This MCQ module is based on: Moment of Inertia
TOPIC 27 OF 33
Moment of Inertia
🎓 Class 11
Physics
CBSE
Theory
Ch 6 – System of Particles and Rotational Motion
⏱ ~14 min
🌐 Language: [gtranslate]
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Moment of Inertia
6.11 Moment of Inertia
The moment of inertia measures a body's resistance to angular acceleration about a chosen axis — the rotational analogue of mass. For a discrete system:
\[I = \sum_i m_i r_i^2\]
For a continuous body:
\[I = \int r^2\,dm\]
where r is the perpendicular distance from the axis. Crucially, I depends on the AXIS — the same body has different I for different axes.
6.11.1 Rotational Kinetic Energy
For a rigid body rotating about a fixed axis with angular speed ω, every particle moves in a circle. Total KE:
\[K = \tfrac{1}{2}\sum m_i v_i^2 = \tfrac{1}{2}\sum m_i (r_i\omega)^2 = \tfrac{1}{2}\omega^2 \sum m_i r_i^2 = \tfrac{1}{2}I\omega^2\]
6.11.2 Newton's Second Law for Rotation
Combining \(\vec{\tau} = d\vec{L}/dt\) with \(L = I\omega\) (for fixed axis, constant I) gives:
\[\vec{\tau} = I\vec{\alpha}\]
| Body | Axis | I |
|---|---|---|
| Thin rod (length L) | Through centre, ⊥ to length | ML²/12 |
| Thin rod (length L) | Through one end, ⊥ to length | ML²/3 |
| Hoop / Thin ring (radius R) | Through centre, ⊥ to plane | MR² |
| Hoop / Thin ring | Through diameter | MR²/2 |
| Solid disc (radius R) | Through centre, ⊥ to plane | MR²/2 |
| Solid disc | Through diameter | MR²/4 |
| Solid sphere (radius R) | Through diameter | 2MR²/5 |
| Spherical shell | Through diameter | 2MR²/3 |
| Solid cylinder (radius R) | Symmetry axis | MR²/2 |
6.12 Theorems of Moment of Inertia
6.12.1 Theorem of Parallel Axes
If \(I_{\text{cm}}\) is the moment of inertia about an axis through the CoM, the moment of inertia about a PARALLEL axis a distance \(d\) away is:
\[I = I_{\text{cm}} + Md^2\]
This makes I about CoM the smallest of all parallel-axis values — a useful design fact.
6.12.2 Theorem of Perpendicular Axes (planar bodies only)
For a planar (2D) body lying in the xy-plane, the moment of inertia about the z-axis (⊥ to the plane through any chosen point) equals the sum of moments about two perpendicular axes (x and y) in the plane through that point:
\[I_z = I_x + I_y\]
6.13 Kinematics of Rotational Motion about a Fixed Axis
For constant angular acceleration α (analogous to constant linear acceleration), with initial angular velocity ω₀ and angular position θ₀:
\[\omega = \omega_0 + \alpha t\]
\[\theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha t^2\]
\[\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)\]
6.14 Dynamics of Rotational Motion about a Fixed Axis
Newton's law for rotation: τ = Iα. The work done by torque is:
\[W = \int \tau\,d\theta\]
And by analogy with translational motion: power P = τω.
6.15 Angular Momentum in Rolling Motion
Rolling without slipping is the most common motion in everyday life — wheels, balls, cylinders. The constraint is:
\[v_{\text{cm}} = R\omega \quad \text{(rolling condition)}\]
The total kinetic energy is the sum of translational + rotational:
\[K = \tfrac{1}{2}M v_{\text{cm}}^2 + \tfrac{1}{2}I_{\text{cm}}\omega^2\]
For a rolling ball or cylinder rolling DOWN an incline, energy conservation:
\[Mgh = \tfrac{1}{2}M v^2 + \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}M v^2\left(1 + \frac{I}{MR^2}\right)\]
Hence different shapes reach different speeds at the bottom — see the table.
| Shape | I/(MR²) | v at bottom (vs height h) | Time to roll down |
|---|---|---|---|
| Sliding (no rotation) | 0 | √(2gh) — fastest | Shortest |
| Solid sphere | 2/5 | √(10gh/7) | Quickest of rolling |
| Solid cylinder/disc | 1/2 | √(4gh/3) | Middle |
| Spherical shell | 2/3 | √(6gh/5) | Slower |
| Hoop/thin ring | 1 | √(gh) — slowest | Longest |
🎯 Interactive Simulation: Rolling-Race Calculator
Pick a shape; it rolls down an incline of height h. Calculate its speed at the bottom and compare with sliding without friction.
v² = 2gh / (1 + I/MR²)
v = 5.29 m/s
Sliding speed: 6.26 m/s ← always faster
Example 6.10: Parallel-Axis Application
Find the moment of inertia of a uniform rod of mass M = 2 kg, length L = 1 m, about an axis perpendicular to the rod and passing through one end.
About the centre: I_cm = ML²/12 = 2 × 1/12 = 1/6 kg·m².
Distance to end-axis: d = L/2 = 0.5 m.
Parallel-axis: I = I_cm + Md² = 1/6 + 2 × 0.25 = 1/6 + 1/2 = 2/3 kg·m².
Direct formula: ML²/3 = 2/3 kg·m². ✓ Match.
Distance to end-axis: d = L/2 = 0.5 m.
Parallel-axis: I = I_cm + Md² = 1/6 + 2 × 0.25 = 1/6 + 1/2 = 2/3 kg·m².
Direct formula: ML²/3 = 2/3 kg·m². ✓ Match.
Example 6.11: Perpendicular-Axis on a Disc
A uniform disc of mass M and radius R has I = MR²/2 about an axis through its centre perpendicular to its plane. Find I about a diameter using the perpendicular-axis theorem.
By symmetry, I_x = I_y = I_diameter. Perpendicular-axis: I_z = I_x + I_y = 2I_diameter ⇒ I_diameter = I_z/2 = MR²/4.
Example 6.12: Solid Cylinder Rolling Down Incline
A solid cylinder of mass 2 kg and radius 0.10 m rolls without slipping down a 3 m high incline. Find its speed at the bottom. (g = 10 m/s²)
For solid cylinder, I = MR²/2, so I/MR² = 0.5.
Energy conservation: Mgh = ½Mv²(1 + I/MR²) = ½Mv² × 1.5
v² = 2gh/1.5 = 4 × 30/3 = 40 ⇒ v ≈ 6.32 m/s.
Sliding (frictionless) gives v = √(2gh) = √60 ≈ 7.75 m/s — faster, as expected.
Energy conservation: Mgh = ½Mv²(1 + I/MR²) = ½Mv² × 1.5
v² = 2gh/1.5 = 4 × 30/3 = 40 ⇒ v ≈ 6.32 m/s.
Sliding (frictionless) gives v = √(2gh) = √60 ≈ 7.75 m/s — faster, as expected.
🏎 Activity 6.4 — Rolling-Race Experiment
L4 Analyse
Materials: Inclined ramp, several rolling objects (solid ball, hollow shell, solid disc, hollow ring), stopwatch.
Procedure:
- Place all objects at the same starting line at the top of the ramp.
- Release simultaneously and observe the order in which they reach the bottom.
- Time each individually if your reflexes can't catch the order.
Predict: Which arrives FIRST — solid sphere, hollow sphere, solid cylinder, or hoop?
Order (first to last): Solid sphere → solid cylinder → hollow sphere → hoop.
Conclusion: Smaller I/MR² ⇒ less of the kinetic energy goes into rotation, more into translation, so faster v_cm. The result is INDEPENDENT of mass and radius — only the shape's I/MR² ratio matters. This is why marbles beat ping-pong balls (which act as shells) on a slide.
🎯 Competency-Based Questions
A 5 kg solid sphere of radius 0.20 m rolls without slipping along a horizontal surface at 4 m/s, then up a frictionless incline. Find the maximum height attained. (g = 10 m/s²)
Q1. Calculate the translational KE.L3 Apply
Answer: (a). ½(5)(4)² = 40 J.
Q2. Calculate the rotational KE.L3 Apply
Answer: I = (2/5)MR² = 0.4 × 5 × 0.04 = 0.08 kg·m². ω = v/R = 20 rad/s. K_rot = ½(0.08)(400) = 16 J.
Q3. Find the height h reached if the incline is frictionless above (so sphere stops rolling — slides).L4 Analyse
Answer: If the incline is FRICTIONLESS, the sphere can't decelerate its rotation — only its translation slows. Translational KE 40 J converts to PE: 40 = mgh ⇒ h = 40/(5 × 10) = 0.80 m. Rotational KE remains; sphere keeps spinning at the highest point.
Q4. State whether TRUE or FALSE: "Hollow cylinder rolls down an incline faster than a solid cylinder of equal mass and radius." L5 Evaluate
Answer: FALSE. Hollow cylinder has I = MR² while solid has I = MR²/2. With higher I/MR², the hollow cylinder has more rotational KE per unit translational KE, so it rolls SLOWER. Solid wins.
Q5. HOT: A flywheel of moment of inertia 50 kg·m² is at rest. A constant torque of 100 N·m is applied. Calculate (a) angular acceleration, (b) angular speed after 10 s, (c) work done by torque in 10 s. L6 Create
(a) α = τ/I = 100/50 = 2 rad/s².
(b) ω = αt = 20 rad/s.
(c) θ = ½αt² = 100 rad. W = τθ = 100 × 100 = 10 000 J. Verify: K_f = ½Iω² = ½(50)(400) = 10 000 J ✓.
(b) ω = αt = 20 rad/s.
(c) θ = ½αt² = 100 rad. W = τθ = 100 × 100 = 10 000 J. Verify: K_f = ½Iω² = ½(50)(400) = 10 000 J ✓.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
Assertion (A): A solid cylinder rolls down an incline faster than a hollow cylinder of equal mass and radius.
Reason (R): Solid cylinder has smaller moment of inertia per unit MR².
Answer: (A). Both true; R explains A. Smaller I/MR² ⇒ less rotational KE share ⇒ more translational ⇒ faster v_cm.
Assertion (A): The perpendicular-axis theorem applies to all 3D rigid bodies.
Reason (R): The theorem states I_z = I_x + I_y.
Answer: (D). Assertion FALSE — the theorem holds ONLY for planar (2D) bodies. Reason TRUE — the formula is correct for planar bodies.
Assertion (A): The moment of inertia of a body is minimum about an axis through its centre of mass (among all parallel axes).
Reason (R): By the parallel-axis theorem, I = I_cm + Md² ≥ I_cm.
Answer: (A). Both true; R explains A. The shift Md² is non-negative, so I_cm is the minimum.
Frequently Asked Questions - Moment of Inertia
What is the main concept covered in Moment of Inertia?
In NCERT Class 11 Physics Chapter 6 (System of Particles and Rotational Motion), "Moment of Inertia" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Moment of Inertia useful in real-life applications?
Real-life applications of Moment of Inertia from NCERT Class 11 Physics Chapter 6 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Moment of Inertia?
Key formulas in Moment of Inertia (NCERT Class 11 Physics Chapter 6 System of Particles and Rotational Motion) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 6?
NCERT Class 11 Physics Chapter 6 (System of Particles and Rotational Motion) is structured so each part builds on the previous one. Moment of Inertia connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Moment of Inertia?
CBSE board questions from Moment of Inertia typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Moment of Inertia lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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