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When a yo-yo is released from a height $h$, the gravitational potential energy is converted to kinetic energy. However, the yo-yo obviously has less acceleration than $g$, $9.8\frac{m}{s}$. This means that the final speed of the yo-yo is less than what it would be if it were in free fall. It seems that the kinetic energy at the end is less than the potential energy at the beginning. What happened to the rest of the energy?

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The "missing" kinetic energy is still there ... it's now in the rotation of the yo-yo.

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    $\begingroup$ Any idea how angular momentum is conserved in the process of spinning up the yo-yo? $\endgroup$ – Brandon Enright Sep 1 '14 at 2:49
  • $\begingroup$ The throw imparts angular momentum (A.M.) to the yo-yo while reducing the A.M. of your fingers. Your fingers are attached to your very-massive body which stands on the very massive Earth, so the loss of A.M. by your body is negligible. While falling: best to view the system in the frame attached to the yo-yo. The yo-yo spins up, while the string is pulled away. Since the string is offset from the center of rotation (by the radius of the axle), it carries A.M. of the opposite sense of the spin. Total A.M. is zero. $\endgroup$ – garyp Sep 1 '14 at 11:48

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