Is Energy Conserved in a Falling Yo-Yo?

[estro]

Suppose that I hold a Yo-Yo and then I leave it to fall [and of course rotate] holding only the string.
I seems to me that the energy [not the whole of the potential energy will be converted to (1/2)mv^2+(1/2)I\omega^2] is not conserved, because of the force the string applies on the Yo-Yo.
Is my intuition is right?

 

[Doc Al]

Suppose that I hold a Yo-Yo and then I leave it to fall [and of course rotate] holding only the string.
I seems to me that the energy [not the whole of the potential energy will be converted to (1/2)mv^2+(1/2)I\omega^2] is not conserved, because of the force the string applies on the Yo-Yo.
Is my intuition is right?

I would say no. Just because the string exerts a force on the yo-yo does not mean that work is being done. Consider the speed of the end of the string as it unravels.

 

[estro]

I would say no. Just because the string exerts a force on the yo-yo does not mean that work is being done. Consider the speed of the end of the string as it unravels.

Regarding what you said about the work:
The yo-yo goes down so the string force does negative work. So the force does not "spend" energy, right?

Regarding speed of the string:
The string has no speed after it unravels, but I don't understand how is this related to work. As work is only F\centerdot \Delta x

 

[Doc Al]

Regarding what you said about the work:
The yo-yo goes down so the string force does negative work. So the force does not "spend" energy, right?

Regarding speed of the string:
The string has no speed after it unravels, but I don't understand how is this related to work. As work is only F\centerdot \Delta x

Work is better thought of as F\centerdot \Delta x, where \Delta x is the displacement of the point of application of that force. But the point of application is not moving.

You can certainly calculate F_{net}\centerdot \Delta x_{cm}, but that's not the same thing as the work done by the forces.

 

[estro]

Work is better thought of as F\centerdot \Delta x, where \Delta x is the displacement of the point of application of that force. But the point of application is not moving.

You can certainly calculate F_{net}\centerdot \Delta x_{cm}, but that's not the same thing as the work done by the forces.

Great explanation [cleared a lot of confusion for me], thanks!