System of Particles and Rotational Motion — Class 11 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Physics chapter "System of Particles and Rotational Motion" — 6 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "System of Particles and Rotational Motion" — 6 important questions with detailed answe…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

Define centre of mass and moment of inertia. Explain their significance in ro…
Calculate the moment of inertia of a solid cylinder of mass 10 kg and radius …
Two particles of masses 3 kg and 5 kg are separated by a distance of 4 m. Fin…
A disc of mass 2 kg and radius 0.4 m rotates about its central axis with angu…
A thin rod of length 2 m and mass 4 kg is pivoted at one end. Calculate its m…
A rotating wheel has moment of inertia 5 kg⋅m² and angular velocity 40 rad/s.…

Solutions Summary:

Question	Status
Define centre of mass and moment of inertia. Explain thei…	✓ Solved
Calculate the moment of inertia of a solid cylinder of ma…	✓ Solved
Two particles of masses 3 kg and 5 kg are separated by a …	✓ Solved
A disc of mass 2 kg and radius 0.4 m rotates about its ce…	✓ Solved
A thin rod of length 2 m and mass 4 kg is pivoted at one …	✓ Solved
A rotating wheel has moment of inertia 5 kg⋅m² and angula…	✓ Solved

Showing 6 of 6 questions

Q1: Define centre of mass and moment of inertia. Explain their significance in rotational motion.

Centre of Mass Definition: The centre of mass of a system is the point where the entire mass of the system can be considered to be concentrated for the purpose of analyzing translational motion. It is the average position of all the mass in the system. For a system of n particles: r_cm = (m₁r₁ + m₂r₂ + ... + m_nr_n)/(m₁ + m₂ + ... + m_n) = Σ(m_i r_i)/Σm_i In component form: x_cm = Σ(m_i x_i)/M y_cm = Σ(m_i y_i)/M z_cm = Σ(m_i z_i)/M Where M = total mass Significance: - The external force act...

Q2: Calculate the moment of inertia of a solid cylinder of mass 10 kg and radius 0.5 m about its central axis. Also calculate its rotational kinetic energy if it rotates at 100 rpm.

Given: Mass of cylinder M = 10 kg Radius R = 0.5 m Rotational speed N = 100 rpm Part 1: Moment of inertia of solid cylinder For a solid cylinder rotating about its central axis: I = (1/2)MR² Substituting values: I = (1/2) × 10 × (0.5)² I = 5 × 0.25 I = 1.25 kg⋅m² Part 2: Convert rpm to rad/s Angular velocity in rpm: N = 100 rpm To convert to rad/s: ω = 2πN/60 = 2π × 100/60 = 200π/60 = 10π/3 rad/s Numerical value: ω = 10π/3 ≈ 10.47 rad/s Part 3: Calculate rotational kinetic energy Rotati...

Q3: Two particles of masses 3 kg and 5 kg are separated by a distance of 4 m. Find the position of their centre of mass from the 3 kg mass.

Given: Mass 1: m₁ = 3 kg Mass 2: m₂ = 5 kg Separation distance: d = 4 m Find: Position of centre of mass from the 3 kg mass Step 1: Set up coordinate system Place the 3 kg mass at the origin (x = 0) Place the 5 kg mass at x = 4 m Step 2: Use centre of mass formula For one dimension: x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂) Where: m₁ = 3 kg at x₁ = 0 m₂ = 5 kg at x₂ = 4 m Step 3: Calculate centre of mass position x_cm = (3 × 0 + 5 × 4)/(3 + 5) x_cm = (0 + 20)/8 x_cm = 20/8 = 2.5 m Step 4: Verify r...

Q4: A disc of mass 2 kg and radius 0.4 m rotates about its central axis with angular velocity 50 rad/s. Calculate (a) angular momentum (b) torque needed to stop it in 5 seconds (c) work done in stopping it.

Given: Mass M = 2 kg Radius R = 0.4 m Initial angular velocity ω_i = 50 rad/s Final angular velocity ω_f = 0 rad/s (stops) Time t = 5 s Part (a): Angular momentum Step 1: Calculate moment of inertia of disc For a solid disc rotating about central axis: I = (1/2)MR² = (1/2) × 2 × (0.4)² = 1 × 0.16 = 0.16 kg⋅m² Step 2: Calculate angular momentum Angular momentum L = Iω L = 0.16 × 50 = 8 kg⋅m²/s Part (b): Torque needed to stop the disc Step 1: Calculate angular acceleration α = (ω_f - ω_i)/...

Q5: A thin rod of length 2 m and mass 4 kg is pivoted at one end. Calculate its moment of inertia about the pivot point and about its centre.

Given: Length of rod L = 2 m Mass M = 4 kg Part 1: Moment of inertia about the pivot point (one end) For a thin rod pivoted at one end, rotating about that end: I_end = (1/3)ML² Substituting values: I_end = (1/3) × 4 × (2)² I_end = (1/3) × 4 × 4 I_end = 16/3 kg⋅m² I_end ≈ 5.33 kg⋅m² Part 2: Moment of inertia about the centre For a thin rod rotating about its centre (perpendicular axis): I_centre = (1/12)ML² Substituting values: I_centre = (1/12) × 4 × (2)² I_centre = (1/12) × 4 × 4 I_centr...

Q6: A rotating wheel has moment of inertia 5 kg⋅m² and angular velocity 40 rad/s. When the brakes are applied, a constant friction torque of 10 N⋅m acts on it. Find (a) angular deceleration (b) time to stop (c) number of revolutions before stopping.

Given: Moment of inertia I = 5 kg⋅m² Initial angular velocity ω_i = 40 rad/s Final angular velocity ω_f = 0 rad/s (stops) Friction torque τ = -10 N⋅m (negative because it opposes motion) Part (a): Angular deceleration Using rotational equation of motion: τ = Iα -10 = 5 × α α = -10/5 = -2 rad/s² Angular deceleration = 2 rad/s² (magnitude) The negative sign indicates deceleration (opposite to direction of rotation). Part (b): Time to stop Using ω_f = ω_i + αt 0 = 40 + (-2)t 2t = 40 t = 20 s ...

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