Yes. See here. I have one on my desk, by the way. Small, but perfectly formed.
A log has central symmetry, a cone doesn't. It also has curves, but we can get rid of them using Cavalieri's principle.
http://en.m.wikipedia.org/wiki/Cavalieri's_principle
So we can take a small block of length h and chop off pieces until it is a pyramid: two half-blocks sliced off on a diagonal, but then we have to add back in another pyramid whose base is the other side. The mass of the block 3P is three times that of the pyramid P, so to find the moment arm x of the pyramid, the torque of the pyramid is
P xh = 3P h/2 - 2 3P/2 2h/3 + P (1-x)h
2P xh = P h/2
x = 1/4
No calculus necessary! Unless you count Cavalieri's principle (which, btw, I do )
I learned calculus (integration) and understand how to do it but I have not practiced the art in a long time so I am re-learning it. I usually shy away from calculus and just do a quick excel file dividing things into many small sections. After cutting the cone into numerous horizontal slices I come to the same answer as wikipedia and the site posted by swampyankee. I am beginning re-realize the ease of using calculus to find solutions (I always relied on 3-D computer models to do math).
So, for the cone it is true that the
center of mass = 0.2063*h
;whereas, the
center of gravity = 0.25*h
Please review for us your explicit definitions of center of mass and center of gravity, in appropriate mathematical detail. You must be using an unorthodox definition for the former. I stand by my finding, having reviewed the calculus myself on the basis of what I understand to be the standard definition.
The usual definition of center of mass is *not* where half the mass is on one side, and half is on the other. Besides, that point .2063h along the axis is not always at the halfway point--just tilt the cone a little and you can see that the halfway plane moves off that point.
As a mental simulation (or if you have the time actual measurement) Consider:
*- When you suspend an object from any point, the center of mass lies along a vertical line
Passing through the point of suspension.
*- Visualize suspending a cone (or any object) from two points and getting the crossing point of the two vertical lines. This will be the center of mass.
BTW, for a cone this is 1/4 of the way from the base along the center line (as already pointed out).
Technically, that describes the center of gravity, but in a uniform gravity field that will also be the center of mass.
Now I am confused, the C of g is the point through which the force of gravity would act, accelerating without rotation, if it (the object) were replaced by a point mass which surely is the same as the c of M? Clearly the OP idea of mass is wrong, the moment of each element of mass is key to the definition. Aha I see you refer to a constant gravity field which I just assumed, no problem
Last edited by profloater; 2012-Mar-21 at 05:40 PM. Reason: afterthought
*- Both centers of gravity and mass lie along a vertical line through the suspension point at the movement of equilibrium
*- Suspension from a different point on the object may alter center-of-gravity but not center-of-mass
*- Crossing point of two such vertical lines through suspension points will be center-of-mass and not (necessarily) center-of-gravity, which is not a constant/single point if the object is moved.
Two incoincident points don't "have to" lie on a vertical line, but they may do so. In case of centers of gravity and mass they both lie on a line that passes through them and the center of the gravity field (Earth) i.e. vertical.
I fail to see any contradiction in the statements. Please be more specific.
Gentle folk, you are spending much time discussing ideas which can be quickly referenced.
From Wiki, 'Center of Mass':
"In physics, the center of mass or barycenter of a body is a point in space where, for the purpose of various calculations, the entire mass of a body is concentrated. A center of gravity is a related point where the gravitational weight of a body acts as if it were concentrated. In a uniform gravitational field, the center of mass is a center of gravity, and in common usage, the two phrases are used as synonyms. In a non-uniform field, the center of mass no longer serves as an exact center of gravity, so one can distinguish the center of gravity as a separate concept." (my italics)
Bang. Done.
Regards, John M.
I'm not a hardnosed mainstreamer; I just like the observations, theories, predictions, and results to match.
"Mainstream isn’t a faith system. It is a verified body of work that must be taken into account if you wish to add to that body of work, or if you want to change the conclusions of that body of work." - korjik
Are there actual, real situations where the center of gravity of
something is *measurably* different from its center of mass?
-- Jeff, in Minneapolis
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
I assumed that more than mere tidal effects were required.
Could you describe an actual case in which tidal differences
in gravity result in a measurable difference in the location of
the center of gravity of something?
-- Jeff, in Minneapolis
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
So, can somebody help me with the missing number near
the bottom of the first (main) section of this web page?
http://www.freemars.org/jeff/Earth/down.htm
It says "0.00 degree." I need to fill in the correct number.
Calculating it requires elliptical trigonometry, I think.
Should be easy for several people here. But I'd like to
see it worked out so I know what I'm talking about.
I made a diagram of the geometry, below. On a monitor
with close pixel spacing, it may be best viewed at 200%.
That's zoomed in 6 times in Firefox! Yuk. Intermediate
sizes make it messy.
I'm assuming a perfectly ellipsoidal Earth.
Earth's equatorial radius is 6,378 km.
Earth's polar radius is 6,357 km.
I seem to have forgotten since I made the diagram, but I think
S is the distance along the surface of a perfectly spherical
Earth with radius R, from the north pole to the 45-degree line.
theta is 45 degrees.
g is the gravitational force at the 45-degree line.
f is the centrifugal force at the 45-degree line..
phi is the angle I want to know.
-- Jeff, in Minneapolis
.
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
When i saw tidal effects, i went "ohhh of course".
The measurable difference seems obvious. The tides cause equal bulges on both sides, so the center of mas is the same. But far enough away from the earth, say from the moon, the center of gravity is closer and offset to the side. Hence the torque which causes the moon to migrate.
Jeff, the gravitational effect of the equatorial bulge is almost as much as the centrifugal force at a latitude of 45°, so the line you've labeled as g does not point to the center of the earth.
By definition, latitude is determined by the horizon angle with the axis of the earth (ETA: in theory, but international and national interests come into play), so it is the tangent of the ellipsoid.
The ellipsoid is and its derivative is which is equal to one (or negative one) at latitudes of 45°, and so --but that is the slope of the angle to the center of the ellipse. Its arctangent will give you an angle that deviates slightly from 45°, and the difference is the angle you seek.
ETA: I plugged "(arctan ( 6357^2/6378^2)) radians in degrees" into wolframalpha, and the result says your number is about 11 minutes. I multiplied it by 6368 and got 21.00 kilometers, which is about how far it misses the center of the earth. Of course, it intersects the plane of the equator much farther than that from the earth center.
EETA: I was almost sure we'd worked on this before.
http://www.bautforum.com/showthread....jor#post702186
No, in fact every instance of the center of gravity not being the same as the center of mass is going to be an instance of a tidal effect. Think about it, the center of gravity equals the center of mass in a uniform gravitational field. What we call a tidal effect is nothing more than saying that you can't consider the gravitational field as uniform over some extended body, for example the tides on earth because the gravitational pull from the moon is not uniform across the earth. So finding examples of measurable differences is the same as finding examples of measurable tidal effects.
Maybe there is another term for what I describe as the center of mass - what I am describing is the point through which the 3 orthogonal planes pass whereby those planes evenly divide the subject mass into two equal masses. So if I melted down either half then I could pour it into a mold of the second and exactly fill it creating two exact replicas.
The center of gravity being the balance point of a uniform object in a uniform gravitational field.
This point you are after is very important when sharing an ice cream cone and want exactly half, ditto all irregular shapes that might be divided equally, sorry I don't know the term for the half mass point.
That's true, cones have so many interesting properties and being part filled is difficult to include in the calculus but I propose a portable balance with attached nomogram, used to find the balance point of the actual ice cream cone and then giving the half mass point. The nomogram would have lines for various standard cone biscuit weights. Named the "Econometer" However I am stumped by the 99 (with chocolate flake) How could you ever find half of that? It's an irregular apparently random cyclinder.
Order of Kilopi