PE of Hanging Cord: Solve 8th Ed. Ch.8 pr.35

[TimH]

Homework Statement

This is Halliday and Resnick 8th Ed. Ch.8 pr.35. You have a cord of mass 15g and length 25cm that is stuck to a ceiling. Later it hangs vertically from one end. What is the change in potential energy of the cord due to the change in orientation? The problem as a hint says to consider a differential slice of the cord and use an integral.

Homework Equations

Potential Gravitational Energy = m g y, where y is the height above the reference point.

The Attempt at a Solution

Okay this is a "three star" problem in the book (=hard), probably because you have to set up the integral, which I can't seem to do for some stupid reason-- there is something (probably basic) I am not getting. We want the change in potential energy so we need to compute the PE in the two orientations and subtract to get the difference.

When the cord is hanging vertically it is .25m long so we will use the bottom of the (vertical) cord as the reference point for zero potential energy. So when the rope is stuck to the ceiling (and I assume they mean it is stuck horizontally to the ceiling, otherwise there is no "change in orientation") the potential energy is m g h, with h being .25m since it is this high above the reference point. This gives a PE of .015kg * 9.8m/s^2 * .25m = .03675 J.

Now for when the rope is hanging vertically. The potential energy of a differential slice of the rope is U= m g x dx. [Yes?] So the integral as we move up the cord is:

mg * [integral from 0 to .25] x dx. This is mg * [(x^2)/2] evaluated at .25 = .00459 J. This value seems to be way too small and leads to a wrong answer when you subtract it from the first PE. I'm computing the area under the curve U=mg x, which is going to be small because I'm squaring fractions of a meter. So I'm confused setting this up somehow.

The correct final answer is .018 J, which is half the PE of the horizontal cord, so the PE of the hanging cord is .018. Now if we imagine that all the mass of the hanging cord were at a point halfway between the bottom (zero PE) point and the ceiling, then we get this correct answer. But what is wrong with my integral? (There's probably something dumb I'm missing). Thanks.

 

[tiny-tim]

Hi TimH!

Now for when the rope is hanging vertically. The potential energy of a differential slice of the rope is U= m g x dx. [Yes?]
…
(There's probably something dumb I'm missing).

That's no … and yes.

U= (m/L) g x dx.

 

[TimH]

Okay I think I get it. Since I want an equation for the gravitational potential of a slice = dx, its got to have a slice of the mass, not the whole thing, so its the total mass divided by the length for a mass-per-length. I did it out with your equation and got the right answer. Thanks! Would it be more correct to say the equation is dU= (m/L) g x dx? I just mean its giving you the differential element of potential for a change dx, yes?

 

[tiny-tim]

Okay I think I get it. Since I want an equation for the gravitational potential of a slice = dx, its got to have a slice of the mass, not the whole thing, so its the total mass divided by the length for a mass-per-length. I did it out with your equation and got the right answer. Thanks! Would it be more correct to say the equation is dU= (m/L) g x dx? I just mean its giving you the differential element of potential for a change dx, yes?

Yes, that's exactly right!