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Center of mass gravity problem

  1. Jan 1, 2007 #1
    1. The problem statement, all variables and given/known data
    Give the X, Y, and Z coordinates of the center of gravty of a half cubical block of uniform density. The diagram below shows how the block was cut.

    The diagram depicts a cube with sides of length 1 cut diagonally in half. the six vertices are as follows:
    (0,0,0) (1,0,0) (0,0,1)
    (0,1,0) (1,1,0) (0,1,1)

    2. Relevant equations
    X = (w1x1 + w2x2 + ... wnxn) / (w1 + w2 + ... wn)
    Y = (w1y1 + w2y2 + ... wnyn) / (w1 + w2 + ... wn)
    Z = (w1z1 + w2z2 + ... wnzn) / (w1 + w2 + ... wn)

    Where each set of points is the center of the shape (x1,y1,z1) and the w represent the weight (w1).

    3. The attempt at a solution
    Center Points of the five shapes:
    1) (1/3, 1, 1/3); W1 = .5
    2) (.5, .5, 0); W2 = 1
    3) (0, .5, .5); W3 = 1
    4) (1/3, 0, 1/3); W4 = .5
    5) (.5, .5, .5); W5 = sqrt(2)

    X= [ (1/3) (.5) + (.5) (1) + (0) (1) + (1/3) (.5) + (.5) (sqrt(2)) ] / (.5 + 1 + 1 + .5 + sqrt(2))
    X= .349

    Y= [ (1) (.5) + (.5) (1) + (.5) (1) + (0) (.5) + (.5) (sqrt(2)) ] / (.5 + 1 + 1 + .5 + sqrt(2))
    Y= .5

    Z= [ (1/3) (.5) + (0) (1) + (.5) (1) + (1/3) (.5) + (.5) (sqrt(2)) ] / (.5 + 1 + 1 + .5 + sqrt(2))
    Z= .349

    The X,Y,Z coordinates of the center of gravity of the half cubical: (.349, .5, .349)

    My only concern is the i took the surface area of the different shapes to be the weight of the different parts. I'm not sure if this is the correct way to go about the problem. (I do know for 2D shapes that the area of the shape can be used as its weight when dealing in the Cartesian plane if the density is uniform since the area is proportional to the weight).
  2. jcsd
  3. Jan 1, 2007 #2
    Is there any way you can post the diagram?

    EDIT: and hey, welcome aboard! :smile:
    Last edited: Jan 1, 2007
  4. Jan 1, 2007 #3

    Umm.... I've used my "1337" (and i use that term EXTREMELY loosely) photoshop skillz to make this figure for you guys....


    I hope this helps!

    (Thank you! Glad to be here!)
  5. Jan 1, 2007 #4
    Another Question

    While i'm at it, I've another question on this topic:

    1. The problem statement, all variables and given/known data
    My teacher gave us a cutout of a pacman shape. Its a circle with the mouth opening which has an angle of 45 degrees. I'll the diameter is about 5.9inches (it really doesn't matter, all i need help with is HOW to go about solving this). The mouth opening starts at the center of the circle and goes out making a 45 degree angle. I need to find the center of mass (the pen and paper way).

    2. Relevant equations
    X = (w1x1 + w2x2 + ... wnxn) / (w1 + w2 + ... wn)
    Y = (w1y1 + w2y2 + ... wnyn) / (w1 + w2 + ... wn)

    Where each set of points is the center of the shape (x1,y1,z1) and the w represent the weight (w1).

    3. The attempt at a solution

    I really have NO idea how to go about solving this.... :confused:


    I'd appreciate any guidance on how to solve this problem. (I was told to think of polar coordinates and to take the integral... but uhhh, i haven't learned how to solve center of mass yet in class, we just got the lab to do as an introductory thing and the pacman assignment for fun, but i really wanna do the pacman thing :rolleyes: )
  6. Jan 2, 2007 #5
    OK, I'll come clean. I am not at all confident about this stuff; the only rule that I feel totally comfortable applying is that of symmetry. The center of mass always lies on an/all axis/axes of symmetry. So your 1/2 cube stuff looks good to me, the y-value is for sure correct, but I can't (don't want to?) follow your reasoning for the other stuff. I do know that just as you use density*area for weight of a 2-D object, you use density*volume for 3-D stuff.

    As far as pacman goes, symmetry can get you an axis along which the COM must lie, but past that, I don't know what to do.

    BTW--love the digram of the dissected cube!
  7. Jan 2, 2007 #6
    hehehe, glad you enjoyed it!

    But I'm not so sure on all of that symmetry stuff, haven't really gone in depth with the subject and my textbook was written for high level physics courses in college (I'm only in High School); it's pretty much gibberish to me... If anything, did u find any problems with the points i got? my weights? And if you say use density *volume for my weight, how would i go about finding EACH shapes volume? (or would i just use the whole blocks volume?)
  8. Jan 2, 2007 #7
    Think about it intuitively--if you wanted to balance a symmetrical object, you'd balance it in the middle, right?

    And I'm in college, on intersession, having told myself that I would spend break boning up on things like center of mass and oscillations. You really don't want me telling you if you're right or wrong; I'm not comfortable enough with the material.
    OK, I just realized that you're throwing around this "five shapes" business. What five shapes? I only see one shape--ONE half of a cube.

    EDIT: I just re-read my post. I am neither grumpy nor averse to helping out. Just wanted to give you fair warning that I am not one of the insanely wonderful people on this site (like Doc Al or OlderDan, for example) that pretty much whatever they post is on the money. I want to go through this, but in the end of the day, I just can't say if it's right or wrong.
    Last edited: Jan 2, 2007
  9. Jan 2, 2007 #8
    well... by five shapes i mean the five different faces i got the center points of... and with 2d figures, i can break down the figure into shapes, but i guess not with 3d... hmmm.... but i'm still a little confused, would i multiply the volume of the half cube for each point? (as in X1V+X2V+.../V) Wouldn't that effectively just give me the the sum of all the points? (I know mbrmbrg that you don't know, so please don't feel obligated to help if you don't want to! But thanks for the though process thus far!)

    Another thought i had was to find the centroid of the shape, which is just averaging ALL the centers i found, giving me (1/3, .5, 1/3). The two answers (from my first post) and this one are off by hundredths, but i need to know which seems to make more sense... and is correct :smile:
  10. Jan 2, 2007 #9
    While googling, i came across this from http://www.gaffer.org/game-physics/physics-in-3d/":

    "The center of mass of the object is the weighted average of all points making up the body by their mass. For objects of uniform density, the center of mass is always the geometric center of the object, for example the center of a sphere or a cube."

    So i'm guessing i have to find the centroid of the shape, but is my prior statement on how to find the center correct?
    Last edited by a moderator: Apr 22, 2017 at 2:55 PM
  11. Jan 2, 2007 #10
    Dude, you can do centroids?! A centroid is a center of mass! I wouldn't worry too much about being off by hundredths, it's probably a rounding error, sig-figs, all that.

    Use centroids for pac-man also, if you like 'em!

    This sounds like it should make sense... Um. If the cube were hollow, you'd definitely be right on with the five facets. Very smart, actually. Only problem is, I don't know how the cube's solidity affects that train of reasoning. I'd like to think that it still holds, but it's 1:53am my time. The brain rebels...
    Have loads of fun! G'night/whatever time of day it is in your bit of the world:zzz:
  12. Jan 2, 2007 #11
    that's funny... its 1:54 am over here... anyway... i don't know how to find the centroid... and the centroid of the pacman shape will be off cause of the missing chunk in its big idiot head... sooooo anybody know the formula for finding the centroid of a 3D object?

    another thought:
    Even if the cube is hollow or not, it shouldn't effect where the center of mass is located... at least from what I'm thinking about.
  13. Jan 2, 2007 #12
    You can easily solve the half-cube problem by thinking two-dimensionally, then expanding it to three dimensions. Imagine looking at the half-cube from the top: it's a triangle, right? Find the triangle's center of mass and you'll be on the right track.
    Last edited: Jan 2, 2007
  14. Jan 2, 2007 #13
    (1/3, 1, 1/3) and (1/3,0,1/3) are the center of masses for the two triangles...

    There must be some mix up in the formulas i am using (either incorrectly or leaving a part out)

    I feel the centroid is at (1/3,.5,1/3) but i have no1 to verify my claim... I'd like to think i'm good at physics, but my test scores beg to differ.

    and any help on pacman is still in need!
  15. Jan 3, 2007 #14
    If any1 is intereseted with the pacman thing, here's how i went about solving it:

    Assumptions made:
    Pacman has a diameter of 15cm,
    Pacman’s mouth opens at a 45-degree angle,
    Pacman’s positioned with his mouth starting on the horizontal opening upwards.

    Center of Mass of circle (center of circle): (7.5 cm, 7.5 cm) (X1, Y1)

    Center of Mass of sector: 2r sin θ / 3θ away from the center on the line bisecting the angle
    = 2 (7.5) (sin (45˚)) / (3 * 45˚) = 4.50158

    X2 = 7.5 + (4.50158) (cos (45˚ / 2))
    = 11.6588 cm
    Y2 = 7.5 + (4.50158) (sin (45˚ / 2))
    = 9.222649 cm
    Center of Mass of sector: (11.6588 cm, 9.222649 cm)

    Area (weight) of circle = 2 r2 = 2 (7.5)2 = 88.65729 cm2
    Area (weight) of sector = r2θ / 2 = (7.5)2 ( /4) / 2 = 22.08932335 cm2

    Center of Mass (Theoretical) of Pacman:
    X-bar = [(7.5cm) (88.65729cm2) – (11.6588) (22.08932335cm2)] /(88.65729cm2 – 22.08932335cm2)
    X-bar = 6.11 cm
    Y-bar = [(7.5cm) (88.65729cm2) – (9.222649) (22.08932335cm2)]/(88.65729cm2 – 22.08932335cm2)
    Y-bar = 6.93 cm

    Center of Mass (Theoretical) of Pacman coordinate: (6.11 cm, 6.93 cm)

    Center of Mass (Experimental) of Pacman (using plumb lines) coordinate: (6.19 cm, 7.25 cm)


    Percent error on value = | 6.19 – 6.11 | / 6.11 = 1.3%
    Percent error on value = | 7.25 – 6.93 | / 6.93 = 4.62%

    Sources of error: Pacman was cut with diameters ranging from 15.1 cm to 15.3 cm. The mouth opening did not start directly in the center of the circle and was not forming an exact 45-degree angle. In short, all sources of error originated from human error.

    And the cubical was (1/3, 1/2, 1/3). Its just a triangle in 3D, so since in 2D you find the centroid for the center of mass, the same rule applies with a 3D triangle, except now you account for the z values when calculating the centroid. (average all the x's, y's, and z's from the six vertices's points)

    Any feed back would be great.... (agree/disagree, i'm using wrong logic, etc)
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