Chapter 6: Systems of Particles and Rotational Motion | Summary

The centre of mass (CM) of a system of particles is a specific point whose position is the mass-weighted mean of all particle positions. For a two-particle system with masses m1 and m2 at positions x1 and x2, the CM is at X = (m1x1 + m2x2)/(m1 + m2). For equal masses, the CM lies exactly midway between them. This concept generalises to n particles in three-dimensional space using the vector relation R = (Σmi ri)/M, where M is total mass.

For homogeneous bodies of regular shape (rings, discs, spheres, rods), the CM coincides with the geometric centre, which can be shown using symmetry arguments. For a continuous body, summation is replaced by integration: R = (1/M) ∫r dm. When the CM is chosen as the origin, Σmi ri = 0.

Differentiating the CM position with respect to time gives MV = Σmivi = P, showing that the total linear momentum of a system equals M times the velocity of the CM. A second differentiation yields MA = Fext, meaning the CM of any system moves as if all the mass were concentrated there and all external forces acted on that point. Internal forces cancel in pairs by Newton's third law and do not affect CM motion.

This result explains why a projectile's CM continues on its original parabolic path even after an internal explosion breaks it into fragments — only external gravity acts on the system. The translational motion of any rigid body (rotating or not) can always be described through its CM using MA = Fext.

The total linear momentum of a system is P = MV, where V is the velocity of the CM. Newton's second law for a system is dP/dt = Fext. If the net external force is zero, P remains constant — this is the law of conservation of linear momentum for a system. Equivalently, the CM moves with constant velocity when no net external force acts.

Examples include radioactive decay: the CM of a radium nucleus continues moving uniformly after it splits into radon and an alpha particle, since the decay forces are internal. In binary star systems, the CM moves in a straight line while the two stars orbit around it in complicated paths.

The vector (cross) product of two vectors a and b is a vector c = a × b with magnitude |c| = ab sinθ, directed perpendicular to the plane of a and b according to the right-hand screw rule. Unlike the scalar product, the vector product is not commutative: a × b = −b × a. It is, however, distributive: a × (b + c) = a × b + a × c.

Key cross products of unit vectors are: i × j = k, j × k = i, k × i = j, and i × i = j × j = k × k = 0. The component form is a × b = (aybz − azby)i + (azbx − axbz)j + (axby − aybx)k, which can be written as a 3×3 determinant. Torque and angular momentum are both defined as vector products.

Every particle of a rigid body rotating about a fixed axis moves in a circle. The instantaneous angular velocity ω = dθ/dt is the same for all particles of the body. For a particle at perpendicular distance r from the axis, its linear velocity magnitude is v = ωr. More precisely, the vector relation is v = ω × r, where ω is directed along the axis by the right-hand rule.

Angular acceleration α = dω/dt is the time rate of change of angular velocity. For a fixed axis, both ω and α are along the axis direction, reducing vector equations to scalar ones. Pure rotation is characterised by all particles having the same angular velocity at any instant, analogous to pure translation where all particles have the same linear velocity.

Torque (moment of force) is τ = r × F, with magnitude τ = rF sinθ. Its SI unit is N m and dimensions are ML2T−2. Torque is the rotational analogue of force. For a single particle, dl/dt = τ, meaning torque equals the time rate of change of angular momentum. Angular momentum of a particle is l = r × p, with magnitude l = rp sinθ.

For a system of particles, the total angular momentum L = Σli and dL/dt = τext. If the total external torque is zero, L is conserved. This law has applications in rotating chairs, acrobats, ballet dancers, and divers who change their angular speed by redistributing their mass (changing moment of inertia) while no external torque acts.

A rigid body is in mechanical equilibrium when both its linear and angular momenta are constant (no linear or angular acceleration). Two conditions must hold: (1) Translational equilibrium — net external force is zero: ΣFi = 0. (2) Rotational equilibrium — net external torque is zero: Στi = 0. For coplanar forces, three scalar conditions (two force components and one torque) must be satisfied.

Partial equilibrium is possible: a body may be in translational equilibrium but not rotational (or vice versa). A couple — two equal and opposite forces with different lines of action — creates pure rotation without translation. The principle of moments for a lever states: load arm × load = effort arm × effort. Mechanical advantage M.A. = F1/F2 = d2/d1 relates the effort and load arms.

The centre of gravity (CG) of a body is the point about which the total gravitational torque on the body is zero. This is defined by Στi = Σ(ri × mi g) = 0. Since g is the same for all particles in a uniform gravitational field, this condition reduces to Σmi ri = 0, which means the CG coincides with the CM in a uniform gravitational field.

For an extended body in a non-uniform gravitational field, the CG and CM do not coincide. For small bodies where g is effectively constant, the CG is the same as the CM. Experimentally, the CG of an irregular body can be found by suspending it from different points — the CG lies on the intersection of the vertical lines through each suspension point.

The moment of inertia I = Σmiri² is the rotational analogue of mass. The kinetic energy of a rotating body is K = (1/2)Iω², analogous to K = (1/2)mv² for translation. Moment of inertia depends on the mass, shape, size, and distribution of mass of the body relative to the axis of rotation. Unlike mass, it is not fixed but depends on the choice of axis.

For a thin ring of radius R and mass M about its central axis, I = MR². For a solid cylinder about its axis, I = MR²/2. For a thin rod of length L about a perpendicular axis through its midpoint, I = ML²/12. The radius of gyration k is defined by I = Mk², giving a geometric measure of mass distribution. A flywheel uses large I to resist sudden changes in rotational speed.

The kinematic equations for uniformly accelerated rotation are exactly analogous to those for linear motion: ω = ω0 + αt; θ = θ0 + ω0t + (1/2)αt²; ω² = ω0² + 2α(θ − θ0). Here θ is angular displacement (replacing x), ω is angular velocity (replacing v), and α is angular acceleration (replacing a). These follow directly from integrating α = dω/dt.

This one-dimensional analogy holds only for fixed-axis rotation, which involves a single degree of freedom. The angular displacement of any particle of the rigid body can represent the angular displacement of the entire body. Directions of ω and α are fixed along the axis, so scalar treatment suffices.

The rotational equation of motion τ = Iα is the analogue of Newton's second law F = ma. Work done by a torque in angular displacement dθ is dW = τ dθ, and instantaneous power is P = τω. These are analogous to dW = F ds and P = Fv in linear motion. The full translational–rotational analogy includes: displacement↔angular displacement, velocity↔angular velocity, mass↔moment of inertia, force↔torque, momentum↔angular momentum.

For rotation about a fixed axis, only force components perpendicular to the axis and position vector components perpendicular to the axis contribute to the effective torque. Constraint forces handle the perpendicular torque components automatically. In a rigid body, work done by external torques entirely converts to rotational kinetic energy.

For a rigid body rotating about a fixed (z) axis, the component of angular momentum along the axis is Lz = Iω. For symmetric bodies (axis is a symmetry axis), the perpendicular component L⊥ vanishes, so L = Iω directed along the axis. For asymmetric bodies, L and ω are not parallel. The equation of motion along the axis is d(Iω)/dt = τext.

If τext = 0, then Iω = constant (conservation of angular momentum). Practical examples: a skater spinning faster by pulling arms in (reducing I increases ω), a diver tucking to spin faster, a girl on a swivel chair stretching or folding arms. Acrobats and classical dancers exploit this principle in their performances.

This chapter establishes the complete framework for analysing rotational and combined motion of rigid bodies. The centre of mass concept unifies the treatment of any system — its motion obeys MA = Fext regardless of internal complexity. The precise analogies between translational and rotational quantities (mass↔moment of inertia, force↔torque, linear momentum↔angular momentum) allow all results of linear dynamics to be mirrored in rotational dynamics, with kinematic equations, Newton's second law, work-energy theorem, and conservation laws carrying over directly.

For CBSE board examinations, this chapter is highly important. Students should master: calculation of CM for discrete and continuous systems, conditions for equilibrium and their application to levers and ladders, derivation of τ = Iα and use of rotational kinematic equations, moment of inertia values from Table 6.1, and conservation of angular momentum with real-world examples. Numericals from equilibrium, MI, kinematics of rotation, and angular momentum conservation appear frequently in board and competitive exams.