{"id":583,"date":"2026-07-16T16:25:34","date_gmt":"2026-07-16T16:25:34","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=583"},"modified":"2026-07-16T16:25:36","modified_gmt":"2026-07-16T16:25:36","slug":"circular-motion-physics","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/circular-motion-physics\/","title":{"rendered":"Circular Motion Physics: Formula, Examples &amp; Uses"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\nCircular motion physics describes an object travelling along a circular path, where the velocity always points along the tangent and a centripetal acceleration of a = v\u00b2\/r points constantly toward the centre. Even at a perfectly constant speed the object is accelerating, because its direction changes every instant, so a net inward force is always required.\n<\/p><\/div>\n<p>Take a roundabout at a steady 36 km\/h and your speedometer will not budge. Your body will. You feel yourself pressed toward the door, your coffee slides across the cup holder, and something deep in you insists you are being flung outwards.<\/p>\n<p>Nothing is flinging you anywhere. You are simply trying to go straight while the car turns out from under you \u2014 and once that clicks, it unlocks everything from a hammer throw to why the Moon has not yet fallen on us.<\/p>\n<h2>What Is Circular Motion?<\/h2>\n<p>Circular motion is the movement of an object along a circular path, in which its velocity stays tangent to the circle while a net force pulls it continuously toward the centre. That definition hides a trap \u2014 and the trap is where most marks are lost.<\/p>\n<p>Speed and velocity are not the same thing. Speed is just a number; velocity is a number <em>and<\/em> a direction. Ride that roundabout at a steady 10 m\/s and your speed is pinned, but your <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/kinematics\/velocity-vs-speed\/\">velocity is being rewritten every instant<\/a>, because the direction keeps turning.<\/p>\n<p>A changing velocity is precisely what <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/kinematics\/acceleration-in-physics\/\">acceleration<\/a> means. So an object in uniform circular motion accelerates without ever going faster.<\/p>\n<p>That is the whole subject in one sentence.<\/p>\n<h3>Uniform vs non-uniform circular motion<\/h3>\n<ul>\n<li><strong>Uniform circular motion<\/strong> \u2014 the speed is constant, so only the direction changes. The acceleration points purely at the centre and its size is fixed at v\u00b2\/r.<\/li>\n<li><strong>Non-uniform circular motion<\/strong> \u2014 the object also speeds up or slows down, adding a <em>tangential<\/em> acceleration along the direction of travel. The total acceleration is the vector sum of the two, and it no longer aims at the centre.<\/li>\n<\/ul>\n<p>A car holding a steady speed round a bend is uniform. The same car braking mid-bend is not \u2014 which is exactly why braking hard in a corner is how you lose grip.<\/p>\n<svg viewBox=\"0 0 640 430\" role=\"img\" aria-label=\"Circular motion physics diagram: the velocity arrow points along the tangent while the centripetal acceleration a equals v squared over r points to the centre of the circle\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\">\n  <defs>\n    <marker id=\"cmG\" markerUnits=\"userSpaceOnUse\" markerWidth=\"15\" markerHeight=\"12\" refX=\"14\" refY=\"6\" orient=\"auto\"><path d=\"M0,0 L14,6 L0,12 Z\" fill=\"#C8932A\"><\/path><\/marker>\n    <marker id=\"cmW\" markerUnits=\"userSpaceOnUse\" markerWidth=\"15\" markerHeight=\"12\" refX=\"14\" refY=\"6\" orient=\"auto\"><path d=\"M0,0 L14,6 L0,12 Z\" fill=\"#7A1F2B\"><\/path><\/marker>\n    <marker id=\"cmI\" markerUnits=\"userSpaceOnUse\" markerWidth=\"11\" markerHeight=\"9\" refX=\"10\" refY=\"4.5\" orient=\"auto\"><path d=\"M0,0 L10,4.5 L0,9 Z\" fill=\"#142139\"><\/path><\/marker>\n  <\/defs>\n  <rect x=\"1\" y=\"1\" width=\"638\" height=\"428\" rx=\"5\" fill=\"#F5F2EA\" stroke=\"#D9CFB8\" stroke-width=\"2\"><\/rect>\n  <text x=\"26\" y=\"34\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" fill=\"#7A1F2B\" letter-spacing=\"1.2\">UNIFORM CIRCULAR MOTION<\/text>\n  <text x=\"26\" y=\"54\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12.5\" fill=\"#142139\">Speed never changes. Velocity changes every instant.<\/text>\n  <circle cx=\"230\" cy=\"245\" r=\"130\" fill=\"none\" stroke=\"#C5D0DC\" stroke-width=\"2\" stroke-dasharray=\"7 5\"><\/circle>\n  <path d=\"M 295 132.4 A 130 130 0 0 0 165 132.4\" fill=\"none\" stroke=\"#142139\" stroke-width=\"2\" marker-end=\"url(#cmI)\"><\/path>\n  <text x=\"230\" y=\"101\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11.5\" fill=\"#142139\" text-anchor=\"middle\">direction of travel<\/text>\n  <line x1=\"230\" y1=\"245\" x2=\"360\" y2=\"245\" stroke=\"#0A1628\" stroke-width=\"1.6\"><\/line>\n  <circle cx=\"230\" cy=\"245\" r=\"4.5\" fill=\"#0A1628\"><\/circle>\n  <text x=\"230\" y=\"272\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12\" fill=\"#142139\" text-anchor=\"middle\">centre<\/text>\n  <text x=\"250\" y=\"235\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"17\" font-style=\"italic\" font-weight=\"700\" fill=\"#0A1628\">r<\/text>\n  <line x1=\"360\" y1=\"245\" x2=\"292\" y2=\"245\" stroke=\"#7A1F2B\" stroke-width=\"4.5\" marker-end=\"url(#cmW)\"><\/line>\n  <line x1=\"360\" y1=\"245\" x2=\"360\" y2=\"140\" stroke=\"#C8932A\" stroke-width=\"4.5\" marker-end=\"url(#cmG)\"><\/line>\n  <line x1=\"360\" y1=\"132\" x2=\"360\" y2=\"76\" stroke=\"#142139\" stroke-width=\"2\" stroke-dasharray=\"6 5\"><\/line>\n  <path d=\"M 360 224 L 339 224 L 339 245\" fill=\"none\" stroke=\"#0A1628\" stroke-width=\"1.4\"><\/path>\n  <circle cx=\"360\" cy=\"245\" r=\"11\" fill=\"#C8932A\" stroke=\"#0A1628\" stroke-width=\"2\"><\/circle>\n  <text x=\"322\" y=\"290\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"16\" font-weight=\"700\" fill=\"#7A1F2B\" text-anchor=\"middle\">a = v\u00b2\/r<\/text>\n  <text x=\"322\" y=\"308\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11.5\" fill=\"#142139\" text-anchor=\"middle\">centripetal \u2014 always toward the centre<\/text>\n  <text x=\"382\" y=\"152\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"17\" font-weight=\"700\" font-style=\"italic\" fill=\"#0A1628\">v<\/text>\n  <text x=\"382\" y=\"171\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11.5\" fill=\"#142139\">tangential speed \u2014<\/text>\n  <text x=\"382\" y=\"187\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11.5\" fill=\"#142139\">constant in size,<\/text>\n  <text x=\"382\" y=\"203\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11.5\" fill=\"#142139\">turning in direction<\/text>\n  <text x=\"382\" y=\"88\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11.5\" font-weight=\"700\" fill=\"#142139\">Cut the inward force and<\/text>\n  <text x=\"382\" y=\"104\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11.5\" fill=\"#142139\">it carries straight on along<\/text>\n  <text x=\"382\" y=\"120\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11.5\" fill=\"#142139\">this tangent \u2014 not outward.<\/text>\n  <text x=\"230\" y=\"405\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12\" fill=\"#142139\" text-anchor=\"middle\">v is always at 90\u00b0 to r \u00b7 a is always along r<\/text>\n<\/svg>\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;margin-top:8px;\">The anatomy of uniform circular motion: velocity along the tangent, acceleration toward the centre, always at right angles.<\/p>\n<h2>The Circular Motion Formula<\/h2>\n<p>The formula for uniform circular motion is a = v\u00b2\/r: the centripetal acceleration equals the square of the tangential speed divided by the radius of the path.<\/p>\n<div class=\"pf-formula\">a = v\u00b2 \/ r<\/div>\n<ul>\n<li><strong>a<\/strong> \u2014 centripetal acceleration, in metres per second squared (m\/s\u00b2). Always directed toward the centre.<\/li>\n<li><strong>v<\/strong> \u2014 tangential speed, in metres per second (m\/s), measured along the path.<\/li>\n<li><strong>r<\/strong> \u2014 radius of the circular path, in metres (m).<\/li>\n<\/ul>\n<p>Multiply by mass and Newton&#8217;s second law hands you the force needed to sustain the turn.<\/p>\n<div class=\"pf-formula\">F = m v\u00b2 \/ r<\/div>\n<ul>\n<li><strong>F<\/strong> \u2014 centripetal force, in newtons (N).<\/li>\n<li><strong>m<\/strong> \u2014 mass of the object, in kilograms (kg).<\/li>\n<\/ul>\n<p>Notice the square. Radius matters, but speed matters twice over: double your speed round the same bend and you do not need twice the grip \u2014 you need <strong>four times<\/strong> it. You can feel that exponent by dragging the speed slider in the lab below, or by running your own numbers through our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/circular-motion\">Circular Motion Calculator<\/a>.<\/p>\n<h3>The related quantities<\/h3>\n<p>Circular motion is usually described with a small family of linked quantities. Any one of them can be traded for another.<\/p>\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n<thead>\n<tr style=\"background:#0A1628;color:#FAF6EE;\">\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Quantity<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Symbol<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Equation<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">SI unit<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">Tangential speed<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">v<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">v = 2\u03c0r \/ T = \u03c9r<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">m\/s<\/td><\/tr>\n<tr style=\"background:#F5F2EA;\"><td style=\"padding:10px;border:1px solid #D9CFB8;\">Period (one full lap)<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">T<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">T = 2\u03c0r \/ v<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">s<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">Frequency<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">f<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">f = 1 \/ T<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Hz<\/td><\/tr>\n<tr style=\"background:#F5F2EA;\"><td style=\"padding:10px;border:1px solid #D9CFB8;\">Angular velocity<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">\u03c9<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">\u03c9 = 2\u03c0f = v \/ r<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">rad\/s<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">Centripetal acceleration<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">a<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">a = v\u00b2\/r = \u03c9\u00b2r = 4\u03c0\u00b2r \/ T\u00b2<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">m\/s\u00b2<\/td><\/tr>\n<tr style=\"background:#F5F2EA;\"><td style=\"padding:10px;border:1px solid #D9CFB8;\">Centripetal force<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">F<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">F = mv\u00b2\/r = m\u03c9\u00b2r<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">N<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>In practice, a common student slip is letting degrees creep in. Every one of these assumes <strong>radians<\/strong>. Leave your calculator in degree mode and the answers will look plausible and be wrong.<\/p>\n<h2>How Circular Motion Works<\/h2>\n<p>Circular motion works because a sideways force bends the object&#8217;s straight-line path into a curve without ever speeding it up. The force acts perpendicular to the motion, so it changes direction only.<\/p>\n<p>That perpendicularity is the key. A force pulling forwards would make the object faster; a force pulling backwards would slow it. A force pulling exactly sideways can do neither, so it has no choice but to steer.<\/p>\n<p>It also explains a tidy result: the centripetal force does <strong>no work<\/strong>. Work needs a force component along the motion, and there isn&#8217;t one. The Moon has orbited for billions of years without gravity spending a single joule on it.<\/p>\n<h3>Where a = v\u00b2\/r comes from<\/h3>\n<p>Take two snapshots of the object a small angle \u0394\u03b8 apart. Because the velocity always sits at 90\u00b0 to the radius, the velocity vector turns through exactly the same angle \u0394\u03b8 that the radius does.<\/p>\n<p>Now draw both velocities tail to tail. You get a triangle with two equal sides \u2014 the speed hasn&#8217;t changed \u2014 separated by that same \u0394\u03b8. It is <em>similar<\/em> to the triangle made by the two radii and the chord between the positions.<\/p>\n<svg viewBox=\"0 0 700 400\" role=\"img\" aria-label=\"Derivation of the circular motion formula a equals v squared over r using similar triangles: the change in velocity delta v points toward the centre of the circle\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\">\n  <defs>\n    <marker id=\"dvG\" markerUnits=\"userSpaceOnUse\" markerWidth=\"13\" markerHeight=\"10\" refX=\"12\" refY=\"5\" orient=\"auto\"><path d=\"M0,0 L12,5 L0,10 Z\" fill=\"#C8932A\"><\/path><\/marker>\n    <marker id=\"dvW\" markerUnits=\"userSpaceOnUse\" markerWidth=\"13\" markerHeight=\"10\" refX=\"12\" refY=\"5\" orient=\"auto\"><path d=\"M0,0 L12,5 L0,10 Z\" fill=\"#7A1F2B\"><\/path><\/marker>\n    <marker id=\"dvI\" markerUnits=\"userSpaceOnUse\" markerWidth=\"9\" markerHeight=\"7\" refX=\"8\" refY=\"3.5\" orient=\"auto\"><path d=\"M0,0 L8,3.5 L0,7 Z\" fill=\"#142139\"><\/path><\/marker>\n  <\/defs>\n  <rect x=\"1\" y=\"1\" width=\"698\" height=\"398\" rx=\"5\" fill=\"#F5F2EA\" stroke=\"#D9CFB8\" stroke-width=\"2\"><\/rect>\n  <text x=\"26\" y=\"34\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" fill=\"#7A1F2B\" letter-spacing=\"1.2\">WHERE a = v\u00b2\/r COMES FROM<\/text>\n  <line x1=\"336\" y1=\"78\" x2=\"336\" y2=\"300\" stroke=\"#D9CFB8\" stroke-width=\"1.5\" stroke-dasharray=\"5 5\"><\/line>\n  <text x=\"175\" y=\"62\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12.5\" font-weight=\"700\" fill=\"#142139\" text-anchor=\"middle\">Two snapshots, \u0394\u03b8 apart<\/text>\n  <text x=\"500\" y=\"62\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12.5\" font-weight=\"700\" fill=\"#142139\" text-anchor=\"middle\">The same two velocities, tail to tail<\/text>\n  <circle cx=\"175\" cy=\"215\" r=\"105\" fill=\"none\" stroke=\"#C5D0DC\" stroke-width=\"1.8\" stroke-dasharray=\"6 5\"><\/circle>\n  <circle cx=\"175\" cy=\"215\" r=\"4\" fill=\"#0A1628\"><\/circle>\n  <line x1=\"175\" y1=\"215\" x2=\"273.7\" y2=\"179.1\" stroke=\"#0A1628\" stroke-width=\"1.4\"><\/line>\n  <line x1=\"175\" y1=\"215\" x2=\"219.4\" y2=\"119.8\" stroke=\"#0A1628\" stroke-width=\"1.4\"><\/line>\n  <path d=\"M 210.7 202 A 38 38 0 0 0 191.1 180.6\" fill=\"none\" stroke=\"#7A1F2B\" stroke-width=\"1.8\"><\/path>\n  <text x=\"216\" y=\"182\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" fill=\"#7A1F2B\">\u0394\u03b8<\/text>\n  <line x1=\"273.7\" y1=\"179.1\" x2=\"219.4\" y2=\"119.8\" stroke=\"#142139\" stroke-width=\"1.8\" stroke-dasharray=\"5 4\"><\/line>\n  <text x=\"224\" y=\"150\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"12.5\" font-weight=\"700\" fill=\"#142139\" text-anchor=\"middle\">\u0394s<\/text>\n  <line x1=\"273.7\" y1=\"179.1\" x2=\"258.3\" y2=\"136.8\" stroke=\"#C8932A\" stroke-width=\"3.4\" marker-end=\"url(#dvG)\"><\/line>\n  <line x1=\"219.4\" y1=\"119.8\" x2=\"178.6\" y2=\"100.8\" stroke=\"#C8932A\" stroke-width=\"3.4\" marker-end=\"url(#dvG)\"><\/line>\n  <circle cx=\"273.7\" cy=\"179.1\" r=\"6\" fill=\"#C8932A\" stroke=\"#0A1628\" stroke-width=\"1.6\"><\/circle>\n  <circle cx=\"219.4\" cy=\"119.8\" r=\"6\" fill=\"#C8932A\" stroke=\"#0A1628\" stroke-width=\"1.6\"><\/circle>\n  <text x=\"266\" y=\"128\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" font-style=\"italic\" fill=\"#0A1628\">v1<\/text>\n  <text x=\"158\" y=\"93\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" font-style=\"italic\" fill=\"#0A1628\">v2<\/text>\n  <text x=\"150\" y=\"272\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11.5\" font-weight=\"700\" fill=\"#142139\" text-anchor=\"middle\">Speed is equal: |v1| = |v2|<\/text>\n  <text x=\"150\" y=\"289\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11.5\" fill=\"#142139\" text-anchor=\"middle\">Only the direction has turned.<\/text>\n  <line x1=\"490\" y1=\"255\" x2=\"450.7\" y2=\"146.9\" stroke=\"#C8932A\" stroke-width=\"3.4\" marker-end=\"url(#dvG)\"><\/line>\n  <line x1=\"490\" y1=\"255\" x2=\"385.8\" y2=\"206.4\" stroke=\"#C8932A\" stroke-width=\"3.4\" marker-end=\"url(#dvG)\"><\/line>\n  <line x1=\"450.7\" y1=\"146.9\" x2=\"385.8\" y2=\"206.4\" stroke=\"#7A1F2B\" stroke-width=\"3.4\" marker-end=\"url(#dvW)\"><\/line>\n  <path d=\"M 475.6 215.5 A 42 42 0 0 0 451.9 237.2\" fill=\"none\" stroke=\"#7A1F2B\" stroke-width=\"1.8\"><\/path>\n  <text x=\"443\" y=\"216\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" fill=\"#7A1F2B\">\u0394\u03b8<\/text>\n  <circle cx=\"490\" cy=\"255\" r=\"4\" fill=\"#0A1628\"><\/circle>\n  <text x=\"458\" y=\"140\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" font-style=\"italic\" fill=\"#0A1628\">v1<\/text>\n  <text x=\"366\" y=\"222\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" font-weight=\"700\" font-style=\"italic\" fill=\"#0A1628\">v2<\/text>\n  <text x=\"398\" y=\"163\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" font-weight=\"700\" fill=\"#7A1F2B\">\u0394v<\/text>\n  <text x=\"500\" y=\"290\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11.5\" font-weight=\"700\" fill=\"#7A1F2B\" text-anchor=\"middle\">\u0394v points toward the centre.<\/text>\n  <text x=\"500\" y=\"308\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"11.5\" fill=\"#142139\" text-anchor=\"middle\">So the acceleration does too.<\/text>\n  <text x=\"350\" y=\"352\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#0A1628\" text-anchor=\"middle\">Same \u0394\u03b8 between equal-length sides, so the triangles are similar:  |\u0394v| \/ v = |\u0394s| \/ r<\/text>\n  <text x=\"350\" y=\"378\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"13\" fill=\"#0A1628\" text-anchor=\"middle\">For a short \u0394t, |\u0394s| \u2248 v\u00b7\u0394t. Divide by \u0394t:  a = |\u0394v|\/\u0394t = v\u00b2 \/ r<\/text>\n<\/svg>\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;margin-top:8px;\">Two similar triangles are all it takes to derive the circular motion formula.<\/p>\n<p>Similar triangles give |\u0394v| \/ v = |\u0394s| \/ r. Over a short interval the chord is nearly the arc, so |\u0394s| \u2248 v\u00b7\u0394t. Substitute and divide by \u0394t, and the result falls out: a = v\u00b2\/r \u2014 pointing wherever \u0394v points, which is toward the centre.<\/p>\n<p>If you want the same derivation with the geometry drawn out step by step, Georgia State&#8217;s <a href=\"http:\/\/hyperphysics.phy-astr.gsu.edu\/hbase\/circ.html\" target=\"_blank\" rel=\"noopener\">HyperPhysics circular motion pages<\/a> work through it carefully.<\/p>\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Circular Motion 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\/circular-motion.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n<h2>Real-World Examples of Circular Motion<\/h2>\n<p>The formula never changes. Only the identity of the inward force does.<\/p>\n<h3>1. A car on a roundabout \u2014 friction<\/h3>\n<p>At 10 m\/s round a 25 m roundabout, a = 10\u00b2\/25 = 4.0 m\/s\u00b2. The road must supply that sideways pull through <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/what-is-friction\/\">friction<\/a>, needing a coefficient of at least 4.0\/9.81 \u2248 0.41.<\/p>\n<p>Dry asphalt offers roughly 0.7\u20130.9, so you sail round without thinking about it. Ice offers perhaps 0.1. The physics did not change \u2014 the supplier defaulted.<\/p>\n<figure style=\"margin:32px auto;max-width:640px;text-align:center;\">\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/07\/track-discipline.avif\"\n       alt=\"Banked velodrome curve demonstrating circular motion physics, where the track tilt supplies the centripetal force\"\n       loading=\"lazy\"\n       style=\"width:100%;height:auto;border-radius:4px;\" \/>\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Bank the track and the normal force tilts inward, supplying the centripetal force without relying on friction at all.<\/figcaption>\n<\/figure>\n<h3>2. A hammer throw \u2014 tension<\/h3>\n<p>An elite thrower whirls a 7.26 kg hammer at a radius near 1.7 m, releasing at roughly 29 m\/s. That demands a = 29\u00b2\/1.7 \u2248 495 m\/s\u00b2, about 50g.<\/p>\n<p>The wire must therefore pull with F = 7.26 \u00d7 495 \u2248 3.6 kN \u2014 comparable to the weight of a small car, held through the arms. Release, and the hammer leaves along the tangent, not radially outward.<\/p>\n<h3>3. A washing machine spin cycle \u2014 the drum wall<\/h3>\n<p>A 0.25 m drum at 1400 rpm gives \u03c9 = 146.6 rad\/s and a = \u03c9\u00b2r \u2248 5,370 m\/s\u00b2. That is roughly <strong>550g<\/strong>.<\/p>\n<p>Water is not &#8220;thrown out&#8221; of your clothes. The drum wall simply stops providing the inward force at the perforations, so the water carries straight on and leaves.<\/p>\n<h3>4. The International Space Station \u2014 gravity<\/h3>\n<p>The ISS orbits about 400 km up, so r \u2248 6.771 \u00d7 10\u2076 m from Earth&#8217;s centre, moving at roughly 7.67 km\/s. Its centripetal acceleration is a = v\u00b2\/r \u2248 8.69 m\/s\u00b2.<\/p>\n<p>Now compare local gravity at that same altitude: also 8.69 m\/s\u00b2, which is about 89% of surface gravity. The match is no coincidence \u2014 gravity <em>is<\/em> the centripetal force, and it is entirely spent on turning. One lap takes 92 minutes.<\/p>\n<h3>5. The Moon \u2014 Newton&#8217;s original sanity check<\/h3>\n<p>The Moon sits 3.844 \u00d7 10\u2078 m away and takes 27.3 days per orbit, giving a = 4\u03c0\u00b2r\/T\u00b2 \u2248 2.72 \u00d7 10\u207b\u00b3 m\/s\u00b2.<\/p>\n<p>Newton&#8217;s inverse-square law independently predicts 2.70 \u00d7 10\u207b\u00b3 m\/s\u00b2 at that distance. Agreement to within about 1%, from two completely separate routes, is what convinced him that the force holding the Moon and the force dropping an apple were the same force.<\/p>\n<h2>Common Misconceptions About Circular Motion<\/h2>\n<h3>&#8220;Constant speed means no acceleration&#8221;<\/h3>\n<p>This is the big one. Acceleration is the rate of change of <em>velocity<\/em>, and velocity carries a direction. Turning at a fixed speed is still accelerating \u2014 hard.<\/p>\n<p>Your body already agrees, even if your intuition doesn&#8217;t: that pressure against the car door is you being accelerated.<\/p>\n<h3>&#8220;Centrifugal force flings you outward&#8221;<\/h3>\n<p>In the ground frame, no outward force acts on you at all. Your body tries to continue in a straight line, the door gets in the way and pushes you <em>inward<\/em>, and you read that squeeze as an outward pull.<\/p>\n<p>Centrifugal force is a bookkeeping device that exists only if you insist on doing the physics in the rotating frame. Useful there \u2014 but it is not what is pushing you.<\/p>\n<h3>&#8220;Cut the string and the ball flies straight out from the centre&#8221;<\/h3>\n<p>It leaves along the <strong>tangent<\/strong>, at 90\u00b0 to the radius, not along it. With the inward force gone, <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/newtons-laws-of-motion\/\">Newton&#8217;s first law<\/a> takes over and the ball simply keeps the velocity it already had.<\/p>\n<h3>&#8220;Centripetal force is a special new force&#8221;<\/h3>\n<p>It is a <strong>job description<\/strong>, not a new entry in the force catalogue. Tension does the job for a hammer, friction for a car, gravity for the ISS, the normal force for a drum wall.<\/p>\n<p>Always ask &#8220;what is providing the centripetal force here?&#8221; \u2014 never &#8220;where do I add the centripetal force?&#8221; It is never an extra arrow on a free-body diagram; it is the name for the resultant that already points inward.<\/p>\n<h2>Circular Motion vs Centripetal Force<\/h2>\n<p>Circular motion is the <em>description<\/em> of the path; centripetal force is the <em>cause<\/em> that keeps the object on it. One is kinematics \u2014 what is happening. The other is dynamics \u2014 why.<\/p>\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n<thead>\n<tr style=\"background:#0A1628;color:#FAF6EE;\">\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Feature<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Circular motion<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Centripetal force<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>What it is<\/strong><\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">The motion itself \u2014 travel along a circular path<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">The net inward force that causes that motion<\/td><\/tr>\n<tr style=\"background:#F5F2EA;\"><td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Branch<\/strong><\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Kinematics \u2014 the description<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Dynamics \u2014 the cause<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Key equation<\/strong><\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">a = v\u00b2\/r<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">F = mv\u00b2\/r<\/td><\/tr>\n<tr style=\"background:#F5F2EA;\"><td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>SI unit<\/strong><\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">m\/s\u00b2 (its acceleration)<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">N<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Direction<\/strong><\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Velocity tangent; acceleration inward<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Always toward the centre<\/td><\/tr>\n<tr style=\"background:#F5F2EA;\"><td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>A force in its own right?<\/strong><\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Not a force at all<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">No \u2014 a role played by tension, friction, gravity or the normal force<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>Put simply: circular motion is the effect, and <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/centripetal-force\/\">centripetal force<\/a> is the reason you get it.<\/p>\n<h2>How Circular Motion Relates to Orbits, Oscillations and Newton&#8217;s Laws<\/h2>\n<p>Circular motion is the bridge between Newton&#8217;s laws, orbital mechanics and oscillations \u2014 the same a = v\u00b2\/r turns up in all three.<\/p>\n<h3>Newton&#8217;s laws<\/h3>\n<p>The first law supplies the tangent: with no net force, the object goes straight. The second law supplies the size: F = ma becomes F = mv\u00b2\/r the moment the acceleration is centripetal.<\/p>\n<h3>Simple harmonic motion<\/h3>\n<p>Shine a light sideways at a ball in uniform circular motion and its shadow on the wall performs <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/simple-harmonic-motion\/\">simple harmonic motion<\/a> exactly. The \u03c9 in circular motion and the \u03c9 in SHM are the same quantity \u2014 which is why one is often taught as the shadow of the other.<\/p>\n<h3>Orbits and planets<\/h3>\n<p>Orbits are genuinely ellipses, not circles. But Earth&#8217;s orbital eccentricity is only 0.0167, so its path is out of round by just 0.014% \u2014 visually, a circle.<\/p>\n<p>The catch is that the Sun sits about 1.67% off-centre, which is enough to swing our distance from 147.1 to 152.1 million km and to make Earth measurably faster in January than in July. So uniform circular motion is an excellent approximation here, and an honest one only if you say so.<\/p>\n<h2>Worked Problems<\/h2>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A stone is whirled on a string of radius 0.80 m at a constant speed of 4.0 m\/s. Find its centripetal acceleration.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Use the definition of centripetal acceleration, a = v\u00b2\/r.<\/p>\n<p>Step 2: Substitute with units: a = (4.0 m\/s)\u00b2 \/ 0.80 m = 16 m\u00b2\/s\u00b2 \/ 0.80 m.<\/p>\n<p>Step 3: Solve: a = 20 m\/s\u00b2, directed toward the centre.<\/p>\n<p><strong>Answer: a = 20 m\/s\u00b2 (2 s.f.), toward the centre<\/strong><\/p>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">A 1200 kg car takes a bend of radius 45 m at a steady 15 m\/s. Find the centripetal acceleration and the force the road must supply.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Acceleration first: a = v\u00b2\/r.<\/p>\n<p>Step 2: a = (15 m\/s)\u00b2 \/ 45 m = 225 \/ 45 = 5.0 m\/s\u00b2.<\/p>\n<p>Step 3: Then Newton&#8217;s second law: F = ma = 1200 kg \u00d7 5.0 m\/s\u00b2 = 6000 N.<\/p>\n<p><strong>Answer: a = 5.0 m\/s\u00b2, F = 6.0 kN supplied by friction, pointing toward the centre of the bend<\/strong><\/p>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">A fairground ride of radius 6.0 m completes one full turn in 4.0 s. Find the tangential speed and the centripetal acceleration.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: One lap is a circumference, so v = 2\u03c0r \/ T.<\/p>\n<p>Step 2: v = 2\u03c0(6.0 m) \/ 4.0 s = 37.70 m \/ 4.0 s = 9.42 m\/s.<\/p>\n<p>Step 3: a = v\u00b2\/r = (9.42)\u00b2 \/ 6.0 = 88.8 \/ 6.0 = 14.8 m\/s\u00b2. (Check with a = 4\u03c0\u00b2r\/T\u00b2 = 14.8 m\/s\u00b2. \u2713)<\/p>\n<p><strong>Answer: v = 9.4 m\/s, a = 15 m\/s\u00b2 \u2014 about 1.5g<\/strong><\/p>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A washing machine drum of radius 0.22 m spins at 1200 rpm. Find the angular velocity and the centripetal acceleration in multiples of g.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Convert rpm to rad\/s: \u03c9 = 1200 \u00d7 2\u03c0 \/ 60.<\/p>\n<p>Step 2: \u03c9 = 125.7 rad\/s. Then use a = \u03c9\u00b2r = (125.7 rad\/s)\u00b2 \u00d7 0.22 m.<\/p>\n<p>Step 3: a = 15790 \u00d7 0.22 = 3474 m\/s\u00b2. Divide by g: 3474 \/ 9.81 = 354.<\/p>\n<p><strong>Answer: \u03c9 = 126 rad\/s, a = 3.47 \u00d7 10\u00b3 m\/s\u00b2, about 354g<\/strong><\/p>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">A roller coaster does a vertical loop of radius 2.5 m. Find the minimum speed at the very top for the car to maintain contact with the track.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: At the minimum speed the track pushes with N = 0, so gravity alone supplies the centripetal force: mg = mv\u00b2\/r.<\/p>\n<p>Step 2: Mass cancels \u2014 the answer is the same for every rider. So v\u00b2 = gr = 9.81 m\/s\u00b2 \u00d7 2.5 m = 24.5 m\u00b2\/s\u00b2.<\/p>\n<p>Step 3: v = \u221a24.5 = 4.95 m\/s.<\/p>\n<p><strong>Answer: v(min) = 5.0 m\/s (2 s.f.), independent of mass<\/strong><\/p>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A car rounds a flat, unbanked curve of radius 60 m. The coefficient of static friction is 0.70. Find the maximum speed before it slides.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Friction supplies the centripetal force, and it maxes out at \u03bc(s)N = \u03bc(s)mg. Set that equal to the requirement: \u03bc(s)mg = mv\u00b2\/r.<\/p>\n<p>Step 2: Mass cancels again, leaving v\u00b2 = \u03bc(s)\u00b7g\u00b7r = 0.70 \u00d7 9.81 m\/s\u00b2 \u00d7 60 m = 412 m\u00b2\/s\u00b2.<\/p>\n<p>Step 3: v = \u221a412 = 20.3 m\/s.<\/p>\n<p><strong>Answer: v(max) = 20 m\/s (about 73 km\/h) \u2014 and note it does not depend on the car&#8217;s mass<\/strong><\/p>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A 0.15 kg ball on a 1.2 m string sweeps a horizontal circle as a conical pendulum, with the string at 30\u00b0 to the vertical. Find the tension and the ball&#039;s speed.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: The radius is not the string length: r = L\u00b7sin \u03b8 = 1.2 m \u00d7 sin 30\u00b0 = 0.60 m. Resolve the tension. Vertically there is no acceleration, so T\u00b7cos \u03b8 = mg.<\/p>\n<p>Step 2: T = mg \/ cos \u03b8 = (0.15 kg \u00d7 9.81 m\/s\u00b2) \/ cos 30\u00b0 = 1.4715 N \/ 0.8660 = 1.70 N.<\/p>\n<p>Step 3: Horizontally, the tension component is the centripetal force: T\u00b7sin \u03b8 = mv\u00b2\/r, so 1.70 N \u00d7 0.5 = (0.15 kg) v\u00b2 \/ 0.60 m, giving v\u00b2 = 3.40 m\u00b2\/s\u00b2 and v = 1.84 m\/s.<\/p>\n<p><strong>Answer: T = 1.70 N, v = 1.84 m\/s. (Sanity check: a = v\u00b2\/r = 5.66 m\/s\u00b2, which equals g\u00b7tan 30\u00b0 = 5.66 m\/s\u00b2. \u2713)<\/strong><\/p>\n<\/div><\/details><\/div>\n<h2>Frequently Asked Questions<\/h2>\n<details class=\"pf-faq-item\"><summary>What is circular motion?<\/summary><div class=\"pf-faq-item-answer\">\nCircular motion is the movement of an object along a circular path, where the velocity stays tangent to the circle and a net force pulls the object toward the centre. It can be uniform, meaning the speed is constant and only direction changes, or non-uniform, where the object also speeds up or slows down as it goes round.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>What is the formula for uniform circular motion?<\/summary><div class=\"pf-faq-item-answer\">\nThe formula for uniform circular motion is a = v\u00b2\/r, where a is the centripetal acceleration in m\/s\u00b2, v is the tangential speed in m\/s, and r is the radius in metres. Multiplying by mass gives the required force, F = mv\u00b2\/r. Equivalent forms are a = \u03c9\u00b2r and a = 4\u03c0\u00b2r\/T\u00b2, using angular velocity or period.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>What is the difference between circular motion and centripetal force?<\/summary><div class=\"pf-faq-item-answer\">\nCircular motion is the motion itself; centripetal force is the inward force that causes it. Circular motion is a kinematic description of the path, measured through a = v\u00b2\/r. Centripetal force is the dynamic cause, measured in newtons through F = mv\u00b2\/r. In short, circular motion is the effect and centripetal force is the reason.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>What is centripetal force?<\/summary><div class=\"pf-faq-item-answer\">\nCentripetal force is the net force directed toward the centre of a circular path, with magnitude F = mv\u00b2\/r. It is not a new kind of force but a role that an ordinary force fills: tension in a hammer wire, friction under a cornering car, gravity on the ISS, or the normal force from a spinning drum wall.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Are planets in circular motion?<\/summary><div class=\"pf-faq-item-answer\">\nNot exactly \u2014 planetary orbits are ellipses, as Kepler&#8217;s first law states, so uniform circular motion is only an approximation. It is a very good one for Earth, whose orbit is out of round by just 0.014%. But the Sun sits about 1.67% off-centre, so Earth&#8217;s distance and orbital speed both vary measurably across the year.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Why do you feel pushed outwards when a car turns?<\/summary><div class=\"pf-faq-item-answer\">\nYou feel pushed outwards because your body is trying to travel in a straight line while the car curves away beneath you. No outward force acts on you in the ground frame. The door or seat pushes you inward, and you interpret that inward squeeze as an outward pull, commonly called centrifugal force.\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>Circular motion physics explains why an object moving in a circle at constant speed is still accelerating: its direction changes every instant. This guide covers the a = v\u00b2\/r formula, five real examples, four common misconceptions and seven worked problems.<\/p>\n","protected":false},"author":1,"featured_media":586,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"class_list":["post-583","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mechanics"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/583","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=583"}],"version-history":[{"count":2,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/583\/revisions"}],"predecessor-version":[{"id":587,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/583\/revisions\/587"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/586"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=583"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=583"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=583"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}