NCERT Plus One (Class 11) Physics • Chapter 6

System of Particles and Rotational Motion

Q: Give one example of a body whose centre of mass lies outside its material body.

Key (1 Mark): A uniform circular ring / bangle , a hollow sphere, or a curved boomerang (COM lies at the empty geometric centre).

Q: A firecracker fired from a cannon explodes mid-flight into fragments. Explain why the centre of mass of the fragments still follows the original parabolic trajectory.

Forces during explosion are strictly internal forces , which cancel in equal and opposite pairs (Newton's 3rd Law). Since M A cm = F ext and gravity is the only external force before and after burst, the COM trajectory stays completely unchanged.

Q: State the two conditions required for a rigid body to be in complete mechanical equilibrium. Does zero net force guarantee equilibrium?

1. Translational equilibrium: Total external force ∑ F ext = 0 . 2. Rotational equilibrium: Total external torque ∑ τ ext = 0 . No! A Couple has ∑ F = 0 yet causes pure rotation because ∑ τ ≠ 0 .

Q: Define Torque and Angular Momentum. Establish the relation τ = dL/dt for a particle, and hence state the law of conservation of angular momentum.

• Torque: τ = r × F . • Angular Momentum: L = r × p . Differentiating: d L /dt = (d r /dt × p ) + ( r × d p /dt) = ( v × m v ) + ( r × F ) = 0 + τ . Thus τ ext = d L /dt . If τ ext = 0 , L = I ω = constant .

Q: A girl rotates on a frictionless turntable with arms stretched out at angular speed ω₁. How does her angular speed change when she folds her arms close to her chest? Explain.

No external torque acts ( τ ext = 0 ), so angular momentum L = Iω is conserved. Folding arms brings mass closer to the axis, decreasing Moment of Inertia I . Hence angular speed ω increases ( I₁ω₁ = I₂ω₂ ).

Q: Write the Moment of Inertia of: (a) a circular disc about an axis perpendicular to its plane through its centre, (b) a thin rod about an axis perpendicular to it at its midpoint, and (c) a solid sphere about its diameter.

• (a) Circular disc: I = ½ MR² • (b) Thin rod at midpoint: I = (1/12) ML² • (c) Solid sphere about diameter: I = (2/5) MR²

Q: A uniform metre stick balances horizontally on a knife edge at the 50 cm mark. When two coins of 5 g each are stacked at the 12 cm mark, the stick balances at 45 cm. Find the mass of the metre stick.

Stick's own weight M acts at its 50 cm midpoint ( 50 − 45 = 5 cm right of new pivot). Two coins = 10 g sit at 45 − 12 = 33 cm left of pivot. By Principle of Moments: 10 × 33 = M × 5 ⇒ M = 330 / 5 = 66 g !

Q: Obtain an expression for the linear velocity and acceleration of a rigid body of radius R and radius of gyration K rolling without slipping down an inclined plane of height h and angle θ.

By energy conservation: mgh = ½mv² + ½Iω² . Substitute I = mK² and pure rolling ω = v/R : mgh = ½mv²(1 + K²/R²) ⇒ v = √[2gh / (1 + K²/R²)] . Using v² = 2a(h / sin θ) gives acceleration a = g sin θ / (1 + K²/R²) .

Centre of mass, vector cross product, torque, angular momentum conservation, rigid-body equilibrium, moment of inertia & rotational dynamics.

📄 Official NCERT Textbook PDF

Topic 6.2 – 6.4: Centre of Mass & Linear Momentum of a System

Try balancing a wooden coconut-leaf broom or a heavy cricket bat across one finger in your Thrissur courtyard! The finger does NOT sit at the geometric middle — it naturally slides closer to the thicker, heavier end until both sides balance perfectly. That magic balance point is the Centre of Mass (COM)!

2-PARTICLE COMx_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂) • Inverse ratio: m₁r₁ = m₂r₂

SYSTEM MOTIONM V_cm = P_total | M A_cm = F_ext

EXPLOSION MAGICInternal forces cancel! If F_ext = 0, V_cm NEVER changes!

🧠 Memory Trick: COM is a Mass-Weighted VIP Vote! Heavier particles pull the Centre of Mass closer to themselves (r₁/r₂ = m₂/m₁)!

💡 Simple Kid-Check: Throw a spinning cricket bat across a field: every corner wobbles wildly in loops, yet one tiny painted dot — its Centre of Mass — glides along a super-smooth parabola as if all mass lived inside that single dot!

1
Position of COM: For particles of masses m₁, m₂, ... at r₁, r₂, ...:
R_cm = ∑ mᵢ rᵢM, where total mass M = ∑ mᵢ.
2
Linear Momentum Conservation: Differentiating gives M V_cm = ∑ mᵢ vᵢ = P, and M A_cm = F_ext! Internal pushes between particles always cancel in pairs by Newton's Third Law.
3
Firecracker / Shell Explosion: If a flying shell explodes mid-air, gravity is the only external force, so even as pieces blast everywhere, the COM calmly finishes the exact original parabolic arc!

🧒 Super-Easy Example: Two balls of 2 kg (at x = 0) and 6 kg (at x = 8 m) sit on a light rod.
x_cm = (2×0 + 6×8) / (2 + 6) = 48 / 8 = 6 m — three times closer to the heavier 6 kg ball!

Major misconception: The Centre of Mass does NOT have to lie inside solid material! For a circular bangle (ring), a hollow bowl, or an L-shaped boomerang, the COM floats right in empty air at the centre!

Topic 6.5 – 6.6: Vector Cross Product & Angular Velocity

Watch a ceiling fan spinning above your desk! Pick two dust specks on one blade — one near the centre hub and one right at the outer tip. Both finish one full circle in the exact same time, yet the outer speck zips through the air way faster! Why? Because same angular speed ω gives bigger linear speed v = rω farther out!

CROSS PRODUCTA × B = AB sin θ n̂ • A × B = −(B × A)!

CYCLIC UNIT TRIADî×ĵ = k̂, ĵ×k̂ = î, k̂×î = ĵ (reverse = minus; î×î = 0)

ANGULAR ↔ LINEARω = dθ/dt (rad/s) | v = ω × r (v = rω)

🧠 Memory Trick: i → j → k → i Circle! Go clockwise = + next letter; go backward = − sign! And |A × B| = Area of parallelogram!

💡 Simple Kid-Check: Unlike dot product (which gives a plain number), a cross product builds a brand-new perpendicular vector pointing along your right thumb when fingers curl from A to B!

1
Right-Hand Thumb Rule: Curl your right fingers from the first vector A to the second vector B; your thumb points along A × B (perpendicular to both!).
2
Determinant Shortcut:
A × B = (A_yB_z − A_zB_y)î + (A_zB_x − A_xB_z)ĵ + (A_xB_y − A_yB_x)k̂.
3
Angular Velocity ω & Linear Velocity v: Every particle of a rigid rotating body shares the same ω (pointing along the axis by right-hand thumb rule) and moves tangentially at v = ω × r.

🧒 Super-Easy Example: A ceiling fan spins at 10 rad/s. A speck 0.1 m from hub moves at v = 0.1 × 10 = 1 m/s, while the tip at 0.6 m speeds along at v = 0.6 × 10 = 6 m/s!

Order is SACRED in cross products! Never write v = r × ω — swapping order flips direction to the exact opposite (A × B = − B × A). Correct order is always v = ω × r!

Topic 6.7: Torque, Angular Momentum & Conservation

Ever tried pushing open a heavy wooden door near its metal hinges? Your hand strains and the door barely budges! Push the exact same lightly near the outer brass handle and it swings wide open! Why? Because rotation doesn't care only about Force — it cares about Force × Lever Arm, called Torque!

TORQUE (ROTATIONAL PUSH)τ = r × F → τ = r F sin θ = r_⊥ F (Unit: N·m)

ANGULAR MOMENTUML = r × p = I ω | Newton's Law: τ_ext = dL/dt

SKATER CONSERVATIONIf τ_ext = 0 ⇒ L = I ω = constant (Arms in → I↓ → ω↑!)

🧠 Memory Trick: Torque turns, Angular Momentum keeps spinning! Just replace F → τ and p → L in F = dp/dt to get τ = dL/dt!

💡 Simple Kid-Check: Long spanner wrenches loosen tight scooter bolts easily because increasing r multiplies your Torque τ = rF sin θ without you pushing any harder!

1
Torque (Moment of Force): τ = r × F, magnitude τ = r F sin θ. Maximum when force pushes at 90° to the handle; zero if pushed directly toward/away from the hinge (θ = 0° or 180°)!
2
Angular Momentum (Moment of Momentum): L = r × p (SI unit kg·m²/s or J·s). Differentiating gives τ_ext = dLdt — internal torques cancel!
3
Conservation of Angular Momentum: When external torque is zero (τ_ext = 0), I₁ ω₁ = I₂ ω₂! Pulling arms inward shrinks Moment of Inertia I, so spin speed ω rockets upward (used by ice skaters, divers & ballet dancers!).

🧒 Super-Easy Example: A dancer spinning on a frictionless stool at 2 rad/s pulls her arms in, halving her moment of inertia from I to I/2.
I × 2 = (I/2) × ω₂ ⇒ ω₂ = 4 rad/s (doubles spin speed)!

Two giant DHSE traps: (1) Even though Torque and Work share dimensions [ML²T⁻²], Work is a scalar in Joules while Torque is a vector in N·m (NEVER write Torque in Joules!). (2) Order is τ = r × F and L = r × p — NEVER reverse them!

Topic 6.8: Equilibrium of a Rigid Body & Centre of Gravity

Watch a vegetable seller at a Kochi market balancing a brass two-pan beam scale (Tharasu), or two kids adjusting their seats on a park seesaw until it floats flat! Even when the kids have different weights, putting the heavier kid closer to the pivot freezes the beam in perfect Equilibrium!

TWO EQUILIBRIUM RULES1) Translational: ∑ F = 0 | 2) Rotational: ∑ τ = 0

PRINCIPLE OF MOMENTSLoad × Load arm = Effort × Effort arm (F₁d₁ = F₂d₂)

COUPLE & CGCouple: ∑F = 0 yet spins! Uniform g ⇒ CG = COM

🧠 Memory Trick: No Slide + No Spin = True Equilibrium! Zero net force stops sliding; zero net torque stops spinning!

💡 Simple Kid-Check: Turn a car steering wheel with one hand pushing up on the right and your other hand pulling down equally on the left: the forces cancel (∑F = 0) so the wheel doesn't fly off the dashboard, yet their Couple torques add up to spin it smoothly!

1
Partial vs Complete Equilibrium: A rigid body is in complete mechanical equilibrium ONLY when BOTH total external force ∑F = 0 (no linear acceleration) AND total external torque ∑τ = 0 about any point (no angular acceleration)!
2
Principle of Moments (Levers): For an ideal balanced beam with pivot reaction R = F₁ + F₂, clockwise moment equals anticlockwise moment: F₁ d₁ = F₂ d₂. Mechanical Advantage = F₁ / F₂ = d₂ / d₁.
3
Centre of Gravity (CG): The point where the entire weight Mg effectively acts so total gravitational torque is zero. In uniform gravity, CG sits right on top of COM (only skyscrapers tall enough for g to fade at the top differ slightly!).

🧒 Super-Easy Example: A 40 kg child sits 1.5 m left of a seesaw pivot. Where must a 30 kg friend sit on the right to balance?
40 × 1.5 = 30 × d₂ ⇒ 60 = 30 d₂ ⇒ d₂ = 2.0 m (and pivot pushes up with R = 70 g)!

Top NCERT conceptual trap: Can a body have ∑F = 0 and still NOT be in equilibrium? YES! A Couple (two equal, opposite, parallel forces along different lines — like turning a water tap or steering wheel) has zero net force yet produces pure rotation!

Topic 6.9: Moment of Inertia & Radius of Gyration of Standard Bodies

Grab a long wooden broomstick: hold it by the exact middle and twist it back and forth — super easy! Now hold the exact same broomstick by one tip and try twisting it! The mass didn't gain a single gram, yet twisting feels way tougher because shifting the mass farther from the axis rockets up its Moment of Inertia!

ROTATIONAL INERTIAI = ∑ mᵢ rᵢ² (SI unit: kg·m²) • Depends on mass AND axis distance!

RADIUS OF GYRATIONI = M K² ⇒ K = √(I/M) (Effective distance of mass)

BIG 4 BODIES (AXIS)Ring MR² • Disc ½MR² • Solid Sphere (2/5)MR² • Rod centre (1/12)ML²

🧠 Memory Trick: Farther Mass = Lazier Spin! Ring puts all mass on rim (1·MR²), Disc spreads it (½MR²), Solid Sphere packs inside (0.4 MR²)!

💡 Simple Kid-Check: Mass measures how lazy an object is when pushed straight; Moment of Inertia I measures how lazy an object is when twisted around an axis!

Rigid Body	Axis of Rotation	Moment of Inertia `I`	Gyration `K`
Thin Circular Ring (`R`)	⊥ to plane, through centre	`MR²`	`R`
Thin Circular Ring (`R`)	Along any diameter	`½ MR²`	`R / √2`
Circular Disc / Solid Cylinder	⊥ to plane, through centre	`½ MR²`	`R / √2`
Circular Disc (`R`)	Along any diameter	`¼ MR²`	`R / 2`
Hollow Cylinder (`R`)	Central symmetry axis	`MR²`	`R`
Thin Rod (length `L`)	⊥ to rod at midpoint	`(1/12) ML²`	`L / √12`
Thin Rod (length `L`)	⊥ to rod at one end	`(1/3) ML²`	`L / √3`
Solid Sphere (`R`)	Along any diameter	`(2/5) MR²`	`R √(2/5)`

Do NOT treat Moment of Inertia as a fixed number like mass! Changing the axis changes I dramatically: a thin rod spun around its centre has (1/12)ML², whereas spun around its end has (1/3)ML² — 4 times harder to spin! Also note flywheels concentrate heavy iron on the outer rim to maximize I for smooth engines!

Topic 6.10 – 6.12: Rotational Kinematics, Dynamics (τ = Iα) & Rolling Motion

Grab a solid glass marble, a flat steel coin (disc), and a metal bangle (ring) and roll all three down a slanted wooden plank at the exact same moment! Even if all three have identical mass and radius, the solid marble ALWAYS crosses the finish line first, followed by the coin, and the bangle finishes last! Why? Because of how each splits energy into spinning!

ROTATIONAL DYNAMICSτ = I α | K_rot = ½ I ω² | W = τ Δθ | P = τ ω

3 KINEMATIC EQUATIONSω = ω₀ + αt • θ = ω₀t + ½αt² • ω² = ω₀² + 2αθ

PURE ROLLING (v_cm = Rω)K_total = ½mv²(1 + K²/R²) • Downhill: Sphere > Disc > Ring!

🧠 Memory Trick: Linear-to-Rotation Translator! Just swap x→θ, v→ω, a→α, m→I, F→τ, p→L and EVERY linear formula turns into a rotational formula!

💡 Simple Kid-Check: When a bicycle wheel rolls smoothly without skidding, the rubber touching the road stops dead for an instant (v_bottom = 0), the axle moves at v, and the top rim zooms forward at 2v!

Translational Motion	Rotational Analog
Displacement `x` \| Velocity `v` \| Accel `a`	Angle `θ` \| Angular vel `ω` \| Angular accel `α`
Mass `m` \| Force `F = ma`	Moment of Inertia `I` \| Torque `τ = Iα`
Momentum `p = mv` \| `K = ½mv²`	Angular Mom `L = Iω` \| `K_rot = ½Iω²`
Work `dW = F dx` \| Power `P = Fv`	Work `dW = τ dθ` \| Power `P = τω`

🧒 Why Solid Sphere Wins Downhill: Rolling from rest at height h: mgh = ½mv²(1 + K²/R²) ⇒ v = √[2gh / (1 + K²/R²)].
Smaller K²/R² wins! Sphere (2/5 = 0.4) beats Disc (1/2 = 0.5) beats Ring (1.0)!

Watch two sneaky PYQ questions: (1) In pure rolling without slipping (v_cm = Rω), what is the speed of the bottom contact point? ZERO! (While the topmost point moves at 2v_cm!). (2) On a frictionless inclined plane, a sphere CANNOT roll cleanly — it just slides without spinning!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Give one example of a body whose centre of mass lies outside its material body.

Key (1 Mark): A uniform circular ring / bangle, a hollow sphere, or a curved boomerang (COM lies at the empty geometric centre).

Q2 (Kerala HSE Say • 2 Marks): A firecracker fired from a cannon explodes mid-flight into fragments. Explain why the centre of mass of the fragments still follows the original parabolic trajectory.

Forces during explosion are strictly internal forces, which cancel in equal and opposite pairs (Newton's 3rd Law). Since M A_cm = F_ext and gravity is the only external force before and after burst, the COM trajectory stays completely unchanged.

Q3 (Kerala HSE March • 2 Marks): State the two conditions required for a rigid body to be in complete mechanical equilibrium. Does zero net force guarantee equilibrium?

1. Translational equilibrium: Total external force ∑F_ext = 0.
2. Rotational equilibrium: Total external torque ∑τ_ext = 0.
No! A Couple has ∑F = 0 yet causes pure rotation because ∑τ ≠ 0.

Q4 (Kerala HSE Model • 3 Marks): Define Torque and Angular Momentum. Establish the relation τ = dL/dt for a particle, and hence state the law of conservation of angular momentum.

• Torque: τ = r × F. • Angular Momentum: L = r × p.
Differentiating: dL/dt = (dr/dt × p) + (r × dp/dt) = (v × mv) + (r × F) = 0 + τ.
Thus τ_ext = dL/dt. If τ_ext = 0, L = Iω = constant.

Q5 (Kerala HSE March • 2 Marks): A girl rotates on a frictionless turntable with arms stretched out at angular speed ω₁. How does her angular speed change when she folds her arms close to her chest? Explain.

No external torque acts (τ_ext = 0), so angular momentum L = Iω is conserved.
Folding arms brings mass closer to the axis, decreasing Moment of Inertia I. Hence angular speed ω increases (I₁ω₁ = I₂ω₂).

Q6 (Kerala HSE Say • 2 Marks): Write the Moment of Inertia of: (a) a circular disc about an axis perpendicular to its plane through its centre, (b) a thin rod about an axis perpendicular to it at its midpoint, and (c) a solid sphere about its diameter.

• (a) Circular disc: I = ½ MR²
• (b) Thin rod at midpoint: I = (1/12) ML²
• (c) Solid sphere about diameter: I = (2/5) MR²

Q7 (Kerala HSE March • 3 Marks): A uniform metre stick balances horizontally on a knife edge at the 50 cm mark. When two coins of 5 g each are stacked at the 12 cm mark, the stick balances at 45 cm. Find the mass of the metre stick.

Stick's own weight M acts at its 50 cm midpoint (50 − 45 = 5 cm right of new pivot).
Two coins = 10 g sit at 45 − 12 = 33 cm left of pivot.
By Principle of Moments: 10 × 33 = M × 5 ⇒ M = 330 / 5 = 66 g!

Q8 (NCERT / HSE Model • 4 Marks): Obtain an expression for the linear velocity and acceleration of a rigid body of radius R and radius of gyration K rolling without slipping down an inclined plane of height h and angle θ.

By energy conservation: mgh = ½mv² + ½Iω².
Substitute I = mK² and pure rolling ω = v/R:
mgh = ½mv²(1 + K²/R²) ⇒ v = √[2gh / (1 + K²/R²)].
Using v² = 2a(h / sin θ) gives acceleration a = g sin θ / (1 + K²/R²).

Translational Motion	Rotational Analog
Displacement `x` \| Velocity `v` \| Accel `a`	Angle `θ` \| Angular vel `ω` \| Angular accel `α`
Mass `m` \| Force `F = ma`	Moment of Inertia `I` \| Torque `τ = Iα`
Momentum `p = mv` \| `K = ½mv²`	Angular Mom `L = Iω` \| `K_rot = ½Iω²`
Work `dW = F dx` \| Power `P = Fv`	Work `dW = τ dθ` \| Power `P = τω`