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how do you find the Center of mass of an object like this?? |
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| Apr21-07, 07:05 PM | #1 |
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how do you find the Center of mass of an object like this??
1. The problem statement, all variables and given/known data
Find the location of the center of mass for the system of particles below. 2. Relevant equations ![]() 3. The attempt at a solution tried to use Xcm= 1/(M1+M2)*(M1+M2) but im stuck. |
| Apr22-07, 01:04 AM | #2 |
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They all equal in value?
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| Apr22-07, 01:10 AM | #3 |
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yes they are all equal sorry about the picture..
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| Apr22-07, 01:24 AM | #4 |
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how do you find the Center of mass of an object like this??
You can join the segments of the object to make diffent sections that you can add to get your center of mass. Also represent each mass as a variable, because you say they are the same, don't they all have the same mass? What is the total mass? Don't you have to find the center of mass in the x and in the y direction?
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| Apr22-07, 10:23 AM | #5 |
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Is the picture a rough diagram? I am presuming that all blocks are supposed to be the same size.
How I would find the centre of mass is hang it from an adjustable piece of string near the centre. When the object is horizontal, the string is at the CoM. This doesn't help you I know, I've always been interested in analog problem-solving. |
| Apr22-07, 02:13 PM | #6 |
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yah im sorry about the diagram, this is the best i could do. They're all the same size and same mass.
So i still dont really understand.. how do i combine them? |
| Apr22-07, 07:06 PM | #7 |
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simple weighted mean, along the x axis, choose the center were the blocks are,
then just multiply the moments by the area. ie, -3*4+-1*2+-1*1+0*3 and same for right hand side, divide by total number of squares. Again weighted average. From symmetry its gonna be in the middle for x, do same for Y, Cm are these coordinates. |
| Apr22-07, 10:00 PM | #8 |
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hmm... still dont quite get it..
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| Apr22-07, 10:02 PM | #9 |
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Here's a hint: There are 5 pieces.
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| Apr22-07, 10:08 PM | #10 |
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ok... my approach is there are 5 pieces, and then 10 pieces... but after that im stuck still...
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| Apr22-07, 10:11 PM | #11 |
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Ok I'll try again, its a torque problem. Imagine two weights on a seesaw. The two kids weigh different, find the point where they balance.
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| Apr22-07, 10:12 PM | #12 |
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distance for each from the point you want to referance from(like an origin on a graph) for each direction of x or y. find the middle of the objects on the x directions and in the y directions. Distances from the orgin to the middle of each object for each dimention. add the multipuls of the mass of the objects times the distance to the center(for each direction). Divide by the total mass for each directional center of mass. You will see the mass cancel out.
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| Apr22-07, 10:27 PM | #13 |
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k so i got CM of the 2 pieces is 3/2L apart.
CM is (5/15)(3/2L)=1/2 above the center of mass of those 5 pieces. what abt the rest of the squares? does this make sense? |
| Apr22-07, 10:32 PM | #14 |
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| Apr22-07, 11:02 PM | #15 |
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Assign an origin as you say, and then assign the center of each block a numerical value that can then be averaged. |
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