Conservation of Energy

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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 [itex](1/2)mv^2+(1/2)I\omega^2[/itex]] is not conserved, because of the force the string applies on the Yo-Yo.
Is my intuition is right?
 
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  • #2
Doc Al
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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 [itex](1/2)mv^2+(1/2)I\omega^2[/itex]] 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.
 
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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 [itex] F\centerdot \Delta x [/itex]
 
  • #4
Doc Al
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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 [itex] F\centerdot \Delta x [/itex]
Work is better thought of as [itex] F\centerdot \Delta x [/itex], where [itex] \Delta x [/itex] is the displacement of the point of application of that force. But the point of application is not moving.

You can certainly calculate [itex] F_{net}\centerdot \Delta x_{cm} [/itex], but that's not the same thing as the work done by the forces.
 
  • #5
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Work is better thought of as [itex] F\centerdot \Delta x [/itex], where [itex] \Delta x [/itex] is the displacement of the point of application of that force. But the point of application is not moving.

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

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

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