Class 11 Physics Notes Chapter 7 (Systems of particles and rotational motion) – Physics Part-I Book
Alright class, let's get straight into Chapter 7: Systems of Particles and Rotational Motion from your NCERT Physics Part-I textbook. This is a crucial chapter, extending our understanding from single-particle motion to the more complex, real-world scenarios involving systems of particles and rigid bodies, especially their rotational behaviour. Pay close attention, as these concepts are frequently tested in government exams.
Chapter 7: Systems of Particles and Rotational Motion - Detailed Notes
1. Introduction
- So far, we've treated objects as point masses. However, real objects have finite size and are composed of many particles.
- This chapter deals with:
- Motion of extended bodies (systems of particles/rigid bodies).
- Concept of Centre of Mass.
- Rotational motion of rigid bodies.
2. Centre of Mass (CM)
- Definition: The Centre of Mass of a system is a hypothetical point where the entire mass of the system is assumed to be concentrated for describing its translational motion.
- Importance: The motion of the CM represents the average translational motion of the system as a whole. Newton's laws apply directly to the motion of the CM.
- Position Vector of CM (Discrete System): For a system of 'n' particles with masses m₁, m₂, ..., m<0xE2><0x82><0x99> and position vectors r₁, r₂, ..., r<0xE2><0x82><0x99>:
R_CM = (m₁r₁ + m₂r₂ + ... + m<0xE2><0x82><0x99>r<0xE2><0x82><0x99>) / (m₁ + m₂ + ... + m<0xE2><0x82><0x99>) = (Σ mᵢrᵢ) / M
where M = Σ mᵢ is the total mass of the system. - Coordinates of CM:
X_CM = (Σ mᵢxᵢ) / M
Y_CM = (Σ mᵢyᵢ) / M
Z_CM = (Σ mᵢzᵢ) / M - CM for 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:
- 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.
- Triangular Lamina: At the centroid.
- Solid Cone: At height h/4 from the base (h = height of cone).
- Motion of CM:
- Velocity of CM: V_CM = dR_CM/dt = (Σ mᵢvᵢ) / M
- Acceleration of CM: A_CM = dV_CM/dt = (Σ mᵢaᵢ) / M
- From Newton's Second Law: M A_CM = Σ F_ext (Internal forces cancel out in pairs). The CM moves as if the entire mass M is concentrated at that point and the net external force acts on it.
- Conservation of Momentum: If Σ F_ext = 0, then A_CM = 0, which means V_CM = constant. The total linear momentum of the system (P_sys = M V_CM = Σ mᵢvᵢ) remains conserved if the net external force is zero.
3. Vector Product (Cross Product) of Two Vectors
- Essential for defining rotational quantities like torque and angular momentum.
- Definition: c = a × b. The magnitude is |c| = |a| |b| sin θ, where θ is the angle between a and b.
- Direction: The direction of c is perpendicular to the plane containing a and b, given by the Right-Hand Screw Rule or Right-Hand Thumb Rule.
- Properties:
- Not Commutative: a × b = - (b × a)
- Distributive: a × (b + c) = (a × b) + (a × c)
- a × a = 0 (zero vector)
- For unit vectors: 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
- In component form: If a = aₓî + a<0xE1><0xB5><0xB3>ĵ + a<0xE2><0x82><0x91>k̂ and b = bₓî + b<0xE1><0xB5><0xB3>ĵ + b<0xE2><0x82><0x91>k̂, then
a × b = det | î ĵ k̂ |
| aₓ a<0xE1><0xB5><0xB3> a<0xE2><0x82><0x91> |
| bₓ b<0xE1><0xB5><0xB3> b<0xE2><0x82><0x91> |
a × b = (a<0xE1><0xB5><0xB3>b<0xE2><0x82><0x91> - a<0xE2><0x82><0x91>b<0xE1><0xB5><0xB3>)î - (aₓb<0xE2><0x82><0x91> - a<0xE2><0x82><0x91>bₓ)ĵ + (aₓb<0xE1><0xB5><0xB3> - a<0xE1><0xB5><0xB3>bₓ)k̂
4. Angular Velocity (ω) and Angular Acceleration (α)
- Angular Displacement (θ): Angle turned by a body about a fixed axis. Unit: radians (rad).
- Angular Velocity (ω): Rate of change of angular displacement. ω = dθ/dt. Unit: rad/s. It's an axial vector, direction along the axis of rotation (Right-Hand Rule).
- Angular Acceleration (α): Rate of change of angular velocity. α = dω/dt = d²θ/dt². Unit: rad/s². Also an axial vector.
- Relation between Linear and Angular Velocity: For a particle at position vector r from the axis, its linear velocity v is v = ω × r. The magnitude is v = ωr sin φ, where φ is the angle between ω and r. For circular motion, φ = 90°, so v = ωr.
5. Torque (Moment of Force) (τ)
- Definition: The rotational analogue of force. It's the turning effect of a force about an axis or point.
- Formula: τ = r × F, where r is the position vector of the point of application of force F relative to the pivot point/axis.
- Magnitude: |τ| = |r| |F| sin θ, where θ is the angle between r and F. |τ| = r (F sin θ) = r F<0xE2><0x8A><0xA5> = (r sin θ) F = r<0xE2><0x8A><0xA5> F. (F<0xE2><0x8A><0xA5> is the component of F perpendicular to r, and r<0xE2><0x8A><0xA5> is the perpendicular distance from the pivot to the line of action of F - called the lever arm).
- Unit: Newton-metre (N m). Dimension: [ML²T⁻²] (same as work, but torque is a vector, work is scalar).
- Direction: Perpendicular to the plane containing r and F, given by the Right-Hand Rule.
6. Angular Momentum (Moment of Linear Momentum) (L)
- Definition: The rotational analogue of linear momentum.
- Formula (Single Particle): L = r × p, where r is the position vector and p = mv is the linear momentum.
- Magnitude: |L| = |r| |p| sin θ = rp sin θ = p r<0xE2><0x8A><0xA5> = r p<0xE2><0x8A><0xA5>.
- Unit: kg m²/s or Joule-second (J s). Dimension: [ML²T⁻¹].
- Direction: Perpendicular to the plane containing r and p, given by the Right-Hand Rule.
- For a System of Particles: L = Σ Lᵢ = Σ (rᵢ × pᵢ)
- For a Rigid Body Rotating about a Fixed Axis: L = Iω, where I is the Moment of Inertia about that axis. (This is analogous to p = mv). Note: L and ω are in the same direction only if the axis is an axis of symmetry. In general, L is not necessarily parallel to ω.
7. Relation between Torque and Angular Momentum
- dL/dt = τ_ext
- This is the rotational analogue of Newton's Second Law (dp/dt = F_ext). The rate of change of angular momentum of a system is equal to the net external torque acting on it.
8. Conservation of Angular Momentum
- Principle: If the net external torque acting on a system is zero (τ_ext = 0), then its total angular momentum L remains constant. dL/dt = 0 ⇒ L = constant.
- If L = Iω, then Iω = constant. If I changes, ω adjusts itself such that Iω remains constant. (e.g., spinning dancer pulling arms in, planet's speed changing as it orbits the sun).
9. Equilibrium of a Rigid Body
- Rigid Body: A body with a perfectly definite and unchanging shape. The distance between any two particles remains constant.
- Translational Equilibrium: The net external force is zero. Σ F_ext = 0. (Prevents linear acceleration).
- Rotational Equilibrium: The net external torque about any point is zero. Σ τ_ext = 0. (Prevents angular acceleration).
- Complete Equilibrium: Both conditions must be satisfied simultaneously.
10. Moment of Inertia (I)
- Definition: The rotational analogue of mass. It measures a body's resistance to changes in its rotational motion.
- Formula (Discrete Particles): I = Σ mᵢrᵢ², where rᵢ is the perpendicular distance of the i-th particle from the axis of rotation.
- Formula (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.
- Position and orientation of the axis of rotation.
- Unit: kg m². Dimension: [ML²].
- Radius of Gyration (K): A measure of mass distribution. Defined by I = MK², where M is the total mass. K is the effective distance from the axis where the entire mass could be concentrated to give the same moment of inertia. 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 (I_CM) about a parallel axis passing through its centre of mass and the product of its mass (M) and the square of the perpendicular distance (a) between the two axes.
I = I_CM + Ma²
(Very useful for finding MI about an axis not passing through CM). - Theorem of Perpendicular Axes: For a planar body (lamina), the moment of inertia about an axis perpendicular to its plane (say, z-axis) is the sum of its moments of inertia about two 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).
12. Moment of Inertia of Some Common Regular Bodies
(Axis passing through CM unless specified)
* Thin Rod (length L, about perpendicular axis through centre): I = ML²/12
* Thin Rod (length L, about perpendicular axis through one end): I = ML²/3
* Circular Ring (radius R, about axis through centre, perpendicular to plane): I = MR²
* Circular Ring (radius R, about a diameter): I = MR²/2
* Circular Disc (radius R, about axis through centre, perpendicular to plane): I = MR²/2
* Circular Disc (radius R, about a diameter): I = MR²/4
* Hollow Cylinder (radius R, about cylinder axis): I = MR²
* Solid Cylinder (radius R, about cylinder axis): I = MR²/2
* Solid Sphere (radius R, about a diameter): I = (2/5)MR²
* Hollow Sphere (radius R, about a diameter): I = (2/3)MR²
13. Kinematics of Rotational Motion (Fixed Axis)
- Analogous to linear kinematics:
- ω = ω₀ + αt
- θ = θ₀ + ω₀t + ½ αt²
- ω² = ω₀² + 2α(θ - θ₀)
(Assuming constant angular acceleration α. θ₀ is initial angular position, ω₀ is initial angular velocity).
14. Dynamics of Rotational Motion (Fixed Axis)
- τ = Iα (Rotational analogue of F = ma). This relates the net external torque about the fixed axis to the angular acceleration and moment of inertia about that axis.
15. Work, Power, and Energy in Rotational Motion
- Work Done by Torque: W = ∫ τ dθ. For constant torque, W = τθ.
- Rotational Kinetic Energy: K_rot = ½ Iω². (Analogue of K_trans = ½ mv²).
- Power Delivered by Torque: P = dW/dt = τ (dθ/dt) = P = τω. (Analogue of P = Fv).
- Work-Energy Theorem (Rotational): Work done by net torque = Change in rotational kinetic energy. W_τ = ΔK_rot = ½ Iω_f² - ½ Iω_i².
16. Rolling Motion
- Definition: Combination of translation (of CM) and rotation (about CM).
- Pure Rolling (without slipping): The point of contact with the surface is instantaneously at rest.
- Condition: v_CM = Rω (where R is the radius).
- Also, a_CM = Rα.
- Kinetic Energy of Rolling: Sum of translational and rotational kinetic energy.
K_roll = K_trans + K_rot = ½ Mv_CM² + ½ I_CMω²
Using v_CM = Rω and I_CM = MK² (K = radius of gyration):
K_roll = ½ Mv_CM² (1 + K²/R²) - Rolling on an Inclined Plane (Angle θ):
- Force causing translation: Mg sinθ - f (f = static friction)
- Torque causing rotation: τ = fR = I_CMα
- Acceleration: a_CM = (g sinθ) / (1 + K²/R²) = (g sinθ) / (1 + I_CM / MR²)
- Friction force: f = (Mg sinθ) / (1 + R²/K²) = (Mg sinθ * I_CM) / (I_CM + MR²)
- Condition for rolling without slipping: f ≤ μ_s N = μ_s Mg cosθ.
⇒ (tanθ) / (1 + R²/K²) ≤ μ_s
Multiple Choice Questions (MCQs)
-
The centre of mass of a uniform solid cone of height 'h' is located at a height of ______ from its base.
(a) h/2
(b) h/3- (c) h/4
(d) 3h/4
- (c) h/4
-
If the net external force on a system of particles is zero, which of the following must be constant?
(a) Velocity of each particle
(b) Acceleration of the centre of mass- (c) Velocity of the centre of mass
(d) Total kinetic energy of the system
- (c) Velocity of the centre of mass
-
A couple produces:
(a) Purely linear motion- (b) Purely rotational motion
(c) Both linear and rotational motion
(d) No motion
- (b) Purely rotational motion
-
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
(b) MR²/2
(c) MR²/4
(d) (2/5)MR²
- (a) MR²
-
A solid sphere is rolling on a horizontal surface without slipping. What fraction of its total kinetic energy is rotational? (I_CM = (2/5)MR²)
(a) 1/2
(b) 2/5- (c) 2/7
(d) 5/7
(Hint: K_rot / K_total = (½ Iω²) / (½ Mv² + ½ Iω²) = (Iω²) / (M(Rω)² + Iω²) = I / (MR² + I) = (2/5)MR² / (MR² + (2/5)MR²) = (2/5) / (7/5) = 2/7)
- (c) 2/7
-
If the radius of the earth contracts to half of its present value without change in mass, the duration of the day will be:
(a) 24 hours
(b) 12 hours- (c) 6 hours
(d) 48 hours
(Hint: Conservation of angular momentum L = Iω = constant. I ∝ R². ω ∝ 1/T. So, R² (1/T) = constant. If R becomes R/2, (R/2)² (1/T') = R² (1/T) ⇒ T' = T/4 = 24/4 = 6 hours)
- (c) 6 hours
-
The angular momentum (L) and torque (τ) are related by:
(a) L = τ × r
(b) τ = L × r- (c) τ = dL/dt
(d) L = dτ/dt
- (c) τ = dL/dt
-
The theorem of parallel axes for moment of inertia is given by: (I_CM is MI about CM axis, M is mass, 'a' is distance between parallel axes)
(a) I = I_CM - Ma²
(b) I = I_CM / Ma²
(c) I = Ma² - I_CM- (d) I = I_CM + Ma²
-
A particle performs uniform circular motion with angular momentum L. If its angular frequency is doubled and kinetic energy is halved, the new angular momentum is:
(a) L/2- (b) L/4
(c) 2L
(d) 4L
(Hint: K = ½ Iω²; L = Iω. So K = ½ (L/ω) ω² = ½ Lω. L' = 2K'/ω'. Given ω' = 2ω, K' = K/2. L' = 2(K/2) / (2ω) = K / (2ω) = (½ Lω) / (2ω) = L/4)
- (b) L/4
-
A wheel starts from rest and accelerates uniformly with an angular acceleration of 2.0 rad/s². After 5 seconds, the angle turned through (in radians) is:
(a) 10 rad
(b) 20 rad- (c) 25 rad
(d) 50 rad
(Hint: Use θ = ω₀t + ½ αt². ω₀ = 0, α = 2.0, t = 5. θ = 0 + ½ (2.0)(5)² = 25 rad)
- (c) 25 rad
Answers to MCQs:
- (c)
- (c)
- (b)
- (a)
- (c)
- (c)
- (c)
- (d)
- (b)
- (c)
Remember to thoroughly understand these concepts and practice numerical problems based on them. Focus on the definitions, formulas, conservation laws, and the theorems of moment of inertia, as they are fundamental to solving problems in rotational mechanics. Good luck with your preparation!