how do I find the center of mass for an object?
103147
Is this a real-world or mathematically-defined object?  Three dimensional?  Not enough info.\nIf a real object: You can do all sorts of hanging and balancing, but 3D objects, and especially those that are not clean geometric shapes (picture a randomly shaped rock) can be tough, requiring the assignment of a coordinate system, attachment of balancing pivot points, etc.  2D real objects are a snap: Hang from any point, draw a vertical line through the point, repeat at another point; where the lines intersect is the CM.  Here's a neat trick I read somewhere for finding the CM of something long & skinny like a stick:  Hold it horizontal, resting on two small sticks that are far enough apart so the CM is in between.  Slowly move the sticks together; they will meet at the CM!  I think you can do this with a flat 2D piece and three pins if you happen to have three hands.
508636
same mass, same weight
909738
It will rotate about the center of mass O at d/2\ncalculate the moment or inertia about  O, include both mass and both rods. see link for formulas if you don't know how\nhttp://en.wikipedia.org/wiki/Moment_of_inertia\nThe formula for KE is also there
227728
Very simple. start out asking yourself what information do i know about the problem. Anything from masses and forces to symmetries of the problem. Next Ask yourself what additional information you need to solve the problem, values, formulas, etc. from there it should all fall out with ease.
332070
Partition this into parts of known sizes ,then calculate weights  of each part and its cent of gravity. Then take a point outside the assy. The moment of each part about this point equals the total weight times  the dist of the CG from this point. Repeat this process to find the vert and horiz position of the CG.Consider a letter L shape.,height l1,cross secton a1,b1 inches.weight=w1=l1*a1*b1*0.284 (steel) lbs.Bottom of L  , length l2, cross sectio a2,b2. weight=w2=l2*a2*b2*0.284 lbs.Then dist of CG from corner of L is w2*l2/(2*(w1+w2)), and dist up from corner is \nw1*l1/(2*(w1+w2)).- I hope ?
997326
Use Newton's universal gravitation equation, then sum the forces.\n\nF = G*m1*m2/r^2\n\nF = force (newtons)\nG = gravitational constant (6.6742 x 10^-11)\nm1 = mass of object 1 (kg)\nm2 = mass of object 2 (kg)\nr = distance between objects (meters) ==> don't forget to convert the units\n\nIn this case, you do it twice:\nF1 = G * 390 * 390 / 0.11^2 = 0.84mN to the left\nF2 = G * 390 * 390 / 0.39^2 = 0.067mN to the right \n\nNet Force = 0.84 - 0.067 = 0.772mN to the left
79440
center of gravity \n\nThe point in any solid where a single applied force could support it; the point where the mass of the object is equally balanced. The center of gravity is also called the center of mass. When a man on a ladder leans sideways so far that his center of gravity is no longer over his feet, he begins to fall.
871968
Here is the math for finding our solar system's barycenter. (It's under the heading "Barycenter").\nhttp://en.wikipedia.org/wiki/Center_of_mass
233618
1 mass\n2 magnitude of separation
952581
http://planetmath.org/?op=getobj&from=objects&id=3808
298529
If you know the density of the material, you can measure it's volume.  D = M / V.  Therefore, the mass equals the density of the object times the volume.  The weight of the material is its mass times gravity.
149911
Density = Mass / Volume\nIf you know any 2 of the above quantities, you can find the other.
981150
a common way of handling this type of problem is the least squares method. Instead of finding the minimum of the distance distance to each point (difficult because of the absolute values) find the minimum of the squares of the distances (always a possitive value) So the formula you want to diferentiate is:\n\nD=(sqrt((x-5)^2+(y-5)^2)-4)^2 + ...\n\nthe more complex problem is what is meant by proximity to the circles. from your equation, it seems that it is only to the nearest point on each circle, but it could also mean to the centers of each circle (significantly easier) or to the area of each circle.\n\none of the first things you should do is plot the circles and eyeball guess. This gives you an idea of what the correct answer should be. With 20+ circles it is going to be difficult to eyeball. It also doesn't make much sense because there will be circles hiden behind other circles.\n\nDepending on how accurate you need to be, there are several ways to guess:\n\na really quick way would be to do a center of mass calculation (this would give you the closest point to the areas of the circle and could be done very easily on a spreadsheet or code.
896953
The net force is the difference between the two weights.\n\nHowever, there is not enough information in the question to answer without making some assumptions.\n\nYou are not given the mass of the man and/or the bricks. One can calculate the mass if given the force of gravity, however you aren't given that either.\n\nOne method of solving is:\n\nASSUMING that this system is on the surface of the Earth, you can find the two masses, using F=mg with g=9.8N/kg. Once you've got that you can find the acceleration using F=ma since you know the net force and the mass of the system.\n\nMake sure you indicate which direction the acceleration is in (in this case, which object is falling, and which is being raised.
981506
a. The force of the gravitational attraction between two objects is given by the formula\n\nF= G*(M1*M2/R^2)\n\nwhere G is the gravitational constant, M1,M2 are the two masses and R is the distance between their center of masses.\n\nSince the distance between your center of mass and the Earth's center of mass is such a large distance, you can approximate this distance as being the mean, average, radius of the Earth.\n\nG can be easily found in any physics book and you are given everything else. Just plug everything in to the equation and that will give you the gravitational force between you and the Earth.\n\nb. How much you weigh is directly related to the force you just calculated. In fact, they are the same thing. Your weight is a measure of how hard the Earth is pulling you towards its center.\n\nMy old physics professor used to joke that if you want to loose weight just walk up some stairs. Since you can't loose any mass, if you are able to decrease the distance between you and the center of the Earth you will lessen the pull of gravity and thus loose weight.
881707
Determining the mass is trivial: just weigh the block.  For the density, you will need to know the volume, which will require some measuring with a ruler.  Then divide the mass by the volume for your answer.
237452
The correct answer is that for equal masses the center of mass changes at 1/2 the SUM of their velocities: i.e., 10.05 m/s is to the right in standard geometry, and -11.6 m/s is to the left (velocities are VECTOR quantities, remember?). So the center of mass changes by 1/2 (10.05 + (-11.6)) = 1/2 (-1.55) m/s = -0.775 m/s, or in lay terms, 0.775 m/s to the left. Just keep your magnitudes (speeds) labelled correctly (+ or -) and your final answer will, by its sign, tell you which way the center of mass is moving.\nFor unequal masses, m1 and m2, moving with velocities v1 and v2, just use (m1 x v1 + m2 x v2)/(m1 + m2). Again, keep your +/- velocity values intact. This is the general equation that applies to the equal mass situation (a) also.