Change in potential energy problem

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
7 replies · 5K views
Cherrybawls
Messages
18
Reaction score
0

Homework Statement


A uniform cord of length 25cm and mass 15g is initially stuck to a ceiling. Later, it hangs vertically from the cieling with only one end still stuck. What is the change in the gravitational potential energy of the cord with this change in orientation? (Hint: Consider a differential slice of the cord and then use integral calculus)


Homework Equations


Ug=mgh


The Attempt at a Solution


Well in calculus we have only recently started integrals so my understanding of integrals is shaky at best, so i am not really sure how to apply them to this problem... any push in the right direction would be appreciated
 
Physics news on Phys.org
A better idea, use centre of mass instead.
 
Hi Cherrybawls! :smile:
Cherrybawls said:
A uniform cord of length 25cm and mass 15g is initially stuck to a ceiling. Later, it hangs vertically from the cieling with only one end still stuck. What is the change in the gravitational potential energy of the cord with this change in orientation? (Hint: Consider a differential slice of the cord and then use integral calculus)
aim1732 said:
A better idea, use centre of mass instead.

Yes (you could lower the whole cord half-way, and then rotate it!), but you'd better do it the way they've hinted, since it's to give you practice (which you need! :wink:) at integration.

The general method is to slice the length (or area or volume) into bits of thickness d(something), treat each bit separately, and then "add" all the bits.

So choose a slice between x and x + dx, decide how much PE it gets, and integrate. :smile:
 
so would I be right in writing:

Ug= ∫0.25mgx dx? If I set the lowest point of the rope equal to 0 and x cooresponds to the height of the rope?
 
Cherrybawls said:
so would I be right in writing:

Ug= ∫0.25mgx dx? If I set the lowest point of the rope equal to 0 and x cooresponds to the height of the rope?

Yes, except the mass of the slice from x to x + dx isn't m, is it? :wink:
 
I'm not quite sure, is it another variable or is it something like m/x?
 
Cherrybawls said:
I'm not quite sure, is it another variable or is it something like m/x?

Well, it's not m/x (where x is the distance from the slice to one end of the cord), but it's something similar.

The mass of the whole cord is 15g, the cord has length 0.25m, so what is the mass of a bit of the cord of length dx ?
 
m/dx? I am just gessing really...