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Class XI · Chapter 06

System of Particles and Rotational Motion

Everything so far treated objects as single points. Real objects have size and can spin — this chapter builds the rotational twin of every law of motion you already know.

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1Centre of Mass

Every previous chapter secretly treated objects as single points — fine for many problems, but real objects have size, shape, and can spin. The centre of mass is the single point that represents an entire system's average position, weighted by mass.

Centre of Mass (Discrete System)
Rcm = Σmiri / Σmi
For two particles: xcm = (m₁x₁ + m₂x₂)/(m₁+m₂). The centre of mass needn't lie inside the object at all — a ring's centre of mass sits in the empty space at its middle.

2Motion of the Centre of Mass

This is what makes the centre of mass genuinely useful, not just a definition: however complicated a system's internal motion (spinning, wobbling, fragments flying apart), the centre of mass itself moves exactly as if all the mass were concentrated there, acted on only by the net external force.

Centre of Mass Motion
Fext = M acm
Internal forces (particles pushing on each other within the system) never affect the centre of mass's motion at all — they always cancel out in pairs, by Newton's third law. A spinning, exploding firework's centre of mass still traces a smooth parabola, exactly as a single projectile would.

3The Vector (Cross) Product

Rotational quantities like torque and angular momentum depend on both a distance and a force or momentum, combined in a direction-sensitive way — exactly what the vector (cross) product is built for.

Cross Product
A × B = AB sin θ n̂
Magnitude is AB sin θ; direction n̂ is perpendicular to both A and B, found by the right-hand rule. Unlike ordinary multiplication, order matters: A × B = −(B × A).

4Torque and Angular Momentum

Torque
τ = r × F, magnitude τ = rF sin θ
Torque is the rotational equivalent of force — it's what actually changes an object's rotational motion, not force alone. The same force applied farther from the pivot, or more perpendicular to the lever arm, produces more torque.
Angular Momentum
L = r × p
The rotational analogue of linear momentum. Just as F = dp/dt, torque and angular momentum are linked by τ = dL/dt — external torque is exactly what changes angular momentum.

5Equilibrium of a Rigid Body

Chapter 4's equilibrium condition (ΣF = 0) is only half the story for an extended object — it guarantees the object won't accelerate linearly, but says nothing about whether it might start spinning. A rigid body is in complete equilibrium only when both conditions hold simultaneously:

Full Equilibrium Conditions
ΣF = 0 (translational) AND Στ = 0 (rotational)
A see-saw can have zero net force yet still rotate if torques aren't balanced too — both conditions are genuinely independent requirements.

6Moment of Inertia

Moment of inertia is rotational motion's version of mass — a measure of how much an object resists a change in its rotational motion. Unlike mass, it isn't fixed for an object; it depends on how that mass is distributed relative to the specific axis of rotation.

Moment of Inertia
I = Σmiri² (discrete), I = ∫r² dm (continuous)
Mass farther from the axis contributes disproportionately more (r² growth) — which is exactly why a figure skater's arm position matters so much (see Section 10).
Common Shapes (axis through centre)
Ring: I=MR² Disc: I=½MR² Solid Sphere: I=(2/5)MR² Rod (about centre, ⊥ to length): I=(1/12)ML²

7Parallel and Perpendicular Axis Theorems

Parallel Axis Theorem
I = Icm + Md²
Relates the moment of inertia about any axis to the moment of inertia about a parallel axis through the centre of mass, plus Md² (d = distance between the two axes). Works for any rigid body, any axis.
Perpendicular Axis Theorem (Planar Bodies Only)
Iz = Ix + Iy
Applies only to flat, two-dimensional (planar) bodies — the moment of inertia about an axis perpendicular to the plane equals the sum of moments of inertia about two perpendicular in-plane axes, all intersecting at the same point.

8Kinematics of Rotational Motion

For rotation about a fixed axis with constant angular acceleration, every linear kinematic equation from Chapter 2 has a direct rotational counterpart — just swap x → θ, v → ω, a → α.

Rotational Kinematic Equations
ω = ω₀ + αt   θ = ω₀t + ½αt²   ω² = ω₀² + 2αθ
Linear-Angular Velocity Relation
v = ω × r, |v| = ωr
Every point on a rotating rigid body shares the same ω, but points farther from the axis move faster linearly — exactly why the outer edge of a spinning disc moves faster than a point near its centre.

9Dynamics of Rotational Motion

Rotational Analogue of Newton's Second Law
τ = I α
Direct parallel to F = ma: net torque produces angular acceleration, with moment of inertia playing the role mass plays for linear motion.
Rotational Kinetic Energy
KErot = ½ I ω²

10Conservation of Angular Momentum

Conservation of Angular Momentum
L = Iω = constant, when τext = 0
If moment of inertia changes while angular momentum stays fixed, angular velocity must change to compensate — this is exactly why a figure skater spins dramatically faster by pulling their arms in (reducing I), and slows back down extending them out again.

11Rolling Motion

A rolling object (like a wheel) combines translational motion of its centre of mass with rotation about that same centre — simultaneously, not one after the other.

Rolling Without Slipping
vcm = R ω
This condition is what "rolling without slipping" actually means mathematically — the contact point is momentarily at rest relative to the ground.
Total Kinetic Energy of a Rolling Body
KE = ½ M vcm² + ½ I ω²
Translational and rotational kinetic energy add — this is exactly why a rolling ball and a sliding (frictionless) block starting with the same speed carry different total kinetic energy, and why objects with different moment-of-inertia distributions (a solid ball vs. a hollow shell) roll down the same slope at different rates.

Formula Summary

Centre of Mass
R_cm = Σmᵢrᵢ/Σmᵢ
COM Motion
F_ext = Ma_cm
Torque
τ = rF sinθ
Angular Momentum
L = Iω
Torque–AM Relation
τ = dL/dt
Parallel Axis
I = I_cm + Md²
Perpendicular Axis
I_z = I_x + I_y
Rotational 2nd Law
τ = Iα
Rotational KE
KE = ½Iω²
Rolling Condition
v_cm = Rω

Solved Examples

Example 1 · Parallel Axis Theorem

A disc of mass 2 kg and radius 0.5 m rotates about an axis tangent to its edge, parallel to its central axis. Find its moment of inertia about this axis.

Solution: Icm = ½MR² = ½×2×(0.5)² = 0.25 kg·m².

I = Icm + Md² = 0.25 + 2×(0.5)² = 0.25 + 0.5 = 0.75 kg·m².

Example 2 · Conservation of Angular Momentum

An ice skater with moment of inertia 4 kg·m² spins at 2 rad/s. She pulls her arms in, reducing her moment of inertia to 1 kg·m². Find her new angular velocity.

Solution: By conservation of angular momentum, I₁ω₁ = I₂ω₂.

4 × 2 = 1 × ω₂ → ω₂ = 8 rad/s — four times faster, matching the fourfold decrease in moment of inertia.

Quick Check

1. The moment of inertia of a rigid body depends on:
Mass alone
✓ How mass is distributed relative to the axis (correct)
Angular velocity
Applied torque
2. Torque is the rotational analogue of which linear quantity?
Momentum
✓ Force (correct)
Velocity
Kinetic energy
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Work, Energy and Power

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Gravitation

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