Centre of Mass & System of Particles Centre of Mass & System of Particles Centre of mass (CoM) is the single point that represents the overall“location” of mass for a system. If you want to track the motion of many particles together, the CoM is the shortcut: apply all external forces to that point and you get the correct translation of the whole system. In everyday terms, when you carry a heavy bag and a light lunchbox on a rod, the balance point shifts towards the heavier bag. That balance point is the CoM. In physics, the CoM is defined as a mass-weighted average of positions: heavier parts pull the average more. For two particles on a line, the CoM lies closer to the heavier one; for a uniform rod, it’s at the centre because mass is symmetrically spread. CoM extends cleanly to 2D and 3D bodies, to discrete particle sets, and to continuous mass distributions (rods, plates, solids). Most importantly for dynamics, internal forces (tensions inside, action–reaction pairs) always cancel in the total, so they cannot accelerate the CoM. Only external forces can change the CoM’s motion, and the CoM obeys Newton’s second law with the system’s total mass. That’s why a firework shell that explodes mid-air has fragments flying all over, but their collective CoM still follows the same parabolic path as if there was no explosion. CoM also connects directly to momentum: total momentum equals total mass times CoM velocity. In rotational problems, choosing the CoM as origin often simplifies torque and angular momentum. This chapter builds your intuition first, then equips you with precise formulas, careful sign conventions, edge cases, and NEET-style strategies. remember Analogy: Think of the CoM as the balancing point on a see-saw with children of different weights sitting at different positions. Heavier child pulls the balance point towards them. In 2D, it’s the point where you could support a uniform plate with one finger without it tipping. Core vocabulary The mass-weighted average position of a system. In vector form: the unique point whose position vector equals total first moment of mass divided by total mass. Centre of mass (CoM) System of particles A collection of particles or rigid bodies considered together so that total mass, total momentum, and internal vs external forces are meaningfully defined. Position vector Vector from the chosen origin to a particle’s location. Choice of origin is arbitrary but must be consistent. The point where the resultant weight can be considered to act. CoG coincides with CoM only in a uniform gravitational field. Centre of gravity (CoG) External vs internal forces External forces come from outside the system boundary and can change the CoM motion. Internal forces are action–reaction pairs within the system and cancel in the total. Discrete systems: definition and coordinates For n particles with masses m 1, m 2, , m n at position vectors r 1, r 2, , r n from a fixed origin O , define total mass M = i=1 n m i . The centre of mass position vector R cm is the mass-weighted average: heavier masses pull the average more strongly. In 1D, it reduces to a weighted mean of x -coordinates: x cm = (m 1 x 1 + m 2 x 2 + )/M . In 2D or 3D, compute each component similarly: x cm , y cm , and z cm . The CoM changes predictably if you translate the origin: shift the origin by a and every position vector changes by a , so R cm also shifts by a . Relative positions (like distances within the system) do not change. Centre of mass of n particles as a mass-weighted average. Vector definition (discrete) Compute each component independently using the same weighted-average idea. Component form Interpretation: R cm is where a single mass M would have to be placed so that the system’s mass distribution has the same total first moment. If all masses lie on a line, the CoM lies on the same line. If the masses are symmetrically placed and equal, the CoM is at the geometric centre. If one mass is much heavier than the others, the CoM lies very close to it. Use x cm = (m 1 x 1 + m 2 x 2)/(m 1 + m 2) . Two-particle CoM on a line CoM lies closer to the heavier mass at 4 m. Masses: m 1 = 2 , kg at x 1 = 0 , m Mass: m 2 = 6 , kg at x 2 = 4 , m easy x cm neet-alert Trap: Do not use a simple average of positions unless masses are equal. Always use the mass-weighted average for x cm . Continuous mass distributions For continuous objects (rods, plates, solids), replace summation by integration. Use appropriate density: linear density = d m/ d for wires, surface density = d m/ d A for plates, and volume density = d m/ d V for solids. Pick a coordinate system that matches symmetry. Many CoM questions are easy if you exploit symmetry: for a uniform semicircular wire, x cm =0 by symmetry about the y -axis; you only need y cm . Vector definition (continuous) Replace sum with integral over the mass distribution. Choose d m in terms of , , or and integrate over geometry. Component integrals Use this formula when the system consists of a few distinct particles, contrasting with the continuous integral approach. CoM of a uniform semicircular wire of radius R lies on the symmetry axis at y cm = 2R from the centre. Uniform linear density Arc angle 0 with origin at circle centre Symmetry about y -axis so x cm = 0 y cm = 2R , x cm = 0 Numerical CoM of a semicircular wire Use y cm = 2R/ for uniform semicircular wire. Distance of CoM from the circle centre along symmetry axis medium Uniform wire, radius R = 0.50 , m forming a semicircle Symmetry first: If a body is symmetric about an axis and density is uniform, the CoM lies on that axis. This can zero out components without any integration. tip Motion of the centre of mass Define the CoM velocity V cm = d R cm d t = 1 M m i v i , and acceleration A cm = d V cm d t = 1 M m i a i . By Newton’s second law for each particle, m i a i = F ,( ext ) i + f ,( int ) i . When you sum over all particles, internal forces cancel pairwise (action–reaction), leaving only external forces. Therefore, the CoM obeys M , A cm = F ext . This means the translation of any complex system can be computed as if all mass were concentrated at the CoM and all external forces acted at that point. CoM velocity and acceleration Mass-weighted average of velocities and accelerations. This calculation determines the initial velocity of the system's center of mass, which dictates the system's overall translation. Newton’s laws hold for each particle Internal forces occur in equal and opposite pairs (Newton’s third law) and act along the line joining the pair Mass M is constant (no mass ejection/accumulation) Newton’s second law for a system: M , A cm = F ext M , A cm = F ext Only external forces can change the motion of the CoM. Internal explosions, springs, or tensions may violently rearrange parts, but the CoM keeps its course unless acted upon from outside. remember Consequences: If F ext = 0 , then A cm = 0 and the CoM moves at constant velocity in a straight line. If F ext is constant, R cm (t) is parabolic in the direction of the force. For projectile fragmentation in uniform gravity, the CoM follows the same parabola as the original projectile because F ext = M g is unchanged. External force is unchanged (uniform gravity). CoM follows the same trajectory as the intact projectile. Compute the range R of the original projectile. Projectile fragments mid-flight Explosion changes internal energy and individual paths, not the CoM path. A projectile is launched with u = 20 , m ,s -1 at 45 above horizontal. Gravitational field is uniform with g = 10 , m ,s -2 . At the top of its trajectory, it explodes into two fragments that land at different points. medium Where does the CoM land relative to the launch point? Composite bodies can often be handled by the add–remove (superposition) method. Treat a missing part (a hole) as negative mass with the same density as the rest. Compute the CoM of the full body and the CoM of the removed part, then take a mass-weighted average including the negative mass. This is an algebraic bookkeeping of first moments. Composite (add–remove) method: steps Choose an origin and axes suited to symmetry. Find total mass M = V (or A , L ) of the original full body. Find mass and CoM of the removed/added part using the same density. Use R cm = (M 1 R 1 + M 2 R 2 + )/ M i with signs (negative for removed). Check direction and magnitude; if you doubled a length, the CoM should move roughly proportionally. neet-alert Trap in composite bodies: Always divide by the net remaining mass, not the original mass. A removed hole contributes negative mass in the average. CoM of a circular disc with an off-centre circular hole Treat the hole as negative mass. M full = R 2 at (0,0) . M hole = r 2 at (d,0) with negative sign. New CoM of the remaining lamina hard Uniform disc of radius R centred at origin O A small disc of radius r is removed with its centre at (d, 0) on the x -axis Surface density is uniform Non-uniform density requires building d m correctly. For a rod along x from 0 to L with linear density (x) , use d m = (x) , d x . Then x cm = 0 L x , (x) , d x 0 L (x) , d x . Example: (x) = 0 x (heavier towards the right) gives x cm = 2L/3 . x cm medium Rod from x=0 to x=L on the x -axis Linear density (x) = 0 x CoM of a rod with linearly varying density Use x cm = 0 L x , (x) , d x 0 L (x) , d x . Thin uniform rod of length L Midpoint Rectangular plate a × b Geometric centre Circular ring / disc Centre of circle Solid sphere / spherical shell Centre of sphere Semicircular wire (radius R) On symmetry axis at 2R/π from centre Semicircular lamina (radius R) On symmetry axis at 4R/(3π) from centre Body (uniform) CoM location Momentum connection: total linear momentum P = m i v i = M V cm . In an isolated system, P is conserved, so V cm is constant. This is the backbone of many collision and explosion questions. In rotational analysis about a point O , the total angular momentum can be split into motion of the CoM and rotation about the CoM, which is why choosing the CoM as origin simplifies torque relations. Total momentum and CoM In the CoM frame, total momentum is zero. 0 to T Time t If external force is zero, the CoM position vs time is a straight line with slope v0. With constant external force (constant a), the curve is a parabola opening upwards or downwards depending on the sign of a. control x0 v0 control control Initial position x0 x0 + v0 t + 0.5 a t 2 General point Position of CoM along x x0 to x0 + v0 T + 0.5 a T 2 st Set a = 0 for force-free motion (straight line); a ≠ 0 gives a parabola. 2D PLOT Centre-of-mass position vs time xcm = x0 + v0 t + 0.5 a t 2 xcm x0 Initial position v0 Initial CoM velocity CoM acceleration CoM can lie outside. Example: a semicircular wire’s CoM is in empty space along the symmetry axis; a boomerang’s CoM can be outside its material. The CoM must lie inside the object. Internal forces cancel in the total. They can change the configuration but not the CoM’s acceleration. Only external forces affect A cm . Internal forces can shift the CoM motion. They coincide only in a uniform gravitational field. In non-uniform fields (e.g., very large objects or near massive bodies), CoG and CoM can differ. Centre of gravity always equals centre of mass. Origin shift: If you shift the origin by a , then R cm shifts by the same a . Relative positions and the physics (like A cm ) are unchanged. tip CoM frame: An inertial frame moving with V cm where the total momentum is zero. In this frame, collisions often become symmetric, and calculations of relative speeds are straightforward. While energies transform between frames, impulse–momentum analyses are often simplest in the CoM frame. A person of mass m = 60 , kg stands at one end of a 6 , m long boat of mass M = 140 , kg floating on still water (no external horizontal force). The person walks to the other end relative to water (slowly, no splashing). Boat’s displacement relative to water when the person reaches the other end hard Person walking on a boat: CoM stays fixed horizontally No external horizontal force, so x cm remains fixed. Let right be positive. Initially, place origin at the boat’s initial left end. Negative means the boat moves 1.8 m left as the person walks right. neet-alert Boat–person trap: Use a fixed ground/water frame. If you compute relative to the moving boat, you’ll mix frames and get the wrong displacement. External horizontal force must be zero for CoM conservation. Draw a clear diagram with a sensible origin and axes. Mark masses and position vectors (or coordinates). Decide: discrete sum or continuous integral? Pick , , appropriately. Exploit symmetry to zero out components whenever possible. For dynamics, isolate external forces; check if F ext = 0 . In composite bodies, include negative masses for holes and divide by net mass. Sanity-check: Is the CoM closer to heavier parts? Does direction make sense? Checklist for CoM problems Choosing origin at the CoM eliminates the M R cm V cm term, simplifying rotational dynamics. Torque–angular momentum split (about O) tip Uniform gravity shortcut: For rigid bodies near Earth’s surface, gravity is effectively uniform, so weight can be taken to act at the CoM (equivalently CoG). This justifies balancing at the CoM. Connection to rotational kinematics: In later topics, we describe rotation about a fixed axis using , , . For a rigid body in general plane motion, the translation of the CoM plus rotation about the CoM fully describes the motion. Thus, once you know R cm (t) from external forces, you can superpose rotation about the CoM. Rotational kinematics ( = 0 + 0 t + 1 2 t 2 ) will later model rotation about the CoM or any fixed axis; here, CoM gives the translational part. M on top of Mean: Weighted MEAN with total MASS on top. x cm = ( m i x i)/M — never forget the M in the denominator. Edge cases and limits: If one mass m k 0 , it has negligible effect on R cm . If one mass dominates ( m 1 others), R cm r 1 . As distances grow large, a far but tiny mass may still matter due to m x in the numerator; always compute the product. In non-uniform gravity, CoG may not coincide with CoM; near Earth’s surface we safely take them equal. Sign convention trap: Always carry coordinate signs into the weighted sums. A mass at x = -2 , m contributes a negative term m(-2) to x cm . neet-alert Worked reasoning example (no numbers): In a dumbbell (two equal masses at ends of a light rod), the CoM is at the midpoint by symmetry. Tilted or rotated configurations do not change R cm relative to the body; only translation of the whole affects it. If one mass is doubled, the CoM shifts to 1/3 of the distance from the heavier mass along the line joining them. Dimensional consistency: x cm has dimensions of length because it is a ratio of a first moment m i x i (units kg ,m ) to mass M (units kg ). In integrals, ensure d m is expressed correctly so units match. This quick check can catch algebra slips. Strategy for NEET numericals: Reduce the system to a few representative points using symmetry. Use CoM conservation only along directions with zero net external force. Convert continuous mass to integrals only if symmetry fails or a formula is unknown. For composite plates, place the origin at an intersection of symmetry lines to simplify signs and arithmetic. m 1 = 1 , kg at (0,0) m 2 = 2 , kg at (2,0) m 3 = 3 , kg at (0,3) easy (x cm , y cm ) Use component averages: x cm = ( m i x i)/M , y cm = ( m i y i)/M . Three masses on a plane: find CoM General method summary: Identify masses and geometry, pick a smart origin, write either a sum or integral for each coordinate, apply symmetry to drop terms, evaluate carefully with signs, and interpret the result (direction, magnitude). For dynamics, reduce the system’s external forces to get A cm and integrate to find R cm (t) . Centre of mass R cm CoM Mass-weighted average position; governs translational motion. External force Force from outside the system; changes A cm . Force between system parts; cancels in total. Internal force Linear density = d m/ d for wires. Surface density = d m/ d A for plates. = d m/ d V for solids. Volume density Key terms recap