Gravitational Potential of Hanging Cord

AI Thread Summary
The discussion focuses on calculating the change in gravitational potential energy of a uniform cord as it transitions from being stuck to the ceiling to hanging vertically. The formula used for gravitational potential energy is Ug = mgh, and the solution involves integrating a differential mass element. The final expression derived for the change in potential energy is mgh/2, which is confirmed as correct. The key point of confusion is the definition of the height variable, h, which needs clarification for completeness. The calculations and reasoning presented are accurate for the problem at hand.
NotZakalwe
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Homework Statement


A uniform cord of length .25 meters and mass .015 kg is initially stuck to a ceiling. Later, it hangs vertically from the ceiling with one end still stuck. What is the change in gravitational potential energy of the cord with this change in orientation?

Homework Equations


Ug = mgh

The Attempt at a Solution


The initial gravitational potential energy of the cord is mgh
Considering a differential mass element of the cord $$\frac{m}{l}$$ the gravitational potential energy of the cord after the change in orientation is

$$\int^{h}_{h_0} \frac{m}{l}gh = \frac {\frac{m}{l}gh^2}{2} - \frac{\frac{m}{l}g{h_0}^2}{2}$$

Taking $$l = h$$ $$h_0 = 0$$ this simplifies to

$$\frac{mgh}{2}$$

Subtracting the final gravitational potential energy from the initial gravitational potential energy yields

$$\frac{mgh}{2}$$

Is this correct?
 
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Not sure as you didn't define what h is?
 
NotZakalwe said:

Homework Statement


A uniform cord of length .25 meters and mass .015 kg is initially stuck to a ceiling. Later, it hangs vertically from the ceiling with one end still stuck. What is the change in gravitational potential energy of the cord with this change in orientation?

Homework Equations


Ug = mgh

The Attempt at a Solution


The initial gravitational potential energy of the cord is mgh
Considering a differential mass element of the cord $$\frac{m}{l}$$ the gravitational potential energy of the cord after the change in orientation is

$$\int^{h}_{h_0} \frac{m}{l}gh = \frac {\frac{m}{l}gh^2}{2} - \frac{\frac{m}{l}g{h_0}^2}{2}$$

Taking $$l = h$$ $$h_0 = 0$$ this simplifies to

$$\frac{mgh}{2}$$

Subtracting the final gravitational potential energy from the initial gravitational potential energy yields

$$\frac{mgh}{2}$$

Is this correct?
I
 
It is correct.
 
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