sasha**, Wednesday, December 1, 2010 at 11:06pmI still don't get it.Moment of inertia of the yo-yo 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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