System of Particles and Rotational Motion Class 11 Notes | NEET Physics<\/title>\n<meta name=\"description\" content=\"Complete System of Particles and Rotational Motion class 11 notes for NEET Physics. 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margin: 0; color: #7c2d12; }\n\n ul.content-list {\n padding-left: 0;\n list-style: none;\n margin: 14px 0;\n display: grid;\n gap: 8px;\n }\n ul.content-list li {\n padding-left: 22px;\n position: relative;\n font-size: 15px;\n }\n ul.content-list li::before {\n content: '';\n position: absolute;\n left: 0;\n top: 9px;\n width: 8px;\n height: 8px;\n background: var(--accent);\n border-radius: 50%;\n }\n\n .two-col { display: grid; grid-template-columns: 1fr 1fr; gap: 20px; margin: 20px 0; }\n @media (max-width: 640px) { .two-col { grid-template-columns: 1fr; } .cta-section { padding: 48px 0; } }\n\n a { color: var(--accent); }\n strong { font-weight: 600; }\n\n .section-divider {\n height: 4px;\n background: linear-gradient(90deg, var(--accent) 0%, transparent 100%);\n border: none;\n margin: 0;\n }\n\n .pyq-badge {\n display: inline-block;\n background: #fef3c7;\n border: 1px solid #fbbf24;\n color: #92400e;\n font-size: 11px;\n font-weight: 700;\n padding: 2px 8px;\n border-radius: 4px;\n font-family: 'Plus Jakarta Sans', sans-serif;\n letter-spacing: 0.5px;\n text-transform: uppercase;\n margin-right: 8px;\n }\n<\/style>\n<\/head>\n<body>\n<article>\n\n\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">01<\/div>\n <h2 class=\"section-title\">Introduction to <span>System of Particles<\/span><\/h2>\n <\/div>\n\n <p>The <strong>System of Particles and Rotational Motion class 11 notes<\/strong> begin with a foundational shift in perspective: instead of treating a body as a single point mass, we now consider an extended body as a collection of particles, each obeying Newton’s laws individually. Together, their collective behaviour determines the motion of the entire system. This chapter is one of the most concept-dense and formula-rich chapters in Class 11 Physics, and it carries consistent weight in NEET.<\/p>\n\n <p>A <strong>system of particles<\/strong> is any group of two or more particles whose interactions and collective motion are analyzed as a whole. The particles may be bound together (rigid body) or free (like gas molecules). For NEET, the focus is on rigid bodies \u2014 bodies where the distance between any two particles remains constant regardless of the forces applied.<\/p>\n\n <div class=\"highlight-grid\">\n <div class=\"highlight-card\">\n <div class=\"hc-title\">Rigid Body<\/div>\n <p>Inter-particle distances are fixed. No deformation under applied forces.<\/p>\n <\/div>\n <div class=\"highlight-card\">\n <div class=\"hc-title\">Types of Motion<\/div>\n <p>Pure translation, pure rotation, or combined rolling motion.<\/p>\n <\/div>\n <div class=\"highlight-card\">\n <div class=\"hc-title\">Why This Matters<\/div>\n <p>Unifies force, momentum, energy concepts under one mechanical framework.<\/p>\n <\/div>\n <\/div>\n\n <p>Key distinction: in <strong>translational motion<\/strong>, every point of the body has the same velocity and acceleration at any instant. In <strong>rotational motion<\/strong>, every point moves in a circular path about the axis of rotation, with angular velocity and angular acceleration defining the state of motion.<\/p>\n\n <div class=\"related-links\">\n <h4>Related Resources<\/h4>\n <ul>\n <li><a href=\"https:\/\/ksquareinstitute.in\/blog\/neet-physics-survival-kit-2026\/\">NEET Physics Survival Kit<\/a><\/li>\n <li><a href=\"https:\/\/ksquareinstitute.in\/blog\/organic-chemistry-strategy-neet\/\">Organic Chemistry Strategy<\/a><\/li>\n <li><a href=\"https:\/\/ksquareinstitute.in\/blog\/neet-biology-tricks-for-exams\/\">NEET Biology Tricks for Exams<\/a><\/li>\n <li><a href=\"https:\/\/ksquareinstitute.in\/blog\/score-340-in-neet-biology\/\">How to Score 340+ in NEET Biology<\/a><\/li>\n <li><a href=\"https:\/\/ksquareinstitute.in\/blog\/top-10-tricky-neet-biology-diagrams\/\">Top 10 Tricky NEET Biology Diagrams<\/a><\/li>\n <li><a href=\"https:\/\/ksquareinstitute.in\/free-study-material\/\">Free Study Materials<\/a><\/li>\n <\/ul>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">02<\/div>\n <h2 class=\"section-title\">Centre of <span>Mass (COM)<\/span><\/h2>\n <\/div>\n\n <p>The <strong>centre of mass<\/strong> is a uniquely defined point in a system where the entire mass of the system can be assumed concentrated for the purpose of analyzing translational motion. It is not necessarily at the geometric centre; it depends entirely on the mass distribution.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">COM for Discrete System<\/div>\n <code>x_cm = (m\u2081x\u2081 + m\u2082x\u2082 + ... + m\u2099x\u2099) \/ (m\u2081 + m\u2082 + ... + m\u2099)<\/code>\n <code>r_cm = \u03a3(m\u1d62 r\u1d62) \/ M where M = total mass<\/code>\n <\/div>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">COM for Continuous Distribution<\/div>\n <code>r_cm = (1\/M) \u222b r dm<\/code>\n <code>x_cm = (1\/M) \u222b x dm , y_cm = (1\/M) \u222b y dm<\/code>\n <\/div>\n\n <p>For symmetric uniform bodies, the COM lies at the geometric centre. Key results to memorize:<\/p>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Body<\/th>\n <th>COM Location<\/th>\n <th>Notes<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td>Uniform Rod<\/td>\n <td>Midpoint (L\/2 from either end)<\/td>\n <td>Symmetric body<\/td>\n <\/tr>\n <tr>\n <td>Uniform Disc \/ Ring<\/td>\n <td>Geometric centre<\/td>\n <td>COM lies at centre even for hollow ring<\/td>\n <\/tr>\n <tr>\n <td>Uniform Solid Sphere<\/td>\n <td>Geometric centre<\/td>\n <td>\u2014<\/td>\n <\/tr>\n <tr>\n <td>Semicircular Ring (radius R)<\/td>\n <td>2R\/\u03c0 from diameter<\/td>\n <td>Asymmetric; COM off centre<\/td>\n <\/tr>\n <tr>\n <td>Semicircular Disc (radius R)<\/td>\n <td>4R\/(3\u03c0) from diameter<\/td>\n <td>Standard NEET result<\/td>\n <\/tr>\n <tr>\n <td>Solid Hemisphere (radius R)<\/td>\n <td>3R\/8 from flat face<\/td>\n <td>High NEET frequency<\/td>\n <\/tr>\n <tr>\n <td>Hollow Hemisphere (radius R)<\/td>\n <td>R\/2 from flat face<\/td>\n <td>\u2014<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n\n <div class=\"callout callout-tip\">\n <span class=\"callout-icon\">Tip<\/span>\n <p>For a system with a cavity (material removed), use the concept of negative mass. Treat the removed portion as a particle of negative mass at its original COM location and apply the COM formula for the remaining system.<\/p>\n <\/div>\n<\/div>\n\n\n<a href=\"https:\/\/courses.ksquare.co.in\/new-courses\/3-mission-180-neet-physics-rankers-batch\" target=\"_blank\" rel=\"nofollow noopener noreferrer\" style=\"display:block; margin-bottom:20px;\">\n <img decoding=\"async\" src=\"https:\/\/ksquareinstitute.in\/blog\/wp-content\/uploads\/2026\/03\/Course-Poromo-Banner-scaled.png\" alt=\"Mission 180 NEET Physics Rankers Batch - KSquare Career Institute\" style=\"width:100%; height:auto; border-radius:10px; display:block;\">\n<\/a>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">03<\/div>\n <h2 class=\"section-title\">Motion of <span>Centre of Mass<\/span><\/h2>\n <\/div>\n\n <p>The most powerful result in the study of a system of particles: <strong>the centre of mass of a system moves as if all external forces act on a single particle of mass equal to the total mass of the system, located at the COM.<\/strong><\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Equation of Motion for COM<\/div>\n <code>F_ext = M \u00d7 a_cm<\/code>\n <code>where a_cm = d\u00b2r_cm\/dt\u00b2 and M = total mass of system<\/code>\n <\/div>\n\n <p><strong>Internal forces<\/strong> \u2014 forces between particles within the system \u2014 always appear in action-reaction pairs and cancel out when summed over the entire system (Newton’s Third Law). They do not affect the motion of the COM; only <strong>external forces<\/strong> determine how the COM accelerates.<\/p>\n\n <div class=\"callout callout-warn\">\n <span class=\"callout-icon\">Warning<\/span>\n <p>In an explosion or collision, even if the object shatters into many pieces, the COM continues on the same parabolic or linear trajectory as before \u2014 because no external horizontal force acts. This is a classic NEET trap question.<\/p>\n <\/div>\n\n <p>Practical application: A bomb thrown as a projectile follows a parabolic path. When it explodes mid-air, the fragments scatter in all directions, but the COM of all fragments continues along the original parabola. If one fragment lands at a known position, the other fragment’s landing position can be located using COM principles.<\/p>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">04<\/div>\n <h2 class=\"section-title\">Linear Momentum of <span>a System<\/span><\/h2>\n <\/div>\n\n <p>The total linear momentum of a system of particles is the vector sum of the momenta of all individual particles:<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Total Linear Momentum<\/div>\n <code>P = \u03a3 p\u1d62 = \u03a3 m\u1d62v\u1d62 = M \u00d7 v_cm<\/code>\n <code>The total momentum equals total mass times velocity of COM.<\/code>\n <\/div>\n\n <p>Differentiating this with respect to time: <strong>dP\/dt = F_ext<\/strong>. This is Newton’s second law for a system of particles. When no external force acts, dP\/dt = 0, which leads directly to conservation of linear momentum.<\/p>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Impulse\u2013Momentum Theorem for System<\/div>\n <code>J = F_ext \u00d7 \u0394t = \u0394P = M(v_cm,f \u2212 v_cm,i)<\/code>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">05<\/div>\n <h2 class=\"section-title\">Conservation of <span>Linear Momentum<\/span><\/h2>\n <\/div>\n\n <p>One of the most fundamental principles in physics: <strong>if the net external force on a system is zero, the total linear momentum of the system remains constant.<\/strong><\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Conservation Condition<\/div>\n <code>F_ext = 0 \u27f9 P_total = constant<\/code>\n <code>m\u2081u\u2081 + m\u2082u\u2082 = m\u2081v\u2081 + m\u2082v\u2082 (for two-particle system)<\/code>\n <\/div>\n\n <p>Critical applications for NEET:<\/p>\n\n <ul class=\"content-list\">\n <li><strong>Recoil of a gun:<\/strong> Before firing, total momentum = 0. After firing, bullet and gun have equal and opposite momenta. m_gun \u00d7 v_gun = m_bullet \u00d7 v_bullet.<\/li>\n <li><strong>Rocket propulsion:<\/strong> Exhaust gases ejected backward give the rocket forward momentum. No external force needed \u2014 conservation applies within the system.<\/li>\n <li><strong>Elastic collision:<\/strong> Both kinetic energy and momentum are conserved. For equal masses, velocities exchange.<\/li>\n <li><strong>Perfectly inelastic collision:<\/strong> Momentum conserved; maximum kinetic energy lost. Bodies stick together: (m\u2081 + m\u2082)v = m\u2081u\u2081 + m\u2082u\u2082.<\/li>\n <\/ul>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Elastic Collision (1D) \u2014 Velocity Formulae<\/div>\n <code>v\u2081 = [(m\u2081 \u2212 m\u2082)\/(m\u2081 + m\u2082)]u\u2081 + [2m\u2082\/(m\u2081 + m\u2082)]u\u2082<\/code>\n <code>v\u2082 = [2m\u2081\/(m\u2081 + m\u2082)]u\u2081 + [(m\u2082 \u2212 m\u2081)\/(m\u2081 + m\u2082)]u\u2082<\/code>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">06<\/div>\n <h2 class=\"section-title\">Introduction to <span>Rotational Motion<\/span><\/h2>\n <\/div>\n\n <p>When a rigid body rotates about a fixed axis, every particle moves in a circle. The axis of rotation may pass through the body (spin) or be external (revolution). Understanding <strong>rotational motion<\/strong> is central to mastering System of Particles and Rotational Motion class 11 notes for NEET.<\/p>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Translational Quantity<\/th>\n <th>Rotational Analogue<\/th>\n <th>Symbol<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td>Displacement (s)<\/td>\n <td>Angular displacement<\/td>\n <td>\u03b8 (radians)<\/td>\n <\/tr>\n <tr>\n <td>Velocity (v)<\/td>\n <td>Angular velocity<\/td>\n <td>\u03c9 = d\u03b8\/dt (rad\/s)<\/td>\n <\/tr>\n <tr>\n <td>Acceleration (a)<\/td>\n <td>Angular acceleration<\/td>\n <td>\u03b1 = d\u03c9\/dt (rad\/s\u00b2)<\/td>\n <\/tr>\n <tr>\n <td>Mass (m)<\/td>\n <td>Moment of inertia<\/td>\n <td>I (kg\u00b7m\u00b2)<\/td>\n <\/tr>\n <tr>\n <td>Force (F)<\/td>\n <td>Torque<\/td>\n <td>\u03c4 = r \u00d7 F (N\u00b7m)<\/td>\n <\/tr>\n <tr>\n <td>Linear momentum (p)<\/td>\n <td>Angular momentum<\/td>\n <td>L = I\u03c9 (kg\u00b7m\u00b2\/s)<\/td>\n <\/tr>\n <tr>\n <td>Kinetic energy (\u00bdmv\u00b2)<\/td>\n <td>Rotational KE<\/td>\n <td>\u00bdI\u03c9\u00b2<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n<\/div>\n\n\n<a href=\"https:\/\/ksquareinstitute.in\/neet-2026-rank-predictor\/\" target=\"_blank\" rel=\"nofollow noopener noreferrer\" style=\"display:block; margin-bottom:20px;\">\n <img decoding=\"async\" src=\"https:\/\/ksquareinstitute.in\/blog\/wp-content\/uploads\/2026\/03\/neet-2026-college-and-rank-predictor-scaled.png\" alt=\"NEET 2026 College and Rank Predictor - KSquare Career Institute\" style=\"width:100%; height:auto; border-radius:10px; display:block;\">\n<\/a>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">07<\/div>\n <h2 class=\"section-title\">Angular Variables and <span>Kinematics<\/span><\/h2>\n <\/div>\n\n <p>The kinematic equations for <strong>uniform angular acceleration<\/strong> are exact analogues of the linear kinematic equations. Mastering the parallel structure makes both sets trivially easy.<\/p>\n\n <div class=\"two-col\">\n <div class=\"formula-dark\">\n <div class=\"f-label\">Linear Kinematics<\/div>\n <code>v = u + at<\/code>\n <code>s = ut + \u00bdat\u00b2<\/code>\n <code>v\u00b2 = u\u00b2 + 2as<\/code>\n <\/div>\n <div class=\"formula-dark\">\n <div class=\"f-label\">Rotational Kinematics<\/div>\n <code>\u03c9 = \u03c9\u2080 + \u03b1t<\/code>\n <code>\u03b8 = \u03c9\u2080t + \u00bd\u03b1t\u00b2<\/code>\n <code>\u03c9\u00b2 = \u03c9\u2080\u00b2 + 2\u03b1\u03b8<\/code>\n <\/div>\n <\/div>\n\n <p>Relationship between linear and angular variables for a particle at radius r from axis:<\/p>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Linear\u2013Angular Relationships<\/div>\n <code>v = r\u03c9 (tangential speed)<\/code>\n <code>a\u209c = r\u03b1 (tangential acceleration)<\/code>\n <code>a\u1d9c = r\u03c9\u00b2 = v\u00b2\/r (centripetal acceleration)<\/code>\n <\/div>\n\n <div class=\"callout callout-tip\">\n <span class=\"callout-icon\">Tip<\/span>\n <p>Angular displacement \u03b8 must always be in radians (not degrees) when using these kinematic equations. Convert degrees to radians: \u03b8(rad) = \u03b8(\u00b0) \u00d7 \u03c0\/180.<\/p>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">08<\/div>\n <h2 class=\"section-title\">Moment of <span>Inertia<\/span><\/h2>\n <\/div>\n\n <p><strong>Moment of inertia (I)<\/strong> is the rotational analogue of mass. It measures a body’s resistance to angular acceleration. Unlike mass, I depends on both the total mass and its distribution relative to the axis of rotation.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Definition<\/div>\n <code>I = \u03a3 m\u1d62r\u1d62\u00b2 (discrete system)<\/code>\n <code>I = \u222b r\u00b2 dm (continuous body)<\/code>\n <code>Unit: kg\u00b7m\u00b2 Dimensional formula: [ML\u00b2]<\/code>\n <\/div>\n\n <p><strong>Radius of gyration (K):<\/strong> An equivalent radius at which the entire mass can be concentrated to give the same moment of inertia: <strong>I = MK\u00b2<\/strong>, so K = \u221a(I\/M).<\/p>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Body<\/th>\n <th>Axis<\/th>\n <th>Moment of Inertia (I)<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td>Thin Ring (mass M, radius R)<\/td>\n <td>Through centre, perpendicular to plane<\/td>\n <td>MR\u00b2<\/td>\n <\/tr>\n <tr>\n <td>Thin Ring<\/td>\n <td>Diameter<\/td>\n <td>MR\u00b2\/2<\/td>\n <\/tr>\n <tr>\n <td>Solid Disc (mass M, radius R)<\/td>\n <td>Through centre, perpendicular<\/td>\n <td>MR\u00b2\/2<\/td>\n <\/tr>\n <tr>\n <td>Solid Disc<\/td>\n <td>Diameter<\/td>\n <td>MR\u00b2\/4<\/td>\n <\/tr>\n <tr>\n <td>Thin Rod (length L)<\/td>\n <td>Through centre, perpendicular to rod<\/td>\n <td>ML\u00b2\/12<\/td>\n <\/tr>\n <tr>\n <td>Thin Rod (length L)<\/td>\n <td>Through one end, perpendicular<\/td>\n <td>ML\u00b2\/3<\/td>\n <\/tr>\n <tr>\n <td>Solid Sphere (radius R)<\/td>\n <td>Through diameter<\/td>\n <td>2MR\u00b2\/5<\/td>\n <\/tr>\n <tr>\n <td>Hollow Sphere (radius R)<\/td>\n <td>Through diameter<\/td>\n <td>2MR\u00b2\/3<\/td>\n <\/tr>\n <tr>\n <td>Solid Cylinder (radius R)<\/td>\n <td>Geometric axis<\/td>\n <td>MR\u00b2\/2<\/td>\n <\/tr>\n <tr>\n <td>Hollow Cylinder (radius R)<\/td>\n <td>Geometric axis<\/td>\n <td>MR\u00b2<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n\n <div class=\"callout callout-warn\">\n <span class=\"callout-icon\">Warning<\/span>\n <p>I is not invariant \u2014 it changes with the axis chosen. The same disc has I = MR\u00b2\/2 about its central axis but I = MR\u00b2\/4 about a diameter. Always specify the axis before quoting a value.<\/p>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">09<\/div>\n <h2 class=\"section-title\">Theorems of <span>Moment of Inertia<\/span><\/h2>\n <\/div>\n\n <p>Two powerful theorems allow calculation of I about any axis if the I about a standard axis is known.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Parallel Axis Theorem<\/div>\n <code>I = I_cm + Md\u00b2<\/code>\n <code>I_cm = MOI about axis through COM (parallel to new axis)<\/code>\n <code>d = perpendicular distance between the two axes<\/code>\n <\/div>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Perpendicular Axis Theorem (Planar Bodies Only)<\/div>\n <code>I_z = I_x + I_y<\/code>\n <code>Applies only to laminar (2D) bodies.<\/code>\n <code>z-axis is perpendicular to the plane of the body.<\/code>\n <\/div>\n\n <div class=\"callout callout-tip\">\n <span class=\"callout-icon\">Tip<\/span>\n <p>The perpendicular axis theorem is restricted to flat (laminar) objects. Do not apply it to 3D bodies like spheres or cylinders. The parallel axis theorem, however, applies universally to all bodies \u2014 2D or 3D.<\/p>\n <\/div>\n\n <p>Classic NEET application: For a disc, I about diameter = MR\u00b2\/4. By perpendicular axis theorem, I about central perpendicular axis = I_x + I_y = MR\u00b2\/4 + MR\u00b2\/4 = MR\u00b2\/2. This confirms the standard result.<\/p>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">10<\/div>\n <h2 class=\"section-title\">Torque and <span>Angular Momentum<\/span><\/h2>\n <\/div>\n\n <p><strong>Torque (\u03c4)<\/strong> is the rotational analogue of force. It is the tendency of a force to produce rotation about an axis. Torque depends not just on the magnitude of the force but also on the point of application and direction.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Torque<\/div>\n <code>\u03c4 = r \u00d7 F (vector cross product)<\/code>\n <code>|\u03c4| = rF sin \u03b8 where \u03b8 = angle between r and F<\/code>\n <code>\u03c4 = I \u00d7 \u03b1 (Newton's 2nd law for rotation)<\/code>\n <\/div>\n\n <p><strong>Angular momentum (L)<\/strong> is the rotational analogue of linear momentum:<\/p>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Angular Momentum<\/div>\n <code>L = r \u00d7 p = r \u00d7 mv (for a particle)<\/code>\n <code>L = I\u03c9 (for a rigid body about fixed axis)<\/code>\n <code>\u03c4 = dL\/dt (analogous to F = dp\/dt)<\/code>\n <\/div>\n\n <div class=\"callout callout-warn\">\n <span class=\"callout-icon\">Warning<\/span>\n <p>Torque and angular momentum are both axial vectors (direction given by right-hand rule). In NEET numerical problems, always confirm that the axis of rotation is fixed before applying L = I\u03c9. If the axis changes, the problem is significantly more complex.<\/p>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">11<\/div>\n <h2 class=\"section-title\">Conservation of <span>Angular Momentum<\/span><\/h2>\n <\/div>\n\n <p>When no external torque acts on a system, the total angular momentum remains constant. This is one of the most beautiful and practically significant principles in classical mechanics.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Conservation of Angular Momentum<\/div>\n <code>\u03c4_ext = 0 \u27f9 L = I\u03c9 = constant<\/code>\n <code>I\u2081\u03c9\u2081 = I\u2082\u03c9\u2082 (before and after any internal change)<\/code>\n <\/div>\n\n <p>Classic applications and NEET examples:<\/p>\n\n <ul class=\"content-list\">\n <li><strong>Ballet dancer \/ spinning skater:<\/strong> Pulls arms inward \u2192 I decreases \u2192 \u03c9 increases (spins faster). Extends arms \u2192 I increases \u2192 \u03c9 decreases.<\/li>\n <li><strong>Diver tucks in mid-air:<\/strong> Reduces I, increases \u03c9 to complete more rotations before hitting water.<\/li>\n <li><strong>Collapsing star (pulsar):<\/strong> As a dying star collapses, its radius decreases enormously \u2192 I decreases \u2192 \u03c9 increases dramatically, producing the characteristic rapid spin of neutron stars.<\/li>\n <li><strong>Planet in elliptical orbit:<\/strong> Closer to sun (perihelion) \u2192 moves faster; farther (aphelion) \u2192 moves slower. L = mvr = constant (Kepler’s second law).<\/li>\n <\/ul>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">12<\/div>\n <h2 class=\"section-title\">Rotational <span>Dynamics<\/span><\/h2>\n <\/div>\n\n <p><strong>Newton’s second law for rotation:<\/strong> The net external torque acting on a rigid body equals the product of its moment of inertia and angular acceleration.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Newton’s 2nd Law (Rotational Form)<\/div>\n <code>\u03c4_net = I \u00d7 \u03b1<\/code>\n <code>This is the exact analogue of F_net = ma<\/code>\n <\/div>\n\n <p>When both translational and rotational motions occur simultaneously (e.g., a cylinder rolling down an incline), two equations are written:<\/p>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Combined Dynamics (e.g., Rolling on Inclined Plane)<\/div>\n <code>Translational: Mg sin \u03b8 \u2212 f = Ma_cm<\/code>\n <code>Rotational: f \u00d7 R = I \u00d7 \u03b1 = I \u00d7 (a_cm\/R)<\/code>\n <code>Solving: a_cm = g sin \u03b8 \/ (1 + I\/MR\u00b2)<\/code>\n <\/div>\n\n <p>This result shows that a solid sphere (I = 2MR\u00b2\/5) accelerates faster down an incline than a hollow sphere (I = 2MR\u00b2\/3), which in turn accelerates faster than a ring (I = MR\u00b2). The more the mass is concentrated at the rim, the slower the acceleration.<\/p>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">13<\/div>\n <h2 class=\"section-title\">Work Done in <span>Rotational Motion<\/span><\/h2>\n <\/div>\n\n <p>The work\u2013energy theorem extends naturally to rotational systems. Work done by a torque over an angular displacement is:<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Work by Torque<\/div>\n <code>W = \u222b \u03c4 d\u03b8<\/code>\n <code>For constant torque: W = \u03c4 \u00d7 \u03b8<\/code>\n <code>Power: P = \u03c4 \u00d7 \u03c9 (analogous to P = F \u00d7 v)<\/code>\n <\/div>\n\n <div class=\"callout callout-tip\">\n <span class=\"callout-icon\">Tip<\/span>\n <p>For NEET problems involving work done against friction torque in rotational motion, apply work-energy theorem directly: Work done by all torques = Change in rotational KE = \u00bdI(\u03c9\u2082\u00b2 \u2212 \u03c9\u2081\u00b2).<\/p>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">14<\/div>\n <h2 class=\"section-title\">Rotational <span>Kinetic Energy<\/span><\/h2>\n <\/div>\n\n <p>A rotating body possesses kinetic energy by virtue of its rotation, distinct from (and additive to) any translational kinetic energy it may have.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Rotational Kinetic Energy<\/div>\n <code>KE_rot = \u00bd I \u03c9\u00b2<\/code>\n <code>Compare with translational: KE_trans = \u00bd mv\u00b2<\/code>\n <code>Both have the same mathematical structure: \u00bd \u00d7 (inertia) \u00d7 (velocity)\u00b2<\/code>\n <\/div>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Work\u2013Energy Theorem (Rotational)<\/div>\n <code>W_net = \u0394KE_rot = \u00bdI(\u03c9_f\u00b2 \u2212 \u03c9_i\u00b2)<\/code>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">15<\/div>\n <h2 class=\"section-title\">Rolling <span>Motion<\/span><\/h2>\n <\/div>\n\n <p><strong>Rolling motion<\/strong> is the superposition of pure translation of the COM and pure rotation about the COM. For pure rolling without slipping, there is a crucial constraint linking linear and angular quantities.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Pure Rolling Condition (No Slipping)<\/div>\n <code>v_cm = R \u00d7 \u03c9 (velocity constraint)<\/code>\n <code>a_cm = R \u00d7 \u03b1 (acceleration constraint)<\/code>\n <code>The contact point P has zero instantaneous velocity.<\/code>\n <\/div>\n\n <p>Velocity at different points on a rolling body:<\/p>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Point on Body<\/th>\n <th>Velocity<\/th>\n <th>Magnitude<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td>Contact point (bottom)<\/td>\n <td>v_cm \u2212 R\u03c9<\/td>\n <td>Zero (pure rolling)<\/td>\n <\/tr>\n <tr>\n <td>Centre of mass<\/td>\n <td>v_cm (horizontal)<\/td>\n <td>v_cm<\/td>\n <\/tr>\n <tr>\n <td>Top point<\/td>\n <td>v_cm + R\u03c9<\/td>\n <td>2v_cm<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n\n <div class=\"callout callout-warn\">\n <span class=\"callout-icon\">Warning<\/span>\n <p>The topmost point of a rolling body moves at twice the speed of the centre of mass. This frequently appears in NEET questions as a conceptual trap \u2014 students often forget the rotational contribution at the top point.<\/p>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">16<\/div>\n <h2 class=\"section-title\">Energy in <span>Rolling Motion<\/span><\/h2>\n <\/div>\n\n <p>The total mechanical energy of a rolling body is the sum of its translational and rotational kinetic energies. This distinction is critical for inclined plane problems.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Total Kinetic Energy in Rolling<\/div>\n <code>KE_total = KE_trans + KE_rot<\/code>\n <code> = \u00bdmv_cm\u00b2 + \u00bdI\u03c9\u00b2<\/code>\n <code> = \u00bdmv_cm\u00b2(1 + I\/mR\u00b2) [using v_cm = R\u03c9]<\/code>\n <code> = \u00bdmv_cm\u00b2(1 + K\u00b2\/R\u00b2) [using I = mK\u00b2]<\/code>\n <\/div>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Rolling Body<\/th>\n <th>I\/MR\u00b2<\/th>\n <th>KE_rot\/KE_total<\/th>\n <th>v at bottom (from height h)<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td>Ring \/ Hollow Cylinder<\/td>\n <td>1<\/td>\n <td>50%<\/td>\n <td>\u221a(gh)<\/td>\n <\/tr>\n <tr>\n <td>Solid Cylinder \/ Disc<\/td>\n <td>1\/2<\/td>\n <td>33%<\/td>\n <td>\u221a(4gh\/3)<\/td>\n <\/tr>\n <tr>\n <td>Hollow Sphere<\/td>\n <td>2\/3<\/td>\n <td>40%<\/td>\n <td>\u221a(6gh\/5)<\/td>\n <\/tr>\n <tr>\n <td>Solid Sphere<\/td>\n <td>2\/5<\/td>\n <td>29%<\/td>\n <td>\u221a(10gh\/7)<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n\n <p>The solid sphere reaches the bottom with the highest velocity because the least fraction of energy goes into rotation. The ring is slowest for the same reason in reverse.<\/p>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">17<\/div>\n <h2 class=\"section-title\">Numerical <span>Framework<\/span><\/h2>\n <\/div>\n\n <p>NEET-level numericals in System of Particles and Rotational Motion class 11 notes fall into predictable categories. Here is the solution strategy for each type:<\/p>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Problem Type<\/th>\n <th>Key Equation(s)<\/th>\n <th>Watch Out For<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td>Finding COM of composite body<\/td>\n <td>r_cm = \u03a3m\u1d62r\u1d62 \/ \u03a3m\u1d62<\/td>\n <td>Use negative mass for cavities<\/td>\n <\/tr>\n <tr>\n <td>COM after explosion\/collision<\/td>\n <td>COM trajectory unchanged (no ext. force)<\/td>\n <td>Internal forces do not shift COM<\/td>\n <\/tr>\n <tr>\n <td>Moment of inertia problems<\/td>\n <td>I = Icm + Md\u00b2 (parallel axis)<\/td>\n <td>Choose correct standard I first<\/td>\n <\/tr>\n <tr>\n <td>Torque and angular acceleration<\/td>\n <td>\u03c4 = I\u03b1<\/td>\n <td>Net torque only; check direction<\/td>\n <\/tr>\n <tr>\n <td>Conservation of angular momentum<\/td>\n <td>I\u2081\u03c9\u2081 = I\u2082\u03c9\u2082<\/td>\n <td>No external torque condition must hold<\/td>\n <\/tr>\n <tr>\n <td>Rolling on incline<\/td>\n <td>a = g sin\u03b8 \/ (1 + I\/MR\u00b2)<\/td>\n <td>Friction does no work in pure rolling<\/td>\n <\/tr>\n <tr>\n <td>Rolling energy conservation<\/td>\n <td>mgh = \u00bdmv\u00b2(1 + K\u00b2\/R\u00b2)<\/td>\n <td>Use K\u00b2\/R\u00b2 for that specific body<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Quick Reference \u2014 Inclined Plane Race<\/div>\n <code>Order of reaching bottom (fastest first):<\/code>\n <code>Solid Sphere > Solid Cylinder > Hollow Sphere > Ring<\/code>\n <code>Smaller I\/MR\u00b2 ratio \u27f9 higher bottom speed \u27f9 wins the race<\/code>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">18<\/div>\n <h2 class=\"section-title\">Conceptual <span>Questions<\/span><\/h2>\n <\/div>\n\n <p>Conceptual questions in System of Particles and Rotational Motion class 11 notes for NEET test the depth of understanding, not just formula recall. These types appear regularly in the NEET MCQ paper:<\/p>\n\n <ul class=\"content-list\">\n <li>Can the COM of a body lie outside the body? Yes \u2014 hollow ring, horseshoe, donut-shaped body.<\/li>\n <li>A body can have angular momentum even without rotating about its own axis \u2014 any particle moving in a straight line has angular momentum about a point not on that line (L = mvr).<\/li>\n <li>Friction is essential for rolling \u2014 without friction, a body would slide, not roll. However, friction does no work during pure rolling.<\/li>\n <li>A spinning top exhibits gyroscopic precession \u2014 the gravitational torque causes the angular momentum vector to precess, not fall.<\/li>\n <li>If a person walks from the rim of a rotating platform toward the centre, the platform speeds up (L conserved, I decreases).<\/li>\n <li>Two identical cylinders (solid and hollow) released simultaneously from the top of an identical incline \u2014 the solid cylinder always reaches the bottom first, regardless of mass or radius.<\/li>\n <\/ul>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">19<\/div>\n <h2 class=\"section-title\">PYQ <span>Trends<\/span><\/h2>\n <\/div>\n\n <p>Analysis of previous year NEET questions reveals consistent high-frequency areas within this chapter:<\/p>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Topic<\/th>\n <th>Frequency<\/th>\n <th>Type<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td><span class=\"pyq-badge\">Very High<\/span>Moment of Inertia (standard bodies + theorems)<\/td>\n <td>3\u20134 questions\/year<\/td>\n <td>Numerical + Conceptual<\/td>\n <\/tr>\n <tr>\n <td><span class=\"pyq-badge\">High<\/span>Rolling Motion (energy, velocity at bottom)<\/td>\n <td>2\u20133 questions\/year<\/td>\n <td>Numerical<\/td>\n <\/tr>\n <tr>\n <td><span class=\"pyq-badge\">High<\/span>Conservation of Angular Momentum<\/td>\n <td>1\u20132 questions\/year<\/td>\n <td>Conceptual + Numerical<\/td>\n <\/tr>\n <tr>\n <td><span class=\"pyq-badge\">High<\/span>Torque and Rotational Dynamics<\/td>\n <td>1\u20132 questions\/year<\/td>\n <td>Numerical<\/td>\n <\/tr>\n <tr>\n <td><span class=\"pyq-badge\">Moderate<\/span>Centre of Mass (COM location, motion)<\/td>\n <td>1 question\/year<\/td>\n <td>Conceptual<\/td>\n <\/tr>\n <tr>\n <td><span class=\"pyq-badge\">Moderate<\/span>Conservation of Linear Momentum (collisions)<\/td>\n <td>1 question\/year<\/td>\n <td>Numerical<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n\n <div class=\"callout callout-tip\">\n <span class=\"callout-icon\">Tip<\/span>\n <p>NEET PYQ pattern shows that rolling motion questions have appeared every single year since 2010 without exception. Master the I\/MR\u00b2 ratios for all standard bodies and the energy partition formula. This alone can secure 8\u201312 marks in NEET Physics.<\/p>\n <\/div>\n\n <div class=\"pdf-btn-wrap\">\n <a href=\"#\" rel=\"nofollow noopener noreferrer\" class=\"pdf-btn\">\n <svg viewBox=\"0 0 24 24\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\"><path d=\"M19 9h-4V3H9v6H5l7 7 7-7zM5 18v2h14v-2H5z\"\/><\/svg>\n Download These Notes as PDF\n <\/a>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">20<\/div>\n <h2 class=\"section-title\">Chapter <span>Summary<\/span><\/h2>\n <\/div>\n\n <p>This chapter builds a unified framework connecting translational and rotational mechanics. Every concept on the translational side has a precise rotational counterpart, and the chapter culminates in rolling motion \u2014 where both operate simultaneously.<\/p>\n\n <div class=\"revision-box\">\n <h3>Quick Revision \u2014 Must-Know Points<\/h3>\n <ul>\n <li>COM of a system: r_cm = \u03a3m\u1d62r\u1d62 \/ M; lies at geometric centre for symmetric uniform bodies.<\/li>\n <li>COM moves as if all external forces act on total mass M at that point.<\/li>\n <li>Total linear momentum P = Mv_cm; conserved when F_ext = 0.<\/li>\n <li>Angular velocity \u03c9 = d\u03b8\/dt; angular acceleration \u03b1 = d\u03c9\/dt; analogues of v and a.<\/li>\n <li>Moment of inertia I = \u03a3m\u1d62r\u1d62\u00b2; depends on axis chosen. Key values: disc = MR\u00b2\/2, ring = MR\u00b2, solid sphere = 2MR\u00b2\/5.<\/li>\n <li>Parallel axis theorem: I = I_cm + Md\u00b2; perpendicular axis theorem: I_z = I_x + I_y (laminar bodies only).<\/li>\n <li>Torque \u03c4 = r \u00d7 F; Newton’s 2nd law for rotation: \u03c4 = I\u03b1.<\/li>\n <li>Angular momentum L = I\u03c9; conserved when \u03c4_ext = 0.<\/li>\n <li>Rolling without slipping: v_cm = R\u03c9; total KE = \u00bdmv\u00b2(1 + K\u00b2\/R\u00b2).<\/li>\n <li>Inclined plane race order: solid sphere wins, ring is last.<\/li>\n <li>Power in rotational motion: P = \u03c4\u03c9; work = \u03c4\u03b8 (constant torque).<\/li>\n <\/ul>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">21<\/div>\n <h2 class=\"section-title\">Common <span>Mistakes<\/span><\/h2>\n <\/div>\n\n <ul class=\"content-list\">\n <li><strong>Using Perpendicular Axis Theorem for 3D bodies:<\/strong> It applies only to flat (laminar) objects. Applying it to a sphere or cylinder is incorrect.<\/li>\n <li><strong>Forgetting the axis when quoting I:<\/strong> Moment of inertia of a disc is MR\u00b2\/2 only about the central axis perpendicular to its plane. About a diameter, it is MR\u00b2\/4.<\/li>\n <li><strong>Confusing rolling and sliding:<\/strong> Pure rolling means v_cm = R\u03c9 exactly. If there is slipping, this constraint does not hold and friction does work.<\/li>\n <li><strong>Applying angular momentum conservation incorrectly:<\/strong> External torque must be zero. Gravity can create torque; verify the condition before applying I\u2081\u03c9\u2081 = I\u2082\u03c9\u2082.<\/li>\n <li><strong>Sign errors in torque:<\/strong> Always define a positive direction of rotation. Torques in the positive direction are positive; opposing torques are negative. Sum them algebraically.<\/li>\n <li><strong>Using translational KE formula for rolling objects:<\/strong> Rolling objects have BOTH \u00bdmv\u00b2 AND \u00bdI\u03c9\u00b2. Omitting the rotational term leads to incorrect energy calculations.<\/li>\n <li><strong>COM trajectory after explosion:<\/strong> Students often assume the COM shifts after a mid-air explosion. It does not \u2014 in the absence of external horizontal forces, the COM trajectory is unchanged.<\/li>\n <li><strong>Radius of gyration vs. radius of body:<\/strong> K (radius of gyration) is defined via I = MK\u00b2 and is not the physical radius R unless I = MR\u00b2 (ring\/hollow cylinder case).<\/li>\n <\/ul>\n<\/div>\n\n\n<div class=\"faq-section\">\n <h2 class=\"faq-title\">Frequently Asked Questions \u2014 <span>Rotational Motion NEET<\/span><\/h2>\n\n <details>\n <summary>\n What is the difference between moment of inertia and mass in rotational motion?\n <span class=\"faq-icon\">+<\/span>\n <\/summary>\n <div class=\"faq-body\">\n Mass is the measure of resistance to linear (translational) acceleration under a net force (F = ma). Moment of inertia (I) is the analogous quantity for rotational motion \u2014 it measures resistance to angular acceleration under a net torque (\u03c4 = I\u03b1). The key distinction is that while mass is an intrinsic scalar property of a body, moment of inertia depends on both the mass and the distribution of that mass relative to a chosen axis of rotation. The same body has different values of I for different axes.\n <\/div>\n <\/details>\n\n <details>\n <summary>\n Why does a solid sphere reach the bottom of an incline before a hollow sphere?\n <span class=\"faq-icon\">+<\/span>\n <\/summary>\n <div class=\"faq-body\">\n When rolling down an incline, a portion of the potential energy converts to translational KE and the rest to rotational KE. The fraction going to rotation is I\/(I + MR\u00b2). A solid sphere has I = 2MR\u00b2\/5, giving a ratio of 2\/7 \u2248 29% to rotation. A hollow sphere has I = 2MR\u00b2\/3, giving 2\/5 = 40% to rotation. Since the solid sphere devotes less energy to spinning, more goes into translational speed \u2014 it reaches the bottom faster. This result is independent of mass and radius.\n <\/div>\n <\/details>\n\n <details>\n <summary>\n How can the centre of mass lie outside the body?\n <span class=\"faq-icon\">+<\/span>\n <\/summary>\n <div class=\"faq-body\">\n The COM is a weighted average of mass positions and is not required to lie within the physical boundaries of the body. For a uniform ring or hollow cylinder, all mass lies at radius R from the centre \u2014 the centre itself is empty space, yet that is exactly where the COM is located. Similarly, a boomerang or horseshoe has its COM in the hollow interior region. Any body with a hole or concave shape where mass is absent near the geometric centre can have its COM in empty space.\n <\/div>\n <\/details>\n\n <details>\n <summary>\n Does friction do work in pure rolling motion?\n <span class=\"faq-icon\">+<\/span>\n <\/summary>\n <div class=\"faq-body\">\n No. In pure rolling without slipping, the contact point between the rolling body and the surface has zero instantaneous velocity. Since the point of application of the static friction force has zero velocity, the work done by friction is zero (W = F \u00d7 displacement of point of application = 0). However, friction is still essential \u2014 it provides the torque necessary to maintain the rolling constraint. Without friction, the body would slide rather than roll.\n <\/div>\n <\/details>\n\n <details>\n <summary>\n How is angular momentum conservation applied in the System of Particles and Rotational Motion class 11 notes?\n <span class=\"faq-icon\">+<\/span>\n <\/summary>\n <div class=\"faq-body\">\n Angular momentum is conserved when the net external torque on the system is zero. The law states L = I\u03c9 = constant in such situations. A classic example is a dancer who spins with arms outstretched (large I, small \u03c9) and then pulls arms in (smaller I). Since I\u03c9 must remain constant, \u03c9 increases \u2014 the dancer spins faster. For NEET, identify that no external torque acts, then write I\u2081\u03c9\u2081 = I\u2082\u03c9\u2082 and solve for the unknown angular velocity or moment of inertia.\n <\/div>\n <\/details>\n\n <details>\n <summary>\n What is the perpendicular axis theorem and when can it be used?\n <span class=\"faq-icon\">+<\/span>\n <\/summary>\n <div class=\"faq-body\">\n The perpendicular axis theorem states that for a laminar (flat, 2D) body, the moment of inertia about an axis perpendicular to its plane equals the sum of moments of inertia about two mutually perpendicular axes lying in the plane: I_z = I_x + I_y. This theorem is strictly limited to planar (2D) objects such as discs, rings, rectangular sheets, and triangular plates. It cannot be applied to three-dimensional bodies like spheres, cylinders, or cones.\n <\/div>\n <\/details>\n<\/div>\n\n\n<div class=\"section\" style=\"padding-top:0;\">\n <div class=\"related-links\">\n <h4>Recommended Courses and Study Materials<\/h4>\n <ul>\n <li><a href=\"https:\/\/courses.ksquare.co.in\/new-courses\/31-umeed-neet-2026\" target=\"_blank\" rel=\"nofollow noopener noreferrer\">Umeed NEET 2026 Study Materials<\/a><\/li>\n <li><a href=\"https:\/\/courses.ksquare.co.in\/new-courses\/29-pc4-29\" target=\"_blank\" rel=\"nofollow noopener noreferrer\">Grip NCERT Biology<\/a><\/li>\n <li><a href=\"https:\/\/courses.ksquare.co.in\/new-courses\/28-grip-ncert-chemistry\" target=\"_blank\" rel=\"nofollow noopener noreferrer\">Grip NCERT Chemistry<\/a><\/li>\n <\/ul>\n <\/div>\n<\/div>\n\n\n<div class=\"cta-section\">\n <h2>Ready to Master NEET Physics?<\/h2>\n <p>System of Particles and Rotational Motion demands consistent practice. Use our rank predictor, join the Mission 180 batch, or test yourself with PYQ practice now.<\/p>\n <div class=\"cta-btns\">\n <a href=\"https:\/\/courses.ksquare.co.in\/new-courses\/3-mission-180-neet-physics-rankers-batch\" target=\"_blank\" rel=\"nofollow noopener noreferrer\" class=\"btn-white\">Join Mission 180 Physics Batch<\/a>\n <a href=\"https:\/\/ksquareinstitute.in\/neet-2026-rank-predictor\/\" target=\"_blank\" rel=\"nofollow noopener noreferrer\" class=\"btn-outline\">Use NEET 2026 Rank Predictor<\/a>\n <a href=\"https:\/\/ksquareinstitute.in\/free-study-material\/\" target=\"_blank\" rel=\"nofollow noopener noreferrer\" class=\"btn-outline\">Download Free Materials<\/a>\n <\/div>\n<\/div>\n\n<\/article>\n<\/body>\n<\/html>\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n <meta charset=\"UTF-8\">\n <meta name=\"viewport\" content=\"width=device-width, initial-scale=1.0\">\n <title>Table of Contents \u2014 Physics Class 11<\/title>\n \n \n <link rel=\"preconnect\" href=\"https:\/\/fonts.googleapis.com\">\n <link rel=\"preconnect\" href=\"https:\/\/fonts.gstatic.com\" crossorigin>\n <link href=\"https:\/\/fonts.googleapis.com\/css2?family=DM+Sans:ital,opsz,wght@0,9..40,100..1000;1,9..40,100..1000&family=Plus+Jakarta+Sans:ital,wght@0,200..800;1,200..800&display=swap\" rel=\"stylesheet\">\n \n <style>\n \/* Scoped wrapper using a unique ID to prevent CSS conflicts. *\/\n #physics-toc-wrapper {\n font-family: 'DM Sans', sans-serif;\n width: 100%;\n margin: 0;\n padding: 60px 0;\n color: #111;\n background: #fff;\n -webkit-font-smoothing: antialiased;\n }\n\n #physics-toc-wrapper .container-inner {\n width: 100%;\n margin: 0 auto;\n padding: 0; \/* Set left\/right padding to 0 *\/\n }\n\n #physics-toc-wrapper h1 {\n font-family: 'Plus Jakarta Sans', sans-serif;\n font-size: 0.85rem;\n font-weight: 700;\n color: #71717a;\n margin: 0 0 8px;\n letter-spacing: 0.1em;\n text-transform: uppercase;\n padding-left: 16px; \/* Keeping a small offset for 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cubic-bezier(0.4, 0, 0.2, 1);\n white-space: nowrap;\n }\n\n #physics-toc-wrapper a.go:hover {\n background: #18181b;\n color: #ffffff;\n transform: translateY(-2px);\n box-shadow: 0 4px 12px rgba(24, 24, 27, 0.15);\n }\n\n \/* Responsive adjustments *\/\n @media (max-width: 768px) {\n #physics-toc-wrapper h2 {\n font-size: 1.75rem;\n margin-bottom: 32px;\n }\n #physics-toc-wrapper td {\n padding: 18px 12px;\n font-size: 0.95rem;\n }\n }\n <\/style>\n<\/head>\n<body>\n\n<div id=\"physics-toc-wrapper\">\n <div class=\"container-inner\">\n <h1>Table of Contents<\/h1>\n <h2>Physics — Class 11<\/h2>\n \n <table>\n <tr><td>01<\/td><td>Units and Measurements<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/units-and-measurements-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>02<\/td><td>Motion in a Straight Line<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/motion-in-a-straight-line-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>03<\/td><td>Motion in a Plane<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/motion-in-a-plane-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>04<\/td><td>Laws of Motion<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/laws-of-motion-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>05<\/td><td>Work, Energy and Power<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/work-energy-and-power-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>06<\/td><td>System of Particles and Rotational Motion<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/system-of-particles-and-rotational-motion-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>07<\/td><td>Gravitation<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/gravitation-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>08<\/td><td>Mechanical Properties of Solids<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/mechanical-properties-of-solids-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>09<\/td><td>Mechanical Properties of Fluids<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/mechanical-properties-of-fluids-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>10<\/td><td>Thermal Properties of Matter<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/thermal-properties-of-matter-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>11<\/td><td>Thermodynamics<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/thermodynamics-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>12<\/td><td>Kinetic Theory<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/kinetic-theory-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>13<\/td><td>Oscillations<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/oscillations-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>14<\/td><td>Waves<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/waves-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <\/table>\n <\/div>\n<\/div>\n\n<\/body>\n<\/html>\n","protected":false},"excerpt":{"rendered":"<p>System of Particles and Rotational Motion Class 11 Notes | NEET Physics 01 Introduction to System of Particles The System of Particles and Rotational Motion class 11 notes begin with a foundational shift in perspective: instead of treating a body as a single point mass, we now consider an extended body as a collection of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[127],"tags":[],"class_list":["post-3940","post","type-post","status-publish","format-standard","hentry","category-free-study-material"],"blocksy_meta":{"page_structure_type":"type-1","styles_descriptor":{"styles":{"desktop":"","tablet":"","mobile":""},"google_fonts":[],"version":6}},"_links":{"self":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/3940","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/comments?post=3940"}],"version-history":[{"count":2,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/3940\/revisions"}],"predecessor-version":[{"id":4209,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/3940\/revisions\/4209"}],"wp:attachment":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/media?parent=3940"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/categories?post=3940"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/tags?post=3940"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

{"id":3940,"date":"2026-03-27T07:03:24","date_gmt":"2026-03-27T07:03:24","guid":{"rendered":"https:\/\/ksquareinstitute.in\/blog\/?p=3940"},"modified":"2026-04-03T12:15:28","modified_gmt":"2026-04-03T12:15:28","slug":"system-of-particles-and-rotational-motion-class-11-notes","status":"publish","type":"post","link":"https:\/\/ksquareinstitute.in\/blog\/system-of-particles-and-rotational-motion-class-11-notes\/","title":{"rendered":"System of Particles and Rotational Motion Class 11 Notes"},"content":{"rendered":"\n\n\n\n\n\nSystem of Particles and Rotational Motion Class 11 Notes | NEET Physics<\/title>\n<meta name=\"description\" content=\"Complete System of Particles and Rotational Motion class 11 notes for NEET Physics. 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margin: 0; color: #7c2d12; }\n\n ul.content-list {\n padding-left: 0;\n list-style: none;\n margin: 14px 0;\n display: grid;\n gap: 8px;\n }\n ul.content-list li {\n padding-left: 22px;\n position: relative;\n font-size: 15px;\n }\n ul.content-list li::before {\n content: '';\n position: absolute;\n left: 0;\n top: 9px;\n width: 8px;\n height: 8px;\n background: var(--accent);\n border-radius: 50%;\n }\n\n .two-col { display: grid; grid-template-columns: 1fr 1fr; gap: 20px; margin: 20px 0; }\n @media (max-width: 640px) { .two-col { grid-template-columns: 1fr; } .cta-section { padding: 48px 0; } }\n\n a { color: var(--accent); }\n strong { font-weight: 600; }\n\n .section-divider {\n height: 4px;\n background: linear-gradient(90deg, var(--accent) 0%, transparent 100%);\n border: none;\n margin: 0;\n }\n\n .pyq-badge {\n display: inline-block;\n background: #fef3c7;\n border: 1px solid #fbbf24;\n color: #92400e;\n font-size: 11px;\n font-weight: 700;\n padding: 2px 8px;\n border-radius: 4px;\n font-family: 'Plus Jakarta Sans', sans-serif;\n letter-spacing: 0.5px;\n text-transform: uppercase;\n margin-right: 8px;\n }\n<\/style>\n<\/head>\n<body>\n<article>\n\n\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">01<\/div>\n <h2 class=\"section-title\">Introduction to <span>System of Particles<\/span><\/h2>\n <\/div>\n\n <p>The <strong>System of Particles and Rotational Motion class 11 notes<\/strong> begin with a foundational shift in perspective: instead of treating a body as a single point mass, we now consider an extended body as a collection of particles, each obeying Newton’s laws individually. Together, their collective behaviour determines the motion of the entire system. This chapter is one of the most concept-dense and formula-rich chapters in Class 11 Physics, and it carries consistent weight in NEET.<\/p>\n\n <p>A <strong>system of particles<\/strong> is any group of two or more particles whose interactions and collective motion are analyzed as a whole. The particles may be bound together (rigid body) or free (like gas molecules). For NEET, the focus is on rigid bodies \u2014 bodies where the distance between any two particles remains constant regardless of the forces applied.<\/p>\n\n <div class=\"highlight-grid\">\n <div class=\"highlight-card\">\n <div class=\"hc-title\">Rigid Body<\/div>\n <p>Inter-particle distances are fixed. No deformation under applied forces.<\/p>\n <\/div>\n <div class=\"highlight-card\">\n <div class=\"hc-title\">Types of Motion<\/div>\n <p>Pure translation, pure rotation, or combined rolling motion.<\/p>\n <\/div>\n <div class=\"highlight-card\">\n <div class=\"hc-title\">Why This Matters<\/div>\n <p>Unifies force, momentum, energy concepts under one mechanical framework.<\/p>\n <\/div>\n <\/div>\n\n <p>Key distinction: in <strong>translational motion<\/strong>, every point of the body has the same velocity and acceleration at any instant. In <strong>rotational motion<\/strong>, every point moves in a circular path about the axis of rotation, with angular velocity and angular acceleration defining the state of motion.<\/p>\n\n <div class=\"related-links\">\n <h4>Related Resources<\/h4>\n <ul>\n <li><a href=\"https:\/\/ksquareinstitute.in\/blog\/neet-physics-survival-kit-2026\/\">NEET Physics Survival Kit<\/a><\/li>\n <li><a href=\"https:\/\/ksquareinstitute.in\/blog\/organic-chemistry-strategy-neet\/\">Organic Chemistry Strategy<\/a><\/li>\n <li><a href=\"https:\/\/ksquareinstitute.in\/blog\/neet-biology-tricks-for-exams\/\">NEET Biology Tricks for Exams<\/a><\/li>\n <li><a href=\"https:\/\/ksquareinstitute.in\/blog\/score-340-in-neet-biology\/\">How to Score 340+ in NEET Biology<\/a><\/li>\n <li><a href=\"https:\/\/ksquareinstitute.in\/blog\/top-10-tricky-neet-biology-diagrams\/\">Top 10 Tricky NEET Biology Diagrams<\/a><\/li>\n <li><a href=\"https:\/\/ksquareinstitute.in\/free-study-material\/\">Free Study Materials<\/a><\/li>\n <\/ul>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">02<\/div>\n <h2 class=\"section-title\">Centre of <span>Mass (COM)<\/span><\/h2>\n <\/div>\n\n <p>The <strong>centre of mass<\/strong> is a uniquely defined point in a system where the entire mass of the system can be assumed concentrated for the purpose of analyzing translational motion. It is not necessarily at the geometric centre; it depends entirely on the mass distribution.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">COM for Discrete System<\/div>\n <code>x_cm = (m\u2081x\u2081 + m\u2082x\u2082 + ... + m\u2099x\u2099) \/ (m\u2081 + m\u2082 + ... + m\u2099)<\/code>\n <code>r_cm = \u03a3(m\u1d62 r\u1d62) \/ M where M = total mass<\/code>\n <\/div>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">COM for Continuous Distribution<\/div>\n <code>r_cm = (1\/M) \u222b r dm<\/code>\n <code>x_cm = (1\/M) \u222b x dm , y_cm = (1\/M) \u222b y dm<\/code>\n <\/div>\n\n <p>For symmetric uniform bodies, the COM lies at the geometric centre. Key results to memorize:<\/p>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Body<\/th>\n <th>COM Location<\/th>\n <th>Notes<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td>Uniform Rod<\/td>\n <td>Midpoint (L\/2 from either end)<\/td>\n <td>Symmetric body<\/td>\n <\/tr>\n <tr>\n <td>Uniform Disc \/ Ring<\/td>\n <td>Geometric centre<\/td>\n <td>COM lies at centre even for hollow ring<\/td>\n <\/tr>\n <tr>\n <td>Uniform Solid Sphere<\/td>\n <td>Geometric centre<\/td>\n <td>\u2014<\/td>\n <\/tr>\n <tr>\n <td>Semicircular Ring (radius R)<\/td>\n <td>2R\/\u03c0 from diameter<\/td>\n <td>Asymmetric; COM off centre<\/td>\n <\/tr>\n <tr>\n <td>Semicircular Disc (radius R)<\/td>\n <td>4R\/(3\u03c0) from diameter<\/td>\n <td>Standard NEET result<\/td>\n <\/tr>\n <tr>\n <td>Solid Hemisphere (radius R)<\/td>\n <td>3R\/8 from flat face<\/td>\n <td>High NEET frequency<\/td>\n <\/tr>\n <tr>\n <td>Hollow Hemisphere (radius R)<\/td>\n <td>R\/2 from flat face<\/td>\n <td>\u2014<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n\n <div class=\"callout callout-tip\">\n <span class=\"callout-icon\">Tip<\/span>\n <p>For a system with a cavity (material removed), use the concept of negative mass. Treat the removed portion as a particle of negative mass at its original COM location and apply the COM formula for the remaining system.<\/p>\n <\/div>\n<\/div>\n\n\n<a href=\"https:\/\/courses.ksquare.co.in\/new-courses\/3-mission-180-neet-physics-rankers-batch\" target=\"_blank\" rel=\"nofollow noopener noreferrer\" style=\"display:block; margin-bottom:20px;\">\n <img decoding=\"async\" src=\"https:\/\/ksquareinstitute.in\/blog\/wp-content\/uploads\/2026\/03\/Course-Poromo-Banner-scaled.png\" alt=\"Mission 180 NEET Physics Rankers Batch - KSquare Career Institute\" style=\"width:100%; height:auto; border-radius:10px; display:block;\">\n<\/a>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">03<\/div>\n <h2 class=\"section-title\">Motion of <span>Centre of Mass<\/span><\/h2>\n <\/div>\n\n <p>The most powerful result in the study of a system of particles: <strong>the centre of mass of a system moves as if all external forces act on a single particle of mass equal to the total mass of the system, located at the COM.<\/strong><\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Equation of Motion for COM<\/div>\n <code>F_ext = M \u00d7 a_cm<\/code>\n <code>where a_cm = d\u00b2r_cm\/dt\u00b2 and M = total mass of system<\/code>\n <\/div>\n\n <p><strong>Internal forces<\/strong> \u2014 forces between particles within the system \u2014 always appear in action-reaction pairs and cancel out when summed over the entire system (Newton’s Third Law). They do not affect the motion of the COM; only <strong>external forces<\/strong> determine how the COM accelerates.<\/p>\n\n <div class=\"callout callout-warn\">\n <span class=\"callout-icon\">Warning<\/span>\n <p>In an explosion or collision, even if the object shatters into many pieces, the COM continues on the same parabolic or linear trajectory as before \u2014 because no external horizontal force acts. This is a classic NEET trap question.<\/p>\n <\/div>\n\n <p>Practical application: A bomb thrown as a projectile follows a parabolic path. When it explodes mid-air, the fragments scatter in all directions, but the COM of all fragments continues along the original parabola. If one fragment lands at a known position, the other fragment’s landing position can be located using COM principles.<\/p>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">04<\/div>\n <h2 class=\"section-title\">Linear Momentum of <span>a System<\/span><\/h2>\n <\/div>\n\n <p>The total linear momentum of a system of particles is the vector sum of the momenta of all individual particles:<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Total Linear Momentum<\/div>\n <code>P = \u03a3 p\u1d62 = \u03a3 m\u1d62v\u1d62 = M \u00d7 v_cm<\/code>\n <code>The total momentum equals total mass times velocity of COM.<\/code>\n <\/div>\n\n <p>Differentiating this with respect to time: <strong>dP\/dt = F_ext<\/strong>. This is Newton’s second law for a system of particles. When no external force acts, dP\/dt = 0, which leads directly to conservation of linear momentum.<\/p>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Impulse\u2013Momentum Theorem for System<\/div>\n <code>J = F_ext \u00d7 \u0394t = \u0394P = M(v_cm,f \u2212 v_cm,i)<\/code>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">05<\/div>\n <h2 class=\"section-title\">Conservation of <span>Linear Momentum<\/span><\/h2>\n <\/div>\n\n <p>One of the most fundamental principles in physics: <strong>if the net external force on a system is zero, the total linear momentum of the system remains constant.<\/strong><\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Conservation Condition<\/div>\n <code>F_ext = 0 \u27f9 P_total = constant<\/code>\n <code>m\u2081u\u2081 + m\u2082u\u2082 = m\u2081v\u2081 + m\u2082v\u2082 (for two-particle system)<\/code>\n <\/div>\n\n <p>Critical applications for NEET:<\/p>\n\n <ul class=\"content-list\">\n <li><strong>Recoil of a gun:<\/strong> Before firing, total momentum = 0. After firing, bullet and gun have equal and opposite momenta. m_gun \u00d7 v_gun = m_bullet \u00d7 v_bullet.<\/li>\n <li><strong>Rocket propulsion:<\/strong> Exhaust gases ejected backward give the rocket forward momentum. No external force needed \u2014 conservation applies within the system.<\/li>\n <li><strong>Elastic collision:<\/strong> Both kinetic energy and momentum are conserved. For equal masses, velocities exchange.<\/li>\n <li><strong>Perfectly inelastic collision:<\/strong> Momentum conserved; maximum kinetic energy lost. Bodies stick together: (m\u2081 + m\u2082)v = m\u2081u\u2081 + m\u2082u\u2082.<\/li>\n <\/ul>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Elastic Collision (1D) \u2014 Velocity Formulae<\/div>\n <code>v\u2081 = [(m\u2081 \u2212 m\u2082)\/(m\u2081 + m\u2082)]u\u2081 + [2m\u2082\/(m\u2081 + m\u2082)]u\u2082<\/code>\n <code>v\u2082 = [2m\u2081\/(m\u2081 + m\u2082)]u\u2081 + [(m\u2082 \u2212 m\u2081)\/(m\u2081 + m\u2082)]u\u2082<\/code>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">06<\/div>\n <h2 class=\"section-title\">Introduction to <span>Rotational Motion<\/span><\/h2>\n <\/div>\n\n <p>When a rigid body rotates about a fixed axis, every particle moves in a circle. The axis of rotation may pass through the body (spin) or be external (revolution). Understanding <strong>rotational motion<\/strong> is central to mastering System of Particles and Rotational Motion class 11 notes for NEET.<\/p>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Translational Quantity<\/th>\n <th>Rotational Analogue<\/th>\n <th>Symbol<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td>Displacement (s)<\/td>\n <td>Angular displacement<\/td>\n <td>\u03b8 (radians)<\/td>\n <\/tr>\n <tr>\n <td>Velocity (v)<\/td>\n <td>Angular velocity<\/td>\n <td>\u03c9 = d\u03b8\/dt (rad\/s)<\/td>\n <\/tr>\n <tr>\n <td>Acceleration (a)<\/td>\n <td>Angular acceleration<\/td>\n <td>\u03b1 = d\u03c9\/dt (rad\/s\u00b2)<\/td>\n <\/tr>\n <tr>\n <td>Mass (m)<\/td>\n <td>Moment of inertia<\/td>\n <td>I (kg\u00b7m\u00b2)<\/td>\n <\/tr>\n <tr>\n <td>Force (F)<\/td>\n <td>Torque<\/td>\n <td>\u03c4 = r \u00d7 F (N\u00b7m)<\/td>\n <\/tr>\n <tr>\n <td>Linear momentum (p)<\/td>\n <td>Angular momentum<\/td>\n <td>L = I\u03c9 (kg\u00b7m\u00b2\/s)<\/td>\n <\/tr>\n <tr>\n <td>Kinetic energy (\u00bdmv\u00b2)<\/td>\n <td>Rotational KE<\/td>\n <td>\u00bdI\u03c9\u00b2<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n<\/div>\n\n\n<a href=\"https:\/\/ksquareinstitute.in\/neet-2026-rank-predictor\/\" target=\"_blank\" rel=\"nofollow noopener noreferrer\" style=\"display:block; margin-bottom:20px;\">\n <img decoding=\"async\" src=\"https:\/\/ksquareinstitute.in\/blog\/wp-content\/uploads\/2026\/03\/neet-2026-college-and-rank-predictor-scaled.png\" alt=\"NEET 2026 College and Rank Predictor - KSquare Career Institute\" style=\"width:100%; height:auto; border-radius:10px; display:block;\">\n<\/a>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">07<\/div>\n <h2 class=\"section-title\">Angular Variables and <span>Kinematics<\/span><\/h2>\n <\/div>\n\n <p>The kinematic equations for <strong>uniform angular acceleration<\/strong> are exact analogues of the linear kinematic equations. Mastering the parallel structure makes both sets trivially easy.<\/p>\n\n <div class=\"two-col\">\n <div class=\"formula-dark\">\n <div class=\"f-label\">Linear Kinematics<\/div>\n <code>v = u + at<\/code>\n <code>s = ut + \u00bdat\u00b2<\/code>\n <code>v\u00b2 = u\u00b2 + 2as<\/code>\n <\/div>\n <div class=\"formula-dark\">\n <div class=\"f-label\">Rotational Kinematics<\/div>\n <code>\u03c9 = \u03c9\u2080 + \u03b1t<\/code>\n <code>\u03b8 = \u03c9\u2080t + \u00bd\u03b1t\u00b2<\/code>\n <code>\u03c9\u00b2 = \u03c9\u2080\u00b2 + 2\u03b1\u03b8<\/code>\n <\/div>\n <\/div>\n\n <p>Relationship between linear and angular variables for a particle at radius r from axis:<\/p>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Linear\u2013Angular Relationships<\/div>\n <code>v = r\u03c9 (tangential speed)<\/code>\n <code>a\u209c = r\u03b1 (tangential acceleration)<\/code>\n <code>a\u1d9c = r\u03c9\u00b2 = v\u00b2\/r (centripetal acceleration)<\/code>\n <\/div>\n\n <div class=\"callout callout-tip\">\n <span class=\"callout-icon\">Tip<\/span>\n <p>Angular displacement \u03b8 must always be in radians (not degrees) when using these kinematic equations. Convert degrees to radians: \u03b8(rad) = \u03b8(\u00b0) \u00d7 \u03c0\/180.<\/p>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">08<\/div>\n <h2 class=\"section-title\">Moment of <span>Inertia<\/span><\/h2>\n <\/div>\n\n <p><strong>Moment of inertia (I)<\/strong> is the rotational analogue of mass. It measures a body’s resistance to angular acceleration. Unlike mass, I depends on both the total mass and its distribution relative to the axis of rotation.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Definition<\/div>\n <code>I = \u03a3 m\u1d62r\u1d62\u00b2 (discrete system)<\/code>\n <code>I = \u222b r\u00b2 dm (continuous body)<\/code>\n <code>Unit: kg\u00b7m\u00b2 Dimensional formula: [ML\u00b2]<\/code>\n <\/div>\n\n <p><strong>Radius of gyration (K):<\/strong> An equivalent radius at which the entire mass can be concentrated to give the same moment of inertia: <strong>I = MK\u00b2<\/strong>, so K = \u221a(I\/M).<\/p>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Body<\/th>\n <th>Axis<\/th>\n <th>Moment of Inertia (I)<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td>Thin Ring (mass M, radius R)<\/td>\n <td>Through centre, perpendicular to plane<\/td>\n <td>MR\u00b2<\/td>\n <\/tr>\n <tr>\n <td>Thin Ring<\/td>\n <td>Diameter<\/td>\n <td>MR\u00b2\/2<\/td>\n <\/tr>\n <tr>\n <td>Solid Disc (mass M, radius R)<\/td>\n <td>Through centre, perpendicular<\/td>\n <td>MR\u00b2\/2<\/td>\n <\/tr>\n <tr>\n <td>Solid Disc<\/td>\n <td>Diameter<\/td>\n <td>MR\u00b2\/4<\/td>\n <\/tr>\n <tr>\n <td>Thin Rod (length L)<\/td>\n <td>Through centre, perpendicular to rod<\/td>\n <td>ML\u00b2\/12<\/td>\n <\/tr>\n <tr>\n <td>Thin Rod (length L)<\/td>\n <td>Through one end, perpendicular<\/td>\n <td>ML\u00b2\/3<\/td>\n <\/tr>\n <tr>\n <td>Solid Sphere (radius R)<\/td>\n <td>Through diameter<\/td>\n <td>2MR\u00b2\/5<\/td>\n <\/tr>\n <tr>\n <td>Hollow Sphere (radius R)<\/td>\n <td>Through diameter<\/td>\n <td>2MR\u00b2\/3<\/td>\n <\/tr>\n <tr>\n <td>Solid Cylinder (radius R)<\/td>\n <td>Geometric axis<\/td>\n <td>MR\u00b2\/2<\/td>\n <\/tr>\n <tr>\n <td>Hollow Cylinder (radius R)<\/td>\n <td>Geometric axis<\/td>\n <td>MR\u00b2<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n\n <div class=\"callout callout-warn\">\n <span class=\"callout-icon\">Warning<\/span>\n <p>I is not invariant \u2014 it changes with the axis chosen. The same disc has I = MR\u00b2\/2 about its central axis but I = MR\u00b2\/4 about a diameter. Always specify the axis before quoting a value.<\/p>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">09<\/div>\n <h2 class=\"section-title\">Theorems of <span>Moment of Inertia<\/span><\/h2>\n <\/div>\n\n <p>Two powerful theorems allow calculation of I about any axis if the I about a standard axis is known.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Parallel Axis Theorem<\/div>\n <code>I = I_cm + Md\u00b2<\/code>\n <code>I_cm = MOI about axis through COM (parallel to new axis)<\/code>\n <code>d = perpendicular distance between the two axes<\/code>\n <\/div>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Perpendicular Axis Theorem (Planar Bodies Only)<\/div>\n <code>I_z = I_x + I_y<\/code>\n <code>Applies only to laminar (2D) bodies.<\/code>\n <code>z-axis is perpendicular to the plane of the body.<\/code>\n <\/div>\n\n <div class=\"callout callout-tip\">\n <span class=\"callout-icon\">Tip<\/span>\n <p>The perpendicular axis theorem is restricted to flat (laminar) objects. Do not apply it to 3D bodies like spheres or cylinders. The parallel axis theorem, however, applies universally to all bodies \u2014 2D or 3D.<\/p>\n <\/div>\n\n <p>Classic NEET application: For a disc, I about diameter = MR\u00b2\/4. By perpendicular axis theorem, I about central perpendicular axis = I_x + I_y = MR\u00b2\/4 + MR\u00b2\/4 = MR\u00b2\/2. This confirms the standard result.<\/p>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">10<\/div>\n <h2 class=\"section-title\">Torque and <span>Angular Momentum<\/span><\/h2>\n <\/div>\n\n <p><strong>Torque (\u03c4)<\/strong> is the rotational analogue of force. It is the tendency of a force to produce rotation about an axis. Torque depends not just on the magnitude of the force but also on the point of application and direction.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Torque<\/div>\n <code>\u03c4 = r \u00d7 F (vector cross product)<\/code>\n <code>|\u03c4| = rF sin \u03b8 where \u03b8 = angle between r and F<\/code>\n <code>\u03c4 = I \u00d7 \u03b1 (Newton's 2nd law for rotation)<\/code>\n <\/div>\n\n <p><strong>Angular momentum (L)<\/strong> is the rotational analogue of linear momentum:<\/p>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Angular Momentum<\/div>\n <code>L = r \u00d7 p = r \u00d7 mv (for a particle)<\/code>\n <code>L = I\u03c9 (for a rigid body about fixed axis)<\/code>\n <code>\u03c4 = dL\/dt (analogous to F = dp\/dt)<\/code>\n <\/div>\n\n <div class=\"callout callout-warn\">\n <span class=\"callout-icon\">Warning<\/span>\n <p>Torque and angular momentum are both axial vectors (direction given by right-hand rule). In NEET numerical problems, always confirm that the axis of rotation is fixed before applying L = I\u03c9. If the axis changes, the problem is significantly more complex.<\/p>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">11<\/div>\n <h2 class=\"section-title\">Conservation of <span>Angular Momentum<\/span><\/h2>\n <\/div>\n\n <p>When no external torque acts on a system, the total angular momentum remains constant. This is one of the most beautiful and practically significant principles in classical mechanics.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Conservation of Angular Momentum<\/div>\n <code>\u03c4_ext = 0 \u27f9 L = I\u03c9 = constant<\/code>\n <code>I\u2081\u03c9\u2081 = I\u2082\u03c9\u2082 (before and after any internal change)<\/code>\n <\/div>\n\n <p>Classic applications and NEET examples:<\/p>\n\n <ul class=\"content-list\">\n <li><strong>Ballet dancer \/ spinning skater:<\/strong> Pulls arms inward \u2192 I decreases \u2192 \u03c9 increases (spins faster). Extends arms \u2192 I increases \u2192 \u03c9 decreases.<\/li>\n <li><strong>Diver tucks in mid-air:<\/strong> Reduces I, increases \u03c9 to complete more rotations before hitting water.<\/li>\n <li><strong>Collapsing star (pulsar):<\/strong> As a dying star collapses, its radius decreases enormously \u2192 I decreases \u2192 \u03c9 increases dramatically, producing the characteristic rapid spin of neutron stars.<\/li>\n <li><strong>Planet in elliptical orbit:<\/strong> Closer to sun (perihelion) \u2192 moves faster; farther (aphelion) \u2192 moves slower. L = mvr = constant (Kepler’s second law).<\/li>\n <\/ul>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">12<\/div>\n <h2 class=\"section-title\">Rotational <span>Dynamics<\/span><\/h2>\n <\/div>\n\n <p><strong>Newton’s second law for rotation:<\/strong> The net external torque acting on a rigid body equals the product of its moment of inertia and angular acceleration.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Newton’s 2nd Law (Rotational Form)<\/div>\n <code>\u03c4_net = I \u00d7 \u03b1<\/code>\n <code>This is the exact analogue of F_net = ma<\/code>\n <\/div>\n\n <p>When both translational and rotational motions occur simultaneously (e.g., a cylinder rolling down an incline), two equations are written:<\/p>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Combined Dynamics (e.g., Rolling on Inclined Plane)<\/div>\n <code>Translational: Mg sin \u03b8 \u2212 f = Ma_cm<\/code>\n <code>Rotational: f \u00d7 R = I \u00d7 \u03b1 = I \u00d7 (a_cm\/R)<\/code>\n <code>Solving: a_cm = g sin \u03b8 \/ (1 + I\/MR\u00b2)<\/code>\n <\/div>\n\n <p>This result shows that a solid sphere (I = 2MR\u00b2\/5) accelerates faster down an incline than a hollow sphere (I = 2MR\u00b2\/3), which in turn accelerates faster than a ring (I = MR\u00b2). The more the mass is concentrated at the rim, the slower the acceleration.<\/p>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">13<\/div>\n <h2 class=\"section-title\">Work Done in <span>Rotational Motion<\/span><\/h2>\n <\/div>\n\n <p>The work\u2013energy theorem extends naturally to rotational systems. Work done by a torque over an angular displacement is:<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Work by Torque<\/div>\n <code>W = \u222b \u03c4 d\u03b8<\/code>\n <code>For constant torque: W = \u03c4 \u00d7 \u03b8<\/code>\n <code>Power: P = \u03c4 \u00d7 \u03c9 (analogous to P = F \u00d7 v)<\/code>\n <\/div>\n\n <div class=\"callout callout-tip\">\n <span class=\"callout-icon\">Tip<\/span>\n <p>For NEET problems involving work done against friction torque in rotational motion, apply work-energy theorem directly: Work done by all torques = Change in rotational KE = \u00bdI(\u03c9\u2082\u00b2 \u2212 \u03c9\u2081\u00b2).<\/p>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">14<\/div>\n <h2 class=\"section-title\">Rotational <span>Kinetic Energy<\/span><\/h2>\n <\/div>\n\n <p>A rotating body possesses kinetic energy by virtue of its rotation, distinct from (and additive to) any translational kinetic energy it may have.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Rotational Kinetic Energy<\/div>\n <code>KE_rot = \u00bd I \u03c9\u00b2<\/code>\n <code>Compare with translational: KE_trans = \u00bd mv\u00b2<\/code>\n <code>Both have the same mathematical structure: \u00bd \u00d7 (inertia) \u00d7 (velocity)\u00b2<\/code>\n <\/div>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Work\u2013Energy Theorem (Rotational)<\/div>\n <code>W_net = \u0394KE_rot = \u00bdI(\u03c9_f\u00b2 \u2212 \u03c9_i\u00b2)<\/code>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">15<\/div>\n <h2 class=\"section-title\">Rolling <span>Motion<\/span><\/h2>\n <\/div>\n\n <p><strong>Rolling motion<\/strong> is the superposition of pure translation of the COM and pure rotation about the COM. For pure rolling without slipping, there is a crucial constraint linking linear and angular quantities.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Pure Rolling Condition (No Slipping)<\/div>\n <code>v_cm = R \u00d7 \u03c9 (velocity constraint)<\/code>\n <code>a_cm = R \u00d7 \u03b1 (acceleration constraint)<\/code>\n <code>The contact point P has zero instantaneous velocity.<\/code>\n <\/div>\n\n <p>Velocity at different points on a rolling body:<\/p>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Point on Body<\/th>\n <th>Velocity<\/th>\n <th>Magnitude<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td>Contact point (bottom)<\/td>\n <td>v_cm \u2212 R\u03c9<\/td>\n <td>Zero (pure rolling)<\/td>\n <\/tr>\n <tr>\n <td>Centre of mass<\/td>\n <td>v_cm (horizontal)<\/td>\n <td>v_cm<\/td>\n <\/tr>\n <tr>\n <td>Top point<\/td>\n <td>v_cm + R\u03c9<\/td>\n <td>2v_cm<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n\n <div class=\"callout callout-warn\">\n <span class=\"callout-icon\">Warning<\/span>\n <p>The topmost point of a rolling body moves at twice the speed of the centre of mass. This frequently appears in NEET questions as a conceptual trap \u2014 students often forget the rotational contribution at the top point.<\/p>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">16<\/div>\n <h2 class=\"section-title\">Energy in <span>Rolling Motion<\/span><\/h2>\n <\/div>\n\n <p>The total mechanical energy of a rolling body is the sum of its translational and rotational kinetic energies. This distinction is critical for inclined plane problems.<\/p>\n\n <div class=\"formula-dark\">\n <div class=\"f-label\">Total Kinetic Energy in Rolling<\/div>\n <code>KE_total = KE_trans + KE_rot<\/code>\n <code> = \u00bdmv_cm\u00b2 + \u00bdI\u03c9\u00b2<\/code>\n <code> = \u00bdmv_cm\u00b2(1 + I\/mR\u00b2) [using v_cm = R\u03c9]<\/code>\n <code> = \u00bdmv_cm\u00b2(1 + K\u00b2\/R\u00b2) [using I = mK\u00b2]<\/code>\n <\/div>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Rolling Body<\/th>\n <th>I\/MR\u00b2<\/th>\n <th>KE_rot\/KE_total<\/th>\n <th>v at bottom (from height h)<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td>Ring \/ Hollow Cylinder<\/td>\n <td>1<\/td>\n <td>50%<\/td>\n <td>\u221a(gh)<\/td>\n <\/tr>\n <tr>\n <td>Solid Cylinder \/ Disc<\/td>\n <td>1\/2<\/td>\n <td>33%<\/td>\n <td>\u221a(4gh\/3)<\/td>\n <\/tr>\n <tr>\n <td>Hollow Sphere<\/td>\n <td>2\/3<\/td>\n <td>40%<\/td>\n <td>\u221a(6gh\/5)<\/td>\n <\/tr>\n <tr>\n <td>Solid Sphere<\/td>\n <td>2\/5<\/td>\n <td>29%<\/td>\n <td>\u221a(10gh\/7)<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n\n <p>The solid sphere reaches the bottom with the highest velocity because the least fraction of energy goes into rotation. The ring is slowest for the same reason in reverse.<\/p>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">17<\/div>\n <h2 class=\"section-title\">Numerical <span>Framework<\/span><\/h2>\n <\/div>\n\n <p>NEET-level numericals in System of Particles and Rotational Motion class 11 notes fall into predictable categories. Here is the solution strategy for each type:<\/p>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Problem Type<\/th>\n <th>Key Equation(s)<\/th>\n <th>Watch Out For<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td>Finding COM of composite body<\/td>\n <td>r_cm = \u03a3m\u1d62r\u1d62 \/ \u03a3m\u1d62<\/td>\n <td>Use negative mass for cavities<\/td>\n <\/tr>\n <tr>\n <td>COM after explosion\/collision<\/td>\n <td>COM trajectory unchanged (no ext. force)<\/td>\n <td>Internal forces do not shift COM<\/td>\n <\/tr>\n <tr>\n <td>Moment of inertia problems<\/td>\n <td>I = Icm + Md\u00b2 (parallel axis)<\/td>\n <td>Choose correct standard I first<\/td>\n <\/tr>\n <tr>\n <td>Torque and angular acceleration<\/td>\n <td>\u03c4 = I\u03b1<\/td>\n <td>Net torque only; check direction<\/td>\n <\/tr>\n <tr>\n <td>Conservation of angular momentum<\/td>\n <td>I\u2081\u03c9\u2081 = I\u2082\u03c9\u2082<\/td>\n <td>No external torque condition must hold<\/td>\n <\/tr>\n <tr>\n <td>Rolling on incline<\/td>\n <td>a = g sin\u03b8 \/ (1 + I\/MR\u00b2)<\/td>\n <td>Friction does no work in pure rolling<\/td>\n <\/tr>\n <tr>\n <td>Rolling energy conservation<\/td>\n <td>mgh = \u00bdmv\u00b2(1 + K\u00b2\/R\u00b2)<\/td>\n <td>Use K\u00b2\/R\u00b2 for that specific body<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n\n <div class=\"formula-orange\">\n <div class=\"f-label\">Quick Reference \u2014 Inclined Plane Race<\/div>\n <code>Order of reaching bottom (fastest first):<\/code>\n <code>Solid Sphere > Solid Cylinder > Hollow Sphere > Ring<\/code>\n <code>Smaller I\/MR\u00b2 ratio \u27f9 higher bottom speed \u27f9 wins the race<\/code>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">18<\/div>\n <h2 class=\"section-title\">Conceptual <span>Questions<\/span><\/h2>\n <\/div>\n\n <p>Conceptual questions in System of Particles and Rotational Motion class 11 notes for NEET test the depth of understanding, not just formula recall. These types appear regularly in the NEET MCQ paper:<\/p>\n\n <ul class=\"content-list\">\n <li>Can the COM of a body lie outside the body? Yes \u2014 hollow ring, horseshoe, donut-shaped body.<\/li>\n <li>A body can have angular momentum even without rotating about its own axis \u2014 any particle moving in a straight line has angular momentum about a point not on that line (L = mvr).<\/li>\n <li>Friction is essential for rolling \u2014 without friction, a body would slide, not roll. However, friction does no work during pure rolling.<\/li>\n <li>A spinning top exhibits gyroscopic precession \u2014 the gravitational torque causes the angular momentum vector to precess, not fall.<\/li>\n <li>If a person walks from the rim of a rotating platform toward the centre, the platform speeds up (L conserved, I decreases).<\/li>\n <li>Two identical cylinders (solid and hollow) released simultaneously from the top of an identical incline \u2014 the solid cylinder always reaches the bottom first, regardless of mass or radius.<\/li>\n <\/ul>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">19<\/div>\n <h2 class=\"section-title\">PYQ <span>Trends<\/span><\/h2>\n <\/div>\n\n <p>Analysis of previous year NEET questions reveals consistent high-frequency areas within this chapter:<\/p>\n\n <div class=\"table-wrap\">\n <table>\n <thead>\n <tr>\n <th>Topic<\/th>\n <th>Frequency<\/th>\n <th>Type<\/th>\n <\/tr>\n <\/thead>\n <tbody>\n <tr>\n <td><span class=\"pyq-badge\">Very High<\/span>Moment of Inertia (standard bodies + theorems)<\/td>\n <td>3\u20134 questions\/year<\/td>\n <td>Numerical + Conceptual<\/td>\n <\/tr>\n <tr>\n <td><span class=\"pyq-badge\">High<\/span>Rolling Motion (energy, velocity at bottom)<\/td>\n <td>2\u20133 questions\/year<\/td>\n <td>Numerical<\/td>\n <\/tr>\n <tr>\n <td><span class=\"pyq-badge\">High<\/span>Conservation of Angular Momentum<\/td>\n <td>1\u20132 questions\/year<\/td>\n <td>Conceptual + Numerical<\/td>\n <\/tr>\n <tr>\n <td><span class=\"pyq-badge\">High<\/span>Torque and Rotational Dynamics<\/td>\n <td>1\u20132 questions\/year<\/td>\n <td>Numerical<\/td>\n <\/tr>\n <tr>\n <td><span class=\"pyq-badge\">Moderate<\/span>Centre of Mass (COM location, motion)<\/td>\n <td>1 question\/year<\/td>\n <td>Conceptual<\/td>\n <\/tr>\n <tr>\n <td><span class=\"pyq-badge\">Moderate<\/span>Conservation of Linear Momentum (collisions)<\/td>\n <td>1 question\/year<\/td>\n <td>Numerical<\/td>\n <\/tr>\n <\/tbody>\n <\/table>\n <\/div>\n\n <div class=\"callout callout-tip\">\n <span class=\"callout-icon\">Tip<\/span>\n <p>NEET PYQ pattern shows that rolling motion questions have appeared every single year since 2010 without exception. Master the I\/MR\u00b2 ratios for all standard bodies and the energy partition formula. This alone can secure 8\u201312 marks in NEET Physics.<\/p>\n <\/div>\n\n <div class=\"pdf-btn-wrap\">\n <a href=\"#\" rel=\"nofollow noopener noreferrer\" class=\"pdf-btn\">\n <svg viewBox=\"0 0 24 24\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\"><path d=\"M19 9h-4V3H9v6H5l7 7 7-7zM5 18v2h14v-2H5z\"\/><\/svg>\n Download These Notes as PDF\n <\/a>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">20<\/div>\n <h2 class=\"section-title\">Chapter <span>Summary<\/span><\/h2>\n <\/div>\n\n <p>This chapter builds a unified framework connecting translational and rotational mechanics. Every concept on the translational side has a precise rotational counterpart, and the chapter culminates in rolling motion \u2014 where both operate simultaneously.<\/p>\n\n <div class=\"revision-box\">\n <h3>Quick Revision \u2014 Must-Know Points<\/h3>\n <ul>\n <li>COM of a system: r_cm = \u03a3m\u1d62r\u1d62 \/ M; lies at geometric centre for symmetric uniform bodies.<\/li>\n <li>COM moves as if all external forces act on total mass M at that point.<\/li>\n <li>Total linear momentum P = Mv_cm; conserved when F_ext = 0.<\/li>\n <li>Angular velocity \u03c9 = d\u03b8\/dt; angular acceleration \u03b1 = d\u03c9\/dt; analogues of v and a.<\/li>\n <li>Moment of inertia I = \u03a3m\u1d62r\u1d62\u00b2; depends on axis chosen. Key values: disc = MR\u00b2\/2, ring = MR\u00b2, solid sphere = 2MR\u00b2\/5.<\/li>\n <li>Parallel axis theorem: I = I_cm + Md\u00b2; perpendicular axis theorem: I_z = I_x + I_y (laminar bodies only).<\/li>\n <li>Torque \u03c4 = r \u00d7 F; Newton’s 2nd law for rotation: \u03c4 = I\u03b1.<\/li>\n <li>Angular momentum L = I\u03c9; conserved when \u03c4_ext = 0.<\/li>\n <li>Rolling without slipping: v_cm = R\u03c9; total KE = \u00bdmv\u00b2(1 + K\u00b2\/R\u00b2).<\/li>\n <li>Inclined plane race order: solid sphere wins, ring is last.<\/li>\n <li>Power in rotational motion: P = \u03c4\u03c9; work = \u03c4\u03b8 (constant torque).<\/li>\n <\/ul>\n <\/div>\n<\/div>\n\n\n<div class=\"section\">\n <div class=\"section-header\">\n <div class=\"badge\">21<\/div>\n <h2 class=\"section-title\">Common <span>Mistakes<\/span><\/h2>\n <\/div>\n\n <ul class=\"content-list\">\n <li><strong>Using Perpendicular Axis Theorem for 3D bodies:<\/strong> It applies only to flat (laminar) objects. Applying it to a sphere or cylinder is incorrect.<\/li>\n <li><strong>Forgetting the axis when quoting I:<\/strong> Moment of inertia of a disc is MR\u00b2\/2 only about the central axis perpendicular to its plane. About a diameter, it is MR\u00b2\/4.<\/li>\n <li><strong>Confusing rolling and sliding:<\/strong> Pure rolling means v_cm = R\u03c9 exactly. If there is slipping, this constraint does not hold and friction does work.<\/li>\n <li><strong>Applying angular momentum conservation incorrectly:<\/strong> External torque must be zero. Gravity can create torque; verify the condition before applying I\u2081\u03c9\u2081 = I\u2082\u03c9\u2082.<\/li>\n <li><strong>Sign errors in torque:<\/strong> Always define a positive direction of rotation. Torques in the positive direction are positive; opposing torques are negative. Sum them algebraically.<\/li>\n <li><strong>Using translational KE formula for rolling objects:<\/strong> Rolling objects have BOTH \u00bdmv\u00b2 AND \u00bdI\u03c9\u00b2. Omitting the rotational term leads to incorrect energy calculations.<\/li>\n <li><strong>COM trajectory after explosion:<\/strong> Students often assume the COM shifts after a mid-air explosion. It does not \u2014 in the absence of external horizontal forces, the COM trajectory is unchanged.<\/li>\n <li><strong>Radius of gyration vs. radius of body:<\/strong> K (radius of gyration) is defined via I = MK\u00b2 and is not the physical radius R unless I = MR\u00b2 (ring\/hollow cylinder case).<\/li>\n <\/ul>\n<\/div>\n\n\n<div class=\"faq-section\">\n <h2 class=\"faq-title\">Frequently Asked Questions \u2014 <span>Rotational Motion NEET<\/span><\/h2>\n\n <details>\n <summary>\n What is the difference between moment of inertia and mass in rotational motion?\n <span class=\"faq-icon\">+<\/span>\n <\/summary>\n <div class=\"faq-body\">\n Mass is the measure of resistance to linear (translational) acceleration under a net force (F = ma). Moment of inertia (I) is the analogous quantity for rotational motion \u2014 it measures resistance to angular acceleration under a net torque (\u03c4 = I\u03b1). The key distinction is that while mass is an intrinsic scalar property of a body, moment of inertia depends on both the mass and the distribution of that mass relative to a chosen axis of rotation. The same body has different values of I for different axes.\n <\/div>\n <\/details>\n\n <details>\n <summary>\n Why does a solid sphere reach the bottom of an incline before a hollow sphere?\n <span class=\"faq-icon\">+<\/span>\n <\/summary>\n <div class=\"faq-body\">\n When rolling down an incline, a portion of the potential energy converts to translational KE and the rest to rotational KE. The fraction going to rotation is I\/(I + MR\u00b2). A solid sphere has I = 2MR\u00b2\/5, giving a ratio of 2\/7 \u2248 29% to rotation. A hollow sphere has I = 2MR\u00b2\/3, giving 2\/5 = 40% to rotation. Since the solid sphere devotes less energy to spinning, more goes into translational speed \u2014 it reaches the bottom faster. This result is independent of mass and radius.\n <\/div>\n <\/details>\n\n <details>\n <summary>\n How can the centre of mass lie outside the body?\n <span class=\"faq-icon\">+<\/span>\n <\/summary>\n <div class=\"faq-body\">\n The COM is a weighted average of mass positions and is not required to lie within the physical boundaries of the body. For a uniform ring or hollow cylinder, all mass lies at radius R from the centre \u2014 the centre itself is empty space, yet that is exactly where the COM is located. Similarly, a boomerang or horseshoe has its COM in the hollow interior region. Any body with a hole or concave shape where mass is absent near the geometric centre can have its COM in empty space.\n <\/div>\n <\/details>\n\n <details>\n <summary>\n Does friction do work in pure rolling motion?\n <span class=\"faq-icon\">+<\/span>\n <\/summary>\n <div class=\"faq-body\">\n No. In pure rolling without slipping, the contact point between the rolling body and the surface has zero instantaneous velocity. Since the point of application of the static friction force has zero velocity, the work done by friction is zero (W = F \u00d7 displacement of point of application = 0). However, friction is still essential \u2014 it provides the torque necessary to maintain the rolling constraint. Without friction, the body would slide rather than roll.\n <\/div>\n <\/details>\n\n <details>\n <summary>\n How is angular momentum conservation applied in the System of Particles and Rotational Motion class 11 notes?\n <span class=\"faq-icon\">+<\/span>\n <\/summary>\n <div class=\"faq-body\">\n Angular momentum is conserved when the net external torque on the system is zero. The law states L = I\u03c9 = constant in such situations. A classic example is a dancer who spins with arms outstretched (large I, small \u03c9) and then pulls arms in (smaller I). Since I\u03c9 must remain constant, \u03c9 increases \u2014 the dancer spins faster. For NEET, identify that no external torque acts, then write I\u2081\u03c9\u2081 = I\u2082\u03c9\u2082 and solve for the unknown angular velocity or moment of inertia.\n <\/div>\n <\/details>\n\n <details>\n <summary>\n What is the perpendicular axis theorem and when can it be used?\n <span class=\"faq-icon\">+<\/span>\n <\/summary>\n <div class=\"faq-body\">\n The perpendicular axis theorem states that for a laminar (flat, 2D) body, the moment of inertia about an axis perpendicular to its plane equals the sum of moments of inertia about two mutually perpendicular axes lying in the plane: I_z = I_x + I_y. This theorem is strictly limited to planar (2D) objects such as discs, rings, rectangular sheets, and triangular plates. It cannot be applied to three-dimensional bodies like spheres, cylinders, or cones.\n <\/div>\n <\/details>\n<\/div>\n\n\n<div class=\"section\" style=\"padding-top:0;\">\n <div class=\"related-links\">\n <h4>Recommended Courses and Study Materials<\/h4>\n <ul>\n <li><a href=\"https:\/\/courses.ksquare.co.in\/new-courses\/31-umeed-neet-2026\" target=\"_blank\" rel=\"nofollow noopener noreferrer\">Umeed NEET 2026 Study Materials<\/a><\/li>\n <li><a href=\"https:\/\/courses.ksquare.co.in\/new-courses\/29-pc4-29\" target=\"_blank\" rel=\"nofollow noopener noreferrer\">Grip NCERT Biology<\/a><\/li>\n <li><a href=\"https:\/\/courses.ksquare.co.in\/new-courses\/28-grip-ncert-chemistry\" target=\"_blank\" rel=\"nofollow noopener noreferrer\">Grip NCERT Chemistry<\/a><\/li>\n <\/ul>\n <\/div>\n<\/div>\n\n\n<div class=\"cta-section\">\n <h2>Ready to Master NEET Physics?<\/h2>\n <p>System of Particles and Rotational Motion demands consistent practice. Use our rank predictor, join the Mission 180 batch, or test yourself with PYQ practice now.<\/p>\n <div class=\"cta-btns\">\n <a href=\"https:\/\/courses.ksquare.co.in\/new-courses\/3-mission-180-neet-physics-rankers-batch\" target=\"_blank\" rel=\"nofollow noopener noreferrer\" class=\"btn-white\">Join Mission 180 Physics Batch<\/a>\n <a href=\"https:\/\/ksquareinstitute.in\/neet-2026-rank-predictor\/\" target=\"_blank\" rel=\"nofollow noopener noreferrer\" class=\"btn-outline\">Use NEET 2026 Rank Predictor<\/a>\n <a href=\"https:\/\/ksquareinstitute.in\/free-study-material\/\" target=\"_blank\" rel=\"nofollow noopener noreferrer\" class=\"btn-outline\">Download Free Materials<\/a>\n <\/div>\n<\/div>\n\n<\/article>\n<\/body>\n<\/html>\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n <meta charset=\"UTF-8\">\n <meta name=\"viewport\" content=\"width=device-width, initial-scale=1.0\">\n <title>Table of Contents \u2014 Physics Class 11<\/title>\n \n \n <link rel=\"preconnect\" 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cubic-bezier(0.4, 0, 0.2, 1);\n white-space: nowrap;\n }\n\n #physics-toc-wrapper a.go:hover {\n background: #18181b;\n color: #ffffff;\n transform: translateY(-2px);\n box-shadow: 0 4px 12px rgba(24, 24, 27, 0.15);\n }\n\n \/* Responsive adjustments *\/\n @media (max-width: 768px) {\n #physics-toc-wrapper h2 {\n font-size: 1.75rem;\n margin-bottom: 32px;\n }\n #physics-toc-wrapper td {\n padding: 18px 12px;\n font-size: 0.95rem;\n }\n }\n <\/style>\n<\/head>\n<body>\n\n<div id=\"physics-toc-wrapper\">\n <div class=\"container-inner\">\n <h1>Table of Contents<\/h1>\n <h2>Physics — Class 11<\/h2>\n \n <table>\n <tr><td>01<\/td><td>Units and Measurements<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/units-and-measurements-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>02<\/td><td>Motion in a Straight Line<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/motion-in-a-straight-line-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>03<\/td><td>Motion in a Plane<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/motion-in-a-plane-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>04<\/td><td>Laws of Motion<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/laws-of-motion-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>05<\/td><td>Work, Energy and Power<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/work-energy-and-power-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>06<\/td><td>System of Particles and Rotational Motion<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/system-of-particles-and-rotational-motion-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>07<\/td><td>Gravitation<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/gravitation-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>08<\/td><td>Mechanical Properties of Solids<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/mechanical-properties-of-solids-class-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>09<\/td><td>Mechanical Properties of Fluids<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/mechanical-properties-of-fluids-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>10<\/td><td>Thermal Properties of Matter<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/thermal-properties-of-matter-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>11<\/td><td>Thermodynamics<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/thermodynamics-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>12<\/td><td>Kinetic Theory<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/kinetic-theory-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>13<\/td><td>Oscillations<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/oscillations-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <tr><td>14<\/td><td>Waves<\/td><td><a class=\"go\" href=\"https:\/\/ksquareinstitute.in\/blog\/waves-11-notes\" target=\"_blank\">Go to page<\/a><\/td><\/tr>\n <\/table>\n <\/div>\n<\/div>\n\n<\/body>\n<\/html>\n","protected":false},"excerpt":{"rendered":"<p>System of Particles and Rotational Motion Class 11 Notes | NEET Physics 01 Introduction to System of Particles The System of Particles and Rotational Motion class 11 notes begin with a foundational shift in perspective: instead of treating a body as a single point mass, we now consider an extended body as a collection of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[127],"tags":[],"class_list":["post-3940","post","type-post","status-publish","format-standard","hentry","category-free-study-material"],"blocksy_meta":{"page_structure_type":"type-1","styles_descriptor":{"styles":{"desktop":"","tablet":"","mobile":""},"google_fonts":[],"version":6}},"_links":{"self":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/3940","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/comments?post=3940"}],"version-history":[{"count":2,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/3940\/revisions"}],"predecessor-version":[{"id":4209,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/3940\/revisions\/4209"}],"wp:attachment":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/media?parent=3940"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/categories?post=3940"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/tags?post=3940"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}