# Calculate the potential energy of a rope hanging

1. Dec 17, 2006

### Gregie666

Hi.
i need to calculate the potential energy of a rope hanging from the cieling. the rope is uniform. the mass of the rope is M. the length of the rope is H. the bottom tip of the rope is the origin of the coordinate system.
i'm *not* supposed to do it by treating the rope as a dimentionless object in the in its center of gravity.
(0.5MgH)

2. Dec 17, 2006

### cristo

Staff Emeritus
Welcome to PF. For homework questions, we ask that you show some working/thoughts about the question, as we can't simply answer it for you! In future try to use the template you were provided with:

3. Dec 17, 2006

### Gregie666

theres obviously an integral involved.. i think that the integral is supposed to be dy (when y is the height above 0).
in every slice of rope the potential energy is y*dy/M*g. but if i integrate it this way it gives (0.5/M)gy^2. which is wrong.

4. Dec 17, 2006

### Ja4Coltrane

is the bottom of the rope touching the floor?

5. Dec 17, 2006

### Ja4Coltrane

I assume so--you are right about the integration but I think you are confusing yourself with it.
the potential energy of each slice of height y is (dm)gy. does that get you started?

6. Dec 17, 2006

7. Dec 17, 2006

### cristo

Staff Emeritus
The bottom tip of the rope is the origin of the coordinate system so, in effect, yes.

8. Dec 17, 2006

### Ja4Coltrane

of course--sorry

9. Dec 17, 2006

### Gregie666

the calculus in the example is too complex for me to follow beyound the initial few steps(this is only my first semester in college).
but still it implies that:
du=gydm=gyM/Ldy (when the differential element of length is dy)
can anyone tell what i am doing wrong?

10. Dec 17, 2006

### cristo

Staff Emeritus
The question in the post is in 2d and so that's why the calculus is harder. You seem to have followed it well though, and the expression you have above, namely, $$dU=\frac{gym}{H}dy$$ is correct.

Now, can you integrate this?

11. Dec 17, 2006

### Gregie666

i think so. but either i do it incorrectly or its just wrong.
$$\int {{{gyM} \over L}dy} = {{gM} \over {2L}}y^2$$
but its supposed to be 0.5MgH

12. Dec 17, 2006

### cristo

Staff Emeritus
You've got the right idea, but note the limits of integration (remember in the above, if you're not using limits of integration, you're missing a constant).

$$U= \int_0^H \frac{gyM}{H} dy$$

So, the limits go from 0 to H (as the length of the rope given in your original question is H, not L)

Can you finish off from here?

13. Dec 17, 2006

### Gregie666

AH! ofcourse.... i get it now. thatnk you very much.

14. Dec 18, 2006

### Ja4Coltrane

I know you get it, but here is another way to do this:
sum up all of the mass elements times g time each respective height h, change dm to (linear density times dh) then change linear density to M/H and then integrate and youve got it.