Gravitational Potential of Hanging Cord

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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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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?
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