Class 11 Physics Notes Chapter 7 (Chapter 7) – Physics Part-II Book

Alright class, let's get straight into Chapter 7: Systems of Particles and Rotational Motion. This is a fundamental chapter that extends our understanding of motion from single particles (as studied earlier) to systems of particles and rigid bodies, focusing particularly on rotational dynamics. Pay close attention, as these concepts are frequently tested in various government examinations.

Chapter 7: Systems of Particles and Rotational Motion - Detailed Notes

1. Introduction

• System of Particles: A collection of a large number of particles interacting with each other.
• Rigid Body: A body with a perfectly definite and unchanging shape. The distance between any two given points on a rigid body remains constant regardless of the external forces applied. In reality, no body is perfectly rigid, but it's a useful idealization for many situations.
• Types of Motion of a Rigid Body:
  • Pure Translation: All particles of the body move with the same velocity at any instant. The orientation of the body remains unchanged.
  • Pure Rotation: All particles of the body move in circles about a fixed axis (axis of rotation). Each particle has the same angular velocity at any instant.
  • Combination of Translation and Rotation: Example: A wheel rolling on a surface.

2. Centre of Mass (CM)

• Definition: The Centre of Mass of a system is a hypothetical point where the entire mass of the system may be assumed to be concentrated for describing its translational motion.
• Position Vector of CM:
  • For a system of n particles with masses m₁, m₂, ..., mₙ and position vectors r₁, r₂, ..., rₙ:
    R_CM = (m₁r₁ + m₂r₂ + ... + mₙrₙ) / (m₁ + m₂ + ... + mₙ) = (Σ mᵢrᵢ) / M
    where M = Σ mᵢ is the total mass of the system.
  • In Cartesian coordinates (rᵢ = xᵢî + yᵢĵ + zᵢk̂):
    X_CM = (Σ mᵢxᵢ) / M
Y_CM = (Σ mᵢyᵢ) / M
Z_CM = (Σ mᵢzᵢ) / M
• CM of Continuous Bodies: For bodies with continuous mass distribution, summation is replaced by integration:
  R_CM = (1/M) ∫ r dm
X_CM = (1/M) ∫ x dm, Y_CM = (1/M) ∫ y dm, Z_CM = (1/M) ∫ z dm
• CM of Some Uniform Bodies: (Important to remember)
  • Uniform Rod: At its geometric centre.
  • Circular Ring: At the centre of the ring.
  • Circular Disc: At the centre of the disc.
  • Solid Sphere / Hollow Sphere: At the geometric centre.
  • Right Circular Cone (height h): At height h/4 from the base on the axis.
  • Semi-circular Ring: At distance 2R/π from the centre on the axis of symmetry.
  • Semi-circular Disc: At distance 4R/(3π) from the centre on the axis of symmetry.
• Motion of the Centre of Mass:
  • Velocity of CM: V_CM = dR_CM/dt = (Σ mᵢvᵢ) / M
  • Acceleration of CM: A_CM = dV_CM/dt = (Σ mᵢaᵢ) / M
  • Newton's Second Law for the System: M * A_CM = Σ Fᵢ = F_ext + F_int. Since internal forces ( F_int) occur in action-reaction pairs (Newton's Third Law), their vector sum is zero (Σ F_int = 0).
    Therefore, M * A_CM = F_ext
    This is a crucial result: The centre of mass of a system moves as if the entire mass of the system were concentrated at the CM and all external forces were applied at that point.
• Conservation of Linear Momentum: If the net external force on the system is zero (F_ext = 0), then A_CM = 0, which means V_CM = constant.
  The total linear momentum of the system (P_sys = M * V_CM) remains conserved if the net external force acting on the system is zero.

3. Vector Product (Cross Product) of Two Vectors

• Needed for defining torque and angular momentum.
• C = A × B
• Magnitude: |C| = |A| |B| sin θ, where θ is the angle between A and B.
• Direction: Perpendicular to the plane containing A and B, given by the Right-Hand Rule or Right-Hand Screw Rule.
• Properties:
  • A × B = - (B × A) (Anti-commutative)
  • A × A = 0
  • î × î = ĵ × ĵ = k̂ × k̂ = 0
  • î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ
  • ĵ × î = -k̂, k̂ × ĵ = -î, î × k̂ = -ĵ
  • A × (B + C) = A × B + A × C (Distributive)
• In determinant form: If A = Aₓî + A<0xE1><0xB5><0xB7>ĵ + A<0xE2><0x82><0x9Ck̂ and B = Bₓî + B<0xE1><0xB5><0xB7>ĵ + B<0xE2><0x82><0x9Ck̂, then
  A × B = | î ĵ k̂ |
| Aₓ A<0xE1><0xB5><0xB7> A<0xE2><0x82><0x9C |
| Bₓ B<0xE1><0xB5><0xB7> B<0xE2><0x82><0x9C |

4. Angular Velocity (ω) and its Relation with Linear Velocity (v)

• Angular Velocity (ω): Rate of change of angular displacement (ω = dθ/dt). It's a vector quantity whose direction is along the axis of rotation (given by the right-hand rule). Unit: rad/s.
• Relation: For a particle P at position vector r from the origin on the axis of rotation, its linear velocity v is related to angular velocity ω by:
  v = ω × r
  The magnitude is v = ωr sinα, where α is the angle between ω and r. If rotation is about a fixed axis and r is the perpendicular distance from the axis, then v = ωr.

5. Torque (τ) or Moment of Force

• Definition: The turning effect of a force about an axis or point. It's the rotational analogue of force.
• Formula: Torque τ due to a force F acting at a point with position vector r relative to the pivot/origin is:
  τ = r × F
• Magnitude: τ = r F sin θ, where θ is the angle between r and F. r sin θ is the perpendicular distance from the axis to the line of action of the force (called the lever arm).
• Direction: Perpendicular to the plane containing r and F, given by the Right-Hand Rule.
• Unit: Newton-meter (Nm).
• Couple: A pair of equal and opposite forces acting along different lines of action. Produces pure rotation. Torque due to a couple is independent of the pivot point. Magnitude = (Force) × (Perpendicular distance between forces).

6. Angular Momentum (L)

• Definition: The rotational analogue of linear momentum.
• For a Particle: Angular momentum L of a particle with linear momentum p at position r relative to the origin is:
  L = r × p = r × (mv)
• Magnitude: L = r p sin θ = mvr sin θ, where θ is the angle between r and p. r sin θ is the perpendicular distance from the origin to the line of motion of the particle.
• Direction: Perpendicular to the plane containing r and p, given by the Right-Hand Rule.
• Unit: kg m²/s or Joule-second (Js).
• For a Rigid Body rotating about a fixed axis:
  L = I ω
  where I is the Moment of Inertia about the axis of rotation and ω is the angular velocity. The direction of L is along the axis of rotation (same as ω).

7. Relation between Torque and Angular Momentum

• dL/dt = d(r × p)/dt = (dr/dt × p) + (r × dp/dt)
• dr/dt = v, p = mv, so dr/dt × p = v × (mv) = m(v × v) = 0.
• dp/dt = F (Newton's Second Law).
• Therefore, dL/dt = r × F = τ
  This is the rotational analogue of F = dp/dt: The rate of change of angular momentum of a particle (or system) is equal to the net external torque acting on it.

8. Conservation of Angular Momentum

• If the net external torque acting on a system is zero (τ_ext = 0), then dL/dt = 0.
• This implies L = constant.
• Statement: If the net external torque on a system is zero, its total angular momentum remains conserved (constant in magnitude and direction).
• For a rigid body rotating about a fixed axis, if τ_ext = 0, then L = Iω = constant.
  I₁ω₁ = I₂ω₂
• Applications: A spinning dancer changing speed by folding/extending arms, a diver curling/straightening body mid-air, orbital motion of planets (Kepler's second law).

9. Equilibrium of a Rigid Body

• Translational Equilibrium: Net external force is zero (Σ F_ext = 0). The linear momentum is conserved (P = constant, often P=0).
• Rotational Equilibrium: Net external torque is zero (Σ τ_ext = 0). The angular momentum is conserved (L = constant, often L=0).
• Complete Equilibrium: Both conditions must be satisfied simultaneously.

10. Moment of Inertia (I)

• Definition: A measure of a body's resistance to changes in its rotational motion. It's the rotational analogue of mass.
• Formula:
  • For a system of n particles: I = Σ mᵢ rᵢ², where rᵢ is the perpendicular distance of the i-th particle (mass mᵢ) from the axis of rotation.
  • For a continuous body: I = ∫ r² dm, where r is the perpendicular distance of the mass element dm from the axis.
• Depends on:
  • Mass of the body.
  • Distribution of mass about the axis of rotation.
  • Position and orientation of the axis of rotation.
• Unit: kg m².
• Radius of Gyration (K): The distance from the axis of rotation where the entire mass of the body could be concentrated without changing its moment of inertia.
  I = M K² => K = √(I/M)

11. Theorems of Moment of Inertia

• Theorem of Parallel Axes: The moment of inertia (I) of a body about any axis is equal to the sum of its moment of inertia about a parallel axis passing through its centre of mass (I_CM) and the product of its total mass (M) and the square of the perpendicular distance (d) between the two parallel axes.
  I = I_CM + M d²
  (Applicable to any rigid body).
• Theorem of Perpendicular Axes: For a planar body (lamina), the moment of inertia about an axis perpendicular to its plane (I_z) is equal to the sum of its moments of inertia about two mutually perpendicular axes (I_x and I_y) lying in the plane of the body and intersecting at the point where the perpendicular axis passes through it.
  I_z = I_x + I_y
  (Applicable only to planar bodies).

12. Moment of Inertia of Some Standard Uniform Bodies: (Crucial for problem-solving)

13. Rotational Kinetic Energy

• K_rot = ½ I ω²
  (Analogue of translational KE = ½ mv²)

14. Work Done by Torque & Power

• Work done by a constant torque τ in rotating a body by angle θ: W = τ θ
• Work done by a variable torque: W = ∫ τ dθ
• Instantaneous Power delivered by torque: P = dW/dt = τ (dθ/dt) = τ ω
  (Analogue of P = F v)

15. Rolling Motion

• Combination of translation (of CM) and rotation (about CM).
• Condition for Pure Rolling (without slipping): The velocity of the point of contact with the surface is zero relative to the surface.
  v_CM = R ω
  where v_CM is the velocity of the centre of mass, R is the radius, and ω is the angular velocity about the CM.
• Kinetic Energy of Rolling: Sum of translational KE of CM and rotational KE about CM.
  K_rolling = K_trans + K_rot = ½ M v_CM² + ½ I_CM ω²
  Using ω = v_CM / R and I_CM = M K² (K = radius of gyration):
  K_rolling = ½ M v_CM² + ½ (M K²) (v_CM / R)² = ½ M v_CM² [1 + (K²/R²)]
• Acceleration on an Inclined Plane (Rolling without slipping):
  Angle of inclination θ.
  a_CM = (g sin θ) / [1 + (I_CM / MR²)] = (g sin θ) / [1 + (K²/R²)]
  • Sphere: K²/R² = 2/5 => a = (5/7) g sin θ
  • Disc/Solid Cylinder: K²/R² = 1/2 => a = (2/3) g sin θ
  • Ring/Hollow Cylinder: K²/R² = 1 => a = (1/2) g sin θ
  • Note: Acceleration is maximum for the sphere and minimum for the ring. The body with the smallest K²/R² rolls down fastest.
• Friction in Rolling: Static friction is necessary for pure rolling on an incline. It provides the torque required for angular acceleration and adjusts itself to prevent slipping. f_s ≤ μ_s N.

Multiple Choice Questions (MCQs)

1. The centre of mass of a system of two particles divides the distance between them:
   (a) In inverse ratio of their masses
   (b) In direct ratio of their masses
   (c) Equally always
   (d) Depends on the forces acting

2. If the net external force acting on a system is zero, which of the following must be constant?
   (a) Velocity of the centre of mass
   (b) Acceleration of the centre of mass
   (c) Angular momentum of the system
   (d) Rotational kinetic energy

3. A couple produces:
   (a) Pure translational motion
   (b) Pure rotational motion
   (c) Both translational and rotational motion
   (d) No motion

4. The moment of inertia of a thin circular ring of mass M and radius R about an axis passing through its centre and perpendicular to its plane is:
   (a) MR²/2
   (b) MR²
   (c) (2/5)MR²
   (d) (2/3)MR²

5. A solid sphere is rotating about a diameter. If its radius increases while its mass remains constant, its angular velocity will:
   (a) Increase
   (b) Decrease
   (c) Remain the same
   (d) Become zero

6. The theorem of parallel axes relates the moment of inertia about an axis through the CM (I_CM) to the moment of inertia (I) about a parallel axis at distance d. The correct relation is:
   (a) I = I_CM - Md²
   (b) I = I_CM + Md²
   (c) I_CM = I + Md²
   (d) I = I_CM / (Md²)

7. For a body undergoing pure rolling on a horizontal surface, the velocity of the point of contact is:
   (a) v_CM
   (b) Rω
   (c) 2 v_CM
   (d) Zero

8. A disc and a solid sphere of the same mass and radius roll down an inclined plane without slipping starting from rest. Which one reaches the bottom first?
   (a) Disc
   (b) Solid Sphere
   (c) Both reach simultaneously
   (d) Cannot be determined

9. The rotational analogue of force in linear motion is:
   (a) Moment of Inertia
   (b) Angular Momentum
   (c) Torque
   (d) Angular Velocity

10. If τ = r × F and L = r × p, the relation between torque τ and angular momentum L is:
    (a) τ = dL/dt
    (b) L = dτ/dt
    (c) τ = L × ω
    (d) L = τ × ω

Answers to MCQs:

1. (a)
2. (a) [Note: Acceleration will be zero. Angular momentum is conserved only if net external torque is zero.]
3. (b)
4. (b)
5. (b) [Assuming no external torque, angular momentum L = Iω is conserved. I = (2/5)MR². If R increases, I increases, so ω must decrease.]
6. (b)
7. (d) [Relative to the surface]
8. (b) [Sphere has smaller K²/R² (2/5) compared to disc (1/2), hence greater acceleration.]
9. (c)
10. (a)

Remember to thoroughly understand the definitions, derivations (especially for theorems), and the analogies between linear and rotational motion. Practice problems involving calculations of CM, MOI, torque, angular momentum conservation, and rolling motion. Good luck with your preparation!