![\"Graph](\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/02\/IMG_0279-300x143.jpg\")
\r\n\r\nCommon sense tells you that the average position of the material making up the two particles is midway between the two particles. Common sense is right. We give the name \u201ccenter of mass\u201d to the average position of the material making up a distribution, and the center of mass of a pair of same-mass particles is indeed midway between the two particles. How about if one of the particles is more massive than the other? One would expect the center of mass to be closer to the more massive particle, and again, one would be right. To determine the position of the center of mass of the distribution of matter in such a case, we compute a weighted sum of the positions of the particles in the distribution, where the weighting factor for a given particle is that fraction of the total mass that the particle\u2019s own mass is. Thus, for two particles on the x axis, one of mass m1<\/sub>, at x1<\/sub>, and the other of mass m2<\/sub>, at x2<\/sub>,\r\n\r\n![\"Graph](\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/02\/IMG_0280-300x121.jpg\")
\r\n\r\nThe position x of the center of mass is given by equation 8-1:\r\n$latex \\bar{x}=\\frac{m_1*x_1}{m_1+m_2}+\\frac{m_2*x_2}{m_1+m_2}$<\/p>\r\n \r\n\r\nNote that each weighting factor is a proper fraction and that the sum of the weighting factors is always 1. Also note that if, for instance, m1<\/sub> is greater than m2<\/sub>, then the position x1<\/sub> of particle 1 will count more in the sum, thus ensuring that the center of mass is found to be closer to the more massive particle (as we know it must be). Further note that if m1<\/sub> = m2<\/sub>, each weighting factor is 1\/2, as is evident when we substitute m for both m1<\/sub> and m2<\/sub> in equation 8-1:\r\n\r\n[latex]\\bar{x}=\\frac{m}{m+m}x_1+\\frac{m}{m+m}x_2\\\\\\bar{x}=\\frac{1}{2}x_1+\\frac{1}{2}x_2\\\\\\bar{x}=\\frac{x_1+x_2}{2}[\/latex]\r\n\r\n \r\n\r\nThe center of mass is found to be midway between the two particles, right where common sense tells us it has to be.\r\n\r\nSource: Calculus-Based Physics 1, Jeffery W. Schnick. p142, <\/span>https:\/\/openlibrary.ecampusontario.ca\/catalogue\/item\/?id=ce74a181-ccde-491c-848d-05489ed182e7<\/a>\r\n\r\n<\/div>\r\n \r\n\r\nBelow is a more visual representation of where the COM would be for two different weighing particles.\r\n\r\n[caption id=\"attachment_557\" align=\"aligncenter\" width=\"285\"] Suppose four point masses are placed on a number line as follows:<\/em><\/p>\r\n\r\n Solution<\/strong><\/p>\r\nFind the moment of the system with respect to the origin and find the center of mass of the system.\r\n\r\nFirst, we need to calculate the moment of the system (the top part of the fraction):\r\n\r\n$latex M =\\sum_{i=1}^4 m_i *x_i \\\\\\qquad \\quad = (30kg)*(-2m) + (5kg)*(3m)+(10kg)*(6m)+(15kg)*(-3m) \\\\\\qquad\\quad = (-60+15+60-45)kg*m \\\\\\qquad\\quad = -30 kg*m $\r\n\r\n \r\n\r\nNow, to find the center of mass, we need the total mass of the system:\r\n\r\n[latex]m = \\sum_{i=1}^4 m_i\u00a0 = (30+5+10+15) kg = 60kg[\/latex]\r\n\r\n \r\n\r\nThen we have $latex \\bar{x} = \\frac{M}{m} = \\frac{-30 kg*m}{60kg} = -0.5 m$\r\n\r\n \r\n\r\nThe center of mass is located 1\/2 m to the left of the origin.\r\n\r\n \r\n\r\nSource: \"Moments and Centers of Mass\" by LibreTexts, <\/a>https:\/\/eng.libretexts.org\/@go\/page\/67237<\/a><\/a>\r\n\r\n<\/div>\r\n $latex m=\\frac{bL^3}{3} $<\/p>\r\nThat can now be used to calculate the mass of a nonlinear rod. The value of L<\/em> is given as 0.890 m, and we defined b<\/em> to be the constant 0.650 kg\/m3<\/sup>, therefore\r\n\r\n[latex]m=\\frac{0.650\\frac{kg}{m^3}(0.890m)^3}{3}\\\\m=0.1527kg[\/latex]\r\n\r\nThat\u2019s a value that will come in handy when we calculate the position of the center of mass.\r\n\r\nNow, when we calculated the center of mass of a set of discrete particles (where a discrete particle is one that is by itself, as opposed, for instance, to being part of a rigid body) we just carried out a weighted sum in which each term was the position of a particle times its weighting factor and the weighting factor was that fraction, of the total mass, represented by the mass of the particle. We carry out a similar procedure for a continuous distribution of mass such as that which makes up the rod in question.\r\n\r\nOnce again, using integration, we find the equation:\r\n $latex \\bar{x}=\\frac{bL^4}{4m}$<\/p>\r\nNow we substitute variables with values; the mass m of the rod that we found earlier, the constant b that we defined to simplify the appearance of the linear density function, and the given length L of the rod:\r\n\r\n[latex]m= \\frac{\\left( 0.650\\frac{kg}{m^3} \\right) (0.890m)^4}{4(0.1527kg)}\\\\\\bar{x}=0.668m[\/latex]\r\n\r\nThis is our final answer for the position of the center of mass. Note that it is closer to the denser end of the rod, as we would expect.\r\n\r\nSource: Calculus-Based Physics 1, Jeffery W. Schnick. p144, <\/span>https:\/\/openlibrary.ecampusontario.ca\/catalogue\/item\/?id=ce74a181-ccde-491c-848d-05489ed182e7<\/a>\r\n\r\n<\/div>\r\n Key Takeaways<\/p>\r\n\r\n<\/header>\r\n To start, let’s calculate the center of mass! This is a weighted function, similar to when we found the location of the resultant force from multiple distributed loads and forces.<\/p>\n [latex]\\bar{x}=\\frac{m_1*x_1}{m_1+m_2}+\\frac{m_2*x_2}{m_1+m_2}[\/latex]<\/p>\n When the density is the same throughout a shape, the center of mass is also the centroid (geometric center).<\/p>\n Consider two particles, having one and the same mass m, each of which is at a different position on the x-axis of a Cartesian coordinate system.<\/p>\n Common sense tells you that the average position of the material making up the two particles is midway between the two particles. Common sense is right. We give the name \u201ccenter of mass\u201d to the average position of the material making up a distribution, and the center of mass of a pair of same-mass particles is indeed midway between the two particles. How about if one of the particles is more massive than the other? One would expect the center of mass to be closer to the more massive particle, and again, one would be right. To determine the position of the center of mass of the distribution of matter in such a case, we compute a weighted sum of the positions of the particles in the distribution, where the weighting factor for a given particle is that fraction of the total mass that the particle\u2019s own mass is. Thus, for two particles on the x axis, one of mass m1<\/sub>, at x1<\/sub>, and the other of mass m2<\/sub>, at x2<\/sub>,<\/p>\n The position x of the center of mass is given by equation 8-1:<\/p>\n [latex]\\bar{x}=\\frac{m_1*x_1}{m_1+m_2}+\\frac{m_2*x_2}{m_1+m_2}[\/latex]<\/p>\n <\/p>\n Note that each weighting factor is a proper fraction and that the sum of the weighting factors is always 1. Also note that if, for instance, m1<\/sub> is greater than m2<\/sub>, then the position x1<\/sub> of particle 1 will count more in the sum, thus ensuring that the center of mass is found to be closer to the more massive particle (as we know it must be). Further note that if m1<\/sub> = m2<\/sub>, each weighting factor is 1\/2, as is evident when we substitute m for both m1<\/sub> and m2<\/sub> in equation 8-1:<\/p>\n [latex]\\bar{x}=\\frac{m}{m+m}x_1+\\frac{m}{m+m}x_2\\\\\\bar{x}=\\frac{1}{2}x_1+\\frac{1}{2}x_2\\\\\\bar{x}=\\frac{x_1+x_2}{2}[\/latex]<\/p>\n <\/p>\n The center of mass is found to be midway between the two particles, right where common sense tells us it has to be.<\/p>\n Source: Calculus-Based Physics 1, Jeffery W. Schnick. p142, <\/span>https:\/\/openlibrary.ecampusontario.ca\/catalogue\/item\/?id=ce74a181-ccde-491c-848d-05489ed182e7<\/a><\/p>\n<\/div>\n <\/p>\n Below is a more visual representation of where the COM would be for two different weighing particles.<\/p>\n A second explanation:<\/p>\n The most common real-life example of a system like this is a playground seesaw, or teeter-totter, with children of different weights sitting at different distances from the center. On a seesaw, if one child sits at each end, the heavier child sinks down and the lighter child is lifted into the air. If the heavier child slides in toward the center, though, the seesaw balances. Applying this concept to the masses on the rod, we note that the masses balance each other if and only if m1<\/sub>d1<\/sub> = m2<\/sub>d2<\/sub>.<\/p>\n This idea is not limited just to two-point masses. In general, if n masses, m1<\/sub>,m2<\/sub>,…, mn<\/sub> are placed on a number line at points x1<\/sub>, x2<\/sub>,…xn<\/sub>,\u00a0respectively, then the center of mass of the system is given by:<\/p>\n [latex]\\bar x=\\frac{\\sum_{i=1}^n m_i x_i}{\\sum_{i=1}^nm_i}[\/latex]<\/p>\n Suppose four point masses are placed on a number line as follows:<\/em><\/p>\n Solution<\/strong><\/p>\n Find the moment of the system with respect to the origin and find the center of mass of the system.<\/p>\n First, we need to calculate the moment of the system (the top part of the fraction):<\/p>\n [latex]M =\\sum_{i=1}^4 m_i *x_i \\\\\\qquad \\quad = (30kg)*(-2m) + (5kg)*(3m)+(10kg)*(6m)+(15kg)*(-3m) \\\\\\qquad\\quad = (-60+15+60-45)kg*m \\\\\\qquad\\quad = -30 kg*m[\/latex]<\/p>\n <\/p>\n Now, to find the center of mass, we need the total mass of the system:<\/p>\n [latex]m = \\sum_{i=1}^4 m_i\u00a0 = (30+5+10+15) kg = 60kg[\/latex]<\/p>\n <\/p>\n Then we have [latex]\\bar{x} = \\frac{M}{m} = \\frac{-30 kg*m}{60kg} = -0.5 m[\/latex]<\/p>\n <\/p>\n The center of mass is located 1\/2 m to the left of the origin.<\/p>\n <\/p>\n Source: “Moments and Centers of Mass” by LibreTexts, <\/a>https:\/\/eng.libretexts.org\/@go\/page\/67237<\/a><\/a><\/p>\n<\/div>\n When we are looking at multiple objects in 2D or 3D, we perform the center of mass equation multiple times in the x, y, and z directions.<\/p>\n [latex]\\bar x=\\frac{\\sum_{i=1}^n m_i x_i}{\\sum_{i=1}^nm_i} \\qquad \\bar y=\\frac{\\sum_{i=1}^n m_i y_i}{\\sum_{i=1}^nm_i} \\qquad \\bar z=\\frac{\\sum_{i=1}^n m_i z_i}{\\sum_{i=1}^nm_i}[\/latex]<\/p>\n <\/p>\n In some sense, one can think about the center of mass of a single object as its “average position.” Let’s consider the simplest case of an “object” consisting of two tiny particles separated along the x<\/span><\/span>-axis, as seen below:<\/span><\/p>\n If the two particles have equal mass, then it’s pretty clear that the “average position” of the two-particle system is halfway between them. If the masses of the two particles are different, would the “average position” still be halfway between them? Perhaps in some sense this is true, but we are not looking for a\u00a0geometric center<\/em><\/span>; we are looking for the average placement of mass. If m1<\/sub>\u00a0has twice the mass of m2<\/sub><\/span>, then when it comes to the average placement of mass, m1<\/sub><\/span>\u00a0gets “two votes.” With more of the mass concentrated at the position x1<\/sub><\/span>\u00a0than at x2<\/sub><\/span>, the center of mass should be closer to x1<\/sub><\/span>\u00a0than x2<\/sub><\/span>. We achieve the perfect balance by “weighting” the positions by the fraction of the total mass that is located there. Accordingly, we define the center of mass:<\/span><\/p>\n [latex]\\bar x_{cm}=(\\frac{m_1}{m_1+m_2})x_1+(\\frac{m_2}{m_1+m_2})x_2=\\frac{m_1x_1+m_2x_2}{M_{system}}[\/latex]<\/p>\n If there are more than two particles, we simply add all of them into the sum in the numerator. To extend this definition of the center of mass into three dimensions, we simply need to do the same things in the y <\/span><\/span>and z<\/span>\u00a0directions. A position vector for the center of mass of a system of many particles would then be:<\/span><\/p>\n [latex]\\vec{r}_{cm}=\\bar x_{cm}\\underline{\\hat{i}}+\\bar y_{cm}\\underline{\\hat{j}}+ \\bar z_{cm}\\underline{\\hat{k}}\\\\=\\frac{[m_1 x_1+m_2 x_2+...]}{M}\\underline{\\hat{i}}+\\frac{[m_1y_1+m_2y_2+...]}{M}\\underline{\\hat{j}}+\\frac{[m_1 z_1+m_2 z_2+...]}{M}\\underline{\\hat{k}}\\\\=\\frac{m_1[x_1\\underline{\\hat{i}}+y_1\\underline{\\hat{j}}+z_1\\underline{\\hat{k}}]+m_2[x_2\\underline{\\hat{i}}+y_2\\underline{\\hat{j}}+z_2\\underline{\\hat{k}}]+...}{M}\\\\=\\frac{m_1\\vec r_1+m_2\\vec r_2+...}{M}[\/latex]<\/p>\n Source: ” Center of Mass” by Tom Weideman, https:\/\/phys.libretexts.org\/Courses\/University_of_California_Davis\/UCD%3A_Physics_9A__Classical_Mechanics\/4%3A_Linear_Momentum\/4.2%3A_Center_of_Mass<\/a><\/p>\n<\/div>\n Suppose three point masses are placed in the x-y plane as follows (assume coordinates are given in meters):<\/em><\/p>\n Find the center of mass of the system.<\/em><\/p>\n Solution<\/strong> [latex]m = \\sum_{i=1}^3 m_i\u00a0 = (2 + 6 + 4) kg = 12 kg[\/latex]<\/p>\n Next, we find the moments with respect to the x- and y-axes:<\/p>\n [latex]M_x =\\sum_{i=1}^3 m_i *x_i \\\\\\qquad \\quad = (2kg)*(-1m) + (6kg)*(1m)+(4kg)*(2m) \\\\\\qquad\\quad = (-2+6+8)kg*m \\\\\\qquad\\quad = 12 kg*m[\/latex]<\/p>\n [latex]M_y =\\sum_{i=1}^3 m_i *y_i \\\\\\qquad \\quad = (2kg)*(3m) + (6kg)*(1m)+(4kg)*(-2m) \\\\\\qquad\\quad = (6+6-8)kg*m \\\\\\qquad\\quad = 4 kg*m[\/latex]<\/p>\n Then we have<\/p>\n [latex]\\bar{x} = \\frac{M_x}{m} = \\frac{12 kgm}{12m} = 1\u00a0 m[\/latex]<\/p>\n [latex]\\bar{y} = \\frac{M_y}{m} = \\frac{4 kgm}{12m} = 0.333 m[\/latex]<\/p>\n <\/p>\n The center of mass of the system is: (1, 0.333)m.<\/p>\n![\"Two](\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/02\/530px-Jacobi_coordinates.svg_.png\")
Source (image): Two_body_jacobi.svg: CWitte, from JPG by Brews O'Hare derivative work: WillowW via Wikimedia Commons https:\/\/zh.wikipedia.org\/wiki\/File:Jacobi_coordinates.svg[\/caption]\r\n\r\nA second explanation:\r\n![\"Two](\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/02\/14.8.3.05-_Moments_and_Centers_of_Mass.jpg\")
\r\n\r\nThis idea is not limited just to two-point masses. In general, if n masses, m1<\/sub>,m2<\/sub>,..., mn<\/sub> are placed on a number line at points x1<\/sub>, x2<\/sub>,...xn<\/sub>,\u00a0respectively, then the center of mass of the system is given by:\r\n\r\n[latex]\\bar x=\\frac{\\sum_{i=1}^n m_i x_i}{\\sum_{i=1}^nm_i}[\/latex]\r\nExample 1: <\/em><\/h2>\r\n
\r\n \t
7.1.2 Center of Mass in 2D & 3D<\/h1>\r\nWhen we are looking at multiple objects in 2D or 3D, we perform the center of mass equation multiple times in the x, y, and z directions.\r\n\r\n[latex]\\bar x=\\frac{\\sum_{i=1}^n m_i x_i}{\\sum_{i=1}^nm_i} \\qquad \\bar y=\\frac{\\sum_{i=1}^n m_i y_i}{\\sum_{i=1}^nm_i} \\qquad \\bar z=\\frac{\\sum_{i=1}^n m_i z_i}{\\sum_{i=1}^nm_i}[\/latex]\r\n\r\n \r\n
![\"Two](\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/02\/IMG_0283.jpg\")
\r\n\r\nIf the two particles have equal mass, then it's pretty clear that the \"average position\" of the two-particle system is halfway between them. If the masses of the two particles are different, would the \"average position\" still be halfway between them? Perhaps in some sense this is true, but we are not looking for a\u00a0geometric center<\/em><\/span>; we are looking for the average placement of mass. If m1<\/sub>\u00a0has twice the mass of m2<\/sub><\/span>, then when it comes to the average placement of mass, m1<\/sub><\/span>\u00a0gets \"two votes.\" With more of the mass concentrated at the position x1<\/sub><\/span>\u00a0than at x2<\/sub><\/span>, the center of mass should be closer to x1<\/sub><\/span>\u00a0than x2<\/sub><\/span>. We achieve the perfect balance by \"weighting\" the positions by the fraction of the total mass that is located there. Accordingly, we define the center of mass:<\/span>\r\n\r\n[latex]\\bar x_{cm}=(\\frac{m_1}{m_1+m_2})x_1+(\\frac{m_2}{m_1+m_2})x_2=\\frac{m_1x_1+m_2x_2}{M_{system}}[\/latex]\r\n\r\nIf there are more than two particles, we simply add all of them into the sum in the numerator. To extend this definition of the center of mass into three dimensions, we simply need to do the same things in the y <\/span><\/span>and z<\/span>\u00a0directions. A position vector for the center of mass of a system of many particles would then be:<\/span>\r\n\r\n[latex]\\vec{r}_{cm}=\\bar x_{cm}\\underline{\\hat{i}}+\\bar y_{cm}\\underline{\\hat{j}}+ \\bar z_{cm}\\underline{\\hat{k}}\\\\=\\frac{[m_1 x_1+m_2 x_2+...]}{M}\\underline{\\hat{i}}+\\frac{[m_1y_1+m_2y_2+...]}{M}\\underline{\\hat{j}}+\\frac{[m_1 z_1+m_2 z_2+...]}{M}\\underline{\\hat{k}}\\\\=\\frac{m_1[x_1\\underline{\\hat{i}}+y_1\\underline{\\hat{j}}+z_1\\underline{\\hat{k}}]+m_2[x_2\\underline{\\hat{i}}+y_2\\underline{\\hat{j}}+z_2\\underline{\\hat{k}}]+...}{M}\\\\=\\frac{m_1\\vec r_1+m_2\\vec r_2+...}{M}[\/latex]\r\n\r\nSource: \" Center of Mass\" by Tom Weideman, https:\/\/phys.libretexts.org\/Courses\/University_of_California_Davis\/UCD%3A_Physics_9A__Classical_Mechanics\/4%3A_Linear_Momentum\/4.2%3A_Center_of_Mass<\/a>\r\n\r\n<\/div>\r\nExample 2:<\/h2>\r\n
\r\n \t
7.1.3 The Center of Mass of a Thin Uniform Rod (Calculus Method)\r\n<\/strong><\/h1>\r\n
![\"Thin](\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/02\/Screen-Shot-2021-08-29-at-11.55.54-PM.png\")
\r\n\r\nSource: Calculus-Based Physics 1, Jeffery W. Schnick. p143, <\/span>https:\/\/openlibrary.ecampusontario.ca\/catalogue\/item\/?id=ce74a181-ccde-491c-848d-05489ed182e7<\/a>\r\n\r\n<\/div>\r\n7.1.4 The Center of Mass of a Non-Uniform Rod<\/strong><\/h1>\r\n
Example 3:<\/em><\/h2>\r\nFind the position of the center of mass of a thin rod that extends from 0 to 0.890 m along the x-axis of a Cartesian coordinate system and has a linear density given by \u00b5 = 0.650 kg\/m3<\/sup><\/em>\r\n\r\nIn order to be able to determine the position of the center of mass of a rod with a given length and a given linear density as a function of position, you first need to be able to find the mass of such a rod. To do that, one might be tempted to use a method that works only for the special case of a uniform rod, namely, to try using m = \u00b5L with L being the length of the rod. The problem with this is that \u00b5 varies along the entire length of the rod. What value would one use for \u00b5? One might be tempted to evaluate the given \u00b5 at x = L and use that, but that would be acting as if the linear density were constant at \u00b5 = \u00b5(L). It is not. In fact, in the case at hand, \u00b5(L) is the maximum linear density of the rod; it only has that value at one point on the rod.\r\n\r\nInstead, using integration, we find the equation:\r\n
7.1.1 Center of Mass of Two Particles<\/strong><\/h1>\n
<\/p>\n<\/p>\n<\/p>\nExample 1: <\/em><\/h2>\n
\n
7.1.2 Center of Mass in 2D & 3D<\/h1>\n
<\/p>\nExample 2:<\/h2>\n
\n
\nFirst, we calculate the total mass of the system:<\/p>\n