Calculate the potential energy of a rope hanging

[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 dimensionless object in the in its center of gravity.
(0.5MgH)

 

[cristo]

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:



So, what are your thoughts about the question?

 

[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.

 

[Ja4Coltrane]

is the bottom of the rope touching the floor?

 

[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?

 

[radou]

This is a similar problem, you should find it useful: http://web.mit.edu/usagi/www/catenary.htm".

 

[cristo]

is the bottom of the rope touching the floor?

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

 

[Ja4Coltrane]

of course--sorry

 

[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?

 

[cristo]

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?

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?

 

[Gregie666]

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

 

[cristo]

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

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?

 

[Gregie666]

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

 

[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.