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

✓ 100% Free ✓ No Login Needed ✓ NCERT / CBSE Aligned ✓ Download as PDF

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…

🤖 Stuck on any question? Ask Syllab's free AI Tutor for a step-by-step explanation — instant, unlimited, no login.

NCERT Solutions for Class 11 Physics — System of Particles and Rotational Motion. Step-by-step answers to all 6 textbook questions from this chapter, written for CBSE board preparation and free to use.

System of Particles and Rotational Motion — All 6 Questions Solved

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 acting on the system equals total mass times acceleration of centre of mass: F_ext = M × a_cm
- The motion of the centre of mass is independent of internal forces
- For a symmetric object, the centre of mass coincides with the geometric centre

Moment of Inertia Definition:
The moment of inertia of a body about an axis is the sum of products of masses of all particles and the squares of their distances from the axis.

For discrete masses:
I = Σ m_i r_i² = m₁r₁² + m₂r₂² + ... + m_nr_n²

For continuous body:
I = ∫ r² dm

Where r is the perpendicular distance from the axis of rotation.

SI Unit: kg⋅m²

Significance in Rotational Motion:

1. Rotational analogue of mass:
- In linear motion: F = ma (force, mass, acceleration)
- In rotational motion: τ = Iα (torque, moment of inertia, angular acceleration)
- Moment of inertia is the rotational analogue of mass

2. Resistance to angular acceleration:
- Greater moment of inertia → greater resistance to change in angular velocity
- A body with large I requires larger torque to produce same angular acceleration

3. Rotational kinetic energy:
KE_rotational = (1/2)Iω²

Comparison:
Linear Motion → Rotational Motion
Mass (m) → Moment of inertia (I)
Force (F) → Torque (τ)
Acceleration (a) → Angular acceleration (α)
Linear KE = (1/2)mv² → Rotational KE = (1/2)Iω²
Linear momentum = mv → Angular momentum = Iω

4. Examples of moments of inertia:
- Thin ring about central axis: I = MR²
- Solid cylinder/disc about central axis: I = (1/2)MR²
- Solid sphere about diameter: I = (2/5)MR²
- Thin rod about perpendicular axis through centre: I = (1/12)ML²

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

Rotational kinetic energy:
KE_rot = (1/2)Iω²

Using exact value with π:
KE_rot = (1/2) × 1.25 × (10π/3)²
KE_rot = 0.625 × (100π²/9)
KE_rot = 62.5π²/9
KE_rot ≈ 68.58 J

Using decimal approximation:
ω ≈ 10.47 rad/s
KE_rot = (1/2) × 1.25 × (10.47)²
KE_rot = 0.625 × 109.62
KE_rot ≈ 68.51 J

Alternative calculation:
KE_rot = (1/2) × 1.25 × (10π/3)²
= 0.625 × 100π²/9
= 62.5π²/9 J
≈ 68.6 J

Part 4: Comparison with linear motion

If the same mass moved linearly at velocity v = ωR:
v = 10.47 × 0.5 = 5.235 m/s

Linear kinetic energy:
KE_linear = (1/2)Mv² = (1/2) × 10 × (5.235)² ≈ 136.8 J

Note: KE_linear ≠ KE_rot because not all parts move with the same velocity in rotation.

Answers:
Moment of inertia = 1.25 kg⋅m²
Angular velocity = 10π/3 rad/s ≈ 10.47 rad/s
Rotational kinetic energy = 62.5π²/9 J ≈ 68.6 J

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 result

Distance from 3 kg mass = 2.5 m
Distance from 5 kg mass = 4 - 2.5 = 1.5 m

Check using principle of moments:
m₁ × d₁ = m₂ × d₂
3 × 2.5 = 5 × 1.5
7.5 = 7.5 ✓

This confirms the centre of mass divides the line joining the two masses in the inverse ratio of their masses.

Step 5: Verify ratio

Ratio of distances: d₁/d₂ = 2.5/1.5 = 5/3 = m₂/m₁

This is correct as per the principle: the centre of mass divides the line segment in the inverse ratio of the masses.

Answer: The position of the centre of mass is 2.5 m from the 3 kg mass (or 1.5 m from the 5 kg mass)

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)/t = (0 - 50)/5 = -10 rad/s²

The negative sign indicates deceleration.
Magnitude of angular acceleration: |α| = 10 rad/s²

Step 2: Calculate torque

Using τ = Iα
τ = 0.16 × (-10) = -1.6 N⋅m

Magnitude of torque required: |τ| = 1.6 N⋅m

Alternative using change in momentum:
ΔL = L_f - L_i = 0 - 8 = -8 kg⋅m²/s
τ = ΔL/t = -8/5 = -1.6 N⋅m
|τ| = 1.6 N⋅m ✓

Part (c): Work done in stopping

The work done equals the change in rotational kinetic energy:

Initial rotational KE:
KE_i = (1/2)Iω_i² = (1/2) × 0.16 × (50)² = 0.08 × 2500 = 200 J

Final rotational KE:
KE_f = (1/2)Iω_f² = (1/2) × 0.16 × 0² = 0 J

Work done:
W = KE_f - KE_i = 0 - 200 = -200 J

Magnitude of work done: |W| = 200 J

The negative sign indicates work is done on the system (energy is removed).

Alternative using work-energy relation:
W = τ × θ

Where θ is the angular displacement:
ω_f² = ω_i² + 2αθ
0 = 50² + 2(-10)θ
20θ = 2500
θ = 125 radians

Work: W = τ × θ = 1.6 × 125 = 200 J
(Or W = -1.6 × 125 = -200 J considering direction)

Answers:
(a) Angular momentum = 8 kg⋅m²/s
(b) Torque required = 1.6 N⋅m (opposing the rotation)
(c) Work done in stopping = 200 J (energy dissipated)

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_centre = 16/12 = 4/3 kg⋅m²
I_centre ≈ 1.33 kg⋅m²

Part 3: Relationship between the two

Using the parallel axis theorem:
I_end = I_centre + Md²

Where d is the distance between the pivot axis and the centre.
For a rod pivoted at one end, d = L/2 = 1 m

I_end = (1/12)ML² + M(L/2)²
I_end = (1/12)ML² + M × L²/4
I_end = (1/12)ML² + (3/12)ML²
I_end = (4/12)ML² = (1/3)ML²

Verification:
I_end = 4/3 + 4 × 1² = 4/3 + 4 = 4/3 + 12/3 = 16/3 ✓

Part 4: Ratio of moments of inertia

I_end/I_centre = (1/3)ML²/(1/12)ML² = (1/3)/(1/12) = 4

Therefore: I_end = 4 × I_centre

Physical interpretation:
- The moment of inertia is 4 times larger about the end than about the centre
- This is because the mass is distributed farther from the pivot at the end
- Greater distance from axis → greater moment of inertia

Answers:
Moment of inertia about the pivot (one end) = 16/3 kg⋅m² ≈ 5.33 kg⋅m²
Moment of inertia about the centre = 4/3 kg⋅m² ≈ 1.33 kg⋅m²
Ratio I_end : I_centre = 4 : 1

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

Verification:
Using ω_f - ω_i = αt
0 - 40 = -2 × 20
-40 = -40 ✓

Part (c): Number of revolutions before stopping

Method 1: Using angular displacement

Using ω_f² = ω_i² + 2αθ
0² = 40² + 2(-2)θ
0 = 1600 - 4θ
4θ = 1600
θ = 400 radians

Number of revolutions:
N = θ/(2π) = 400/(2π) = 200/π revolutions
N ≈ 63.66 revolutions

Method 2: Using average angular velocity

Average angular velocity:
ω_avg = (ω_i + ω_f)/2 = (40 + 0)/2 = 20 rad/s

Angular displacement:
θ = ω_avg × t = 20 × 20 = 400 radians ✓

Number of revolutions = 400/(2π) ≈ 63.66 revolutions

Method 3: Using kinetic energy and work

Initial rotational KE:
KE_i = (1/2)Iω_i² = (1/2) × 5 × (40)² = 2.5 × 1600 = 4000 J

Work done by friction:
W = τ × θ = -10 × θ

By work-energy theorem:
W = KE_f - KE_i
-10θ = 0 - 4000
-10θ = -4000
θ = 400 radians ✓

Part (d): Summary table

Linear motion analogy:

Linear | Rotational
---|---
v = u + at | ω = ω_i + αt
v² = u² + 2as | ω² = ω_i² + 2αθ
s = ut + (1/2)at² | θ = ω_i t + (1/2)αt²
F = ma | τ = Iα

Answers:
(a) Angular deceleration = 2 rad/s²
(b) Time to stop = 20 s
(c) Number of revolutions = 200/π ≈ 63.66 revolutions (or approximately 64 revolutions)

How to Use These Solutions

Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.

Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.

All NCERT Solutions, Class 6–12 →