mechanics: why does treating things as point particles and the usage of center of mass work?

[Darien]

topic.

I've always wondered why you could treat objects as points...and get the "right" answer to what you're looking for...

 

[f95toli]

Because you can. Center of mass is defined that way.

By using the defintion of center of mass you actually mathematicaly prove that you will get the right answer.
To mathematicaly find the center of mass is not easy if you have a complicated shape (mathematicaly it involves a 3 triple intergrals over x,y and z).

And it only works if the object is not rotating (although you can always split the problem inte two, one dealing with the rotation and one with the motion of the center of mass).

 

[Darien]

i thought center of mass involved riemann sums, not integrals.

you wouldn't happen to know why everything is treated as a point particle would you?

 

[ClueLis]

Originally posted by: Darien
i thought center of mass involved riemann sums, not integrals.

you wouldn't happen to know why everything is treated as a point particle would you?

A Riemann sum is one way of representing an integral.

It makes calculations much easier if you can simplify an object down to a single point.

 

[f95toli]

I think you missed my point. You do not ´HAVE to treat the motion of an extended object like the motion of a point particle, but it is just simpler to do it that way.
Unless you know the center of mass of a given shape (there are tables) you have to perform a rather complicated calculation where you acually integrate over the whole object. Using "´center of mass" is just a mathematical technique used to simplify the calculations.

 

[DrPizza]

If you want a fun problem... try calculating the force of gravity someplace inside an object.
(i.e. you dig a tunnel from the north pole through the center of earth to the south pole)

edit: and even more fun, calculating what the air pressure would be at any point in the hole.

 

[Goosemaster]

Originally posted by: DrPizza
If you want a fun problem... try calculating the force of gravity someplace inside an object.
(i.e. you dig a tunnel from the north pole through the center of earth to the south pole)

edit: and even more fun, calculating what the air pressure would be at any point in the hole.

That just gave me a headache:|

What confuses me is the gravity of the surounding mass at the point of observation.

 

[DrPizza]

Originally posted by: Goosemaster

Originally posted by: DrPizza
If you want a fun problem... try calculating the force of gravity someplace inside an object.
(i.e. you dig a tunnel from the north pole through the center of earth to the south pole)

edit: and even more fun, calculating what the air pressure would be at any point in the hole.

That just gave me a headache:|

What confuses me is the gravity of the surounding mass at the point of observation.

Sorry to give you a headache... It's actually a problem that's been bothering me for a year or two... I've never sat down and devoted the many hours it would take to completely solve this problem: With that hole from N to S pole, if you jumped into the hole, when would you reach the center of the earth. From some quick simplifying assumptions, I've figured that either you would never reach the center, (the center would be the limit as t -> infinity) OR, you would just pass the center, but never make it quite back to the center (again, center is the limit as t -> infinity. The air resistance was the killer for me, I was having trouble with the air pressure at the center of the earth.
edit: I think this question was a topic of discussion here sometime before last summer.

 

[Matthias99]

The answer I recall from engineering classes: if you neglect air resistance, you'll yo-yo back and forth through the center. Your gravitational potential never changes -- the energy just converts from potential to kinetic and back again, over and over. With air resistance, eventually you'll end up exactly in the center (where the gravitational force is equal on all sides). However, calculating the exact times and speeds is a huge pain.

You can calculate the gravitational pull on you at any particular point by taking the triple integral over the volume of the earth and using the formula for gravitational attraction (G * m_1 * m_2/ d^2, where d is the distance to the point in question). Then you have to integrate *that* over time, using F=MA to determine your acceleration and velocity at any given point. Air pressure is a whole other can of worms, as then you have to calculate drag, etc based on speed...

This is one of those torturous mechanics problems they make college engineering students do, and as I'm not one (anymore), I'm not going to do it.

 

[DrPizza]

And, as soon as you factor in air resistance, you'll discover that your speed is approximately 0 when you're near the center. And, as you approach the center, the pull of gravity approaches 0. (assuming a symmetrical Earth... which it isn't). I understood the yo-yo part of it, but I tried to solve the problem more in the real world.

 

[Goosemaster]

So you would be weightless in the center?

Would the mass above produce any gravitational pull that would affect you?

 

[MadRat]

If the earth was a perfect sphere then theoretically you would have no weight at the center, true. Which at this point it makes me wonder if the core has positive, negative, or neutral pressure...

 

[uart]

Is it really that hard to calculate the gravitational pull at any point within the hole. I thought that for a system with sphearical symmetry that you could ignore all the mass at radii greater than your current position and only need consider that within the sphear defined by your current position. I thinks its Gausses thm or something that says that the volume integral of the mass enclosed is a equal to the surface integral of the gravitational flux (and likewise with charge/electric flux). So mass (or charge) that's not enclosed doesnt count for Jack in the net gravitational flux over the entire surface. And since you have sphearical symmetry the flux is the same magnetude and radial direction everywhere on the surface.

 

[Pulsar]

Gravity inside the earth is not exactly trivial.

http://stommel.tamu.edu/~baum/reid/book1/book/node31.html

 

[kevinthenerd]

First, take every single quark in an object and define its location and mass. Then average all the locations with the masses to arrive at a weighted average of where the center of mass really is. You'll end up with approximately the same thing as the aformentioned point method as long as the object isn't very flexible, and you won't have to make measurements on the order of Avagadro's number for every mole of particles.