{"id":660,"date":"2026-07-27T22:55:39","date_gmt":"2026-07-27T22:55:39","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=660"},"modified":"2026-07-27T22:55:40","modified_gmt":"2026-07-27T22:55:40","slug":"moment-of-inertia","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/moment-of-inertia\/","title":{"rendered":"Moment of Inertia: Formula for Common Shapes"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\nMoment of inertia is the rotational equivalent of mass: it measures how strongly an object resists a change to its spin about a chosen axis. It equals the sum of each mass element times the square of its distance from that axis, giving I = k m r squared for common shapes, in kilogram metres squared.\n<\/p><\/div>\n\n<p>Hold a hammer by its handle and swing it. Now flip it round, grip the head, and swing again. Same hammer, same mass, same muscles \u2014 yet one swing feels sluggish and the other whips through the air.<\/p>\n\n<p>Nothing about the hammer changed except where the heavy end sat relative to your wrist. That single observation is the whole of this topic: when things rotate, <em>where<\/em> the mass sits matters more than <em>how much<\/em> there is.<\/p>\n\n<h2>What Is Moment of Inertia?<\/h2>\n\n<p>Moment of inertia is a measure of how much an object resists being spun up or slowed down about a particular axis. Push a shopping trolley and its mass fights you; twist a heavy door and something else fights you \u2014 that something is its moment of inertia.<\/p>\n\n<p>Mass tells you how stubborn an object is when you try to move it in a straight line. Moment of inertia tells you how stubborn it is when you try to turn it. The two are cousins, not twins.<\/p>\n\n<p>The critical difference is distance. Every scrap of material in a rotating body sweeps out a circle, and the wider that circle, the faster that scrap has to travel for a given rate of spin. Material near the rim therefore costs far more effort to accelerate than material near the hub.<\/p>\n\n<p>That cost does not grow in proportion to distance. It grows with the <strong>square<\/strong> of distance \u2014 move a bolt twice as far from the axis and it contributes four times as much.<\/p>\n\n<h3>Why the axis is part of the answer<\/h3>\n\n<p>Here is the point that trips up more students than any other: an object does not <em>have<\/em> a moment of inertia. It has one for each axis you might spin it about.<\/p>\n\n<p>A metre rule spun about its middle and the same rule spun about its end are two completely different problems, with answers differing by a factor of four. Quote a moment of inertia without naming the axis and you have said almost nothing.<\/p>\n\n<h2>The Moment of Inertia Formula<\/h2>\n\n<p>For a single point mass, the definition is as simple as physics gets.<\/p>\n\n<div class=\"pf-formula\">I = m r\u00b2<\/div>\n\n<ul>\n<li><strong>I<\/strong> \u2014 moment of inertia, in kilogram metres squared (kg\u00b7m\u00b2)<\/li>\n<li><strong>m<\/strong> \u2014 mass of the particle, in kilograms (kg)<\/li>\n<li><strong>r<\/strong> \u2014 perpendicular distance from the particle to the axis, in metres (m)<\/li>\n<\/ul>\n\n<p>Real objects are not points, so you chop them into countless tiny pieces and add up every piece&#8217;s contribution.<\/p>\n\n<div class=\"pf-formula\">I = \u03a3 m r\u00b2<\/div>\n\n<p>For a uniform shape that sum collapses into something far friendlier: a single number times the mass times a characteristic length squared.<\/p>\n\n<div class=\"pf-formula\">I = k m r\u00b2<\/div>\n\n<ul>\n<li><strong>k<\/strong> \u2014 the shape factor, a pure number with no units, set entirely by the geometry and the axis<\/li>\n<li><strong>m<\/strong> \u2014 total mass of the body, in kilograms (kg)<\/li>\n<li><strong>r<\/strong> \u2014 the defining length: radius for round shapes, length for rods, in metres (m)<\/li>\n<\/ul>\n\n<p>Everything difficult about this topic lives inside <strong>k<\/strong>. It runs from about 0.083 for a rod spun about its centre up to 1.0 for a hoop \u2014 a twelve-fold spread for the same mass. Once you have the right k, you can read the shape table below or check yourself against our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/moment-of-inertia\">Moment of Inertia Calculator<\/a>, which handles the substitution and the units for you.<\/p>\n\n<p>A quick sanity check worth remembering: k can never exceed 1 for a solid body rotating about an axis through it, because no part of the object sits further out than the defining radius.<\/p>\n\n<h2>Moment of Inertia Formulas for 9 Common Shapes<\/h2>\n\n<p>These nine cover almost everything a first-year course or A-level paper will throw at you. Read the axis column first \u2014 it decides which row you actually need.<\/p>\n\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;\">Shape<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Axis<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Formula<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Shape factor k<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Point mass<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">At distance r from the axis<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">I = m r\u00b2<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">1<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Thin hoop or ring<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Central axis, perpendicular to the plane<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">I = m r\u00b2<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">1<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Thin-walled hollow cylinder<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Central long axis<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">I = m r\u00b2<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">1<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Solid disc or solid cylinder<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Central axis<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">I = m r\u00b2 \/ 2<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">0.5<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Thick-walled (annular) cylinder<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Central axis<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">I = m (r<sub>1<\/sub>\u00b2 + r<sub>2<\/sub>\u00b2) \/ 2<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">0.5 to 1<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Solid sphere<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Any diameter<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">I = 2 m r\u00b2 \/ 5<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">0.4<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Thin spherical shell<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Any diameter<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">I = 2 m r\u00b2 \/ 3<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">0.667<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Thin rod (length L)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Through the centre, perpendicular to the rod<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">I = m L\u00b2 \/ 12<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">0.083<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Thin rod (length L)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Through one end, perpendicular to the rod<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">I = m L\u00b2 \/ 3<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">0.333<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>Two rows deserve a second look. The hoop and the solid disc share the same mass and the same radius, yet the hoop&#8217;s moment of inertia is exactly double \u2014 because a hoop keeps every gram out at the rim, while a disc buries most of its material close to the axis.<\/p>\n\n<p>Two more worth knowing, though they appear less often: a solid disc spun about a diameter rather than its central axis gives I = m r\u00b2 \/ 4, and a rectangular plate spun about an axis through its centre and perpendicular to its face gives I = m (a\u00b2 + b\u00b2) \/ 12, where a and b are the side lengths.<\/p>\n\n<h2>How Moment of Inertia Works<\/h2>\n\n<p>Moment of inertia works by weighting every piece of mass according to the square of its distance from the axis, which is why mass sitting far out dominates the total. The reasoning behind that squared term is worth following once, because it explains everything else.<\/p>\n\n<p>Picture a rigid body turning at angular velocity \u03c9. A small piece of mass m sitting at distance r moves in a circle at speed v = \u03c9 r. Its kinetic energy is therefore m v\u00b2 \/ 2, which becomes m r\u00b2 \u03c9\u00b2 \/ 2 once you substitute.<\/p>\n\n<p>Now add up every piece. The \u03c9\u00b2 factor is the same for all of them \u2014 a rigid body turns as one \u2014 so it comes outside the sum, leaving \u03c9\u00b2 \/ 2 multiplied by the sum of all the m r\u00b2 terms.<\/p>\n\n<p>That leftover sum <em>is<\/em> the moment of inertia. It appears not because someone invented it, but because it is exactly what falls out when you add up the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/kinetic-energy-formula\/\">kinetic energy<\/a> of a spinning object.<\/p>\n\n<div class=\"pf-formula\">KE = I \u03c9\u00b2 \/ 2<\/div>\n\n<ul>\n<li><strong>KE<\/strong> \u2014 rotational kinetic energy, in joules (J)<\/li>\n<li><strong>I<\/strong> \u2014 moment of inertia about the spin axis, in kilogram metres squared (kg\u00b7m\u00b2)<\/li>\n<li><strong>\u03c9<\/strong> \u2014 angular velocity, in radians per second (rad\/s)<\/li>\n<\/ul>\n\n<p>Compare that with the straight-line version, KE = m v\u00b2 \/ 2, and the family resemblance is unmistakable. Swap mass for moment of inertia, swap speed for angular velocity, and the whole of linear mechanics reappears in rotational form.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Moment of Inertia 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\/moment-of-inertia.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n\n<h3>Why the squared term changes everything<\/h3>\n\n<p>The squaring is not a mathematical detail \u2014 it is the reason flywheels, gymnasts and satellites behave the way they do. Doubling an object&#8217;s radius while keeping its mass fixed quadruples its moment of inertia.<\/p>\n\n<p>Engineers exploit this ruthlessly. A flywheel designed to store energy puts its metal as far out as the material strength allows, because every extra centimetre of radius buys energy storage on the cheap.<\/p>\n\n<svg viewBox=\"0 0 700 320\" role=\"img\" aria-label=\"Diagram comparing the moment of inertia of a thin rod about its centre and about its end\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" style=\"width:100%;height:auto;max-width:700px;display:block;margin:28px auto;\">\n<rect x=\"0\" y=\"0\" width=\"700\" height=\"320\" fill=\"#F5F2EA\"><\/rect>\n<text x=\"350\" y=\"34\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"17\" font-weight=\"700\" fill=\"#0A1628\">Same rod, same mass \u2014 two different answers<\/text>\n<line x1=\"350\" y1=\"52\" x2=\"350\" y2=\"268\" stroke=\"#D9CFB8\" stroke-width=\"2\" stroke-dasharray=\"5 5\"><\/line>\n<text x=\"175\" y=\"78\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"14\" font-weight=\"700\" fill=\"#7A1F2B\">AXIS THROUGH CENTRE<\/text>\n<rect x=\"55\" y=\"128\" width=\"240\" height=\"16\" rx=\"3\" fill=\"#142139\"><\/rect>\n<line x1=\"175\" y1=\"100\" x2=\"175\" y2=\"172\" stroke=\"#C8932A\" stroke-width=\"3\"><\/line>\n<circle cx=\"175\" cy=\"136\" r=\"9\" fill=\"#C8932A\"><\/circle>\n<circle cx=\"175\" cy=\"136\" r=\"3.5\" fill=\"#F5F2EA\"><\/circle>\n<path d=\"M 128 108 A 58 58 0 0 1 222 108\" stroke=\"#C8932A\" stroke-width=\"2.5\" fill=\"none\"><\/path>\n<path d=\"M 222 108 l -3 -11 l 12 4 z\" fill=\"#C8932A\"><\/path>\n<line x1=\"55\" y1=\"192\" x2=\"295\" y2=\"192\" stroke=\"#0A1628\" stroke-width=\"1.5\"><\/line>\n<line x1=\"55\" y1=\"186\" x2=\"55\" y2=\"198\" stroke=\"#0A1628\" stroke-width=\"1.5\"><\/line>\n<line x1=\"295\" y1=\"186\" x2=\"295\" y2=\"198\" stroke=\"#0A1628\" stroke-width=\"1.5\"><\/line>\n<text x=\"175\" y=\"212\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"13\" fill=\"#0A1628\">length L<\/text>\n<text x=\"175\" y=\"248\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"19\" font-weight=\"700\" fill=\"#0A1628\">I = m L\u00b2 \/ 12<\/text>\n<text x=\"175\" y=\"274\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"13\" fill=\"#7A1F2B\">mass is close in \u2014 easy to spin<\/text>\n<text x=\"525\" y=\"78\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"14\" font-weight=\"700\" fill=\"#7A1F2B\">AXIS THROUGH END<\/text>\n<rect x=\"405\" y=\"128\" width=\"240\" height=\"16\" rx=\"3\" fill=\"#142139\"><\/rect>\n<line x1=\"405\" y1=\"100\" x2=\"405\" y2=\"172\" stroke=\"#C8932A\" stroke-width=\"3\"><\/line>\n<circle cx=\"405\" cy=\"136\" r=\"9\" fill=\"#C8932A\"><\/circle>\n<circle cx=\"405\" cy=\"136\" r=\"3.5\" fill=\"#F5F2EA\"><\/circle>\n<path d=\"M 358 108 A 58 58 0 0 1 452 108\" stroke=\"#C8932A\" stroke-width=\"2.5\" fill=\"none\"><\/path>\n<path d=\"M 452 108 l -3 -11 l 12 4 z\" fill=\"#C8932A\"><\/path>\n<line x1=\"405\" y1=\"192\" x2=\"645\" y2=\"192\" stroke=\"#0A1628\" stroke-width=\"1.5\"><\/line>\n<line x1=\"405\" y1=\"186\" x2=\"405\" y2=\"198\" stroke=\"#0A1628\" stroke-width=\"1.5\"><\/line>\n<line x1=\"645\" y1=\"186\" x2=\"645\" y2=\"198\" stroke=\"#0A1628\" stroke-width=\"1.5\"><\/line>\n<text x=\"525\" y=\"212\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"13\" fill=\"#0A1628\">length L<\/text>\n<text x=\"525\" y=\"248\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"19\" font-weight=\"700\" fill=\"#0A1628\">I = m L\u00b2 \/ 3<\/text>\n<text x=\"525\" y=\"274\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"13\" fill=\"#7A1F2B\">mass is far out \u2014 4\u00d7 harder<\/text>\n<rect x=\"55\" y=\"288\" width=\"590\" height=\"22\" rx=\"3\" fill=\"#0A1628\"><\/rect>\n<text x=\"350\" y=\"303\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"13\" font-weight=\"700\" fill=\"#FAF6EE\">Move the axis, change the answer \u2014 by a factor of four<\/text>\n<\/svg>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">Moment of inertia of a thin rod depends entirely on where the axis sits.<\/p>\n\n<h2>The Parallel Axis Theorem<\/h2>\n\n<p>The parallel axis theorem lets you find the moment of inertia about any axis, provided you already know it about a parallel axis through the centre of mass. It is the single most useful shortcut in rotational mechanics.<\/p>\n\n<div class=\"pf-formula\">I = I(cm) + m d\u00b2<\/div>\n\n<ul>\n<li><strong>I<\/strong> \u2014 moment of inertia about the new, offset axis, in kg\u00b7m\u00b2<\/li>\n<li><strong>I(cm)<\/strong> \u2014 moment of inertia about a parallel axis through the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/center-of-mass\/\">centre of mass<\/a>, in kg\u00b7m\u00b2<\/li>\n<li><strong>m<\/strong> \u2014 total mass, in kilograms (kg)<\/li>\n<li><strong>d<\/strong> \u2014 perpendicular distance between the two parallel axes, in metres (m)<\/li>\n<\/ul>\n\n<p>Test it on the rod. About its centre, I = m L\u00b2 \/ 12; shift the axis to the end, a distance d = L\/2 away, and the theorem predicts m L\u00b2 \/ 12 + m L\u00b2 \/ 4, which is m L\u00b2 \/ 3. That is exactly the value in the table.<\/p>\n\n<p>Notice that the correction term m d\u00b2 is always positive. The centre-of-mass axis is therefore always the easiest axis to spin about \u2014 every other parallel axis costs you more.<\/p>\n\n<svg viewBox=\"0 0 700 300\" role=\"img\" aria-label=\"Diagram of the parallel axis theorem showing a disc with a centre of mass axis and an offset parallel axis separated by distance d\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" style=\"width:100%;height:auto;max-width:700px;display:block;margin:28px auto;\">\n<rect x=\"0\" y=\"0\" width=\"700\" height=\"300\" fill=\"#F5F2EA\"><\/rect>\n<text x=\"350\" y=\"34\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"17\" font-weight=\"700\" fill=\"#0A1628\">The parallel axis theorem<\/text>\n<ellipse cx=\"270\" cy=\"160\" rx=\"118\" ry=\"70\" fill=\"#142139\"><\/ellipse>\n<ellipse cx=\"270\" cy=\"160\" rx=\"118\" ry=\"70\" fill=\"none\" stroke=\"#C5D0DC\" stroke-width=\"1.5\"><\/ellipse>\n<line x1=\"270\" y1=\"62\" x2=\"270\" y2=\"258\" stroke=\"#C8932A\" stroke-width=\"3\" stroke-dasharray=\"8 5\"><\/line>\n<circle cx=\"270\" cy=\"160\" r=\"8\" fill=\"#C8932A\"><\/circle>\n<circle cx=\"270\" cy=\"160\" r=\"3\" fill=\"#0A1628\"><\/circle>\n<text x=\"270\" y=\"52\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"13\" font-weight=\"700\" fill=\"#C8932A\">centre-of-mass axis<\/text>\n<line x1=\"470\" y1=\"62\" x2=\"470\" y2=\"258\" stroke=\"#7A1F2B\" stroke-width=\"3\"><\/line>\n<circle cx=\"470\" cy=\"160\" r=\"6\" fill=\"#7A1F2B\"><\/circle>\n<text x=\"470\" y=\"52\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"13\" font-weight=\"700\" fill=\"#7A1F2B\">new parallel axis<\/text>\n<line x1=\"270\" y1=\"160\" x2=\"470\" y2=\"160\" stroke=\"#0A1628\" stroke-width=\"2\"><\/line>\n<line x1=\"270\" y1=\"152\" x2=\"270\" y2=\"168\" stroke=\"#0A1628\" stroke-width=\"2\"><\/line>\n<line x1=\"470\" y1=\"152\" x2=\"470\" y2=\"168\" stroke=\"#0A1628\" stroke-width=\"2\"><\/line>\n<rect x=\"352\" y=\"146\" width=\"36\" height=\"26\" rx=\"3\" fill=\"#F5F2EA\"><\/rect>\n<text x=\"370\" y=\"166\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"17\" font-weight=\"700\" fill=\"#0A1628\">d<\/text>\n<rect x=\"120\" y=\"244\" width=\"460\" height=\"42\" rx=\"4\" fill=\"#0A1628\"><\/rect>\n<text x=\"350\" y=\"271\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"20\" font-weight=\"700\" fill=\"#FAF6EE\">I = I(cm) + m d\u00b2<\/text>\n<text x=\"350\" y=\"228\" text-anchor=\"middle\" font-family=\"Manrope,Arial,sans-serif\" font-size=\"13\" fill=\"#7A1F2B\">the offset always adds \u2014 never subtracts<\/text>\n<\/svg>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">The parallel axis theorem shifts a known moment of inertia to any parallel axis.<\/p>\n\n<h3>One condition people forget<\/h3>\n\n<p>The theorem only works when your starting value is measured about an axis through the centre of mass. You cannot hop from the rod&#8217;s end-axis to some other axis in one step.<\/p>\n\n<p>In practice, always travel back through the centre of mass. Go from the known axis to the centre, then out to the axis you actually want.<\/p>\n\n<h2>Real-World Examples of Moment of Inertia<\/h2>\n\n<p>Rotational inertia is not an exam abstraction \u2014 it is the reason several very different machines and bodies are built the way they are.<\/p>\n\n<h3>1. The spinning skater<\/h3>\n\n<p>A skater starts a spin with arms outstretched, then pulls them in and accelerates dramatically. No one pushes them. Angular momentum, the product of I and \u03c9, stays constant, so cutting I roughly in half must roughly double \u03c9.<\/p>\n\n<figure style=\"margin:32px auto;max-width:640px;text-align:center;\">\n\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/07\/A_handbook_of_figure_skating_arranged_for_use_on_the_ice_with_over_six_hundred_diagrams_and_illustrations_1907_14596023919.jpg\"\n\n       alt=\"Figure skater pulling arms in to reduce moment of inertia and spin faster\"\n\n       loading=\"lazy\"\n\n       style=\"width:100%;height:auto;border-radius:4px;\" \/ width=\"1640\" height=\"1244\">\n\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Pulling the arms in cuts the skater&#8217;s moment of inertia, so the spin rate rises to conserve angular momentum.<\/figcaption>\n\n<\/figure>\n\n<h3>2. Flywheels and engine crankshafts<\/h3>\n\n<p>A car engine delivers power in violent pulses, not a smooth stream. A heavy flywheel bolted to the crankshaft has a large moment of inertia, so it absorbs each pulse and releases it gradually \u2014 which is why the engine idles smoothly instead of shuddering.<\/p>\n\n<h3>3. Spacecraft that turn without fuel<\/h3>\n\n<p>The Hubble Space Telescope has no thrusters for routine pointing. Instead, <a href=\"https:\/\/science.nasa.gov\/mission\/hubble\/observatory\/design\/\" target=\"_blank\" rel=\"noopener\">four 45 kg reaction wheels<\/a> sit near its centre of gravity; spinning one up forces the telescope to rotate the other way, conserving total angular momentum.<\/p>\n\n<h3>4. Tightrope walkers and their poles<\/h3>\n\n<p>The long pole is not for balance in the intuitive sense \u2014 it is a moment-of-inertia amplifier. By putting mass a long way from the walker&#8217;s centre line, it makes any unwanted rotation build up slowly enough to correct.<\/p>\n\n<h3>5. The rolling race<\/h3>\n\n<p>Release a solid sphere and a hoop together from the top of a ramp and the sphere always wins. Not sometimes \u2014 always, regardless of their masses or radii, which cancel out of the problem completely.<\/p>\n\n<p>The reason is that a rolling object has to split the gravitational energy it gains between moving forwards and spinning. A hoop, with its k of 1, diverts half that energy into rotation; a solid sphere, with k of 0.4, spends only about 29 per cent and arrives faster.<\/p>\n\n<h2>Common Misconceptions About Moment of Inertia<\/h2>\n\n<h3>Myth 1: an object has one moment of inertia<\/h3>\n\n<p>It has one for every axis. The 1.8 m, 3.0 kg rod in the worked problems below scores 0.81 kg\u00b7m\u00b2 about its centre and 3.24 kg\u00b7m\u00b2 about its end \u2014 the same object, four times the resistance.<\/p>\n\n<p>Always name the axis before you quote a number. An answer without an axis is not wrong so much as incomplete.<\/p>\n\n<h3>Myth 2: heavier always means harder to spin<\/h3>\n\n<p>Mass matters, but placement can easily overwhelm it. A 5 kg hoop of radius 0.4 m has I = 0.80 kg\u00b7m\u00b2, while a 10 kg solid disc of radius 0.3 m has only 0.45 kg\u00b7m\u00b2 \u2014 double the mass, barely half the rotational resistance.<\/p>\n\n<h3>Myth 3: moment of inertia and moment of a force are the same thing<\/h3>\n\n<p>They share a word and nothing else. A <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/moment-of-a-force\/\">moment of a force<\/a> is a turning effect measured in newton metres; moment of inertia is a resistance to turning measured in kilogram metres squared.<\/p>\n\n<p>The units alone should settle it. If your working produces N\u00b7m where you expected kg\u00b7m\u00b2, you have mixed the two up.<\/p>\n\n<h3>Myth 4: the parallel axis theorem works from any axis<\/h3>\n\n<p>It starts from the centre-of-mass axis and nowhere else. Applying it from an arbitrary axis is one of the most common ways to lose marks on a rotational dynamics question, and the error is invisible in the final answer \u2014 it just quietly comes out wrong.<\/p>\n\n<h2>How Moment of Inertia Relates to Torque, Energy and Angular Momentum<\/h2>\n\n<p>Moment of inertia is the bridge between the linear mechanics you already know and the rotational version. Every familiar equation has a twin.<\/p>\n\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;\">Linear motion<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Rotational motion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Resistance to change<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">mass, m (kg)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">moment of inertia, I (kg\u00b7m\u00b2)<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Newton&#8217;s second law<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">F = m a<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">\u03c4 = I \u03b1<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Kinetic energy<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">KE = m v\u00b2 \/ 2<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">KE = I \u03c9\u00b2 \/ 2<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Momentum<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">p = m v<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">L = I \u03c9<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>The rotational form of <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/newtons-second-law\/\">Newton&#8217;s second law<\/a> is the workhorse. Apply a <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/torque-physics\/\">torque<\/a> \u03c4 to a body of moment of inertia I and it picks up angular acceleration \u03b1, exactly as a force produces linear acceleration.<\/p>\n\n<p>Angular momentum L = I \u03c9 behaves just as linear momentum does, and it obeys its own <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/conservation-of-momentum\/\">conservation law<\/a>. That is the principle running quietly behind the skater, the gyroscope and every satellite that turns itself without spending fuel.<\/p>\n\n<h2>Worked Problems<\/h2>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A 0.50 kg ball is fixed to the end of a light rod of length 1.2 m, which rotates about its far end. Find the moment of inertia, treating the ball as a point mass and ignoring the rod&#039;s mass.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: For a point mass, I = m r\u00b2.\nStep 2: Substitute m = 0.50 kg and r = 1.2 m, so I = 0.50 \u00d7 (1.2)\u00b2.\nStep 3: (1.2)\u00b2 = 1.44 m\u00b2, so I = 0.50 \u00d7 1.44 = 0.72 kg\u00b7m\u00b2.\n<strong>Answer: I = 0.72 kg\u00b7m\u00b2<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">A solid disc flywheel has mass 12 kg and radius 0.25 m. Calculate its moment of inertia about its central axis.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: For a solid disc about its central axis, I = m r\u00b2 \/ 2.\nStep 2: Substitute m = 12 kg and r = 0.25 m, so I = 12 \u00d7 (0.25)\u00b2 \/ 2.\nStep 3: (0.25)\u00b2 = 0.0625 m\u00b2, giving I = 12 \u00d7 0.0625 \/ 2 = 0.75 \/ 2.\n<strong>Answer: I = 0.375 kg\u00b7m\u00b2<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">A thin hoop has the same mass (12 kg) and radius (0.25 m) as the disc in Problem 2. Find its moment of inertia and compare the two.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: For a thin hoop about its central axis, I = m r\u00b2, since all the mass sits at radius r.\nStep 2: Substitute: I = 12 \u00d7 (0.25)\u00b2 = 12 \u00d7 0.0625.\nStep 3: I = 0.75 kg\u00b7m\u00b2. Comparing with the disc, 0.75 \/ 0.375 = 2.\n<strong>Answer: I = 0.75 kg\u00b7m\u00b2, exactly twice the disc&#8217;s value<\/strong>\n<\/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 has mass 3.0 kg and length 1.8 m. Find its moment of inertia (a) about a perpendicular axis through its centre and (b) about a perpendicular axis through one end.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: About the centre, I = m L\u00b2 \/ 12. About the end, I = m L\u00b2 \/ 3.\nStep 2: L\u00b2 = (1.8)\u00b2 = 3.24 m\u00b2.\nStep 3 (a): I = 3.0 \u00d7 3.24 \/ 12 = 9.72 \/ 12 = 0.81 kg\u00b7m\u00b2.\nStep 4 (b): I = 3.0 \u00d7 3.24 \/ 3 = 9.72 \/ 3 = 3.24 kg\u00b7m\u00b2.\n<strong>Answer: (a) 0.81 kg\u00b7m\u00b2 (b) 3.24 kg\u00b7m\u00b2 \u2014 a factor of 4<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">A solid disc of mass 2.0 kg and radius 0.30 m rotates about an axis through a point on its rim, parallel to its central axis. Use the parallel axis theorem to find its moment of inertia.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: The parallel axis theorem gives I = I(cm) + m d\u00b2, with d = r = 0.30 m here.\nStep 2: I(cm) = m r\u00b2 \/ 2 = 2.0 \u00d7 (0.30)\u00b2 \/ 2 = 2.0 \u00d7 0.09 \/ 2 = 0.09 kg\u00b7m\u00b2.\nStep 3: m d\u00b2 = 2.0 \u00d7 0.09 = 0.18 kg\u00b7m\u00b2.\nStep 4: I = 0.09 + 0.18 = 0.27 kg\u00b7m\u00b2.\n<strong>Answer: I = 0.27 kg\u00b7m\u00b2 (equal to 3 m r\u00b2 \/ 2)<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A solid cylinder of mass 5.0 kg and radius 0.20 m is free to rotate about its central axis. A torque of 2.5 N\u00b7m is applied. Find the angular acceleration.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: Use the rotational form of Newton&#8217;s second law, \u03c4 = I \u03b1, so \u03b1 = \u03c4 \/ I.\nStep 2: I = m r\u00b2 \/ 2 = 5.0 \u00d7 (0.20)\u00b2 \/ 2 = 5.0 \u00d7 0.04 \/ 2 = 0.10 kg\u00b7m\u00b2.\nStep 3: \u03b1 = 2.5 \/ 0.10.\n<strong>Answer: \u03b1 = 25 rad\/s\u00b2<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">The flywheel from Problem 2 (I = 0.375 kg\u00b7m\u00b2) spins at 1500 revolutions per minute. Calculate the rotational kinetic energy stored.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: Convert to rad\/s: \u03c9 = 1500 \u00d7 2\u03c0 \/ 60.\nStep 2: \u03c9 = 1500 \u00d7 0.10472 = 157.08 rad\/s.\nStep 3: KE = I \u03c9\u00b2 \/ 2 = 0.375 \u00d7 (157.08)\u00b2 \/ 2.\nStep 4: (157.08)\u00b2 = 24674 rad\u00b2\/s\u00b2, so KE = 0.375 \u00d7 24674 \/ 2 = 4626 J.\n<strong>Answer: KE \u2248 4.63 kJ (about 4600 J)<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 8<\/div><div class=\"pf-problem-question\">A solid sphere and a thin hoop are released from rest and roll without slipping down a slope of vertical height 1.5 m. Find each one&#039;s speed at the bottom and show that mass and radius do not matter. Take g = 9.81 m\/s\u00b2.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: Energy conservation gives m g h = m v\u00b2 \/ 2 + I \u03c9\u00b2 \/ 2, with I = k m r\u00b2 and \u03c9 = v \/ r for rolling.\nStep 2: Substituting, I \u03c9\u00b2 \/ 2 = k m v\u00b2 \/ 2, so m g h = m v\u00b2 (1 + k) \/ 2. The mass m cancels, and r never appears.\nStep 3: Rearranging, v = the square root of 2 g h \/ (1 + k).\nStep 4 (sphere, k = 0.4): v = sqrt(2 \u00d7 9.81 \u00d7 1.5 \/ 1.4) = sqrt(21.02) = 4.58 m\/s.\nStep 5 (hoop, k = 1): v = sqrt(29.43 \/ 2) = sqrt(14.72) = 3.84 m\/s.\n<strong>Answer: sphere 4.58 m\/s, hoop 3.84 m\/s \u2014 independent of mass and radius<\/strong>\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What is moment of inertia in simple terms?<\/summary><div class=\"pf-faq-item-answer\">\nMoment of inertia is how hard it is to start or stop something spinning about a chosen axis. It plays the same role in rotation that mass plays in straight-line motion. The difference is that it depends not just on how much mass an object has, but on how far that mass sits from the axis.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the SI unit of moment of inertia?<\/summary><div class=\"pf-faq-item-answer\">\nThe SI unit of moment of inertia is the kilogram metre squared (kg\u00b7m\u00b2). It follows directly from the definition I = m r\u00b2, since mass is in kilograms and distance in metres. Note that this is different from the newton metre (N\u00b7m), which is the unit of torque, or moment of a force.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the formula for moment of inertia?<\/summary><div class=\"pf-faq-item-answer\">\nFor a point mass the formula is I = m r\u00b2, where r is the perpendicular distance to the axis. For an extended body you add up every mass element, which for uniform shapes simplifies to I = k m r\u00b2. Here k is a shape factor: 0.5 for a solid disc, 0.4 for a solid sphere, and 1 for a thin hoop.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Does moment of inertia depend on mass or shape?<\/summary><div class=\"pf-faq-item-answer\">\nIt depends on both, plus the axis you choose. Moment of inertia rises in direct proportion to mass but with the square of distance from the axis, so shape usually has the bigger effect. A light hoop can be harder to spin than a much heavier solid disc of similar size.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the difference between moment of inertia and moment of a force?<\/summary><div class=\"pf-faq-item-answer\">\nMoment of inertia is a body&#8217;s resistance to a change in rotation, measured in kg\u00b7m\u00b2. A moment of a force, also called torque, is a turning effect produced by a force, measured in N\u00b7m. One is a property of the object; the other is an action applied to it. They are linked by the equation torque equals I times angular acceleration.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Why does a solid sphere roll down a slope faster than a hoop?<\/summary><div class=\"pf-faq-item-answer\">\nA solid sphere has a smaller shape factor, k = 0.4 against the hoop&#8217;s k = 1, so it diverts less of its energy into spinning and more into moving forwards. From a 1.5 m drop the sphere reaches 4.58 m\/s and the hoop only 3.84 m\/s. Mass and radius cancel out entirely, so they make no difference.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the parallel axis theorem?<\/summary><div class=\"pf-faq-item-answer\">\nThe parallel axis theorem states that I = I(cm) + m d\u00b2, where I(cm) is the moment of inertia about an axis through the centre of mass and d is the perpendicular distance to a parallel axis. It only works starting from the centre-of-mass axis. Because m d\u00b2 is always positive, the centre-of-mass axis is always the easiest to spin about.\n<\/div><\/details>\n\n<h2>Key Takeaways<\/h2>\n\n<ul>\n<li>Moment of inertia is rotational mass: it measures resistance to a change in spin, in kg\u00b7m\u00b2.<\/li>\n<li>It always depends on the axis \u2014 the same object has many different values.<\/li>\n<li>Distance from the axis counts twice over, since the formula squares it.<\/li>\n<li>For uniform shapes, I = k m r\u00b2, with k running from 0.083 for a rod about its centre to 1 for a hoop.<\/li>\n<li>The parallel axis theorem, I = I(cm) + m d\u00b2, moves a known value to any parallel axis.<\/li>\n<\/ul>\n\n<p>For a full university-level treatment, including video derivations of the rod, disc and sphere results, MIT OpenCourseWare&#8217;s <a href=\"https:\/\/ocw.mit.edu\/courses\/8-01sc-classical-mechanics-fall-2016\/pages\/week-10-rotational-motion\/\" target=\"_blank\" rel=\"noopener\">Classical Mechanics rotational motion unit<\/a> covers the same ground with calculus.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Moment of inertia measures how strongly an object resists a change in its spin about a chosen axis. This guide gives the formula, a table for nine common shapes, the parallel axis theorem, eight worked problems and the four mistakes that cost exam marks.<\/p>\n","protected":false},"author":1,"featured_media":662,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"class_list":["post-660","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\/660","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=660"}],"version-history":[{"count":1,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/660\/revisions"}],"predecessor-version":[{"id":663,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/660\/revisions\/663"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/662"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=660"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=660"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=660"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}