# Yo yo motion and energy conservation I am struggling with balancing energy in a yo-yo. So, we have a yo-yo (massless string wrapped around solid cylinder). It is allowed to fall through a distance $$h$$ without rotation. The loss in potential energy $$= mgh$$ will be converted into kinetic energy (KE) and we can find the KE after it has fallen through a distance $$h$$.

Now, we allow the yo-yo to fall like a yo-yo so that it moves the same distance, but also rotates. So, again, we apply the conservation of energy. Now, the KE of the yo-yo will have 2 parts (i.e. due to the rotation around the center of mass and due to the translation of the center of mass). We find that the final KE in this case is 1.5 times than that of the previous case.

My question is, how come the 2nd case has more final energy when the initial energy in both the cases was $$mgh$$?

• How did you get this answer ? Can you show it and include it in the question? – Stratiev Jun 12 '20 at 12:10
• @Stratiev added thx – user31058 Jun 12 '20 at 12:37

## 1 Answer

Obviously, the yo-yo's energy has to be conserved. The potential energy of the yo-yo at the stationary starting point where it starts to move down ($$mgh$$, $$h$$ being the length of the rope) is converted in a linear and a rotational kinetic energy part (as you wrote). The linear acceleration downwards though is less than in free fall. This means that the yo-yo's linear kinetic energy when it's rolled off completely is less than $$mgh$$. The difference with $$mgh$$ is contained in the yo-yo's rotational energy.
Because you took the kinetic energy to be the same in both cases (free fall vs constrained fall), you took $$g$$ as the downward acceleration in both cases, which isn't the case.
In fact, the equipartition theorem implies the potential energy is distributed in equal amounts of kinetic energy among the two degrees of freedom of the motion (vertical linear motion and the rotation around one axis).
All this means the KE in the first case (constrained motion with rotation) is equal to the KE in the second case (free fall without rotation) (the kinetic energy you get in the second case becomes half that value in the first case, i.e. $$\frac{1}{4} m{v_{cm}}^2$$). So the only mistake you made is assuming the linear kinetic energy at the end of the roll is equal in both cases.