Chapter 7 — System Of Particles And Rotational Motion

Fig 1 — Educational Diagram: System of Particles — basic definitions

System of particles: A collection of particles that may interact with each other and with external agents. You can treat the collection as a single entity by studying quantities like total mass, total momentum and the position of the centre of mass.

Particle and mass: Each element of the system is treated as a particle with mass m_i and position vector r_i (measured from a chosen origin). The total mass M of a system of N particles is M = Σ_i m_i.

Centre of mass (COM): The COM is a weighted average of the positions of all particles, weighted by their masses. For discrete particles, the position vector of the centre of mass r_cm is

• r_cm = (1/M) Σ_{i=1}^N m_i r_i

• r_cm = (1/M) ∫ r ρ(r) dV, with M = ∫ ρ(r) dV.

Velocity and acceleration of COM: Differentiate r_cm with respect to time to get the velocity and acceleration of the COM:

• v_cm = (1/M) Σ_i m_i v_i
• a_cm = (1/M) Σ_i m_i a_i

Total momentum: The total linear momentum of the system is P = Σ_i m_i v_i. Using v_cm, we get P = M v_cm. Thus the motion of the whole system (as far as linear momentum is concerned) is equivalent to a single particle of mass M located at the COM and moving with velocity v_cm.

Newton's laws for a system: Internal forces between particles cancel pairwise by Newton’s third law (for central forces). As a result, only external forces change the total momentum. The resultant external force equals the total mass times acceleration of the COM:

• Σ F_ext = dP/dt = M a_cm

Conservation of linear momentum: If Σ F_ext = 0 (isolated system), then dP/dt = 0 and total momentum P is constant. This fundamental result follows directly from the previous relation.

Useful special cases:

• Two-particle system (one-dimensional): x_cm = (m1 x1 + m2 x2)/(m1 + m2).
• Symmetric rigid bodies: COM lies at geometric centre (e.g., uniform rod, sphere) — useful for simplifying motion.

Physical meaning: The COM is the point where the mass of the system can be considered to be concentrated for analyzing translational motion. Even if parts of the system move relative to each other (internal motion), the COM motion depends only on external forces.

Fig 2 — Educational Diagram: Center of Mass (CoM)

Definition: The center of mass (CoM) of a system of particles or a rigid body is the unique point that moves as if the total mass of the system were concentrated there and all external forces were applied there. For uniform gravitational field, CoM coincides with center of gravity.

Discrete system (N particles): If particles of masses m_i are at position vectors r_i (i = 1..N), the position vector of the CoM is

R = (1/M) Σ_{i=1}^N m_i r_i, where M = Σ_{i=1}^N m_i.

Continuous distribution: For a body with mass density described by dm,

R = (1/M) ∫ r dm, with M = ∫ dm.

Cartesian components:

x_{CM} = (1/M) Σ m_i x_i , y_{CM} = (1/M) Σ m_i y_i , z_{CM} = (1/M) Σ m_i z_i

Two-particle special case (on x-axis): x_{CM} = (m_1 x_1 + m_2 x_2) / (m_1 + m_2). Useful rearrangement: x_{CM} measured from m_1 is x_1 + [m_2/(m_1+m_2)](x_2-x_1).

Physical properties and consequences:

• The total linear momentum of a system equals the total mass times velocity of the CoM: P = M V_{CM}.
• Newton's 2nd law for the system: M a_{CM} = Σ F_{ext}. Internal forces cancel and do not affect motion of CoM.
• If ΣF_{ext} = 0, then V_{CM} is constant (linear momentum conserved) even if parts move internally.
• Kinetic energy splits as K_total = (1/2) M V_{CM}^2 + K_relative (motion about CoM).

Finding CoM in practice: Use symmetry where possible (e.g., center of a uniform sphere, midpoint of uniform rod). For composite bodies, treat each piece as a point mass at its own CoM and apply R = (1/M) Σ m_i r_i.

Notes for CBSE problems: Emphasize coordinate method, use integrals for non-uniform continuous bodies, and exploit symmetry. Remember the distinction between CoM motion (affected only by external forces) and internal motion (rotations about the CoM).

Fig 3 — Educational Diagram: Linear momentum of a system

Definition: The total (linear) momentum of a system of particles is the vector sum of the momenta of all the particles in the system. If the system has N particles with masses m_i and velocities v_i, the total momentum is P = Σ m_i v_i.

Relation to centre of mass: Let total mass M = Σ m_i and the position of the centre of mass be R = (1/M) Σ m_i r_i. Differentiating R gives the centre-of-mass velocity V_cm = (1/M) Σ m_i v_i. Hence the total momentum P relates to the centre of mass as

P = M V_cm.

Time rate of change and external forces: Differentiating P with respect to time gives

dP/dt = Σ m_i a_i = Σ F_i^{(ext)} + Σ F_{ij}^{(int)}

By Newton's third law internal forces between particle pairs cancel in the vector sum (F_{ij} = −F_{ji}), so only external forces remain. Therefore

dP/dt = Σ F_i^{(ext)} = M a_cm.

Conservation of momentum: If the net external force on the system is zero (Σ F_ext = 0), then dP/dt = 0 and the total momentum P is constant in time. This is the conservation of linear momentum for a system.

Impulse: The impulse delivered by external forces over a time interval Δt is J = ∫_{t1}^{t2} Σ F_ext dt. Impulse equals the change in total momentum: J = ΔP.

Vector nature: Momentum is a vector. Conservation holds separately for each component: Σ p_x (before) = Σ p_x (after), Σ p_y (before) = Σ p_y (after).

Remarks: Internal interactions can change individual particle momenta but cannot change the total momentum of an isolated system. For variable-mass systems (e.g., rockets) care is needed; the momentum principle still applies if all mass flow and external forces are correctly accounted for.

Fig 4.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

Overview
A system of particles is a collection of masses that may interact with each other (internal forces) and with bodies outside the system (external forces). Newton's laws apply to each particle; when combined for the whole system they give compact and useful results about total momentum, motion of the centre of mass (CM), and conservation laws.

Centre of mass (CM)
For N particles with masses m_i and position vectors r_i, total mass M = Σ_i m_i and the position of the centre of mass is
r_cm = (1/M) Σ_{i=1}^N m_i r_i.
The velocity and acceleration of the CM are v_cm = dr_cm/dt and a_cm = d^2 r_cm/dt^2.

Total momentum and its rate of change
Total linear momentum of the system is
P = Σ_i p_i = Σ_i m_i v_i.
Differentiating, dP/dt = Σ_i m_i a_i = Σ_i F_i (by Newton's second law for each particle), where F_i = external_i + Σ_{j≠i} F_{ij} (internal forces on i from j).

Cancellation of internal forces
If internal forces obey Newton's third law (F_{ij} = −F_{ji}, i.e. equal, opposite and along the line joining particles), then Σ_i Σ_{j≠i} F_{ij} = 0, so internal forces sum to zero. Thus only external forces contribute to the rate of change of total momentum:

F_ext,net ≡ Σ_i F_{ext,i} = dP/dt.

Motion of the centre of mass
Since P = M v_cm, for a closed system with constant total mass M we get
F_ext,net = dP/dt = M a_cm.
Interpretation: the net external force on a system produces acceleration of its centre of mass as if all mass were concentrated there.

Conservation of momentum
If the net external force is zero (isolated system), dP/dt = 0 ⇒ P = constant. Equivalently, velocity of CM is constant. This gives the powerful result that momentum of a system is conserved in collisions, explosions, or interactions when external forces are negligible.

Variable mass systems
For systems whose total mass changes (e.g., rockets expelling fuel), the general law still holds: F_ext,net = dP/dt. But P = M v_cm and dP/dt = M a_cm + v_cm dM/dt, so F_ext,net ≠ M a_cm in general. One must explicitly account for mass flow when applying Newton's laws.

Energy and angular momentum
Total kinetic energy can be split into motion of CM plus motion relative to CM:
K_total = 1/2 M v_cm^2 + Σ_i (1/2 m_i u_i^2), where u_i is velocity of particle i relative to CM.
Total angular momentum about an origin is L = Σ_i r_i × p_i and, when internal forces are central, dL/dt = τ_ext,net (net external torque). If τ_ext,net = 0 then total angular momentum is conserved.

Summary of physical consequences
- External forces change the system's total momentum and accelerate the CM.
- Internal forces do not change total momentum (they only redistribute momentum among particles).
- Conservation of momentum and angular momentum follow when external influences vanish.

Fig 5 — Educational Diagram: Kinetic energy of a system

Definition
The kinetic energy of a system of particles is the sum of the kinetic energies of all its particles: K = Σ_i (1/2 m_i v_i²).

Decomposition into motion of the centre of mass and motion about the centre of mass
Let M = Σ_i m_i be the total mass and V_cm the velocity of the centre of mass. For each particle, write its velocity as v_i = V_cm + v'_i, where v'_i is the velocity of the particle relative to the centre of mass. Then

K = Σ_i (1/2 m_i v_i²) = 1/2 M V_cm² + 1/2 Σ_i m_i v'_i².

Derivation (sketch)
v_i² = V_cm² + v'_i² + 2 V_cm·v'_i. Multiplying by m_i and summing over i gives Σ m_i v_i² = M V_cm² + Σ m_i v'_i² + 2 V_cm·Σ m_i v'_i. But by definition of centre of mass Σ m_i v'_i = 0, so the cross term vanishes.

Physical meaning
- The first term, 1/2 M V_cm², is the kinetic energy of the whole mass moving as if concentrated at the centre of mass (translational KE).
- The second term, 1/2 Σ m_i v'_i², is the internal kinetic energy (motion about the CM) — e.g., rotation, vibration, random thermal motion.

Special case: Rigid body
For a rigid body moving with translation of its CM plus rotation about an axis through the CM with angular speed ω, the internal kinetic energy reduces to rotational kinetic energy about the CM: K = 1/2 M V_cm² + 1/2 I_cm ω², where I_cm is the moment of inertia about the CM.

Energy changes and forces
External work changes the total kinetic energy of the system. Internal forces (between particles) can convert energy between translational and internal forms (for example, inelastic collisions convert translational KE into internal energy like heat) but their net work on the whole system's CM motion is zero.

Summary formula
K_total = K_translation + K_internal = 1/2 M V_cm² + 1/2 Σ m_i v'_i².

Fig 6 — Educational Diagram: Angular momentum

Definition (single particle): Angular momentum of a particle of mass m having linear momentum p and position vector r (measured from a chosen origin) is the vector L = r × p. Its direction is perpendicular to the plane of r and p given by the right-hand rule.

Magnitude: |L| = m v r sinθ, where v is the speed and θ is the angle between r and v. For motion in a circle of radius r with speed v, |L| = m v r = m r^2 ω.

System of particles and rigid body: For a system of particles, total angular momentum about the origin is L_total = Σ r_i × p_i. For a rigid body rotating about a fixed axis (or about a principal axis through the origin) with angular speed ω, the angular momentum is L = I ω, where I is the moment of inertia about that axis.

Relation with torque (rotational analogue of Newton’s second law): Torque about the origin is τ = r × F. For a particle (or a system), the time rate of change of angular momentum equals the net external torque: dL/dt = τ_ext. This gives the condition for conservation: if τ_ext = 0, then L is constant in time.

Conservation of angular momentum: If no external torque acts on a system, its angular momentum remains constant. This explains many phenomena where rotation speed changes when distribution of mass changes (moment of inertia changes) but L stays constant.

Rotational kinetic energy: For a rigid body rotating with angular speed ω, K_rot = (1/2) I ω^2. Using L = I ω, this can be written as K_rot = L^2/(2I).

Vector nature and components: Angular momentum is a vector; both magnitude and direction matter. Under a torque perpendicular to L, only the direction of L changes (precession) while its magnitude may remain nearly constant.

Important conceptual points for Class 11:

• Angular momentum depends on the choice of origin — L is conserved only if external torque about that same origin is zero.
• For central forces (force directed along r), torque is zero and angular momentum is conserved (explains Kepler’s second law).
• For rigid bodies, L = I ω holds when ω is along a principal axis. Otherwise L and ω may not be parallel.

Fig 7.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

Definition: Torque (also called moment of a force) is a measure of the tendency of a force to rotate an object about a point or axis. It depends on the magnitude of the force, the point of application (lever arm) and the angle between the force and the lever arm.

Scalar (magnitude) form: If a force F acts at a point whose position vector from the axis/point is r and the angle between r and F is θ, the magnitude of the torque is

τ = r F sinθ

Often this is written using the perpendicular (shortest) distance from the axis to the line of action of the force, called the lever arm (or moment arm) b. Then

τ = F b

Vector form: Torque is a vector given by the cross product

τ⃗ = r⃗ × F⃗

The direction of τ⃗ is perpendicular to the plane containing r⃗ and F⃗; its sense is given by the right-hand rule (thumb points in direction of τ⃗ when fingers curl from r⃗ to F⃗).

Units and sign: SI unit of torque is newton–metre (N·m). Positive/negative sign is a convention indicating the sense of rotation (e.g. counterclockwise positive).

Couple (pure moment): A couple is formed by two equal and opposite parallel forces separated by distance d. The net force is zero but the couple produces a torque of magnitude

τ = F d

Couples are free vectors: their effect does not depend on the choice of origin.

Net torque and equilibrium: For a system of forces, the net torque about a point is the vector sum τ⃗_net = Σ r⃗_i × F⃗_i. For static equilibrium of a rigid body, ΣF⃗ = 0 and Στ⃗ = 0.

Relation to rotational dynamics: The net torque on a rigid body about a fixed axis produces angular acceleration α according to

τ_net = I α

where I is the moment of inertia about that axis. This is the rotational analogue of Newton's second law.

Work and power: When a torque τ acts through a small angular displacement dθ, the work done is dW = τ dθ. If the body rotates with angular speed ω, the mechanical power is P = τ ω.

Important points to remember:

• Torque depends on where the force is applied: same force at different distances gives different torque.
• A force through the axis (r = 0) produces zero torque.
• Maximum torque for given r and F occurs when F is perpendicular to r (θ = 90°).
• Couples produce rotation without net translation.

Fig 8 — Educational Diagram: Moment of inertia (rotational inertia)

Definition: The moment of inertia (rotational inertia) of a system about a given axis is a scalar measure of how the system's mass is distributed relative to that axis. For a discrete system,

I = Σ m_i r_i²,

and for a continuous body,

I = ∫ r² dm,

where r (perpendicular) is the distance of each mass element dm from the axis. Unit: kg·m².

Physical meaning and role:

• Moment of inertia is the rotational analogue of mass in linear motion: it resists angular acceleration just as mass resists linear acceleration.
• Rotational dynamics: torque τ and angular acceleration α are related by τ = I α (for rigid bodies about a fixed axis).
• Rotational kinetic energy: K_rot = (1/2) I ω², where ω is angular speed.
• I depends on the axis chosen — moving the axis changes I (see Parallel-Axis Theorem).

How to compute:

• Choose an axis, express dm in terms of a density (linear/area/volume) and a coordinate variable.
• Write r (distance to axis) in that coordinate and evaluate I = ∫ r² dm over the body.
• For composite bodies, moments add: I_total = Σ I_parts (about the same axis).

Useful theorems:

• Parallel-axis theorem: I_about axis parallel to one through centre of mass (CM): I = I_CM + M d², where d is distance between axes.
• Perpendicular-axis theorem (planar lamina in xy-plane): I_z = I_x + I_y, for axis z perpendicular to plane.

Worked-sketch (slender rod about centre) — length L, mass M, axis perpendicular through midpoint:

Take x from -L/2 to +L/2, dm = (M/L) dx, r = x, so I = ∫_-L/2^L/2 x² (M/L) dx = (M/L) [2 (L/2)³/3] = (1/12) M L².

Notes:

• Different shapes and axis choices give different I even for the same mass and size.
• Because I involves r², moving mass further from the axis increases I rapidly.

Fig 9 — Educational Diagram: Theorems for moments of inertia

Definition and physical meaning
Moment of inertia (I) of a rigid body about an axis is the measure of the body's resistance to angular acceleration about that axis. For a system of point masses I = sum m_i r_i^2, and for a continuous body I = ∫ r^2 dm, where r is the perpendicular distance of the mass element from the axis.

Why theorems are useful
Direct integration to find I can be hard for many axes. The two main theorems (perpendicular axis theorem and parallel axis theorem) let you obtain moments of inertia about new axes from known ones, greatly simplifying problems.

Perpendicular axis theorem (for planar lamina)
Statement: For a flat lamina lying in the xy plane, the moment of inertia about an axis perpendicular to the plane (z axis) equals the sum of moments about two perpendicular axes in the plane that intersect at the same point: I_z = I_x + I_y.

Outline of proof
For an element dm at coordinates (x,y), its distance squared from the z axis is r^2 = x^2 + y^2. So I_z = ∫(x^2 + y^2) dm = ∫ x^2 dm + ∫ y^2 dm = I_x + I_y. This applies only to planar bodies (thin lamina) and axes that intersect at the same point and are mutually perpendicular.

Parallel axis theorem
Statement: If I_cm is the moment of inertia of a body about an axis through its center of mass, then the moment of inertia I about any axis parallel to it and at a distance d from it is I = I_cm + M d^2, where M is the total mass and d is the perpendicular distance between the axes.

Outline of proof
Place origin at the center of mass. For a mass element dm at position vector r', its distance to the shifted axis (parallel axis displaced by R) satisfies r^2 = |r' + R|^2 = r'^2 + R^2 + 2 R·r'. Integrating, the cross term 2 R·∫ r' dm = 0 because the centroid is at origin. Hence I = ∫ r'^2 dm + M R^2 = I_cm + M d^2. The theorem holds for any 3D body and any pair of parallel axes.

Important remarks

• The perpendicular axis theorem is valid only for thin planar bodies (lamina) and for axes that meet at the same point.
• The parallel axis theorem holds for any rigid body and any pair of parallel axes.
• Using I = ∫ r^2 dm together with symmetry and these theorems makes finding moments for common shapes straightforward.

Common moments of inertia (about central axes)

• Thin ring or hoop about central axis: I = M R^2
• Solid disc or cylinder about central axis: I = (1/2) M R^2
• Thin rod about axis through center perpendicular to length: I = (1/12) M L^2
• Thin rod about axis through one end perpendicular to length: I = (1/3) M L^2
• Solid sphere about diameter: I = (2/5) M R^2
• Thin spherical shell about diameter: I = (2/3) M R^2

How to apply in problems

1. Choose an axis about which you know I (often center of mass axis or an axis of symmetry).
2. Use parallel axis theorem to shift to the required parallel axis by adding M d^2.
3. Use perpendicular axis theorem to relate planar axes when needed (I_z = I_x + I_y).
4. Exploit symmetry to reduce integrals or identify zero cross-terms.

Summary: The perpendicular axis theorem relates three mutually perpendicular axes in a plane for thin lamina, while the parallel axis theorem shifts moment of inertia from a centroidal axis to any parallel axis by adding M d^2. These theorems reduce calculation effort and are widely used in rotational dynamics.

Fig 10 — Educational Diagram: Rigid body rotation — kinematics

Definition and basic idea: A rigid body is an object in which the distance between any two particles remains constant. In pure rotation about a fixed axis every particle of the body moves in a circle whose centers lie on the axis. All particles have the same angular displacement, angular velocity and angular acceleration.

Angular quantities:

• Angular displacement θ (radian): measure of rotation; s = r θ (arc length s on a circle of radius r).
• Angular velocity ω = dθ/dt (rad s⁻¹): same for every particle of the body; vector direction given by the right-hand rule along the rotation axis.
• Angular acceleration α = dω/dt (rad s⁻²): rate of change of angular velocity.

Relationship with linear motion (for a particle at distance r from axis):

• Linear (tangential) speed: v = ω r. (Vector form: v = ω × r.)
• Tangential acceleration: a_t = α r (along the tangent, responsible for change in speed).
• Centripetal (normal) acceleration: a_n = ω² r (directed toward the axis, responsible for change in direction).
• Net acceleration magnitude: a = sqrt(a_t² + a_n²) when a_t and a_n are perpendicular.

Kinematic equations for constant angular acceleration (analogous to linear kinematics):

• ω = ω_0 + α t
• θ = θ_0 + ω_0 t + 1/2 α t²
• ω² = ω_0² + 2 α (θ − θ_0)

Other important points:

• Instantaneous axis (or center) of rotation: at any instant a rigid body moving general planar motion can be considered as rotating about an instantaneous center; for pure rotation this is the fixed axis itself.
• Sign convention and right-hand rule: choose positive rotation direction (usually by right-hand rule); signs of ω and α follow this convention.
• Units: angle in radians (rad), ω in rad s⁻¹, α in rad s⁻².

Short derivations (key ideas): s = r θ (by arc length). Differentiate: v = ds/dt = r dθ/dt = r ω. Differentiate again tangentially to get a_t = r α. Centripetal acceleration comes from v²/r = (ω r)² / r = ω² r.

Fig 11 — Educational Diagram: Rigid body rotation — dynamics

What is a rigid body rotation (dynamics)?
A rigid body is an ideal object whose internal distances do not change. Rigid body rotation (dynamics) studies how such bodies respond to forces and torques when they rotate. It is the rotational analogue of Newtonian dynamics for particles.

Key physical quantities

• Angular displacement θ (rad), angular velocity ω = dθ/dt (rad/s), angular acceleration α = dω/dt (rad/s²).
• Moment of inertia I — measure of mass distribution about an axis; for discrete masses I = Σ m_ir_i², for continuous bodies I = ∫ r² dm (units kg·m²).
• Torque τ = r × F (vector); scalar about a chosen axis τ = rF⊥ = rF sinφ (units N·m). It measures tendency of a force to produce rotation.
• Angular momentum L. For rotation about a fixed axis of a rigid body L = I ω (units kg·m²/s).

Equation of rotational motion (rotational Newton’s second law)
For a rigid body rotating about a fixed axis, the net external torque about the axis equals moment of inertia times angular acceleration:

τ_net = I α

This is directly analogous to F_net = ma for translation. When τ_net is constant, α is constant and ω, θ follow kinematic relations analogous to linear motion.

Work, energy and power in rotation

• Infinitesimal work done by a torque: dW = τ dθ.
• Rotational kinetic energy of a rigid body rotating about a fixed axis: K_rot = (1/2) I ω².
• Power supplied by torque: P = τ ω.

Angular impulse and momentum
Angular impulse is integral of torque over time: ∫τ dt = ΔL. If no external torque acts, total angular momentum is conserved: L_initial = L_final.

Related linear–rotational relations (for a point at distance r from axis)

• Linear speed v = r ω
• Tangential acceleration a_t = r α
• Normal (centripetal) acceleration a_n = r ω²

Special results used often

• Parallel–axis theorem: I_about axis = I_cm + M d², where d is distance between axes.
• For rolling without slipping (wheel of radius R): v_cm = ω R and total kinetic energy = (1/2) M v_cm² + (1/2) I_cm ω².

How the main equation τ = I α is obtained (sketch)
For a rigid body of many mass elements, torque about axis is Σ r_i × F_i. Using F_i = m_i a_i and expressing tangential accelerations a_ti = r_i α, one gets Σ τ_i = Σ m_i r_i² α = (Σ m_i r_i²) α = I α.

When to use conservation of angular momentum
Use when net external torque about the chosen axis is zero or negligible (e.g., ice skater pulling in arms, isolated spinning satellite). Then I_1 ω_1 = I_2 ω_2.

Important remarks

• Moments of inertia depend on axis orientation and mass distribution.
• Sign conventions: angular quantities and torques can be treated as vectors (right-hand rule) or as signed scalars about a chosen axis.
• For extended motion without fixed axis, one can use instantaneous axis of rotation or decompose motion into translation of center of mass + rotation about CM.

Fig 12.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

What is rolling motion?
Rolling motion of a rigid body (wheel, sphere, cylinder) on a surface is a combination of translation of its centre of mass (CM) and rotation about the CM. If the point of contact does not slip relative to the surface, the motion is called pure rolling (rolling without slipping).

Condition for pure rolling
For pure rolling the linear speed of the center and the angular speed about the center are related by
v_cm = R ω , where R is the radius and ω is angular speed. Equivalently the instantaneous velocity of the contact point relative to ground is zero.

Instantaneous axis and velocities of points
In pure rolling the instantaneous axis of rotation is the point of contact (instantaneous rest). Velocities of points on rim combine translation and rotation. For a particle on rim at angle θ measured from the forward direction, its instantaneous velocity = v_cm + ω × r (vector sum). The locus of a point on the rim as the body rolls without slipping is a cycloid.

Kinetic energy
Total kinetic energy = translational KE of CM + rotational KE about CM:
KE_total = (1/2) M v_cm^2 + (1/2) I_cm ω^2. Using v_cm = R ω this can be written as KE_total = (1/2) M v_cm^2 (1 + I_cm/(M R^2)).

Role of friction
Static friction is required to prevent slipping and to provide the torque that changes rotational motion. For pure rolling on a horizontal surface with constant v_cm, static friction does no work (point of contact is instantaneously at rest). When a body rolls down an incline, static friction (usually up the plane) provides torque that increases ω. In rolling with slipping, kinetic friction acts and dissipates mechanical energy.

Equation of motion for rolling down an incline (no slip)
Consider a rigid body of mass M and moment of inertia I_cm rolling without slipping down an incline of angle θ. Using Newton's second law for translation and rotation together with v_cm = R ω gives:

• Translational: M a_cm = M g sin θ - f (f = friction)
• Rotational: I_cm α = f R, and a_cm = α R

Eliminating f and α yields

a_cm = (g sin θ) / (1 + I_cm/(M R^2)).

Common special cases
Substituting I_cm for common shapes gives their accelerations down an incline (no slip):

• Solid sphere (I = 2/5 M R^2): a = (5/7) g sin θ
• Solid cylinder / disc (I = 1/2 M R^2): a = (2/3) g sin θ
• Thin cylindrical shell (I = M R^2): a = (1/2) g sin θ
• Hollow sphere (I = 2/3 M R^2): a = (3/5) g sin θ

Energy viewpoint
When a body of mass M rolls down height h without slipping, gravitational potential energy M g h converts into translational and rotational KE:
M g h = (1/2) M v_cm^2 + (1/2) I_cm (ω)^2 = (1/2) M v_cm^2 (1 + I_cm/(M R^2)).

Rolling with slipping
If v_cm ≠ R ω there is relative motion at the contact and kinetic friction acts, giving both a tangential force and dissipating mechanical energy. Equations of motion must include kinetic friction (f_k = μk N) and the slip relation v_rel = v_cm - R ω.

Practical notes
- Static friction provides torque but does not necessarily do work in pure rolling.
- Objects with smaller I_cm/(M R^2) accelerate faster when rolling down the same incline (because more of the potential energy goes into translation).
- Rolling resistance (deformation of surfaces) can cause energy loss and effective torque opposing motion.

Summary
Rolling is a combined translation-rotation motion governed by v_cm = R ω for no slip, total KE = translational + rotational parts, and dynamics combine Newton’s laws for translation and rotation. Moment of inertia determines how energy splits and thus the linear acceleration for a given driving force (e.g., gravity on an incline).

Fig 13.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

Overview: Conservation laws state that certain physical quantities of an isolated system remain constant in time. In Class 11 mechanics the main conserved quantities are linear momentum, mechanical energy (under conservative forces), and angular momentum. These laws follow from Newton's laws and symmetry principles and are widely used to analyze collisions, explosions, rotations and motion of the centre of mass.

1. Conservation of linear momentum:

• Definition: For an isolated system (net external force = 0) the total linear momentum P = Σ m_iv_i is constant: dP/dt = Σ F_ext = 0 ⇒ P = constant.
• Impulse: Change of momentum equals impulse: J = Δp = ∫ F dt. During a short collision, internal forces cause large impulses while external forces are negligible.
• Applications: Collisions (elastic, inelastic, perfectly inelastic), explosions, recoil (gun and bullet), motion of centre of mass after internal interactions.

2. Conservation of mechanical energy:

• When only conservative forces act, total mechanical energy E = K + U is constant. Kinetic energy K = 1/2 mv^2 (or 1/2 Iω^2 for rotation). Potential energy depends on the force (e.g., U_g = mgh, U_s = 1/2 kx^2).
• If non-conservative forces (friction, inelastic deformation) act, mechanical energy is not conserved; some mechanical energy converts to internal energy (heat, sound).
• Use in problems: relate heights, speeds and spring compression without solving equations of motion directly.

3. Conservation of angular momentum:

• Angular momentum of a particle about a point: L = r × p. For a rigid body rotating about a fixed axis, L = Iω.
• If net external torque about the chosen point is zero, total angular momentum is conserved: dL/dt = τ_ext = 0 ⇒ L = constant.
• Applications: Spinning ice skater (pulling arms in increases ω), planetary motion, gyroscope, collision/interaction of rotating bodies, fragmentation of rotating bodies.

4. Centre of mass and relation to conservation laws:

• Position of centre of mass R_cm = (Σ m_i r_i)/M. The motion of the centre of mass is governed by external forces: M a_cm = Σ F_ext. If Σ F_ext = 0 then P_total = M v_cm = constant.
• Even if internal forces change individual momenta, the centre of mass motion is unaffected when external force is zero.

5. Collisions — practical classification and consequences:

• Elastic collision: momentum and kinetic energy both conserved. Useful for ideal billiard-ball problems.
• Inelastic collision: momentum conserved but kinetic energy is not; some KE converted to internal energy.
• Perfectly inelastic collision: bodies stick together after collision. Final common velocity v = (m1 v1 + m2 v2)/(m1 + m2) from momentum conservation.
• Coefficient of restitution e = (relative speed after)/(relative speed before) in 1D collisions; 0 ≤ e ≤ 1.

How to choose which conservation law to use: If external force/torque on the system is zero → use corresponding conservation (momentum or angular momentum). If only conservative forces act → use conservation of mechanical energy. For collisions, always use momentum conservation; use energy conservation only for elastic collisions (or to relate speeds and heights when no non-conservative forces act).

Notes on combined use: Many problems require combining laws: e.g., an object falling onto a rotating turntable (angular momentum about axis may be conserved if no external torque), followed by energy considerations for frictional losses; or an explosion where momentum conservation gives fragment velocities while energy tells about conversion to internal energy.

Fig 14 — Educational Diagram: Computational methods and examples

Computational methods in the topic System of Particles and Rotational Motion are systematic techniques used to calculate quantities such as center of mass (CM), total momentum, angular momentum, moment of inertia and rotational kinetic energy for systems of discrete particles and continuous bodies. The key idea is to reduce a complex body to simpler elements, apply basic formulas to each element, then sum or integrate those contributions. Symmetry, the parallel-axis and perpendicular-axis theorems, and conservation laws (momentum, angular momentum, energy) are used to simplify computations.

Common methods

• Discrete summation: For a system of point masses m_i at positions r_i compute sums: e.g., r_cm = (Σ m_i r_i)/Σ m_i. Useful for a small number of particles.
• Continuous integration: Replace the body by infinitesimal elements dm with known density (linear λ, surface σ, volume ρ) and integrate: r_cm = (1/M) ∫ r dm, I = ∫ r_perp^2 dm.
• Divide-and-conquer (composite bodies): Break a complex body into simple shapes with known centers of mass and moments of inertia; combine using mass-weighted sums and parallel-axis theorem.
• Use symmetry: Symmetry often gives zero for some coordinates of CM or simple expressions for I without full integration.
• Numerical methods: For irregular distributions, approximate integrals by discrete sums or use numerical integration techniques (trapezoid, Simpson) and computational tools (spreadsheet, Python) to evaluate sums.
• Vector methods for rotation: Use r x p sums to compute total angular momentum L = Σ (r_i x p_i) or for rigid bodies L = I ω (about principal axes). Compute torques τ = Σ (r_i x F_i) and use τ = dL/dt.

Practical steps for a typical problem

1. Identify whether the body is discrete or continuous and choose appropriate density (λ, σ, ρ) if continuous.
2. Choose coordinate axes using symmetry to simplify integrals/sums.
3. Write dm in terms of the density and a small geometric element (dm = λ dx, dm = ρ dV, etc.).
4. Set up integrals for required quantities (r_cm, I about given axis) and evaluate analytically or numerically.
5. When shifting axes, apply parallel-axis theorem: I_axis = I_cm + M d^2.
6. Check limiting cases and units to validate results.

Tips

• Always exploit symmetry first to reduce work.
• Use small-slice elemental approach for rods, plates, disks: choose the slice perpendicular to symmetry axis.
• For rigid body kinetics, decompose kinetic energy: K_total = 1/2 M v_cm^2 + 1/2 I_cm ω^2.
• Use computational tools when integrals are complex or data is discrete.