Class 11 Physics Notes Chapter 7 (Chapter 7) – Examplar Problems (English) Book

Alright class, let's delve into Chapter 7, 'Systems of Particles and Rotational Motion' from the NCERT Exemplar. This is a crucial chapter, bridging the gap between the motion of point objects and real, extended bodies. Understanding these concepts thoroughly is vital for many government exams including NEET, JEE, and others based on the Class 11/12 syllabus.

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

1. Centre of Mass (CM)

• Concept: The centre of mass of a system of particles 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:
    R_CM = (m₁r₁ + m₂r₂ + ... + m_nr_n) / (m₁ + m₂ + ... + m_n) = (Σ m_ir_i) / M
    where M = Σ m_i is the total mass of the system.
  • In Cartesian coordinates:
    X_CM = (Σ m_ix_i) / M
    Y_CM = (Σ m_iy_i) / M
    Z_CM = (Σ m_iz_i) / M
  • For a continuous body:
    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 Common Uniform Bodies:
  • Rod: Midpoint
  • Ring/Disc/Solid Sphere/Hollow Sphere: Geometric centre
  • Triangle (Lamina): Centroid (Intersection of medians)
  • Cone (Solid): At height h/4 from the base (h = height)
  • Hemisphere (Solid): At height 3R/8 from the base (R = radius)
• Motion of CM:
  • Velocity of CM: V_CM = (Σ m_iv_i) / M
  • Acceleration of CM: A_CM = (Σ m_ia_i) / M
  • Newton's Second Law for the System: M A_CM = F_ext
    (The total external force on the system equals the total mass times the acceleration of the centre of mass. Internal forces cancel out in pairs.)
• Conservation of Linear Momentum: If F_ext = 0, then M A_CM = 0, which implies V_CM = constant. The total linear momentum of the system (P_sys = M V_CM) remains conserved if the net external force is zero.

2. Vector Product (Cross Product) of Two Vectors

• Definition: A x B = |A| |B| sinθ n̂
  where θ is the angle between A and B, and n̂ is a unit vector perpendicular to the plane containing A and B (direction given by the Right-Hand Screw Rule).
• Properties:
  • A x B = - (B x A) (Anti-commutative)
  • A x (B + C) = A x B + A x C (Distributive)
  • A x A = 0 (Zero vector)
  • î x ĵ = k̂, ĵ x k̂ = î, k̂ x î = ĵ
  • ĵ x î = -k̂, k̂ x ĵ = -î, î x k̂ = -ĵ
  • î x î = ĵ x ĵ = k̂ x k̂ = 0
• In Cartesian Components: If A = A_xî + A_yĵ + A_zk̂ and B = B_xî + B_yĵ + B_zk̂, then
  A x B = det | î ĵ k̂ |
  | A_x A_y A_z |
  | B_x B_y B_z |
  = (A_yB_z - A_zB_y)î - (A_xB_z - A_zB_x)ĵ + (A_xB_y - A_yB_x)k̂
• Application: Used to define torque, angular momentum, etc.

3. Rotational Motion Variables

• Angular Position (θ): Angle swept by a position vector from a reference line. (Unit: radians)
• Angular Displacement (Δθ): Change in angular position.
• Angular Velocity (ω): Rate of change of angular position.
  • Average: ω_avg = Δθ / Δt
  • Instantaneous: ω = dθ / dt (Unit: rad/s)
  • Vector form: v = ω x r (where v is linear velocity, r is position vector)
• Angular Acceleration (α): Rate of change of angular velocity.
  • Average: α_avg = Δω / Δt
  • Instantaneous: α = dω / dt = d²θ / dt² (Unit: rad/s²)
  • Relation to tangential acceleration (a_t): a_t = α x r ; Magnitude: a_t = rα
  • Centripetal acceleration (a_c or a_r): Magnitude a_c = v²/r = ω²r (directed towards the centre)
  • Total linear acceleration: a = a_t + a_c

4. Torque (Moment of Force)

• Concept: The rotational analogue of force; the turning effect of a force.
• Definition: Torque (τ) about a point O due to a force F acting at a point P with position vector r relative to O is:
  τ = r x F
• Magnitude: τ = |r| |F| sinθ = r (F sinθ) = F (r sinθ) = r F_⊥ = F r_⊥
  where θ is the angle between r and F, F_⊥ is the component of F perpendicular to r, and r_⊥ is the perpendicular distance from O to the line of action of F (lever arm).
• Direction: Perpendicular to the plane containing r and F, given by the Right-Hand Screw Rule.
• Unit: Newton-metre (N m)
• Dimensions: [M L² T⁻²] (Same as work/energy, but it's a vector quantity)

5. Angular Momentum (Moment of Linear Momentum)

• Concept: The rotational analogue of linear momentum.
• Definition: Angular momentum (L) of a particle with linear momentum p about a point O, having position vector r relative to O is:
  L = r x p = r x (mv)
• Magnitude: L = |r| |p| sinθ = r p sinθ = r p_⊥ = p r_⊥
  where θ is the angle between r and p, p_⊥ is the component of p perpendicular to r, and r_⊥ is the perpendicular distance from O to the line of action of p.
• Direction: Perpendicular to the plane containing r and p, given by the Right-Hand Screw Rule.
• Unit: kg m²/s or Joule-second (J s)
• Dimensions: [M L² T⁻¹]
• For a Rigid Body rotating about a fixed axis: L = I ω
  where I is the Moment of Inertia about the axis and ω is the angular velocity. (Note: L and ω are along the axis of rotation for symmetric bodies).
• For a System of Particles: L_sys = Σ L_i = Σ (r_i x p_i)

6. Relation between Torque and Angular Momentum

• Newton's Second Law in Rotational Motion: The time rate of change of angular momentum of a particle (or system) is equal to the net external torque acting on it.
  τ_ext = dL / dt
• Conservation of Angular Momentum: If the net external torque on a system is zero (τ_ext = 0), then dL/dt = 0, which implies L = constant.
  • L = Iω = constant => I₁ω₁ = I₂ω₂ (if I changes, ω adjusts)
  • Examples: Skater pulling arms in, planet orbiting the sun (approximately), rotating stool experiment.

7. Equilibrium of a Rigid Body

• A rigid body is in equilibrium if it is not accelerating, either linearly or rotationally.
• Translational Equilibrium: The net external force on the body is zero.
  Σ F_ext = 0 (Vector sum)
  Σ F_x = 0 ; Σ F_y = 0 ; Σ F_z = 0
• Rotational Equilibrium: The net external torque on the body about any point is zero.
  Σ τ_ext = 0 (Vector sum)
• Complete Equilibrium: Both conditions must be satisfied simultaneously.

8. Moment of Inertia (I)

• Concept: Rotational analogue of mass; a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation.
• Definition:
  • For a system of particles: I = Σ m_i r_i²
    where r_i is the perpendicular distance of the i-th particle 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 of rotation.
• Unit: kg m²
• Dimensions: [M L²]
• 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 axes.
    I = I_CM + M d² (Applicable to any rigid body)
  • Theorem of Perpendicular Axes: The moment of inertia of a planar body (lamina) about an axis perpendicular to its plane (say, z-axis) is equal to the sum of its moments of inertia about two mutually perpendicular axes lying in the plane of the body and intersecting at the point where the perpendicular axis passes through it (say, x and y axes).
    I_z = I_x + I_y (Applicable only to planar bodies)
• Moment of Inertia of Some Common Regular Bodies: (Memorize these!)
  • Thin Rod (length L, about perpendicular axis through centre): ML²/12
  • Thin Rod (length L, about perpendicular axis through one end): ML²/3
  • Ring (radius R, about axis through centre, perpendicular to plane): MR²
  • Ring (radius R, about a diameter): MR²/2
  • Disc (radius R, about axis through centre, perpendicular to plane): MR²/2
  • Disc (radius R, about a diameter): MR²/4
  • Solid Cylinder (radius R, about its symmetry axis): MR²/2
  • Hollow Cylinder (radius R, about its symmetry axis): MR²
  • Solid Sphere (radius R, about a diameter): (2/5)MR²
  • Hollow Sphere (radius R, about a diameter): (2/3)MR²

9. Kinematics of Rotational Motion (for constant α)

• Analogous to linear motion equations:
  • ω = ω₀ + α t
  • θ = θ₀ + ω₀t + ½ α t²
  • ω² = ω₀² + 2 α (θ - θ₀)
  • θ - θ₀ = ½ (ω₀ + ω) t

10. Dynamics of Rotational Motion

• Relation between Torque and Angular Acceleration: For rotation about a fixed axis:
  τ = I α (Analogue of F = ma)
• Work Done by Torque:
  • Work done by a constant torque τ in rotating a body through an angle θ: W = τ θ
  • Work done by a variable torque: W = ∫ τ dθ
• Rotational Kinetic Energy (K_rot): Energy possessed by a body due to its rotation.
  K_rot = ½ I ω² (Analogue of K_trans = ½ mv²)
• Power Delivered by Torque: Rate at which work is done by the torque.
  P = dW/dt = τ (dθ/dt) = τ ω (Analogue of P = F v)

11. Rolling Motion

• Concept: Combination of translational motion of the centre of mass and rotational motion about the centre of mass.
• Pure Rolling (Without Slipping): The point of contact between the rolling body and the surface is instantaneously at rest relative to the surface.
  • Condition: v_CM = R ω (where R is the radius)
  • Acceleration condition: a_CM = R α
• Kinetic Energy of Rolling: Sum of translational and rotational kinetic energy.
  K_total = K_trans + K_rot = ½ M v_CM² + ½ I_CM ω²
  Using v_CM = Rω and I_CM = M k² (where k is the radius of gyration):
  K_total = ½ M v_CM² (1 + k²/R²)
• Acceleration on an Inclined Plane (Pure Rolling):
  a_CM = g sinθ / (1 + k²/R²)
• Friction in Rolling: Static friction is generally required for pure rolling (especially during acceleration/deceleration or on an incline). It provides the necessary torque (f * R = I_CM α) or prevents slipping. If rolling is pure on a horizontal surface at constant velocity, friction is not needed (ideally). Kinetic friction acts if slipping occurs.

Key Analogies: Linear vs. Rotational Motion

Multiple Choice Questions (MCQs)

1. The centre of mass of a uniform L-shaped lamina (thin flat plate) with dimensions shown will lie:
   (Assume vertices are (0,0), (2L,0), (2L,L), (L,L), (L,2L), (0,2L))
   (a) At the point (L, L)
   (b) Somewhere in the first quadrant but not at (L, L)
   (c) On the y-axis
   (d) On the x-axis

2. A particle moves on a circular path with decreasing speed. Choose the correct statement.
   (a) Angular momentum is conserved.
   (b) Only direction of angular momentum L is conserved.
   (c) Acceleration is directed towards the centre.
   (d) The net torque on the particle about the centre is non-zero.

3. A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?
   (a) Angular velocity
   (b) Moment of inertia
   (c) Rotational kinetic energy
   (d) Angular momentum

4. A disc of moment of inertia I₁ is rotating with angular velocity ω₁ about an axis perpendicular to its plane and passing through its centre. If another disc of moment of inertia I₂ is gently placed over it coaxially, the new angular velocity of the combined system will be:
   (a) (I₁ω₁) / (I₁ + I₂)
   (b) (I₁ + I₂)ω₁ / I₁
   (c) (I₂ω₁) / (I₁ + I₂)
   (d) ω₁

5. A solid sphere rolls down two different inclined planes of the same height but different angles of inclination. It will reach the bottom:
   (a) With the same speed in both cases.

   • (b) With larger speed on the steeper incline.
   • (c) With larger speed on the less steep incline.
   • (d) Information insufficient.

6. The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is I₀. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is:
   (a) I₀ + ML²/4
   (b) I₀ + ML²/2
   (c) I₀ + ML²
   (d) 2 I₀

7. For a body to be in complete equilibrium, which condition(s) must be satisfied?
   (a) The vector sum of all forces acting on it must be zero.
   (b) The vector sum of all torques acting on it about any point must be zero.
   (c) Both (a) and (b) must be satisfied.
   (d) Either (a) or (b) must be satisfied.

8. A wheel has angular acceleration of 3.0 rad/s². If its initial angular speed is 2.0 rad/s, the angular displacement (in radians) after 2.0 s is:
   (a) 6
   (b) 10
   (c) 4
   (d) 8

9. The rotational kinetic energy of a body is K_rot and its moment of inertia is I. The angular momentum L is given by:
   (a) √(2 I K_rot)
   (b) 2 I K_rot
   (c) K_rot / (2I)
   (d) √(K_rot / (2I))

10. The ratio of the radius of gyration of a solid sphere of radius R about its own axis to the radius of gyration of a thin hollow sphere of same radius R about its axis is:
    (a) √(3/5)
    (b) √(2/5)
    (c) √(3/2)
    (d) √(2/3)

Answers to MCQs:

1. (b) [Hint: Treat it as two rectangles; find CM of each, then combine.]
2. (d) [Hint: Speed is decreasing, so there must be tangential acceleration, hence a tangential force, causing a torque.]
3. (d) [Hint: No external torque acts, so angular momentum (L = Iω) is conserved. As R increases, I increases, so ω must decrease. K_rot = L²/2I also changes.]
4. (a) [Hint: Apply conservation of angular momentum: L_initial = L_final => I₁ω₁ = (I₁ + I₂)ω_final.]
5. (a) [Hint: Speed at the bottom depends only on height (v² = 2gh / (1 + k²/R²)), not the angle, assuming pure rolling starts from rest.]
6. (a) [Hint: Use Parallel Axis Theorem. I₀ = ML²/12. Distance d = L/2. I = I₀ + Md² = ML²/12 + M(L/2)² = ML²/12 + ML²/4 = 4ML²/12 = ML²/3. So I = I₀ + ML²/4.]
7. (c) [Hint: Definition of complete equilibrium.]
8. (b) [Hint: Use θ = ω₀t + ½ α t² = (2.0)(2.0) + ½ (3.0)(2.0)² = 4 + 6 = 10 rad.]
9. (a) [Hint: K_rot = ½ I ω² = ½ (Iω)² / I = L² / 2I => L = √(2 I K_rot).]
10. (a) [Hint: For solid sphere, I = (2/5)MR² = Mk₁². So k₁ = R√(2/5). For hollow sphere, I = (2/3)MR² = Mk₂². So k₂ = R√(2/3). Ratio k₁/k₂ = √(2/5) / √(2/3) = √(3/5).]

Study these notes carefully, focusing on the definitions, formulas, theorems, and conservation laws. Practice problems applying these concepts, especially those involving moment of inertia calculations, conservation of angular momentum, and rolling motion. Good luck with your preparation!