A yo-yo modeled as a uniform disk (I = \tfrac{1}{2}MR^2) descends as the string unwinds from its rim. What is the linear (downward) acceleration of its center?

\tfrac{2}{3}g Newton's 2nd: Mg - T = Ma. Torque: TR = I\alpha = \tfrac{1}{2}MR^2(a/R), so T = \tfrac{1}{2}Ma. Then a = \tfrac{2}{3}g.

Rolling

AP Physics 1· difficulty 4/5

I = \tfrac{1}{2}MR^2

Explanation

Mg - T = Ma

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