Angular Momentum & Conservation Angular Momentum & Conservation Angular momentum is to rotation what linear momentum is to straight motion. If a moving mass is hard to stop because it has momentum p = m v , a spinning or orbiting object is hard to stop because it has angular momentum L . You feel this when you try to stop a spinning ceiling fan with a stick, or when you rotate on a swivel chair and pull in your arms to spin faster. The central theme is simple: if no external torque acts on a system about a chosen axis, its total angular momentum about that axis remains constant. This single rule explains a skater’s spin-up, why planets sweep equal areas in equal times, and how a bullet–disc collision is solved cleanly if the right axis is chosen. The mathematics is compact and powerful. For a particle, L = r p . For a rigid body about a fixed symmetry axis, L = I . Torque = r F changes angular momentum through d L dt = ext . Over a short time, the area under torque–time gives the angular impulse and the jump in L . The art in NEET problems is picking the axis so that unknown forces exert either no torque or zero moment arm about that axis. For example, hinge forces often do no torque about the hinge axis, and vertical impulses through a vertical axle have no moment arm about the axle. When external torque is truly zero, angular momentum is locked; other quantities like mechanical energy may still change because internal forces can do work while keeping L fixed. Always attend to vectors: direction follows the right-hand rule, and signs come from your chosen axis. With a steady structure—definitions first, then laws, then edge cases—you can handle everything from stool-and-dumbbell questions to clay-on-disc impacts and planetary motion analogies. remember Real-world anchor: A figure skater spins faster by pulling in her arms. Her moment of inertia decreases, and with no external torque about the spin axis, angular momentum stays fixed, so angular speed increases. L = r p about a chosen origin. Magnitude L = r p ; direction by right-hand rule. Angular momentum (particle) Angular momentum (rigid body) About a fixed symmetry axis, L = I . In general L need not be parallel to unless the axis is a principal axis. Torque = r F ; changes angular momentum as d L dt = ext . J = ,dt = L ; useful for short-duration forces. Angular impulse Moment of inertia I = m i r i 2 (discrete) or I = r 2 dm (continuous) about the axis; rotational inertia. External vs internal torque Only external torques change the total L of a system. Internal torques occur in equal and opposite pairs and cancel in the total. Total L can be written as orbital about the center of mass plus spin about each body’s own center. Spin and orbital parts Central force A force always along the line joining body and center ( F r ). Then = 0 and L is conserved. Right-hand rule Curl fingers in direction of rotation; thumb points along L and . Sign convention and axis choice: pick a fixed axis (usually +z out of the page) and treat anticlockwise as positive. Always compute torque and angular momentum about the same axis and origin. For hinged or axle systems, the clean axis is the hinge/axle line because unknown reaction forces pass through it and give zero torque about that line. In collisions, choose the axis so that dominant unknown impulses have zero moment arm. The direction of L and follows the right-hand rule; be explicit with signs to avoid algebra mistakes. Angular momentum of a particle about a chosen origin Definition for a particle Interpretation: L = r p = m v r = m r v tan . For uniform circular motion with radius r and speed v = r , we get L = m r 2 , along the axis given by the right-hand rule. If r and p are collinear (pure radial motion), = 0 and L = 0 . Rigid body about fixed axis For rotation about a principal (symmetry) axis This relationship is fundamental to the study of rigid body dynamics when analyzing rotational motion about a fixed axis. For a rigid body with a fixed symmetric axis, all mass elements have velocities perpendicular to radii, so their contributions add to yield L = I . If the axis is not a principal axis, the relation is more general and L may not align with . In NEET questions, most axes are symmetry axes, so you can safely use L = I as a scalar with sign. External torque changes total angular momentum Torque and rate of change of L This theorem applies to a rigid body or system where the net external torque is the only factor changing the angular momentum. Total angular momentum of a system of particles Product rule for cross products Because ( v i v i = 0 ) Split net force on each particle into external and internal parts Internal forces occur in equal and opposite pairs along the same line; their torques cancel in total Sum of external torques about the same origin Newton’s second law holds: d p dt = F net Rigid bodies or particles; differentiability in time Torques and angular momentum computed about the same fixed origin d L dt = ext d L dt = ext Conservation rule: If the net external torque about a chosen axis is zero at all instants, d L dt = 0 and L is constant in both magnitude and direction. This can happen because either no external forces act or their lines of action pass through the axis so that moment arms are zero. Remember that conservation is always about a specified axis and origin. Angular impulse Area under torque–time graph equals change in angular momentum Angular impulse is handy in collisions and short pushes. If a clay blob lands on a disc, the contact force is large but brief; integrate torque over the short time to get the jump in L . If you choose the disc’s axle as axis, the bearing force has zero torque and drops out, leaving only the clay’s impulse torque. Orbital about CM plus spin about CM (with primed variables measured from CM) Total angular momentum of a system Decomposition: The total L equals orbital angular momentum of the center of mass plus the spin angular momentum about the center of mass. This split is useful: a rolling body has both spin (about its CM) and orbital (motion of CM about the chosen origin). Use the split that simplifies torque calculations for your chosen axis. Boundary conditions: Angular momentum conservation holds only if net external torque about the chosen axis is zero. It can be zero because forces pass through the axis (zero lever arm) or because external forces are central (parallel to radius). It fails if friction at an axle exerts a torque or if the axis itself accelerates in a way that injects external torque. tip Problem-solving flow for angular momentum Choose a clean axis (hinge/axle or center of mass). Declare positive sense. List external forces and check their lever arms about that axis. If net external torque is zero (or negligible) during the interval, write L i = L f about that axis. If a short-time interaction occurs, use angular impulse: ,dt = L . For extended bodies with changing shape, use I 1 1 = I 2 2 only if ext =0 . After finding f , check energy; rotational KE is 1 2 I 2 (not conserved in inelastic processes). Translational vs Rotational Analogs Linear Quantity Rotational Analog Symbol Transition Formula Mapping SI Unit To master rotation, simply swap mass m for inertia I and force F for torque while keeping the mathematical structure identical. Moment of Inertia m I I = m i r i 2 kg m 2 Torque (Moment of Force) = I N m Angular Momentum p L L = I kg m 2 s -1 Rotational Kinetic Energy K trans K rot K rot = 1 2 I 2 Angular Velocity = d dt rad s -1 Angular Acceleration = d dt rad s -2 Angular Displacement = s r rad Rotational Work W = Fs W = W = d Rotational Power P = Fv P = P = Angular Impulse J J rot J rot = L N m s Rotational Equilibrium F = 0 = 0 net = 0 N m translational vs rotational analogs For constant : = 0 + 0 t + 1 2 t 2 . Applies to fixed-axis rotation with uniform angular acceleration; use radians. = 0 + 0 t + 1 2 t 2 = 0 + 0 t + 1 2 t 2 Angular acceleration is constant in time Rotation about a fixed axis Angles measured in radians Integrate to get angular velocity Linear growth in time for constant Relate angular velocity to angle Integrate again with respect to time Final expression for angular position Kinematics vs dynamics: – – equations describe how angle changes when is known, but they do not tell you why has a certain value. Dynamics comes from = I and d L dt = ext . In conservation problems, ext =0 locks L , and may be nonzero only if I changes with time. Angular momentum L about the center Mass m = 0.20 , kg Radius r = 0.50 , m Speed v = 6.0 , m/s A 0.20 , kg stone moves in a horizontal circle of radius 0.50 , m at 6.0 , m/s . Find its angular momentum about the center. easy kg m 2 /s For circular motion, r p Compute magnitude; direction is out of page by RHR for anticlockwise motion The stone’s L points along the axis of rotation by the right-hand rule. If no external torque acted about the center (ideal string, no friction), the magnitude of L would remain constant even if the speed changed due to internal, constraint forces that exert no torque about the center. medium Angular momentum conserved Initial rotational KE Final rotational KE Energy increases because internal work is done while conserving L Initial I 1 = 3.0 , kg ,m 2 Initial 1 = 2.0 , rad/s Final I 2 = 1.2 , kg ,m 2 No external torque about spin axis rad/s, J Conservation of L : I 1 1 = I 2 2 ; rotational KE K = 1 2 I 2 An ice skater pulls her arms in. Initially I 1 = 3.0 , kg ,m 2 and 1 = 2.0 , rad/s . Finally I 2 = 1.2 , kg ,m 2 . Find 2 and the change in rotational KE. 2 and K Key insight: Angular momentum conservation does not imply energy conservation. Internal forces can rearrange mass and do work, changing I and the rotational KE while keeping L fixed because no external torque acts about the spin axis. About the axle; incoming momentum is tangential After sticking, treat as a single rigid body Angular momentum conserved about axle Disc initially at rest Most energy is lost as heat and deformation About 91% energy loss; L conserved but K not hard Disc: M=2.0 , kg , R=0.20 , m Clay: m=0.10 , kg , v=8.0 , m/s Axle is smooth (reaction through axis, zero torque about axis) Impact is short; use angular impulse about axle About the axle: I disc = 1 2 MR 2 , I ball = mR 2 (point mass at rim). Angular impulse from ball equals L . rad/s, dimensionless A uniform disc ( M=2.0 , kg , R=0.20 , m ) is at rest on a smooth vertical axle. A small 0.10 , kg clay ball moving horizontally at 8.0 , m/s sticks at the rim. The ball’s line of motion is tangential at impact. Find the final angular speed and the fractional loss of mechanical energy. f and energy loss fraction custom Time Before impact: L = 0 0.005 Jump due to angular impulse 0.16 0.1 0.16 After impact: L conserved Angular impulse causes a step change in L; with zero external torque after, L is constant. kg m 2 /s Angular momentum about axle L rises sharply from 0 to 0.16 at impact (near t ≈ 0), then stays constant at 0.16 afterwards. dependent This torque–time impulse view shows why angular momentum can jump during a short interaction but then remain fixed. It also clarifies why choosing the axle as axis eliminates the unknown bearing reaction: it has zero lever arm about its own line. Stool+student moment of inertia I s=1.5 , kg ,m 2 Bag mass m=2.0 , kg caught at radius r=0.80 , m Initial 0=1.2 , rad/s Bag’s impulse line passes through axis (no external torque about axis during catch) Frictionless vertical axle hard Add point mass at radius r Initial angular momentum Reduced angular speed due to increased I Final f A student stands on a frictionless rotating stool holding a 2.0 , kg bag at arm’s length ( r=0.80 , m ). Student+stool has I s=1.5 , kg ,m 2 and spins initially at 0=1.2 , rad/s . The student catches the bag (thrown tangentially by a friend so that, at the instant of catch, its line of motion passes through the stool’s axis). The bag sticks at r=0.80 , m . Find the final f just after the catch. rad/s About the axle: L conserved. Before catch: L i = I s 0 . After catch: I f = I s + mr 2 and L f = I f f . Why is L conserved here? The impulsive force exerted by the bag acts through the vertical axis at the instant of catch, so its lever arm about the axis is zero. The axle’s reaction also acts through the axis. Hence ext =0 about that axis during the impulse. neet-alert Do not apply I 1 1 = I 2 2 blindly. It holds only if the net external torque about the chosen axis is zero during the change. In hinge-collision problems, pick the hinge axis so that hinge impulses have zero torque. About the center of mass, hinge impulses usually do exert torque and L is not conserved. Conservation depends on external torques about the specific axis. Even with external forces present, L can be conserved if their lines of action pass through the axis (zero torque). With no external forces at all, L is conserved about any fixed origin. If there are no external forces, angular momentum is always conserved about any axis. Only if ext =0 about the axis during the change. If an external torque acts, L changes and need not increase. When I decreases, must increase in all situations. remember Right-hand rule mnemonic: Curl your fingers in the direction of rotation; your thumb gives the direction of and L . Thumb-up spin: Fingers curl with rotation, thumb points along angular momentum. Think “Curl to Spin, Thumb for L.” right-hand-rule List all large unknown forces (hinge reactions, contact impulses). Pick the axis through their lines of action so their torques vanish. For central forces (gravity by a central body), choose that body as origin. If shape changes but no external torque, pick the spin axis to conserve I . How to choose the axis like a pro Planets and satellites move under central gravitational force directed along r toward the Sun/Earth. Since = r F and F - r , the torque about the Sun/Earth is zero. Thus the orbital angular momentum about that center is conserved, and the areal velocity is constant (Kepler’s second law). Consequence: constant areal velocity Central force implies zero torque Rolling without slipping mixes translation and rotation. About the center of mass, L = I CM (spin) and the external torque from friction may be zero or nonzero depending on the scenario. About the instantaneous contact point, results can be elegant but beware: that point accelerates and requires care if using energy or angular momentum. Safest is to compute about a fixed axis or about the CM with well-identified external torques. Trap alert in rolling: The instantaneous contact point is not an inertial pivot; using it as a fixed axis for conservation can mislead. Use CM or a fixed ground axis unless you are using the well-known rolling relations carefully. tip A 1000 , kg satellite moves in an elliptical orbit about Earth. At a point where its distance from Earth’s center is 7.0 10 6 , m and speed is 7.5 , km/s , find the areal velocity A . Areal velocity A For central force, A = L 2m . Here L = mrv with v = v since motion is tangential to radius at that instant (true at periapsis/apoapsis; otherwise use component). m 2 /s Mass m = 1000 , kg (drops out) Position r = 7.0 10 6 , m Speed v = 7.5 , km/s = 7.5 10 3 , m/s Central gravitational force (zero torque about Earth’s center) Angular momentum about Earth’s center Constant everywhere along the orbit medium Areal velocity has units of area per time and remains constant for motion under any central force. It is a direct restatement of angular momentum conservation because A = 1 2 r v = L/(2m) . Advanced note: For an arbitrary rigid body about a non-principal axis, L need not be parallel to . The simple scalar L = I is safe only for rotation about a principal (symmetry) axis. NEET typically keeps to symmetry axes, but keep this in mind for conceptual clarity. Units and dimensions Use these to sanity-check answers Precession glimpse: If a torque acts perpendicular to L (like gravity on a spinning top’s CM offset from the pivot), L changes direction at nearly constant magnitude, causing the axis to precess. While beyond our present scope, it reinforces that torque changes L ’s vector, not just its size. Scenario Clean axis Is L conserved? Why Skater pulls in arms Spin axis Yes No external torque about spin axis; I changes internally Clay hits disc on axle Axle axis Yes (during impact) Axle reaction through axis; only clay’s impulse torque matters Bullet strikes hinged rod Hinge axis Yes (during impact) Hinge impulse through axis, zero torque there Planet orbits Sun Sun as origin Yes Central gravitational force; =0 Rolling sphere on rough plane CM axis Not generally Friction can exert external torque about CM depending on motion Radians only: In = 0 + 0 t + 1 2 t 2 , use radians. Degrees will give wrong magnitudes. Angles are dimensionless in SI, so keep track of units explicitly. neet-alert Dimensional sense-checks prevent errors. If you compute L and do not get units of kg ,m 2/s , re-check your lever arm. If torque comes out without m , you likely missed the perpendicular distance to the axis. L = r p ; for symmetric rigid rotation L = I Angular momentum rotational momentum moment of force = r F ; d L dt = ext Torque Rotational analogue of mass: I = r 2 dm about the chosen axis Moment of inertia Rate of rotation: = d dt Angular speed Angular impulse ,dt = L Conservation condition L is conserved when ext = 0 about the chosen axis Right-hand rule Direction rule for L and using curled fingers and thumb Angular Momentum Recap