{"id":1953,"date":"2026-09-20T10:09:33","date_gmt":"2026-09-20T10:09:33","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=1953"},"modified":"2026-09-20T10:09:35","modified_gmt":"2026-09-20T10:09:35","slug":"angular-momentum","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/angular-momentum\/","title":{"rendered":"Angular Momentum Explained"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\n\nAngular momentum is the rotational counterpart of linear momentum: an object&#8217;s moment of inertia multiplied by its angular velocity, L = I\u03c9, or for a single mass circling an axis, mass times speed times radius, L = mvr. It is measured in kg\u00b7m\u00b2\/s and stays constant while no net external torque acts about the axis.\n\n<\/p><\/div>\n\n<p>Sit on an office chair, get someone to spin you with your legs stuck out, then tuck them in. You did not push against the floor, the walls or anybody in the room, and you sped up anyway.<\/p>\n\n<p>Nothing about that is a quirk of the chair. It is the one thing a spinning system refuses to give up, and once you can put a number on it you can predict your new spin rate before you tuck.<\/p>\n\n<h2>What Is Angular Momentum?<\/h2>\n\n<p>Angular momentum is the amount of rotation a body carries about a chosen axis: its moment of inertia multiplied by its angular velocity. Write it down and you have L = I\u03c9.<\/p>\n\n<p>The phrase &#8220;about a chosen axis&#8221; is doing real work in that sentence. There is no such thing as the angular momentum of an object full stop \u2014 only its angular momentum about some stated line in space, and choosing a different line gives a different number for exactly the same motion.<\/p>\n\n<p>Three things fix the value, and only three.<\/p>\n\n<ul>\n<li><strong>How much mass there is.<\/strong> Double the mass, double the angular momentum, all else held still.<\/li>\n<li><strong>How far that mass sits from the axis.<\/strong> This one counts twice over, because the distance enters the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/moment-of-inertia\/\">moment of inertia<\/a> squared. Moving a mass from 0.20 m out to 0.80 m multiplies its contribution by sixteen.<\/li>\n<li><strong>How fast it is turning.<\/strong> The <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/angular-velocity-formula\/\">angular velocity<\/a> enters once, in radians per second.<\/li>\n<\/ul>\n\n<p>Why bother defining such a thing? Because it survives. Left alone by any outside twist, a spinning system can rearrange itself as violently as it likes and this one number will not move, which makes it the rotational member of the same family as energy and linear momentum \u2014 the trio that <a href=\"https:\/\/ocw.mit.edu\/courses\/8-01sc-classical-mechanics-fall-2016\/\" target=\"_blank\" rel=\"noopener\">MIT&#8217;s OpenCourseWare classical mechanics course<\/a> introduces together as the conservation approach to solving problems.<\/p>\n\n<p>There is something unusual about this topic worth flagging early. Nothing in the angular momentum relations needs a constant of nature: every number below is built from the definitions of mass, length, time and angle alone, so there is next to nothing to look up and nothing to get out of date. The one exception is problem 6, where g = 9.81 m\/s\u00b2 appears only to turn a force in newtons into the weight of a familiar mass.<\/p>\n\n<h2>The Angular Momentum Formula, and the Two Ways to Write It<\/h2>\n\n<p>The standard form is the moment of inertia about the axis multiplied by the angular velocity about that same axis.<\/p>\n\n<div class=\"pf-formula\">L = I\u03c9<\/div>\n\n<p>For one mass travelling round a circle, the same quantity is usually written a different way: mass, times speed along the circle, times the radius from the axis.<\/p>\n\n<div class=\"pf-formula\">L = mvr<\/div>\n\n<p>Those are not two laws. They are one quantity wearing two hats, and a single line of substitution shows it. A mass on a circle has v = \u03c9r, so mvr becomes m(\u03c9r)r, which regroups as (mr\u00b2)\u03c9 \u2014 and mr\u00b2 is precisely what &#8220;moment of inertia&#8221; means for a point mass at radius r.<\/p>\n\n<p>That regrouping is worth staring at for a moment, because it explains the whole topic. A real spinning body is a crowd of little masses at a crowd of different radii; add up every mr\u00b2 and you get the single number I. Shift material outward and I climbs as the square of the distance, which is why a small movement of your arms changes your spin so much more than you expect.<\/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>\n<th style=\"text-align:left;padding:8px;border-bottom:2px solid #D9CFB8;\">Symbol<\/th>\n<th style=\"text-align:left;padding:8px;border-bottom:2px solid #D9CFB8;\">What it is<\/th>\n<th style=\"text-align:left;padding:8px;border-bottom:2px solid #D9CFB8;\">SI unit<\/th>\n<th style=\"text-align:left;padding:8px;border-bottom:2px solid #D9CFB8;\">Worth knowing<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">L<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">angular momentum, about one stated axis<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">kg\u00b7m<sup>2<\/sup>\/s<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">the same unit may be written N\u00b7m\u00b7s or J\u00b7s; all three are identical combinations of the base units<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">I<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">moment of inertia about that same axis<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">kg\u00b7m<sup>2<\/sup><\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">depends on shape as much as on mass; a point mass at radius r has I = mr<sup>2<\/sup><\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">\u03c9<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">angular velocity, the spin rate<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">rad\/s<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">convert revolutions per minute or degrees per second before multiplying anything<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">m<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">the mass of the one object going round<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">kg<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">only the point-mass form uses it; the I form has already swallowed every mass in the body<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">v<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">its speed along the circle<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">m\/s<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">tied to the spin rate by v = \u03c9r, which is what makes the two formulae one<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">r<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">radius from the axis out to that mass<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">m<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">measured from the axis you are quoting, never from anywhere else<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>The unit is where a lot of marks go missing. Angular momentum comes out in kg\u00b7m\u00b2\/s, built from nothing but the kilogram, the metre and the second, three of the <a href=\"https:\/\/physics.nist.gov\/cuu\/Units\/units.html\" target=\"_blank\" rel=\"noopener\">seven SI base units<\/a>. It is also, written differently, N\u00b7m\u00b7s and J\u00b7s \u2014 which is a nice piece of bookkeeping rather than a deep truth, so do not read a quantum claim into the fact that it happens to be the unit of the Planck constant.<\/p>\n\n<p>If you would rather not grind through the arithmetic twice, the <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/angular-momentum\">angular momentum calculator<\/a> takes either set of inputs \u2014 a moment of inertia and a spin rate, or a mass, a speed and a radius \u2014 and returns L. Enter a known L instead, alongside one or two of the others, and it works backwards to give you \u03c9, I or v.<\/p>\n\n<p>The lab below puts that algebra in motion. It carries two equal masses on a turntable and shows, as you slide them in or out, how the total splits between what the turntable carries and what the masses carry.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Angular Momentum 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\/angular-momentum.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n\n<h2>Why a Spinning Skater Speeds Up<\/h2>\n\n<p>A skater speeds up because pulling the arms in cuts the moment of inertia, and with nothing outside exerting a torque the product I\u03c9 has to stay where it was, so \u03c9 rises in exactly the proportion that I fell.<\/p>\n\n<p>Numbers make it concrete. Take a turntable of moment of inertia 2.00 kg\u00b7m\u00b2 about its axis, carrying two 3.00 kg masses, one on each side, at a radius of 0.80 m, turning at 2.50 rad\/s.<\/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>\n<th style=\"text-align:left;padding:8px;border-bottom:2px solid #D9CFB8;\">Reading<\/th>\n<th style=\"text-align:left;padding:8px;border-bottom:2px solid #D9CFB8;\">Arms out, masses at 0.80 m<\/th>\n<th style=\"text-align:left;padding:8px;border-bottom:2px solid #D9CFB8;\">Arms in, masses at 0.20 m<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">Moment of inertia I<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">5.840 kg\u00b7m<sup>2<\/sup><\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">2.240 kg\u00b7m<sup>2<\/sup><\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">Spin rate \u03c9<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">2.50 rad\/s<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">6.518 rad\/s<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\"><strong>Angular momentum L (the one that holds still)<\/strong><\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\"><strong>14.60 kg\u00b7m<sup>2<\/sup>\/s<\/strong><\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\"><strong>14.60 kg\u00b7m<sup>2<\/sup>\/s<\/strong><\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">Rotational energy<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">18.25 J<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">47.58 J<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">Speed of each mass<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">2.000 m\/s<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">1.304 m\/s<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">m v r for one mass alone<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">4.800 kg\u00b7m<sup>2<\/sup>\/s<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">0.7821 kg\u00b7m<sup>2<\/sup>\/s<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">Both masses together<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">9.600 kg\u00b7m<sup>2<\/sup>\/s<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">1.564 kg\u00b7m<sup>2<\/sup>\/s<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">The turntable&#8217;s own share<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">5.000 kg\u00b7m<sup>2<\/sup>\/s<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">13.04 kg\u00b7m<sup>2<\/sup>\/s<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">Inward pull needed on each mass<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">15.00 N<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">25.49 N<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>Read down the third row and nothing happens. Read down any other row and something does \u2014 the inertia falls by a factor of 2.607, the spin rate rises by the same 2.607, and the two cancel in the product.<\/p>\n\n<p>One caution about arithmetic on that table, because it catches people out. Every computed cell is rounded to four significant figures for display \u2014 only the 2.50 rad\/s you set is printed as typed \u2014 and the shares behave accordingly: 5.000 + 9.600 comes to exactly 14.60, while 13.04 + 1.564 comes to 14.604. The physics is not drifting; the display is.<\/p>\n\n<figure style=\"margin:32px auto;max-width:720px;text-align:center;\">\n  <img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/angular-momentum-diagram-01-768x529.jpg\" srcset=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/angular-momentum-diagram-01-768x529.jpg 768w, https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/angular-momentum-diagram-01.jpg 1240w\" sizes=\"auto, (max-width: 760px) 100vw, 720px\" width=\"1240\" height=\"854\" alt=\"Two drawings of the same turntable seen from above, drawn the same size because it is the same turntable. On the left the two masses sit on spokes 0.80 metres from the hub; on the right the same two masses sit 0.20 metres from the hub, both radii plotted on one shared scale, so the four to one you see is 0.80 metres against 0.20 metres. A curved gold arrow round each disc shows the direction of the spin and not its rate, with the rate labelled 2.50 radians per second on the left and 6.518 radians per second on the right. The moment of inertia is 5.840 kilogram metre squared on the left and 2.240 on the right, and one line under both reads L equals 14.60 kilogram metre squared per second.\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;border-radius:4px;\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Figure 1. The same turntable drawn twice, at the same drawn size, because it is the same turntable \u2014 only the masses moved. Both mass radii are plotted on one shared scale, so the 4:1 you see is 0.80 m against 0.20 m and nothing else. The two spin arrows show the direction of the spin, not its rate; the rates are the gold labels above the discs.<\/figcaption>\n<\/figure>\n\n<p>It helps to watch where the angular momentum is actually stored. With the masses out at 0.80 m they carry 9.600 kg\u00b7m\u00b2\/s of the total and the turntable carries 5.000 kg\u00b7m\u00b2\/s. Bring them in and the masses carry only 1.564 kg\u00b7m\u00b2\/s, while the faster-turning turntable now carries 13.04 kg\u00b7m\u00b2\/s.<\/p>\n\n<figure style=\"margin:32px auto;max-width:720px;text-align:center;\">\n  <img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/angular-momentum-diagram-02-768x424.jpg\" srcset=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/angular-momentum-diagram-02-768x424.jpg 768w, https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/angular-momentum-diagram-02.jpg 1240w\" sizes=\"auto, (max-width: 760px) 100vw, 720px\" width=\"1240\" height=\"684\" alt=\"Two horizontal bars, both plotted on one shared scale. The upper bar, labelled before, is the turntable's share 5.000 followed by the two masses' share 9.600. The lower bar, labelled after, is the turntable's share 13.04 followed by the two masses' share 1.564. Each bar was built from its own state, and both reach the same dashed line at 14.60 kilogram metre squared per second. A scale underneath is ticked 0, 4, 8, 12 and 16, labelled angular momentum in kilogram metre squared per second.\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;border-radius:4px;\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Figure 2. Each bar is summed from its own state \u2014 the turntable&#8217;s share I\u03c9, then the two masses&#8217; share 2mvr \u2014 and both reach 14.60 kg\u00b7m\u00b2\/s. Read where the bars end rather than adding the printed segment labels, which are rounded to four figures.<\/figcaption>\n<\/figure>\n\n<p>That is the whole trick, drawn. The split moves wholesale from the masses to the turntable, and the far end of the bar does not budge.<\/p>\n\n<h2>Where the Extra Energy Comes From<\/h2>\n\n<p>The extra rotational energy is work done on the system. Something has to haul each mass in towards the axis, and hauling it inward while it strains outward on you is work by any definition. Nothing is created and nothing is free.<\/p>\n\n<p>The rotational <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/kinetic-energy-formula\/\">kinetic energy<\/a> of a spinning body is \u00bdI\u03c9\u00b2, which can be rewritten as \u00bdL\u03c9. Hold L still and the energy simply tracks \u03c9 \u2014 so the 2.607-fold speed-up costs a 2.607-fold rise in energy, from 18.25 J to 47.58 J.<\/p>\n\n<figure style=\"margin:32px auto;max-width:720px;text-align:center;\">\n  <img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/angular-momentum-diagram-03-768x396.jpg\" srcset=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/angular-momentum-diagram-03-768x396.jpg 768w, https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/angular-momentum-diagram-03.jpg 1240w\" sizes=\"auto, (max-width: 760px) 100vw, 720px\" width=\"1240\" height=\"640\" alt=\"Two horizontal bars, both plotted on one shared scale, showing the rotational energy of the same event as the previous figure. The upper bar, arms out, runs to 18.25 joules. The lower bar, arms in, runs to 47.58 joules on that same scale. A bracket between the two bar ends marks the 29.33 joule difference and is labelled work done by the arms. A scale underneath is ticked 0, 10, 20, 30, 40 and 50, labelled rotational energy in joules.\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;border-radius:4px;\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Figure 3. The same event as Figure 2, and the quantity that behaves differently: the angular momentum held at 14.60 kg\u00b7m\u00b2\/s while the rotational energy climbed from 18.25 J to 47.58 J, a gain of 29.33 J.<\/figcaption>\n<\/figure>\n\n<p>Where did 29.33 J come from? Each mass was being held on its 0.80 m circle by an inward pull of 15.00 N, and something had to drag it in to 0.20 m against a pull that never once turned outward. Force through a distance is work, and that work went straight into the rotation.<\/p>\n\n<p>That pull is not steady on the way in, and it does not simply grow. Each mass needs F = m\u03c9\u00b2r to stay on its circle, and as it comes in the spin rate climbs, which asks for more force, while the radius shrinks, which asks for less. Put \u03c9 = L \/ I into that, with I = 2.00 + 2mr\u00b2, and the two effects balance where 6mr\u00b2 matches the turntable&#8217;s own 2.00 kg\u00b7m\u00b2 \u2014 that is, at r = sqrt(2.00\/18) = 0.3333 m, where the two masses between them hold a third of the turntable&#8217;s inertia.<\/p>\n\n<p>So the pull climbs from 15.00 N at 0.80 m to a peak of 29.98 N at that radius, then eases back to 25.49 N by 0.20 m \u2014 still well above where it started, and still pointing inward the whole way. Every stage of the journey takes work in, which is why the 29.33 J came out positive.<\/p>\n\n<p>Push the masses back out and the accounting runs backwards: the spin falls, the energy falls, and your arms take the energy out again. Problem 5 below works that case through and the answer comes out negative, which is the sign telling you the direction of the transaction.<\/p>\n\n<p>So angular momentum is conserved here and energy is not \u2014 and there is no contradiction in that. Conservation of angular momentum is a statement about torques; conservation of energy is a statement about a closed system, and a pair of arms doing work is not one.<\/p>\n\n<h2>Common Misconceptions About Angular Momentum<\/h2>\n\n<p>Four beliefs cause most of the trouble on this topic, and all four are worth naming.<\/p>\n\n<h3>&#8220;It is just momentum, going round in a circle&#8221;<\/h3>\n\n<p>It is not. Linear momentum is kg\u00b7m\/s and angular momentum is kg\u00b7m\u00b2\/s, so the two cannot even be added, let alone equated. They are analogues, not the same quantity, and the parallel between them is the most useful map in rotational mechanics.<\/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>\n<th style=\"text-align:left;padding:8px;border-bottom:2px solid #D9CFB8;\">Straight-line quantity<\/th>\n<th style=\"text-align:left;padding:8px;border-bottom:2px solid #D9CFB8;\">Rotational partner<\/th>\n<th style=\"text-align:left;padding:8px;border-bottom:2px solid #D9CFB8;\">Conserved when?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">mass m (kg)<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">moment of inertia I (kg\u00b7m<sup>2<\/sup>)<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">never: I changes the instant the shape changes, which is the whole point<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">velocity v (m\/s)<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">angular velocity \u03c9 (rad\/s)<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">never: it is free to move, and it does<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">momentum p = mv (kg\u00b7m\/s)<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">angular momentum L = I\u03c9 (kg\u00b7m<sup>2<\/sup>\/s)<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">p with no net external force; L with no net external torque about the axis<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">force F = ma (N)<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">torque \u03c4 = I\u03b1 (N\u00b7m)<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">neither: these are the things that change the momentum<\/td><\/tr>\n<tr><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">kinetic energy \u00bdmv<sup>2<\/sup> (J)<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">rotational energy \u00bdI\u03c9<sup>2<\/sup> (J)<\/td><td style=\"padding:8px;border-bottom:1px solid #D9CFB8;\">only while no work is done on the system<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>Read that third row across and the two conservation rules fall out side by side. <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/conservation-of-momentum\/\">Conservation of linear momentum<\/a> needs no net external force; its rotational twin needs no net external torque about the axis you chose.<\/p>\n\n<h3>&#8220;Angular momentum and torque are the same thing&#8221;<\/h3>\n\n<p>They are related the way momentum and force are related, which is to say they are not the same at all. <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/torque-physics\/\">Torque<\/a> is the rate at which angular momentum changes, so a torque of zero means L is flat, not that L is zero.<\/p>\n\n<p>A common slip follows from that. Students see &#8220;no torque&#8221; and conclude &#8220;nothing is happening&#8221;, when in fact a great deal may be happening \u2014 as the skater case shows, the system can be tearing through a dramatic rearrangement while the conserved quantity sits perfectly still.<\/p>\n\n<h3>&#8220;Angular momentum is always conserved&#8221;<\/h3>\n\n<p>Only when the net external torque about the axis is zero. Your office chair has bearings that rub, a skater has a blade on ice and a body moving through air, and in both cases an outside torque is quietly bleeding angular momentum away the entire time.<\/p>\n\n<p>The honest version is conditional: L holds still <em>because<\/em> nothing outside twists the system about that axis. Assume the condition and you can predict the new spin rate; forget it and you will be surprised when a real wheel coasts to a stop.<\/p>\n\n<h3>&#8220;The heavier object must have more angular momentum&#8221;<\/h3>\n\n<p>Not necessarily, because where the mass sits matters more than how much of it there is. Mass enters once and radius enters squared, so a light rim at a large radius can easily beat a heavy hub at a small one.<\/p>\n\n<p>In practice this is why a bicycle wheel, with nearly all its metal out at the rim, is so much harder to spin up than a solid disc of the same mass and radius. The material that counts is the material furthest from the axle.<\/p>\n\n<h2>Angular Momentum Depends on the Axis You Measure About<\/h2>\n\n<p>Quote an angular momentum without naming an axis and you have said nothing. The same motion, measured about two different lines, gives two different numbers \u2014 and one of them may well be zero.<\/p>\n\n<p>Take the cleanest case there is. A ball sails past you in a perfectly straight line at constant speed, with no rotation at all. About a point off to one side of its path it has a real, non-zero angular momentum of mvr, where r is the perpendicular distance from that point to the line of travel. About any point actually on the line, r is zero, and so is L.<\/p>\n\n<p>Nothing about the ball changed between those two statements. Only the axis did. That is why every worked answer on this page names the axis it belongs to, and why a problem that hands you a moment of inertia about the centre of mass and a radius from the pivot is a trap rather than a question.<\/p>\n\n<p>The same discipline applies in <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/circular-motion-physics\/\">circular motion<\/a> generally: pick the axis first, measure every radius from it, and do not change your mind halfway through.<\/p>\n\n<h2>Where You Meet Angular Momentum<\/h2>\n\n<p>The pattern to look for is a system that changes its own shape, or spins one part to steer another, with nothing outside twisting it.<\/p>\n\n<ul>\n<li><strong>A skater or a diver.<\/strong> Tuck in mid-rotation and the spin rises; open out and it falls. The shape is the control, and the conserved quantity does the rest.<\/li>\n<li><strong>A helicopter&#8217;s tail rotor.<\/strong> It exists because the airframe would otherwise take up the main rotor&#8217;s angular momentum and turn the other way underneath it.<\/li>\n<li><strong>A spacecraft&#8217;s reaction wheel.<\/strong> Spin a heavy wheel inside the craft one way and the craft turns the other, with no fuel spent and no external torque involved \u2014 the principle behind the <a href=\"https:\/\/science.nasa.gov\/mission\/hubble\/observatory\/design\/pointing-control\/\" target=\"_blank\" rel=\"noopener\">reaction wheels used for spacecraft pointing control<\/a>.<\/li>\n<li><strong>A flywheel.<\/strong> Here the interest is the energy stored in the spin rather than the angular momentum itself, but the same I and the same \u03c9 set both.<\/li>\n<li><strong>The planets.<\/strong> Gravity pulls a planet straight towards the Sun, which exerts no torque about the Sun, which is why orbital angular momentum is conserved and the orbit keeps its plane.<\/li>\n<\/ul>\n\n<p>Notice what all five have in common. None of them adds rotation from outside; each one only moves rotation about, between parts of a system that was already turning.<\/p>\n\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\/09\/a-figure-skater-in-a-spin-with-the-arms-drawn-in.webp\"\n       alt=\"Figure skater spinning with the arms pulled in, the classic everyday example of angular momentum conservation\"\n       loading=\"lazy\"\n       style=\"width:100%;height:auto;border-radius:4px;\" width=\"1600\" height=\"931\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Arms in, moment of inertia down, spin rate up. Nothing outside the skater supplied the extra rotation.<\/figcaption>\n<\/figure>\n\n<h2>Where Conservation Breaks Down<\/h2>\n\n<p>Conservation of angular momentum is exact under one condition and approximate under real ones, and five limits are worth stating plainly.<\/p>\n\n<ul>\n<li><strong>Friction and drag.<\/strong> A real chair, a real wheel and a real skater all slow down, because bearings rub and air pushes back. Treat a conserved L as the idealisation it is, and check whether the outside torque is genuinely negligible over the time you care about.<\/li>\n<li><strong>L = I\u03c9 is the scalar shortcut.<\/strong> In general angular momentum is a vector, defined as r \u00d7 p, it need not point along the rotation axis at all, and I is a tensor rather than a number. The scalar form used throughout this page is the special case of rotation about a symmetry axis, which covers most school and first-year problems and not much beyond them.<\/li>\n<li><strong>The axis is part of the answer.<\/strong> As above: change the axis and you change the number. An L quoted without its axis is not wrong so much as meaningless.<\/li>\n<li><strong>Point masses are an idealisation.<\/strong> Treating a held mass as a point at radius r ignores the moment of inertia it has about its own centre. For a compact object at a large radius that is a small error; for a long bar held close in, it is not.<\/li>\n<li><strong>Extreme arrangements describe machines, not people.<\/strong> Push the numbers hard enough and the inward pull runs into the thousands of newtons, as problem 6 shows. The arithmetic stays correct; what it is describing has stopped being a person on a chair.<\/li>\n<\/ul>\n\n<p>One more caution about where the numbers on this page come from. The turntable&#8217;s own moment of inertia is a figure chosen for the worked cases, not a measurement of any real chair, stool or skater \u2014 nothing here measures a person, and no page should attach a number to one.<\/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 turntable with a moment of inertia of 2.00 kg\u00b7m\u00b2 about its axis carries two 3.00 kg masses, one on each side, at a radius of 0.80 m, and the whole thing turns at 2.50 rad\/s. Find the moment of inertia and the angular momentum, then pull the masses in to 0.20 m and find the new spin rate.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Build the moment of inertia. Each mass is a point mass at radius r, so it contributes mr<sup>2<\/sup>, and there are two of them: I<sub>1<\/sub> = 2.00 + 2 \u00d7 3.00 \u00d7 0.80<sup>2<\/sup> = 2.00 + 3.84 = 5.84 kg\u00b7m<sup>2<\/sup>.\n\nStep 2: Multiply by the spin rate. L = I<sub>1<\/sub>\u03c9<sub>1<\/sub> = 5.84 \u00d7 2.50 = 14.60 kg\u00b7m<sup>2<\/sup>\/s, about the turntable&#8217;s spin axis.\n\nStep 3: Rebuild the inertia at the new radius. I<sub>2<\/sub> = 2.00 + 2 \u00d7 3.00 \u00d7 0.20<sup>2<\/sup> = 2.00 + 0.24 = 2.24 kg\u00b7m<sup>2<\/sup>.\n\nStep 4: No external torque acts about the axis, so L is unchanged and \u03c9<sub>2<\/sub> = L \/ I<sub>2<\/sub> = 14.60 \/ 2.24 = 6.517857 rad\/s.\n\nStep 5: Check the factor two ways. 6.517857 \/ 2.50 = 2.607143, and I<sub>1<\/sub> \/ I<sub>2<\/sub> = 5.84 \/ 2.24 = 2.607143. The inertia fell by that factor and the spin rate rose by it.\n\n<strong>Answer: I = 5.840 kg\u00b7m<sup>2<\/sup> out and 2.240 kg\u00b7m<sup>2<\/sup> in, L = 14.60 kg\u00b7m<sup>2<\/sup>\/s at both radii, and the spin rate rises to 6.518 rad\/s.<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">Take the same turntable from problem 1. Find the rotational energy before and after the masses are pulled in from 0.80 m to 0.20 m, and say how much work the arms did.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Use the rotational energy formula on the starting state. KE<sub>1<\/sub> = \u00bdI<sub>1<\/sub>\u03c9<sub>1<\/sub><sup>2<\/sup> = 0.5 \u00d7 5.84 \u00d7 2.50<sup>2<\/sup> = 18.25 J.\n\nStep 2: For the finishing state, the tidier route is KE = \u00bdL\u03c9, since L is already known. KE<sub>2<\/sub> = 0.5 \u00d7 14.60 kg\u00b7m<sup>2<\/sup>\/s \u00d7 6.517857 rad\/s = 47.5804 J.\n\nStep 3: The work done on the system is the change in its energy. W = 47.58 &#8211; 18.25 = 29.33 J, and the positive sign says energy went in.\n\nStep 4: Compare the ratios. \u03c9<sub>2<\/sub>\/\u03c9<sub>1<\/sub> = 6.517857 \/ 2.50 = 2.607143, and KE<sub>2<\/sub>\/KE<sub>1<\/sub> = 47.5804 \/ 18.25 = 2.607145 \u2014 the same 2.607 to four figures, because KE = \u00bdL\u03c9 and L never moved.\n\n<strong>Answer: 18.25 J before, 47.58 J after, and the arms did 29.33 J of work. The angular momentum stayed at 14.60 kg\u00b7m<sup>2<\/sup>\/s throughout, which is a different quantity in a different unit and is under no obligation to follow the energy.<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">A light frame, its own moment of inertia taken as zero, carries two 3.00 kg masses at 0.30 m and turns at 2.50 rad\/s. Show that the inertia-times-spin-rate form and the mass-times-speed-times-radius form give the same angular momentum, then pull the masses in to 0.20 m and show that they still agree.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Work the I form first. With the frame contributing nothing, I<sub>1<\/sub> = 2 \u00d7 3.00 \u00d7 0.30<sup>2<\/sup> = 0.54 kg\u00b7m<sup>2<\/sup>, so L = 0.54 \u00d7 2.50 = 1.35 kg\u00b7m<sup>2<\/sup>\/s.\n\nStep 2: Now the mvr form. Each mass travels at v<sub>1<\/sub> = \u03c9<sub>1<\/sub>r<sub>1<\/sub> = 2.50 \u00d7 0.30 = 0.75 m\/s, so mvr = 3.00 \u00d7 0.75 \u00d7 0.30 = 0.675 kg\u00b7m<sup>2<\/sup>\/s each, and two of them give 2 \u00d7 0.675 = 1.35 kg\u00b7m<sup>2<\/sup>\/s. Identical.\n\nStep 3: Pull them in. I<sub>2<\/sub> = 2 \u00d7 3.00 \u00d7 0.20<sup>2<\/sup> = 0.24 kg\u00b7m<sup>2<\/sup>, so \u03c9<sub>2<\/sub> = 1.35 \/ 0.24 = 5.625 rad\/s and v<sub>2<\/sub> = 5.625 \u00d7 0.20 = 1.125 m\/s.\n\nStep 4: Re-run the mvr form at the new radius. 3.00 \u00d7 1.125 \u00d7 0.20 = 0.675 kg\u00b7m<sup>2<\/sup>\/s each again, and 2 \u00d7 0.675 = 1.35 kg\u00b7m<sup>2<\/sup>\/s. The speed went up, the radius went down, and the product held.\n\n<strong>Answer: L = 1.350 kg\u00b7m<sup>2<\/sup>\/s by either formula, at both radii. The two forms are one quantity, and with the frame&#8217;s own inertia at zero the point-mass form is exact rather than approximate.<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A 2.00 kg\u00b7m\u00b2 turntable carries two 3.00 kg masses at 1.00 m and turns at 3.00 rad\/s. The masses are hauled right in to 0.15 m. Find the new spin rate and the work done, and watch what rounding does to the energy subtraction.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Inertia before and after. I<sub>1<\/sub> = 2.00 + 2 \u00d7 3.00 \u00d7 1.00<sup>2<\/sup> = 8.00 kg\u00b7m<sup>2<\/sup>, and I<sub>2<\/sub> = 2.00 + 2 \u00d7 3.00 \u00d7 0.15<sup>2<\/sup> = 2.00 + 0.135 = 2.135 kg\u00b7m<sup>2<\/sup>.\n\nStep 2: Angular momentum, then the new spin rate. L = 8.00 \u00d7 3.00 = 24.00 kg\u00b7m<sup>2<\/sup>\/s, so \u03c9<sub>2<\/sub> = 24.00 \/ 2.135 = 11.2412178 rad\/s.\n\nStep 3: Energies. KE<sub>1<\/sub> = 0.5 \u00d7 8.00 \u00d7 3.00<sup>2<\/sup> = 36.00 J, and KE<sub>2<\/sub> = \u00bdL\u03c9<sub>2<\/sub> = 0.5 \u00d7 24.00 kg\u00b7m<sup>2<\/sup>\/s \u00d7 11.2412178 rad\/s = 134.8946 J.\n\nStep 4: Subtract, carrying the unrounded energy. W = 134.8946 &#8211; 36.00 = 98.8946 J, which is 98.89 J to four significant figures. Round both energies to four figures <em>first<\/em> and the subtraction lands a unit out in the last place \u2014 always subtract before you round.\n\n<strong>Answer: the spin rate rises to 11.24 rad\/s, a factor of 3.747, and the work done is 98.89 J. The angular momentum stays at 24.00 kg\u00b7m<sup>2<\/sup>\/s.<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">The same 2.00 kg\u00b7m\u00b2 turntable carries two 3.00 kg masses at 0.20 m and spins at 8.00 rad\/s. This time the masses are pushed back out to 0.80 m. Find the new spin rate and the work done, and say who does the work.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Start state. I<sub>1<\/sub> = 2.00 + 2 \u00d7 3.00 \u00d7 0.20<sup>2<\/sup> = 2.24 kg\u00b7m<sup>2<\/sup>, so L = 2.24 \u00d7 8.00 = 17.92 kg\u00b7m<sup>2<\/sup>\/s.\n\nStep 2: Finish state. I<sub>2<\/sub> = 2.00 + 2 \u00d7 3.00 \u00d7 0.80<sup>2<\/sup> = 5.84 kg\u00b7m<sup>2<\/sup>, and \u03c9<sub>2<\/sub> = 17.92 \/ 5.84 = 3.068493 rad\/s. Moving mass outward raises I, so the spin must fall.\n\nStep 3: Energies. KE<sub>1<\/sub> = 0.5 \u00d7 2.24 \u00d7 8.00<sup>2<\/sup> = 71.68 J, and KE<sub>2<\/sub> = 0.5 \u00d7 17.92 kg\u00b7m<sup>2<\/sup>\/s \u00d7 3.068493 rad\/s = 27.4937 J.\n\nStep 4: The change. W = 27.4937 &#8211; 71.68 = -44.1863 J, which is -44.19 J to four figures.\n\n<strong>Answer: the spin falls to 3.068 rad\/s while L holds at 17.92 kg\u00b7m<sup>2<\/sup>\/s, and W = -44.19 J. The minus sign says the transfer runs the other way \u2014 the masses push outward as they go, the arms resist that push, and the system does work on the arms.<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A light turntable of 0.40 kg\u00b7m\u00b2 carries two 8.00 kg masses at 1.00 m and turns at 3.00 rad\/s. The masses come in to 0.20 m. Find the spin-up factor and the inward pull each mass then needs to stay on its circle.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Inertias. I<sub>1<\/sub> = 0.40 + 2 \u00d7 8.00 \u00d7 1.00<sup>2<\/sup> = 16.40 kg\u00b7m<sup>2<\/sup>, and I<sub>2<\/sub> = 0.40 + 2 \u00d7 8.00 \u00d7 0.20<sup>2<\/sup> = 0.40 + 0.64 = 1.04 kg\u00b7m<sup>2<\/sup>. A light turntable means the masses dominate, so I swings hard.\n\nStep 2: Angular momentum and the new spin rate. L = 16.40 \u00d7 3.00 = 49.20 kg\u00b7m<sup>2<\/sup>\/s, and \u03c9<sub>2<\/sub> = 49.20 \/ 1.04 = 47.3076923 rad\/s.\n\nStep 3: The factor. 47.30769 \/ 3.00 = 15.76923, so the spin rate goes up more than fifteenfold.\n\nStep 4: The pull each mass needs. A mass on a circle needs a centripetal force m\u03c9<sup>2<\/sup>r, so F = 8.00 \u00d7 47.30769<sup>2<\/sup> \u00d7 0.20 = 3580.8 N.\n\n<strong>Answer: a spin-up factor of 15.77, to 47.31 rad\/s, needing 3581 N on each mass. Divide by g = 9.81 m\/s<sup>2<\/sup> and that is 3580.8 \/ 9.81 = 365.0, so each hand would be holding the weight of a 365 kg mass \u2014 this arrangement describes a machine, not a person on a chair.<\/strong>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A heavy 5.00 kg\u00b7m\u00b2 turntable carries two 1.00 kg masses at 1.00 m and turns at 3.00 rad\/s. The masses come in to 0.20 m. Find the new spin rate and the work done. Then find the angular momentum the two masses alone would have about the spin axis, and about an axis lying along the line each mass is travelling at that instant.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Inertias. I<sub>1<\/sub> = 5.00 + 2 \u00d7 1.00 \u00d7 1.00<sup>2<\/sup> = 7.00 kg\u00b7m<sup>2<\/sup>, and I<sub>2<\/sub> = 5.00 + 2 \u00d7 1.00 \u00d7 0.20<sup>2<\/sup> = 5.00 + 0.08 = 5.08 kg\u00b7m<sup>2<\/sup>. This time the turntable dominates, so I barely moves.\n\nStep 2: Spin rate. L = 7.00 \u00d7 3.00 = 21.00 kg\u00b7m<sup>2<\/sup>\/s, and \u03c9<sub>2<\/sub> = 21.00 \/ 5.08 = 4.133858 rad\/s, a factor of only 4.133858 \/ 3.00 = 1.378.\n\nStep 3: Work. KE<sub>1<\/sub> = 0.5 \u00d7 7.00 \u00d7 3.00<sup>2<\/sup> = 31.50 J and KE<sub>2<\/sub> = 0.5 \u00d7 21.00 kg\u00b7m<sup>2<\/sup>\/s \u00d7 4.133858 rad\/s = 43.41 J, so W = 43.41 &#8211; 31.50 = 11.91 J. Problem 4&#8217;s pull-in was worth 98.89 J, but that case differs from this one in three ways at once \u2014 a lighter turntable, heavier masses and a tighter finishing radius \u2014 so the drop is not the heavy turntable&#8217;s doing alone.\n\nStep 4: Now the axis. The two masses alone, out at 1.00 m, each move at v = 3.00 \u00d7 1.00 = 3.00 m\/s, so about the spin axis they carry 2 \u00d7 1.00 \u00d7 3.00 \u00d7 1.00 = 6.00 kg\u00b7m<sup>2<\/sup>\/s. About an axis lying along the line a mass is travelling at that instant, the perpendicular distance r is zero, so that same mass carries none at all. The other mass is running back along a parallel line, so its momentum points straight down that axis and carries nothing about it either.\n\n<strong>Answer: 4.134 rad\/s and 11.91 J of work, with L held at 21.00 kg\u00b7m<sup>2<\/sup>\/s. The two masses alone carry 6.00 kg\u00b7m<sup>2<\/sup>\/s about the spin axis and zero about a line running along the instantaneous path of either one of them \u2014 the same motion, two axes, two answers.<\/strong>\n\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What is the formula for angular momentum?<\/summary><div class=\"pf-faq-item-answer\">\n\nThe formula is L = I\u03c9: the moment of inertia about an axis multiplied by the angular velocity about that same axis. For a single mass travelling round a circle it is usually written L = mvr, with r measured from the axis. Both give the same number, because v = \u03c9r turns mvr into (mr\u00b2)\u03c9, and mr\u00b2 is the moment of inertia of a point mass.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What are the units of angular momentum?<\/summary><div class=\"pf-faq-item-answer\">\n\nThe SI unit is kg\u00b7m\u00b2\/s, built from the kilogram, the metre and the second. The same combination can be written N\u00b7m\u00b7s or J\u00b7s, and all three are the identical unit rather than three different ones. Linear momentum, by contrast, is kg\u00b7m\/s, with one power of length instead of two, which is why the two quantities can never be added together.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Why does a skater spin faster when they pull their arms in?<\/summary><div class=\"pf-faq-item-answer\">\n\nPulling the arms in moves mass closer to the spin axis, which cuts the moment of inertia. With no outside torque the product I\u03c9 cannot change, so the spin rate has to rise by exactly the factor that the inertia fell. In the worked case above, an inertia of 5.840 kg\u00b7m\u00b2 dropping to 2.240 kg\u00b7m\u00b2 raises 2.50 rad\/s to 6.518 rad\/s.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Is angular momentum always conserved?<\/summary><div class=\"pf-faq-item-answer\">\n\nOnly while the net external torque about the axis is zero. That condition is an idealisation: real bearings rub, real blades scrape ice and real bodies push through air, so a real spinning object always loses angular momentum eventually. Conservation is a powerful tool precisely because it tells you what would happen with no outside twist, which is often close enough over a few seconds.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the difference between angular momentum and linear momentum?<\/summary><div class=\"pf-faq-item-answer\">\n\nLinear momentum measures motion in a straight line, mass times velocity, in kg\u00b7m\/s. Angular momentum measures rotation about a stated axis, moment of inertia times angular velocity, in kg\u00b7m\u00b2\/s. They are analogues with separate conservation rules: linear momentum is conserved with no net external force, angular momentum with no net external torque about the chosen axis.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Where does the extra energy come from when a spinning object speeds up?<\/summary><div class=\"pf-faq-item-answer\">\n\nFrom work done by whatever pulls the mass inward. Each mass is being held on its circle by an inward force, and dragging it closer means pulling against that force through a distance, which is work by definition. In the worked case, 29.33 J goes in and the rotational energy climbs from 18.25 J to 47.58 J while the angular momentum, in kg\u00b7m\u00b2\/s, does not move.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Does angular momentum depend on where you measure it from?<\/summary><div class=\"pf-faq-item-answer\">\n\nYes, completely, which is why an axis must always be quoted with the value. A mass moving in a straight line has angular momentum mvr about any point off that line, where r is the perpendicular distance, and exactly zero about any point on it. Nothing about the motion changed between those two answers; only the reference axis did.\n\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>Angular momentum is an object&#8217;s moment of inertia multiplied by its spin rate, L = I\u03c9, and for one mass on a circle the same quantity is L = mvr. Why a spinning skater speeds up when nothing pushes on them, where the extra 29.33 J of energy actually comes from, why the answer changes when you change the axis, seven worked problems, three diagrams and an interactive lab.<\/p>\n","protected":false},"author":1,"featured_media":1957,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"class_list":["post-1953","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\/1953","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=1953"}],"version-history":[{"count":3,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/1953\/revisions"}],"predecessor-version":[{"id":2033,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/1953\/revisions\/2033"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/1957"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=1953"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=1953"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=1953"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}