System of Particles and Rotational Motion

The centre of mass (CM) of a system of particles is the point at which the entire mass of the system may be considered to be concentrated for the purpose of describing the translational motion of the system as a whole. It is purely a geometric/mass concept and need not coincide with any actual particle of the body — for example, the centre of mass of a uniform ring lies at its centre, where there is no mass at all.

For two particles of masses m₁ and m₂ situated at positions x₁ and x₂ on the x-axis, the x-coordinate of the centre of mass is:

• X_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂)

For a system of n particles, this generalises to:

• X_cm = Σm_ix_i / M, Y_cm = Σm_iy_i / M, Z_cm = Σm_iz_i / M, where M = Σm_i is the total mass.

In vector form, R_cm = (1/M) Σ m_i r_i. For a continuous rigid body, the sum is replaced by an integral over infinitesimal mass elements: R_cm = (1/M) ∫ r dm.

For homogeneous bodies possessing a point, line, or plane of symmetry (a uniform rod, ring, disc, sphere, or cube), the centre of mass lies at the geometric centre, on the axis of symmetry, or on the symmetry plane respectively — a very useful shortcut for board-level problems.

Worked Example.

• Given: A particle of mass m₁ = 1 kg is at x = 0, and a particle of mass m₂ = 3 kg is at x = 4 m on the x-axis.
• Formula: X_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂)
• Substitution: X_cm = (1 × 0 + 3 × 4) / (1 + 3) = 12 / 4
• Result: X_cm = 3 m from the origin, closer to the heavier 3 kg mass, as expected.

Once the centre of mass of a system is known, its overall motion — however complicated the internal motion of the individual particles — can be studied by treating the CM as though it were a single particle of mass M acted upon by the net external force.

The velocity of the centre of mass is obtained by differentiating its position vector with respect to time:

• V_cm = dR_cm/dt = (1/M) Σ m_i v_i

Multiplying by M gives the total linear momentum of the system:

• P = M V_cm = Σ p_i

Differentiating P with respect to time, and using the fact that internal action-reaction force pairs (Newton's third law) cancel out in the sum, gives the equation of motion of the system:

• F_ext = M A_cm = dP/dt

Only the external forces determine how the centre of mass moves; internal forces (mutual forces between particles of the system, such as in an explosion or collision) can never change the total momentum or the motion of the centre of mass.

If the net external force on a system is zero, dP/dt = 0, so the total linear momentum P stays constant — the law of conservation of linear momentum. This holds even when the particles of the system exert complicated internal forces on one another.

Worked Example.

• Given: An object of mass 20 kg moving at 5 m/s explodes into two fragments of mass 8 kg and 12 kg. After the explosion, the 8 kg fragment continues in the same direction with speed 10 m/s.
• Formula: Conservation of momentum: Mv = m₁v₁ + m₂v₂
• Substitution: 20 × 5 = 8 × 10 + 12 × v₂ ⟹ 100 = 80 + 12v₂
• Result: v₂ = 20/12 = 1.67 m/s, in the same direction as the original motion — the internal explosive forces redistribute momentum but never change its total.

Several rotational quantities are defined using the vector (cross) product of two vectors. For vectors A and B inclined at angle θ:

• A × B = AB sinθ n̂, where n̂ is a unit vector perpendicular to the plane containing A and B, its direction given by the right-hand rule.

Key properties: A × B = −(B × A) (not commutative); A × A = 0; and A × B = 0 if A is parallel or anti-parallel to B.

Angular velocity ω describes the rate of change of angular position: ω = dθ/dt, directed along the axis of rotation. It relates to the linear velocity v of a particle at position r from a point on the axis by:

• v = ω × r, so speed v = ωr sinθ, reducing to v = ωr when r is measured perpendicular to the axis.

Torque (moment of force) τ about a point O, due to a force F acting at position r relative to O:

• τ = r × F, magnitude τ = rF sinθ.

Torque is the rotational analogue of force — it produces angular acceleration exactly as force produces linear acceleration.

The angular momentum l of a particle of momentum p = mv at position r relative to O is:

• l = r × p, magnitude l = rp sinθ.

For a system of particles, total angular momentum L = Σ l_i and total torque τ = Σ τ_i. Just as force is the rate of change of linear momentum, torque is the rate of change of angular momentum:

• τ_net = dL/dt

This is the rotational analogue of Newton's second law, valid for a system of particles about a fixed point when internal torques (from Newton's third-law force pairs acting along the line joining particles) cancel in pairs.

Worked Example.

• Given: A force F = (3i + 4j) N acts at a point of position vector r = 2i m relative to the origin.
• Formula: τ = r × F
• Substitution: τ = (2i) × (3i + 4j) = 6(i × i) + 8(i × j) = 0 + 8k, using i × i = 0 and i × j = k
• Result: τ = 8k N·m — a torque of magnitude 8 N·m directed along the positive z-axis.

From τ_net = dL/dt, if the net external torque on a system about a given point (or axis) is zero, then dL/dt = 0, so the total angular momentum L remains constant. This is the law of conservation of angular momentum.

A rigid body rotating about a fixed axis has angular momentum L = Iω, where I is its moment of inertia about that axis. If no external torque acts, L = Iω stays constant even if the body's mass redistributes about the axis, changing I. Hence:

• I₁ω₁ = I₂ω₂

A classic illustration is an ice skater (or a person on a rotating stool) spinning with arms outstretched. Pulling the arms inward decreases the moment of inertia I about the spin axis; since L = Iω must remain constant, the angular speed ω increases. The same principle explains why a diver curls into a tighter tuck to spin faster mid-air, then straightens out to slow the spin before entering the water.

Worked Example.

• Given: A spinning skater has moment of inertia I₁ = 6 kg m² with arms outstretched, spinning at ω₁ = 2 rad/s. She pulls her arms in, reducing her moment of inertia to I₂ = 2 kg m².
• Formula: I₁ω₁ = I₂ω₂
• Substitution: ω₂ = I₁ω₁ / I₂ = (6 × 2) / 2
• Result: ω₂ = 6 rad/s — three times faster than before, purely because the moment of inertia was reduced.

The moment of inertia (I) of a rigid body about a given axis measures its rotational inertia — its reluctance to change its state of rotation about that axis — playing the same role in rotational motion that mass plays in linear motion. For a system of particles, I = Σ m_ir_i², where r_i is the perpendicular distance of the i-th particle from the axis; for a continuous body, I = ∫ r² dm.

Unlike mass, moment of inertia depends on the choice of axis as well as on how mass is distributed relative to it. It is often expressed via the radius of gyration K, defined by I = MK².

Perpendicular axes theorem (for a planar lamina only): if x and y axes lie in the plane of the lamina and intersect at O, and z is the axis perpendicular to the plane through O, then:

• I_z = I_x + I_y

Parallel axes theorem (for any rigid body): the moment of inertia I about any axis equals the moment of inertia I_cm about a parallel axis through the centre of mass, plus Md², where d is the perpendicular distance between the two axes:

• I = I_cm + Md²

Standard moments of inertia used at this level:

• Thin uniform rod, about a perpendicular axis through its centre: I = ML²/12
• Thin uniform rod, about a perpendicular axis through one end: I = ML²/3
• Ring, radius R, about the central axis perpendicular to its plane: I = MR²
• Ring, about a diameter: I = MR²/2
• Disc, radius R, about the central axis perpendicular to its plane: I = MR²/2
• Disc, about a diameter: I = MR²/4
• Solid sphere, about a diameter: I = (2/5)MR²
• Solid cylinder, about its own central longitudinal axis: I = MR²/2

Worked Example.

• Given: A uniform disc of mass M = 2 kg and radius R = 0.5 m spins about an axis through its centre, perpendicular to its plane.
• Formula: I_cm = MR²/2 (central axis); I = I_cm + Md² for a parallel axis shifted to the rim, with d = R.
• Substitution: I_cm = 2 × (0.5)² / 2 = 0.25 kg m²; then I = 0.25 + 2 × (0.5)² = 0.25 + 0.5
• Result: I_cm = 0.25 kg m² about the centre, and I = 0.75 kg m² about a parallel axis through the rim, perpendicular to the disc's plane.

Rolling motion combines translation of the centre of mass with rotation about an axis through the centre of mass. For a wheel, disc, ring, or sphere of radius R rolling without slipping on a surface, the point of contact is instantaneously at rest, giving the essential rolling condition:

• v_cm = Rω, where v_cm is the speed of the centre of mass and ω is the angular speed about the centre.

Because a rolling body simultaneously translates and rotates, its total kinetic energy is the sum of translational and rotational parts:

• KE = (1/2)Mv_cm² + (1/2)Iω²

Writing I = Mk² (k = radius of gyration about the rolling axis) and using ω = v_cm/R, this becomes:

• KE = (1/2)Mv_cm²(1 + k²/R²)

For the same v_cm, a body with larger k²/R² (mass concentrated farther from the axis, e.g. a ring) has more kinetic energy and a smaller linear acceleration when rolling down an incline than a body with smaller k²/R² (mass concentrated near the axis, e.g. a solid sphere). Rolling from rest down an incline of angle θ, energy considerations give acceleration a = g sinθ / (1 + k²/R²), so a solid sphere (k²/R² = 2/5) always reaches the bottom before a disc (k²/R² = 1/2), which beats a ring (k²/R² = 1), regardless of mass or radius.

Worked Example.

• Given: A solid sphere of mass M = 1 kg rolls without slipping with centre-of-mass speed v_cm = 2 m/s. For a solid sphere, k²/R² = 2/5.
• Formula: KE = (1/2)Mv_cm²(1 + k²/R²)
• Substitution: KE = (1/2) × 1 × (2)² × (1 + 2/5) = (1/2) × 1 × 4 × 1.4
• Result: KE = 2 × 1.4 = 2.8 J, of which 2 J is translational and 0.8 J is rotational kinetic energy.