Eng. Mechanics I Bonus question

[CJSGrailKnigh]

Homework Statement

A human with mass “m” lives at the flat end
of a huge cylinder of mass “M”,
diameter “D” and length “L”
An asteroid collides with this cylinder
destroying one half of its length.
Does the human weigh less or more
after the asteroid impact? How much?

sorry about set up of question I had to copy from a bad .pdf file

The Attempt at a Solution

Assume the following:
1) The cylinder has an even distribution of mass M
2) The asteroid causes enough damage to thoroughly destroy and scatter the matter that the second half of the cylinder comprised of.
3) The system is far enough away from any other body that any force of gravity is essentially 0.
4) As in the diagram the human lives on the flat end of the cylinder


F = (GmM)/(0.5L)^2

f = [Gm(0.5M)]/[0.5 * (0.5L)]^2
2f = (GmM)/[0.25 * (0.5L)^2]
2f = 4 * (GmM)/(0.5L)^2
0.5f = F
therefore
f = 2F

The force acting on the person or his weight is doubled because the both the length L and the mass M are halved if the distance from the center of gravity was not squared in the universal gravitation equation this would result in no change however since the change due to the halving of the distance is squared the result is that the force doubles.

 

[Delphi51]

Knigh, that is an interesting problem indeed - congrats to your teacher!

Your calculation is correct in all its details. But the calculation is actually for a different problem because F = GmM/r^2 applies to points of mass - and with 3D calculus you can show it also applies to spheres of mass. Your solution describes a planet that has half its mass taken away and the other half formed into a sphere with half the original radius. Density?

It appears that you would need 3D calculus (triple integral) to do the cylinder problem.
A computer calculation is awkward because splitting the cylinder into 1000 slices would not be very accurate, especially for the slices close to the person because the distance from the person to different parts of the slice would vary quite a bit. This kind of thing is done for models of underground salt domes and other geological features in order to understand maps of surface gravity. Oil and mineral exploration.

An armchair solution with a qualitative answer may also come to mind.

 

[Delphi51]

the force of gravity must be of the form C times G m M / L^2 where C is a messy constant

Oh, interesting thought! I haven't done 3D calculus for more than 30 years so I can't say for sure. My instinct says the C would depend strongly on the relative length and radius of the cylinder - which changes drastically in this problem.

 

[nrqed]

Oh, interesting thought! I haven't done 3D calculus for more than 30 years so I can't say for sure. My instinct says the C would depend strongly on the relative length and radius of the cylinder - which changes drastically in this problem.

Hi!

After thinking about it for a second, I realized that you are right, the C may depend on the length L (so I earsed my previous post in order not to confuse the OP). The point I had missed is that there is a second dimesnionful quantity involved in the calculation, which is the radius R. This makes the whole thing much more complicated because the general form is

F = C G mM/(a L^2 + b LR + c R^2)

with C possibly a function of the dimensionless ratio R/L.

Therefore dimensional analysis does not help here. Thanks for pointing out my mistake.

Regards

Patrick

 

[CJSGrailKnigh]

Wow I don't think I've ever seen the equation with C in it. This is only a first year mechanics course so i don't know if they want us to do triple integrals and such as we've only really started diff. eqns and infinite series this term. Either way my answer seems to be the general consensus.

And you I must congratulate my prof as well, some of my fellow students are giving really complicated answers involving concepts that I didn't even think applied.

Thanks for looking over my algebra and offering your thoughts on the problem

 

[Delphi51]

Either way my answer seems to be the general consensus.

Oh, dear - not the double force answer, I hope. If you think about the slice model - with your mind instead of a computer - I think the answer will become clear. What happens to the gravitational force on the person as you take away slices of the cylinder?

If the prof gave you the 2F idea from F = GmM/r^2, he or she will be looking for an explanation of why that is the answer for the sphere problem but not the cylinder problem. I know I just couldn't rest until I understood why you get double the force when you take half the mass away! The key is to calculate the density of the new planet after half the mass is taken away. (I'm feeling really guilty over giving you such a fabulous hint on how to earn those bonus marks, but of course you learn nothing unless you try it!)

 

[nrqed]

Wow I don't think I've ever seen the equation with C in it. This is only a first year mechanics course so i don't know if they want us to do triple integrals and such as we've only really started diff. eqns and infinite series this term. Either way my answer seems to be the general consensus.

And you I must congratulate my prof as well, some of my fellow students are giving really complicated answers involving concepts that I didn't even think applied.

Thanks for looking over my algebra and offering your thoughts on the problem

It is clear that 2F cannot be the correct answer, though. Think about it: initialy the whol;e cylinder was pulling on the person. Then half the cylinder is removed. The force cannot be larger than before!

 

[microguy]

@ nrqed: But you are leaving out half the puzzle, which is the person is half the distance from half the mass.

If half the Mass only: f/F = 1/2
If half the d only: f/F = 4
If both: ...

 

[CJSGrailKnigh]

I have to agree with microguy given my assumptions

Assume the following:
1) The cylinder has an even distribution of mass M
2) The asteroid causes enough damage to thoroughly destroy and scatter the matter that the second half of the cylinder comprised of.
3) The system is far enough away from any other body that any force of gravity is essentially 0.
4) As in the diagram the human lives on the flat end of the cylinder

The loss of half the length of the cylinder does in fact result in the loss of half the mass. The loss of half the mass does indeed lose half the force but the loss of distance to the center of gravity by half will quadruple the force.

Also quick question to Delphi51 i don't understand where this sphere your talking about is... maybe I just don't understand the triple integral thing I have never even heard of this kind of math. My experience with calculus is with basic differential equations, differential and integral calculus. I mean I know how to find volume and such with various methods. The only other 3d stuff (and more dimensions) I know is from linear algebra. Could you please explain this sphere to me please.

Also i realize that wasn't a quick question now so my bad.

 

[microguy]

Forum, please correct me where I am wrong in my thinking.

Forget about cylinders, spheres, slices, and asteroids. If I am distance D1 from point mass M1 I weigh F1. If I am distance D2 from point mass M2 I weigh F2. The ratio F2/F1 invokes only D1, M1, D2 and M2.

We can treat the "cylinder" as a point mass because it has uniform density. That let's us evaluate D very easily. It doesn't matter if the "cylinder" is a sphere, tetrahedron, soccer ball, or a pear.

 

[Delphi51]

You cannot use the F = GmM/r^2 formula with a cylinder.
You CAN use it for spheres. Thus the calculation GrailNigh did applies ONLY to spheres. He did the planet problem I described, but NOT the cylinder problem.

The conditions such as "half the distance" cook up a very curious situation with those planets - work out the planet densities to see it - which in no way exist in the cylinder problem.

Sorry I mentioned 3D calculus - only needed to get a quantitative answer to the problem, not for a qualitative one.

We can treat the "cylinder" as a point mass because it has uniform density.

No. Spherical symmetry is required.

Imagine mass being chipped away from the opposite side of the cylinder. You can't use the formula to find the force of gravity on the person but you can use common sense. In fact, you can imagine chipping away the mass until there is nothing left. According to the 2F reasoning, the person should then feel an infinite force. In fact he feels no force.

 

[microguy]

I am quite flummoxed by this problem. Quite possibly, I do not recall where the gravitational force equation is supposed to work. Also, where are you going with the density argument? The density of the planet is not changed (?), only its mass and size.

On one hand, indeed, it doesn't make sense that the limit of F goes to infinity as we "chip away" at M. On the other hand, What is the difference between chipping away at a planet, versus positioning ourselves at various planets where we are half the distance from half the mass? Maybe that's what you are trying to say. If not F=GmM/(r^2), then what?

 

[Delphi51]

The density of the planet is not changed (?), only its mass and size.

Yes it is! And drastically. Work it out! You'll see the real reason why the guy feels double the gravitational force.

What is the difference between chipping away at a planet...

The density doesn't change when chipping away. They guy remains the same distance as before from the part of the mass that remains so it has the same effect it originally did. The only change is that the effect of the chipped away mass is GONE. Think what effect it originally had, then subtract it away.

 

[microguy]

Not trying to be bull-headed but I am feeling lost here now...

(a) How does density change when M/V = (M/2)/(V/2)

(b) OK now you say density does not change. Huh? But this I agree with! But what I do not follow is "They [sic] guy remains the same distance as before from the part of the mass that remains". I contend he is half the distance from the center of mass that remains.

Are we just getting to the same answer in different ways?

 

[Delphi51]

How does density change when M/V = (M/2)/(V/2)

The V varies with the radius, which is reduced to half according to the original question. You must use 4/3*pi*r^3 in place of the V. You only want to compare the original planet with the one reduced to half mass and half the radius, so take the ratio of new density to original density and everything will cancel out except for a number, showing by what factor the density has increased.

now you say density does not change. Huh?

The density changes in the sphere problem but not in the cylinder problem. That is why the problems are fundamentally different.

 

[nrqed]

@ nrqed: But you are leaving out half the puzzle, which is the person is half the distance from half the mass.

If half the Mass only: f/F = 1/2
If half the d only: f/F = 4
If both: ...

I am not following. The person is on the flat end of a huge cylinder. Let's call this side the "top" of the cylinder. Now remove the lower half of the cylinder. There is less mass pulling on the person now so the force of gravity will certainly have decreased. Are we talking about the same problem?



Now, it won't be halved because the half of the cylinder taken away was farther from the person was farther than the half that remains. So the answer has to be some value between F/2 and F .

 

[nrqed]

The V varies with the radius, which is reduced to half according to the original question. You must use 4/3*pi*r^3 in place of the V. You only want to compare the original planet with the one reduced to half mass and half the radius, so take the ratio of new density to original density and everything will cancel out except for a number, showing by what factor the density has increased.


The density changes in the sphere problem but not in the cylinder problem. That is why the problems are fundamentally different.

The spherical case is not what the OP was asking about so I feel that this is only confusing the issue. But I just wanted to point out that even in the spherical case, if part of the palnet is removed, the density won't change (unless one is not treating the matter as incompressible).

 

[Delphi51]

So the answer has to be some value between F/2 and F .

Well, that's the answer for sure, Nrged! (Energy Ed?)

It wasn't hard - just picture a bunch of little Fg vectors acting on the man and pointing toward the little chunks of mass in the cylinder that cause them. As you chip away those chunks of mass, some of the Fg vectors disappear, reducing the total force of gravity on the man.

But that incredible 2F solution that was given with the question is much more interesting! It doesn't make sense, yet comes straight out of the Fg = GMm/r^2 formula. I don't know how anyone could sleep last night without figuring out why! That formula only applies to spheres, so you have to think about a sphere (planet?) "whose mass and radius are reduced to half". The density of the new planet compared to the original is
.5*m/(1.33*pi*(.5*r)^3)
------------------------- = .5/(.5)^3 = 4
m/(1.33*pi*r^3)

So, the 2F solution says that if a planet loses half its mass and the rest is compressed by a factor of 4 into a smaller sphere, then its surface gravity is doubled. It makes sense now - the surface is closer to the mass because it is so dense. But this math doesn't apply to the cylinder problem because it is not compressed. The person on the end is indeed only half as far away from the new center of mass, but is not any closer to this half of the mass than before (no increase in density) so no increase in Fg.

Note that if the whole mass is compressed into half the radius, the density is multiplied by 8 and of course is due to the r^3 in the volume formula. All volume formulas go as the cube of the linear size. And that is why elephant legs look fatter than ant legs relative to the size of the beast. If any of you didn't get that story in high school physics, figure it out for yourselves or look it up.

Physics is fascinating!

 

[microguy]

Whew! I'm not trying to beat this problem to death, but it was beating me to death, so I had to pursue it.

I re-read the Gravity chapter in my old textbook and realize--no--I can't approach the cylinder as a point mass/sphere (although the spherical result is rather interesting).

The results of Googling "gravitational attraction of a cylinder" suggest the proper computation of gravity in this problem is horrifically difficult.

So, what do to? The OP said complex math was probably not involved.

Falling back on the point-mass approach, how about this as an approximation: Think of the cylinder as 2 halves stacked lengthwise. Fg before the asteroid is the sum of two Fg components: one from a point 1/4 along the total length, the other from a point 3/4 along the total length. After the asteroid, there is only the first component. The ratio of Fg after/before works out to between 100% and 50%, as Nrged initially stated.

 

[CJSGrailKnigh]

I agree the 2F=f answer is counter intuitive so I shall be listening quite intently at my mechanics lecture tomorrow night (if I can get to it. Bus drivers suck in Ottawa and are on strike.). Either way I shall post my prof's answer tomorrow around 7 pm. I found all your replies interesting. Some of the equations and math you guys were talking about I've never seen. Anyway I want to thank you all for your posts.

 

[nrqed]

Well, that's the answer for sure, Nrged! (Energy Ed?)

It's actually nrQed which stands for nonrelativistic quantum electrodynamics.

It wasn't hard - just picture a bunch of little Fg vectors acting on the man and pointing toward the little chunks of mass in the cylinder that cause them. As you chip away those chunks of mass, some of the Fg vectors disappear, reducing the total force of gravity on the man.

But that incredible 2F solution that was given with the question is much more interesting! It doesn't make sense, yet comes straight out of the Fg = GMm/r^2 formula. I don't know how anyone could sleep last night without figuring out why! That formula only applies to spheres, so you have to think about a sphere (planet?) "whose mass and radius are reduced to half".

Ah! Of course if you reduce BOTH the mass and radius by 2, the density changes since the mass scales like the radius cube. But that does not make much physical sense. If part of a planet was removed (in a spherically symmetric way!) then the mass and radius would NOT be decreased by the same factor.

There is no way an impact could reduce both the mass and radius by the same factor.

 

[Carid]

Let's try another way to look at this problem. (If someone implicitly suggested this already, my apologies)

In our imagination divide the cylinder into two half cylinders - the near half and the far half.
Part of the man's weight is due to the near half, and part of his weight is due to the far half.
If we can estimate the relative influence of each half, then we know how much of his weight remains when the asteroid whacks the far half.

 

[Delphi51]

That makes good sense, Carid.

So the answer has to be some value between F/2 and F .

I just HAD TO work it out! See attachment. This work is way above my level (though I could have done it when I was a student), so I hope someone will check it. I noticed that a 2D integral works because you can do a ring of mass on one slice of the cylinder because only the axial component of g doesn't cancel out by symmetry. And for you first year young people, 2D integrals just mean you do the inner one and then you do the outer one. I evaluated it for a cylinder 1000 m long, 100 m radius. When the length is halved the g field at the end is reduced to 96% of its original value. Rather surprising, isn't it? I would have guessed something like 75% - instinct just doesn't work for inverse square laws.

Wow, NrQed, it must be exciting to be carrying on Feynman's work! I tried to emulate his high school teacher and found one or two students in my career who were interested in his path integral view of projectile motion.

 

[djeitnstine]

That makes good sense, Carid.


I just HAD TO work it out! See attachment. This work is way above my level (though I could have done it when I was a student), so I hope someone will check it. I noticed that a 2D integral works because you can do a ring of mass on one slice of the cylinder because only the axial component of g doesn't cancel out by symmetry. And for you first year young people, 2D integrals just mean you do the inner one and then you do the outer one. I evaluated it for a cylinder 1000 m long, 100 m radius. When the length is halved the g field at the end is reduced to 96% of its original value. Rather surprising, isn't it? I would have guessed something like 75% - instinct just doesn't work for inverse square laws.

Wow, NrQed, it must be exciting to be carrying on Feynman's work! I tried to emulate his high school teacher and found one or two students in my career who were interested in his path integral view of projectile motion.

Your result seems correct. I just presented this problem to my professor and he did indeed say that you would have to integrate over the cylindrical region meaning take the force of gravity at every point on the solid that is interacting with the man!

Your conclusion certainly seems to agree with this notion.

 

[CJSGrailKnigh]

apparently my prof. wanted to know weight change on earth. This of course would result in no significant change in weight... the assumptions I made were what made the question interesting I guess. although I disagree with my prof's answer because it says nothing about still being on Earth for the weighing.

The most logical reasoning to why the weighing is on Earth is because my prof is a civil engineer and because I'm an astronautical engineering student I assumed space and got the question wrong.

 

[Delphi51]

Thanks for those comments. CJSGrailKnigh, you came up with one heck of an interesting question - I suggest you give it to a physics or astro prof for further use. I certainly enjoyed it!