{"id":644,"date":"2026-07-21T02:06:15","date_gmt":"2026-07-21T02:06:15","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=644"},"modified":"2026-07-21T02:06:16","modified_gmt":"2026-07-21T02:06:16","slug":"center-of-mass","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/center-of-mass\/","title":{"rendered":"Center of Mass: Formula, Examples &amp; How to Find It"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\n\nCenter of mass is the point where a system&#8217;s whole mass can be treated as concentrated: the mass-weighted average of every particle&#8217;s position, found by summing each mass times its position and dividing by the total mass. In uniform gravity it coincides with the center of gravity, but the two are conceptually different.\n\n<\/p><\/div>\n\n<p>Toss a hammer across a room and watch it tumble. Every part of it seems to fly on its own wild path \u2014 head over handle, spinning end over end. Yet one invisible point drifts through the air in a smooth, boring arc, as calm as a gently thrown ball.<\/p> <p>That point is the center of mass. Divers rotate around it, dancers leap around it, and the Moon and Earth swing around it. Pin it down and a messy, spinning object suddenly obeys one clean rule \u2014 which is exactly why physicists reach for it first.<\/p> <h2>What Is Center of Mass?<\/h2> <p>The center of mass is the average position of all the mass in a system, weighted so that heavier parts count for more. It is the single balance point that stands in for the whole object when you care about how it moves.<\/p> <p>Think of a mobile hanging over a cot. Slide a heavy shape outward and the balance point shifts toward it; add a feather-light one and barely anything changes. The center of mass leans toward mass, always.<\/p> <h3>Mass-weighted, not just the middle<\/h3> <p>Here is the key idea students often miss: the center of mass is <em>not<\/em> simply the geometric middle. It is an average of positions, but each position is weighted by how much mass sits there. Put more mass on one side and the average slides that way.<\/p> <p>Two children of equal weight on a see-saw balance at the centre. Swap one for a heavier friend and the pivot must move toward the heavier child. Same physics, different weighting.<\/p> <h3>Center of mass and center of gravity<\/h3> <p>In everyday gravity these two points sit in the same place, so people use the terms interchangeably. Strictly, though, they answer different questions: the center of mass depends only on how mass is spread out, while the center of gravity also depends on the gravitational field. Near Earth&#8217;s surface, where gravity is effectively uniform, they coincide exactly.<\/p><figure style=\"margin:32px auto;max-width:640px;text-align:center;\"> <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/07\/blog-hintz-eric-2021-04-08-oregon-digital-df70ch25w-c-oregon-state-banner-edit-1.jpg\" alt=\"High jumper arched over the bar with the center of mass passing below it\" loading=\"lazy\" style=\"width:100%;height:auto;border-radius:4px;\" \/ width=\"1470\" height=\"670\"> <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">A Fosbury-flop jumper arches so their center of mass can pass under the bar while their body goes over it.<\/figcaption> <\/figure> <h2>The Center of Mass Formula<\/h2> <p>For a set of point masses along a line, the center of mass is the sum of each mass times its position, divided by the total mass.<\/p>\n\n<div class=\"pf-formula\">x<sub>cm<\/sub> = (m<sub>1<\/sub>x<sub>1<\/sub> + m<sub>2<\/sub>x<sub>2<\/sub> + \u2026 + m<sub>n<\/sub>x<sub>n<\/sub>) \/ (m<sub>1<\/sub> + m<sub>2<\/sub> + \u2026 + m<sub>n<\/sub>)<\/div>\n\n<p>Written compactly with the summation sign \u03a3 (meaning &#8220;add up over every particle&#8221;), the same relation is:<\/p>\n\n<div class=\"pf-formula\">x<sub>cm<\/sub> = ( \u03a3 m<sub>i<\/sub>x<sub>i<\/sub> ) \/ ( \u03a3 m<sub>i<\/sub> )<\/div>\n\n<p>Real objects live in three dimensions, so the position becomes a vector <strong>r<\/strong> and you compute each axis the same way:<\/p>\n\n<div class=\"pf-formula\">r<sub>cm<\/sub> = (1\/M) \u03a3 m<sub>i<\/sub> r<sub>i<\/sub><\/div>\n\n<p>Every symbol, with its SI unit:<\/p> <ul> <li><strong>x<sub>cm<\/sub><\/strong> (or <strong>r<sub>cm<\/sub><\/strong>) \u2014 position of the center of mass, in metres (m)<\/li> <li><strong>m<sub>i<\/sub><\/strong> \u2014 mass of the <em>i<\/em>-th particle, in kilograms (kg)<\/li> <li><strong>x<sub>i<\/sub><\/strong> (or <strong>r<sub>i<\/sub><\/strong>) \u2014 position of that particle, in metres (m)<\/li> <li><strong>M = \u03a3 m<sub>i<\/sub><\/strong> \u2014 total mass of the whole system, in kilograms (kg)<\/li> <li><strong>\u03a3<\/strong> \u2014 instruction to add the quantity over every particle in the system<\/li> <\/ul> <p>For a solid body the sum becomes an integral over tiny mass elements dm, written r<sub>cm<\/sub> = (1\/M) \u222b r dm \u2014 but the recipe is identical: weight each bit of mass by where it is, then average.<\/p> <p>The arithmetic is quick for two masses and tedious for many, so you can also skip it and drop your values straight into our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/center-of-mass\">Center of Mass Calculator<\/a>, which applies the mass weighting and returns the coordinates for you.<\/p> <svg viewBox=\"0 0 640 250\" role=\"img\" aria-label=\"Diagram of a 2 kilogram mass and a 6 kilogram mass on a beam, with the center of mass marked closer to the heavier mass\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" style=\"width:100%;height:auto;max-width:640px;display:block;margin:24px auto;\"> <rect x=\"0\" y=\"0\" width=\"640\" height=\"250\" rx=\"8\" fill=\"#F5F2EA\"><\/rect> <text x=\"320\" y=\"34\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"17\" font-weight=\"700\" fill=\"#0A1628\">The center of mass is a mass-weighted average<\/text> <rect x=\"70\" y=\"122\" width=\"500\" height=\"8\" rx=\"4\" fill=\"#0A1628\"><\/rect> <circle cx=\"150\" cy=\"126\" r=\"22\" fill=\"#7A1F2B\"><\/circle> <text x=\"150\" y=\"131\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" fill=\"#FAF6EE\">2 kg<\/text> <text x=\"150\" y=\"176\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#0A1628\">m<tspan dy=\"4\" font-size=\"10\">1<\/tspan><\/text> <circle cx=\"500\" cy=\"126\" r=\"38\" fill=\"#C8932A\"><\/circle> <text x=\"500\" y=\"131\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"14\" font-weight=\"700\" fill=\"#0A1628\">6 kg<\/text> <text x=\"500\" y=\"188\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#0A1628\">m<tspan dy=\"4\" font-size=\"10\">2<\/tspan><\/text> <polygon points=\"412,130 396,166 428,166\" fill=\"#0A1628\"><\/polygon> <circle cx=\"412\" cy=\"126\" r=\"7\" fill=\"#C8932A\" stroke=\"#0A1628\" stroke-width=\"2\"><\/circle> <text x=\"412\" y=\"206\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" fill=\"#7A1F2B\">center of mass<\/text> <line x1=\"150\" y1=\"96\" x2=\"412\" y2=\"96\" stroke=\"#0A1628\" stroke-width=\"1\"><\/line> <line x1=\"150\" y1=\"90\" x2=\"150\" y2=\"102\" stroke=\"#0A1628\" stroke-width=\"1\"><\/line> <line x1=\"412\" y1=\"90\" x2=\"412\" y2=\"102\" stroke=\"#0A1628\" stroke-width=\"1\"><\/line> <text x=\"281\" y=\"88\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"12\" fill=\"#0A1628\">d<tspan dy=\"3\" font-size=\"9\">1<\/tspan><\/text> <line x1=\"412\" y1=\"96\" x2=\"500\" y2=\"96\" stroke=\"#0A1628\" stroke-width=\"1\"><\/line> <line x1=\"500\" y1=\"90\" x2=\"500\" y2=\"102\" stroke=\"#0A1628\" stroke-width=\"1\"><\/line> <text x=\"456\" y=\"88\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"12\" fill=\"#0A1628\">d<tspan dy=\"3\" font-size=\"9\">2<\/tspan><\/text> <text x=\"320\" y=\"236\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"14\" fill=\"#0A1628\">The bigger the mass, the closer the center of mass sits.<\/text> <\/svg> <p style=\"text-align:center;font-style:italic;font-size:13px;color:#1F2E47;\">The center of mass of two masses obeys m<sub>1<\/sub>d<sub>1<\/sub> = m<sub>2<\/sub>d<sub>2<\/sub>, so it always lands closer to the heavier mass.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Center of Mass Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame{width:100%;border:none;height:600px}@media(max-width:760px){.pf-sim-frame{height:1000px}}<\/style><iframe src=\"\/labs\/center-of-mass.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n\n<h2>How to Find the Center of Mass<\/h2> <p>To find the center of mass, multiply each mass by its position, add the products, and divide by the total mass \u2014 then repeat for every axis you need. That single procedure handles everything from two blocks on a track to a whole galaxy.<\/p> <h3>The five-step method<\/h3> <ol> <li><strong>Choose an origin and axis.<\/strong> Any point works; a clever choice (one mass at zero) kills half the arithmetic.<\/li> <li><strong>List each mass and its position<\/strong> along that axis.<\/li> <li><strong>Form \u03a3 m<sub>i<\/sub>x<sub>i<\/sub><\/strong> \u2014 multiply, then add.<\/li> <li><strong>Divide by the total mass<\/strong> M = \u03a3 m<sub>i<\/sub>.<\/li> <li><strong>Repeat for y (and z)<\/strong> if the problem is two- or three-dimensional.<\/li> <\/ol> <p>Notice that the answer never depends on where you put the origin \u2014 shift it, and every coordinate shifts by the same amount. The center of mass is a real physical point, not an artefact of your axes.<\/p> <h3>Shortcut: let symmetry do the work<\/h3> <p>For a uniform, symmetric object, the center of mass sits at the geometric centre \u2014 no calculation required. A uniform rod balances at its midpoint; a uniform disc balances at its hub; a uniform sphere balances dead centre.<\/p> <p>Spot the symmetry first. It turns a scary integral into a one-line answer.<\/p> <h3>Composite shapes and the subtraction trick<\/h3> <p>Awkward objects yield to a simple move: split them into easy pieces, treat each piece as a single mass sitting at <em>its own<\/em> center of mass, then average those. An L-bracket becomes two rectangles; a hammer becomes a rod plus a block.<\/p> <p>Got a hole? Treat the missing material as <strong>negative mass<\/strong>. Compute the solid shape, subtract the cut-out weighted at its own centre, and the center of mass shifts away from the gap. It is the same formula, run in reverse.<\/p> <h2>Real-World Examples of Center of Mass<\/h2> <p>Real examples of center of mass run from a wobbling spanner to the Earth\u2013Moon system pivoting around a point buried inside our planet. Once you look for it, it is everywhere.<\/p> <h3>The high jumper who beats their own body<\/h3> <p>An elite high jumper using the Fosbury flop arches backward over the bar. Done well, the body clears the bar while the center of mass \u2014 the average of all that mass \u2014 actually passes <em>beneath<\/em> it. The athlete raises less &#8220;average height,&#8221; so the same leap clears a higher bar.<\/p> <h3>The thrown hammer<\/h3> <p>Launch a hammer and its head, handle and everything between tumble chaotically. But the center of mass ignores the spin and traces a clean parabola, exactly like a point projectile. That is why <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/projectile-motion-guide\/\">projectile motion<\/a> works even for objects that are wildly rotating in flight.<\/p> <h3>The Earth\u2013Moon barycenter<\/h3> <p>We say the Moon orbits Earth, but both bodies actually orbit their common center of mass, called the <em>barycenter<\/em>. Because Earth is about 81 times heavier, that point lies roughly 4,670 km from Earth&#8217;s centre \u2014 beneath the surface, since Earth&#8217;s radius is 6,371 km. Earth doesn&#8217;t sit still; it wobbles around this buried point, a fact <a href=\"https:\/\/spaceplace.nasa.gov\/barycenter\/en\/\" target=\"_blank\" rel=\"noopener\">NASA<\/a> uses to hunt for planets around distant stars.<\/p> <h3>The ring with no centre<\/h3> <p>A uniform ring, boomerang or doughnut carries its center of mass in empty air. There is no material at the point at all \u2014 proof that the center of mass is a location in space, not a lump you can touch.<\/p> <svg viewBox=\"0 0 640 300\" role=\"img\" aria-label=\"A uniform ring with its center of mass marked at the geometric centre, located in the empty hole where there is no material\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" style=\"width:100%;height:auto;max-width:560px;display:block;margin:24px auto;\"> <rect x=\"0\" y=\"0\" width=\"640\" height=\"300\" rx=\"8\" fill=\"#F5F2EA\"><\/rect> <text x=\"320\" y=\"34\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"17\" font-weight=\"700\" fill=\"#0A1628\">The center of mass can lie in empty space<\/text> <circle cx=\"230\" cy=\"172\" r=\"95\" fill=\"#7A1F2B\"><\/circle> <circle cx=\"230\" cy=\"172\" r=\"52\" fill=\"#F5F2EA\"><\/circle> <line x1=\"214\" y1=\"172\" x2=\"246\" y2=\"172\" stroke=\"#C8932A\" stroke-width=\"3\"><\/line> <line x1=\"230\" y1=\"156\" x2=\"230\" y2=\"188\" stroke=\"#C8932A\" stroke-width=\"3\"><\/line> <circle cx=\"230\" cy=\"172\" r=\"7\" fill=\"#C8932A\" stroke=\"#0A1628\" stroke-width=\"2\"><\/circle> <line x1=\"238\" y1=\"165\" x2=\"430\" y2=\"112\" stroke=\"#0A1628\" stroke-width=\"1\"><\/line> <text x=\"435\" y=\"108\" font-family=\"Arial, sans-serif\" font-size=\"15\" font-weight=\"700\" fill=\"#7A1F2B\">center of mass<\/text> <text x=\"435\" y=\"132\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#0A1628\">There is no material here \u2014<\/text> <text x=\"435\" y=\"152\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#0A1628\">it sits in the hole of the ring.<\/text> <text x=\"230\" y=\"292\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#0A1628\">A uniform ring: center of mass at the geometric centre.<\/text> <\/svg> <p style=\"text-align:center;font-style:italic;font-size:13px;color:#1F2E47;\">For a uniform ring, the center of mass lies exactly at the middle of the hole, where there is no material at all.<\/p> <h2>Common Misconceptions About Center of Mass<\/h2> <p>The most common misconception is that the center of mass must lie inside the object. As the ring and the arched high jumper show, it frequently sits in empty space. Here are the slips worth unlearning.<\/p> <h3>&#8220;It has to be inside the object&#8221;<\/h3> <p>It does not. Rings, horseshoes, boomerangs and arched gymnasts all place their center of mass off the material. The point is defined by an average, and averages don&#8217;t have to land on anything solid.<\/p> <h3>&#8220;It&#8217;s the same as the geometric centre&#8221;<\/h3> <p>Only for uniform density. Load one end of a bar with lead and the center of mass slides toward the lead, even though the geometric middle hasn&#8217;t moved. Mass distribution is what counts, not shape alone.<\/p> <h3>&#8220;Center of mass and center of gravity are always identical&#8221;<\/h3> <p>They match in a uniform gravitational field \u2014 which covers essentially every laboratory and sports field. In a field that varies across the object (a very tall structure, or an object near a strong source), the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/center-of-gravity-physics\/\">center of gravity<\/a> shifts slightly toward the region of stronger gravity while the center of mass stays put.<\/p> <h3>&#8220;It just sits where most of the material is&#8221;<\/h3> <p>The center of mass is a weighted average, not the address of the biggest chunk. Two unequal masses always place it <em>between<\/em> them, closer to the heavier one but never on top of it, unless the other mass is zero.<\/p> <h2>How Center of Mass Relates to Momentum, Gravity and Rotation<\/h2> <p>The center of mass ties straight to momentum: a system&#8217;s total momentum equals its total mass times the velocity of its center of mass, p = M v<sub>cm<\/sub>. Track that one point and you have captured the motion of the entire system.<\/p> <h3>Momentum and Newton&#8217;s second law<\/h3> <p>Because internal forces between the parts cancel in pairs, only external forces move the center of mass. That gives a clean system-wide version of <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/newtons-second-law\/\">Newton&#8217;s second law<\/a>: F<sub>ext<\/sub> = M a<sub>cm<\/sub>. It also means that with no net external force, the center of mass glides at constant velocity \u2014 the everyday face of the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/conservation-of-momentum\/\">conservation of momentum<\/a>.<\/p> <p>This is why an exploding firework&#8217;s fragments still average out along the original path, and why studying <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/momentum-and-impulse\/\">momentum<\/a> and <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/elastic-inelastic-collisions\/\">collisions<\/a> becomes far simpler when you switch to the center-of-mass frame.<\/p> <h3>Rotation happens about it<\/h3> <p>A free object that isn&#8217;t pushed off-centre spins about its center of mass, not about some other point. Divers, thrown phones and tumbling satellites all rotate around it, which is why it anchors the study of rotational motion and moment of inertia.<\/p> <div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\"> <table style=\"width:100%;border-collapse:collapse;word-break:break-word;\"> <thead> <tr style=\"background:#142139;color:#FAF6EE;\"> <th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Property<\/th> <th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Center of mass<\/th> <th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Center of gravity<\/th> <th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Centroid (geometric centre)<\/th> <\/tr> <\/thead> <tbody> <tr> <td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Averages positions weighted by<\/strong><\/td> <td style=\"padding:10px;border:1px solid #D9CFB8;\">mass<\/td> <td style=\"padding:10px;border:1px solid #D9CFB8;\">weight (mass \u00d7 local g)<\/td> <td style=\"padding:10px;border:1px solid #D9CFB8;\">nothing \u2014 equal weighting<\/td> <\/tr> <tr style=\"background:#F5F2EA;\"> <td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Depends on<\/strong><\/td> <td style=\"padding:10px;border:1px solid #D9CFB8;\">mass distribution only<\/td> <td style=\"padding:10px;border:1px solid #D9CFB8;\">mass distribution + gravitational field<\/td> <td style=\"padding:10px;border:1px solid #D9CFB8;\">shape only<\/td> <\/tr> <tr> <td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Coincides with center of mass when<\/strong><\/td> <td style=\"padding:10px;border:1px solid #D9CFB8;\">\u2014<\/td> <td style=\"padding:10px;border:1px solid #D9CFB8;\">gravity is uniform<\/td> <td style=\"padding:10px;border:1px solid #D9CFB8;\">density is uniform<\/td> <\/tr> <tr style=\"background:#F5F2EA;\"> <td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Mainly used for<\/strong><\/td> <td style=\"padding:10px;border:1px solid #D9CFB8;\">motion, momentum, collisions<\/td> <td style=\"padding:10px;border:1px solid #D9CFB8;\">balance and stability<\/td> <td style=\"padding:10px;border:1px solid #D9CFB8;\">geometry and design<\/td> <\/tr> <\/tbody> <\/table> <\/div> <h2>Worked Problems<\/h2>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">Two masses sit on a straight track: m\u2081 = 2 kg at x = 0 m and m\u2082 = 6 kg at x = 4 m. Where is the center of mass?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\n\n<p>Step 1: Apply x<sub>cm<\/sub> = (m<sub>1<\/sub>x<sub>1<\/sub> + m<sub>2<\/sub>x<sub>2<\/sub>) \/ (m<sub>1<\/sub> + m<sub>2<\/sub>).<\/p> <p>Step 2: Substitute with units: x<sub>cm<\/sub> = (2 kg \u00d7 0 m + 6 kg \u00d7 4 m) \/ (2 kg + 6 kg).<\/p> <p>Step 3: Solve: x<sub>cm<\/sub> = (0 + 24) \/ 8 = 3 m.<\/p> <strong>Answer: 3 m from the origin \u2014 three-quarters of the way toward the heavier 6 kg mass.<\/strong> <\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">Masses of 1 kg and 3 kg are 8 m apart. Use the inverse-ratio shortcut to place the center of mass, then confirm it with the formula.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\n\n<p>Step 1: Inverse-ratio rule \u2014 the center of mass divides the gap in the inverse ratio of the masses, 1 : 3, so the 3 kg mass claims 3 of every 4 parts of the &#8220;pull.&#8221; Distance from the 1 kg mass = (3\/4) \u00d7 8 m = 6 m.<\/p> <p>Step 2: Check with the formula, taking the 1 kg mass at x = 0 and the 3 kg mass at x = 8 m: x<sub>cm<\/sub> = (1 \u00d7 0 + 3 \u00d7 8) \/ (1 + 3) = 24 \/ 4 = 6 m.<\/p> <p>Step 3: Both methods agree.<\/p> <strong>Answer: 6 m from the 1 kg mass (2 m from the 3 kg mass) \u2014 closer to the heavier one, as expected.<\/strong> <\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">Three masses lie in a plane: 1 kg at (0, 0), 2 kg at (4, 0), and 3 kg at (2, 3), all in metres. Find the center of mass.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\n\n<p>Step 1: Total mass M = 1 + 2 + 3 = 6 kg. Handle each axis separately.<\/p> <p>Step 2: x<sub>cm<\/sub> = (1\u00d70 + 2\u00d74 + 3\u00d72) \/ 6 = (0 + 8 + 6) \/ 6 = 14\/6 \u2248 2.33 m.<\/p> <p>Step 3: y<sub>cm<\/sub> = (1\u00d70 + 2\u00d70 + 3\u00d73) \/ 6 = 9\/6 = 1.5 m.<\/p> <strong>Answer: The center of mass is at approximately (2.33 m, 1.5 m).<\/strong> <\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A uniform rod of mass 3 kg and length 2 m has a 2 kg point mass fixed to one end. Taking the rod from x = 0 to x = 2 m, find the center of mass of the combination.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\n\n<p>Step 1: Replace the rod with its own center of mass \u2014 a uniform rod balances at its midpoint, x = 1 m. The point mass sits at x = 2 m.<\/p> <p>Step 2: x<sub>cm<\/sub> = (m<sub>rod<\/sub>x<sub>rod<\/sub> + m<sub>point<\/sub>x<sub>point<\/sub>) \/ (m<sub>rod<\/sub> + m<sub>point<\/sub>) = (3 kg \u00d7 1 m + 2 kg \u00d7 2 m) \/ (3 kg + 2 kg).<\/p> <p>Step 3: x<sub>cm<\/sub> = (3 + 4) \/ 5 = 7\/5 = 1.4 m.<\/p> <strong>Answer: 1.4 m from the rod&#8217;s free end \u2014 shifted toward the loaded end.<\/strong> <\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">A uniform disc of radius R = 12 cm has a circular hole of radius 6 cm cut from it, the hole&#039;s centre lying 6 cm from the disc&#039;s centre. Where is the center of mass of what remains?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\n\n<p>Step 1: Treat the hole as negative mass. For uniform thickness, mass is proportional to area. Full disc: area \u221d \u03c0R\u00b2 at x = 0. Hole: area \u221d \u03c0(R\/2)\u00b2 = \u03c0R\u00b2\/4 at x = R\/2.<\/p> <p>Step 2: x<sub>cm<\/sub> = [\u03c0R\u00b2(0) \u2212 (\u03c0R\u00b2\/4)(R\/2)] \/ [\u03c0R\u00b2 \u2212 \u03c0R\u00b2\/4] = [\u2212\u03c0R\u00b3\/8] \/ [3\u03c0R\u00b2\/4].<\/p> <p>Step 3: Simplify: x<sub>cm<\/sub> = \u2212(R\u00b3\/8) \u00d7 (4 \/ 3R\u00b2) = \u2212R\/6 = \u2212(12 cm)\/6 = \u22122 cm.<\/p> <strong>Answer: 2 cm from the disc&#8217;s centre, on the side opposite the hole.<\/strong> <\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">Estimate the Earth\u2013Moon barycenter. Use Earth&#039;s mass 5.97 \u00d7 10\u00b2\u2074 kg, the Moon&#039;s mass 7.35 \u00d7 10\u00b2\u00b2 kg, and a centre-to-centre distance of 3.84 \u00d7 10\u2078 m. Measure from Earth&#039;s centre.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\n\n<p>Step 1: Put Earth at x = 0 and the Moon at x = d. The center of mass from Earth&#8217;s centre is x<sub>cm<\/sub> = (M<sub>Moon<\/sub> \u00d7 d) \/ (M<sub>Earth<\/sub> + M<sub>Moon<\/sub>).<\/p> <p>Step 2: Substitute: x<sub>cm<\/sub> = (7.35 \u00d7 10\u00b2\u00b2 \u00d7 3.84 \u00d7 10\u2078) \/ (5.97 \u00d7 10\u00b2\u2074 + 7.35 \u00d7 10\u00b2\u00b2) = (2.82 \u00d7 10\u00b3\u00b9) \/ (6.04 \u00d7 10\u00b2\u2074).<\/p> <p>Step 3: x<sub>cm<\/sub> \u2248 4.67 \u00d7 10\u2076 m \u2248 4,670 km. Earth&#8217;s radius is 6,371 km.<\/p> <strong>Answer: About 4,670 km from Earth&#8217;s centre \u2014 roughly 1,700 km below the surface, so the barycenter lies inside the Earth.<\/strong> <\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A 60 kg person stands at one end of a 40 kg boat, 4 m long, floating on frictionless water. They walk to the other end. How far does the boat move?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\n\n<p>Step 1: No horizontal external force acts, so the center of mass cannot move. Let the boat slide a distance \u0394 opposite to the walk; the person&#8217;s displacement over the ground is (4 \u2212 \u0394).<\/p> <p>Step 2: Keep the center of mass fixed: m<sub>person<\/sub>(4 \u2212 \u0394) = m<sub>boat<\/sub>\u0394, i.e. 60(4 \u2212 \u0394) = 40\u0394.<\/p> <p>Step 3: Solve: 240 \u2212 60\u0394 = 40\u0394 \u2192 240 = 100\u0394 \u2192 \u0394 = 2.4 m. (Check: 60 \u00d7 1.6 m = 96 = 40 \u00d7 2.4 m.)<\/p> <strong>Answer: The boat slides 2.4 m; the person moves 1.6 m over the ground.<\/strong> <\/div><\/details><\/div> <h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>Is the center of mass the same as the center of gravity?<\/summary><div class=\"pf-faq-item-answer\">\nThey are the same point whenever gravity is uniform, which holds for almost every everyday object near Earth&#8217;s surface. The distinction only appears in a non-uniform gravitational field: the center of mass is a pure mass average, while the center of gravity is weighted by the local gravitational pull, so a very tall structure can have them a small distance apart.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Can the center of mass be outside an object?<\/summary><div class=\"pf-faq-item-answer\">\nYes. Any object with a gap or a curve can have its center of mass in empty space. A ring&#8217;s center of mass sits in the central hole, a boomerang&#8217;s lies off the material in the crook, and a high jumper arched over a bar can have a center of mass that passes beneath it. Nothing has to be physically present at the point.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>How do you calculate the center of mass of two objects?<\/summary><div class=\"pf-faq-item-answer\">\nMultiply each mass by its position, add the two products, and divide by the combined mass: x<sub>cm<\/sub> = (m<sub>1<\/sub>x<sub>1<\/sub> + m<sub>2<\/sub>x<sub>2<\/sub>) \/ (m<sub>1<\/sub> + m<sub>2<\/sub>). A fast check is the inverse-ratio rule \u2014 the center of mass divides the line between them in the inverse ratio of their masses, so it always sits closer to the heavier object.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the center of mass of a system of particles?<\/summary><div class=\"pf-faq-item-answer\">\nIt is the mass-weighted average position of all the particles: sum each mass times its position, then divide by the total mass. The result is a single point that moves as though every external force acted there and all the mass were gathered at it, which is what makes it so useful for tracking complicated systems.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Does the center of mass move if there is no external force?<\/summary><div class=\"pf-faq-item-answer\">\nNo. With zero net external force, the center of mass keeps a constant velocity \u2014 staying still if it began still \u2014 no matter how the parts push on one another internally. This follows from conservation of momentum, and it is why an astronaut drifting in space cannot shift their center of mass just by waving their arms.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the difference between the center of mass and the centroid?<\/summary><div class=\"pf-faq-item-answer\">\nThe centroid is the purely geometric centre of a shape, weighting every point equally, while the center of mass weights each point by how much mass is there. For an object of uniform density the two coincide exactly. They part ways only when the material is denser in some regions than others, which pulls the center of mass toward the heavier side.\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>Center of mass is the mass-weighted average position of a system \u2014 the point where all its mass acts as one. 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