{"id":11300,"date":"2026-07-09T14:28:03","date_gmt":"2026-07-09T14:28:03","guid":{"rendered":"https:\/\/www.myengineeringbuddy.com\/blog\/angular-momentum\/"},"modified":"2026-07-12T04:24:31","modified_gmt":"2026-07-12T04:24:31","slug":"angular-momentum","status":"publish","type":"post","link":"https:\/\/www.myengineeringbuddy.com\/blog\/angular-momentum\/","title":{"rendered":"Angular Momentum Explained: Definition, Proof and Examples"},"content":{"rendered":"\n<div style=\"background-color:#f8f8f8; border-left:4px solid #d0d0d0; padding:12px 16px; margin-bottom:20px;\"><strong>Key Takeaways<\/strong>\n<ul>\n<li>Angular momentum equals the moment of linear momentum about an axis of rotation.<\/li>\n<li>For a rigid body, angular momentum L equals moment of inertia times angular velocity (L = I\u03c9).<\/li>\n<li>Angular momentum is a vector quantity derived from the cross product of position and linear momentum.<\/li>\n<li>The perpendicular distance from the axis determines the magnitude of angular momentum.<\/li>\n<\/ul><\/div>\n\n<h2>What is Angular Momentum?<\/h2>\n\n<p><strong>Angular momentum is the moment of linear momentum of a body with respect to an axis of rotation.<\/strong><\/p>\n\n<p>Angular momentum of a particle about O is defined as ${bf{L}} = {bf{r}} times {bf{P}}$<\/p>\n\n<p>Where <strong>P<\/strong> is the linear momentum and <strong>r<\/strong> is the position vector of the particle from the given point O. The angular momentum of a system of particles is the vector sum of the angular momenta of the particles of the system. So,<\/p>\n\n<p>${bf{L}} = sumlimits_{i = 1}^n {{{bf{r}}_i} times {{bf{P}}_i}} $<\/p>\n\n<p style=\"text-align: center;\"><img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/figure-1-4-292x300.png\" alt=\"Diagram showing angular momentum of a particle about point O with perpendicular distance OA\"><\/p>\n\n<p>$sin theta = frac{{OA}}{{OP}}quad Rightarrow quad OA = OPsin theta $<\/p>\n\n<p>Let a particle P of mass m moves at velocity <strong>v<\/strong>. Its angular momentum about a point O can be written as<\/p>\n\n<p>$ell = {bf{OP}} times (m{bf{v}})$<\/p>\n\n<p>$ Rightarrow quad ell = (OP),(mv)sin theta ,hat n$<\/p>\n\n<p>$ Rightarrow quad ell = (OA),(mv),hat n$<\/p>\n\n<p>$ Rightarrow quad ell = {{bf{r}}_ bot } times m{bf{v}} = {bf{r}} times m{bf{v}}$<\/p>\n\n<p>Where $r = OA = (OP)sin theta $ is the perpendicular distance of the line of motion from O.<\/p>\n\n<p>If you want to learn Physics from an <a href=\"https:\/\/www.myengineeringbuddy.com\/online-tutoring\/online-physics-tutor\/\">online Physics tutor<\/a>, feel free to contact us on WhatsApp at +91 8971 383660 or email meb@myengineeringbuddy.com.<\/p>\n\n<p>Students who find physics concepts challenging often benefit from structured support in related quantitative fields. If you are working through data-heavy coursework, connecting with an <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/econometrics\/\">econometrics tutor<\/a> can help build the analytical foundations that underpin many applied sciences.<\/p>\n\n<h2>Proof of L = I \u00d7 \u03c9<\/h2>\n\n<p>Suppose a particle is going in a circle of radius r, and at some instant, the particle&#8217;s speed is v.<\/p>\n\n<p style=\"text-align: center;\"><img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/figure-2-2-300x135.png\" alt=\"Diagram of a particle moving in a circle of radius r with speed v\"><\/p>\n\n<p>The origin may be chosen anywhere on the axis. We choose it at the center of the circle.<\/p>\n\n<p>Now, angular momentum of particle about centre,<\/p>\n\n<p>$ell = {bf{r}} times m{bf{v}} = r,hat i times mv,hat j = rmv,hat k$<\/p>\n\n<p>Also, ${bf{r}} times {bf{P}}$ is perpendicular to <strong>r<\/strong> and <strong>P<\/strong> and hence is along the axis. So, component of ${bf{r}} times {bf{P}}$ along the axis is mvr itself.<\/p>\n\n<p>Now, consider a rigid body rotating about an axis AB. Let the angular velocity of the rigid body be $omega $.<\/p>\n\n<p style=\"text-align: center;\"><img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/figure-3-2-300x242.png\" alt=\"Rigid body rotating about axis AB with angular velocity omega\"><\/p>\n\n<p>Consider the i<sup>th<\/sup> particle going in a circle of radius ${r_i}$ with its plane perpendicular to AB. The linear velocity of this particle at this instant is ${v_i} = {r_i}omega $.<\/p>\n\n<p>Now, the angular momentum about AB (for the considered particle):<\/p>\n\n<p>$ell = {{bf{r}}_i} times {m_i}{{bf{v}}_i},,;,,{{bf{v}}_i} = {bf{omega }} times {{bf{r}}_i}$<\/p>\n\n<p>$ell = {m_i}r_1^2{bf{omega }}$<\/p>\n\n<p>$sum ell = sum {m_i}r_1^2{bf{omega }}$<\/p>\n\n<p>${bf{L}} = {I_{sy}},,,axis,of,,{bf{omega }},,rotation$<\/p>\n\n<p>${bf{L}} = {I_{sy,,AB}},{bf{omega }}$<\/p>\n\n<p>Where I is the moment of inertia of the rigid body about axis AB.<\/p>\n\n<p>Understanding rotational dynamics connects naturally to computational modelling tools used in engineering. Students working with simulation software may find it useful to work with an <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/ansys\/\">ANSYS tutor<\/a> to apply these physical principles in a practical context.<\/p>\n\n<p>For a broader look at how online tutoring platforms support students across technical subjects, see this overview of the <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/top-10-online-tutoring-websites-worldwide\/\">top 10 online tutoring websites worldwide<\/a>.<\/p>\n\n<h2>Problems Based on Angular Momentum<\/h2>\n\n<h3>Example 1: Angular Momentum of a Rotating Disk<\/h3>\n\n<p><strong>The diameter of a disc is 1 m. It has a mass of 20 kg. It is rotating about its axis with a speed of 120 rotations in one minute. Its angular momentum is $kg &#8211; {m^2}{rm{\/}}s$ is __________.<\/strong><\/p>\n\n<p><strong>Ans.<\/strong> Here, the body rotates about an axis, so we will use $L = Iomega $.<\/p>\n\n<p>Here, $r = frac{1}{2}m$; mass = m = 20 kg;<\/p>\n\n<p>$omega = frac{{120 times 2pi r}}{{60 times r}}rad{rm{\/}}squad Rightarrow quad omega = 4pi ,rad{rm{\/}}s$<\/p>\n\n<p>$left| {bf{L}} right| = frac{1}{2}m{r^2}omega = frac{1}{2} times 20 times {left( {frac{1}{2}} right)^2} times 4pi kg &#8211; {m^2}{rm{\/}}s$<\/p>\n\n<p>$ Rightarrow quad L = 10pi = 31.4,kg &#8211; {m^2}{rm{\/}}s$<\/p>\n\n<h3>Example 2: Particle Moving in a Circle<\/h3>\n\n<p><strong>A particle of mass m is moving along a circle of radius r with a time period T. Its angular momentum is _____________.<\/strong><\/p>\n\n<p><strong>Ans.<\/strong> Since the particle is moving along a circle:<\/p>\n\n<p style=\"text-align: center;\"><img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/figure-4-1-300x300.png\" alt=\"Diagram of a particle of mass m moving in a circle of radius r with time period T\"><\/p>\n\n<p>$left| {{{bf{L}}_C}} right| = Iomega = m{r^2}omega = m{r^2}frac{{2pi }}{T}$<\/p>\n\n<p>$ Rightarrow quad {L_C} = frac{{2pi m{r^2}}}{T}$<\/p>\n\n<p>Quantitative reasoning skills developed through physics problems also transfer well to financial modelling. Students who want to strengthen those skills might consider working with a <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/finance\/\">finance tutor<\/a> to see how mathematical rigour applies across disciplines.<\/p>\n\n<h3>Example 3: Angular Momentum of a Projectile<\/h3>\n\n<p><strong>A particle is projected at time t = 0 from a point O with a speed u at an angle ${45^o}$ to horizontal. Find the angular momentum of the particle at time $frac{u}{g}$.<\/strong><\/p>\n\n<p><strong>Ans.<\/strong> Velocity of particle at time t is ${bf{v}} = {v_x}hat i + {v_y}hat j$. Position vector of particle at time t is ${bf{r}} = x,hat i + y,hat j$.<\/p>\n\n<p style=\"text-align: center;\"><img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/figure-5-1-300x225.png\" alt=\"Projectile motion diagram showing particle projected at 45 degrees from point O\"><\/p>\n\n<p>Hence, angular momentum of the particle about the origin,<\/p>\n\n<p>${{{bf{L}}_0} = {bf{r}} times m{bf{v}} = m({bf{r}} times {bf{v}}) = m(x,hat i + y,hat j) times ({v_x}hat i + {v_y}hat j)}$<\/p>\n\n<p>$ Rightarrow quad {{bf{L}}_0} = m(x{v_y} &#8211; y{v_x}),hat k$ \u2026(i)<\/p>\n\n<p>Here, ${v_x} = frac{u}{{sqrt 2 }},;$<\/p>\n\n<p>$x = {u_x}t + frac{1}{2}{a_x}{t^2} = frac{u}{{sqrt 2 }} times frac{u}{g} + 0 = frac{{{u^2}}}{{gsqrt 2 }}$<\/p>\n\n<p>${v_y} = frac{u}{{sqrt 2 }} &#8211; gfrac{u}{g} = left( {frac{{1 &#8211; sqrt 2 }}{{sqrt 2 }}} right)u$<\/p>\n\n<p>$y = usin {45^o}t &#8211; frac{1}{2}g{t^2} = frac{u}{{sqrt 2 }} times frac{u}{g} &#8211; frac{1}{2}g times frac{{{u^2}}}{{{g^2}}}$<\/p>\n\n<p>Now putting values of $x,,{v_y},,y$ and ${v_x}$ into equation (i),<\/p>\n\n<p>${{{bf{L}}_0} = frac{{m{u^3}}}{{2sqrt 2 g}}( &#8211; hat k)}$<\/p>\n\n<p>Applying physics principles to real-world data analysis is a skill that extends into fields like geographic information systems. Students exploring spatial modelling may find it helpful to work with an <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/arcgis-and-arcgis-pro\/\">ArcGIS tutor<\/a> to connect quantitative thinking with applied tools.<\/p>\n\n<p>For students navigating complex academic subjects, structured tutoring support can make a significant difference. This guide on <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/unlocking-your-potential-how-business-administration-tutoring-can-shape-your-future\/\">how business administration tutoring can shape your future<\/a> explores how personalised learning applies across disciplines.<\/p>\n\n<p>Those interested in how cognitive science intersects with learning may also find value in this <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/unlocking-the-brains-secrets-a-deep-dive-into-the-world-of-biopsychology\/\">deep dive into the world of biopsychology<\/a>.<\/p>\n\n<p>If you are comparing tutoring platforms, this <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/extramarks-reviews-alternatives-pricing-offerings-in-2025\/\">Extramarks review covering alternatives, pricing, and offerings<\/a> provides a useful reference.<\/p>\n\n<p>Students who want to strengthen their understanding of financial reporting alongside physics and engineering coursework can also explore working with a <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/financial-accounting\/\">financial accounting tutor<\/a> to build complementary analytical skills.<\/p>\n\n<h2>Related Reading<\/h2>\n\n<ul>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/cracking-the-code-a-medical-students-ultimate-guide-to-conquering-the-usmle\/\">A Medical Student&#8217;s Ultimate Guide to Conquering the USMLE<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/cbt-nuggets-reviews-alternatives-pricing-offerings\/\">CBT Nuggets: Reviews, Alternatives, Pricing and Offerings<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/linkedin-learning-reviews-alternatives-pricing-offerings\/\">LinkedIn Learning: Reviews, Alternatives, Pricing and Offerings<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/benefits-of-getting-calculus-tutor\/\">Benefits of Getting a Calculus Tutor<\/a><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Key Takeaways Angular momentum equals the moment of linear momentum  [&#8230;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-11300","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11300","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/comments?post=11300"}],"version-history":[{"count":1,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11300\/revisions"}],"predecessor-version":[{"id":12176,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11300\/revisions\/12176"}],"wp:attachment":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/media?parent=11300"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/categories?post=11300"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/tags?post=11300"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}