This MCQ module is based on: NCERT Exercises and Solutions: System of Particles and Rotational Motion
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NCERT Exercises and Solutions: System of Particles and Rotational Motion
🎓 Class 11
Physics
CBSE
Theory
Ch 6 – System of Particles and Rotational Motion
⏱ ~8 min
🌐 Language: [gtranslate]
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NCERT Exercises and Solutions: System of Particles and Rotational Motion
Chapter 6 — Summary
Key Equations & Concepts
- Centre of Mass: \(\vec{R}_{cm} = \frac{1}{M}\sum m_i \vec{r}_i\); \(M\vec{a}_{cm} = \vec{F}_{ext}\)
- Vector product: \(\vec{A}\times\vec{B} = AB\sin\theta\,\hat{n}\); anti-commutative.
- Linear-Angular relations: \(\vec{v} = \vec{\omega}\times\vec{r}\); \(v = \omega r\) (perpendicular distance)
- Torque: \(\vec{\tau} = \vec{r}\times\vec{F}\); \(\tau = rF\sin\theta\)
- Angular momentum: \(\vec{L} = \vec{r}\times\vec{p}\) (particle); \(\vec{L} = I\vec{\omega}\) (rigid body)
- Newton's 2nd for rotation: \(\vec{\tau} = I\vec{\alpha} = d\vec{L}/dt\)
- Conservation of L: If \(\vec{\tau}_{ext} = 0\), \(\vec{L}\) is constant
- Equilibrium: \(\sum \vec{F} = 0\) AND \(\sum \vec{\tau} = 0\)
- Parallel-axis: \(I = I_{cm} + Md^2\)
- Perpendicular-axis (planar): \(I_z = I_x + I_y\)
- Rotational KE: \(K = \tfrac{1}{2}I\omega^2\)
- Rolling without slipping: \(v_{cm} = R\omega\); \(K = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2\)
| Quantity | SI Unit | Dimensions |
|---|---|---|
| Angular position (θ) | rad | dimensionless |
| Angular velocity (ω) | rad/s | [T⁻¹] |
| Angular acceleration (α) | rad/s² | [T⁻²] |
| Moment of inertia (I) | kg·m² | [ML²] |
| Torque (τ) | N·m | [ML²T⁻²] |
| Angular momentum (L) | kg·m²/s | [ML²T⁻¹] |
🎯 Quick Self-Check Calculator
Compute the moment of inertia of a uniform rod given mass, length, and axis choice.
I = 0.167 kg·m²
NCERT Exercises — Worked Solutions
Exercise 6.1: CoM of HCl Molecule
Find the position of the centre of mass of an HCl molecule with H–Cl distance ~1.27 Å. (m_H ≈ 1, m_Cl ≈ 35.5 atomic units; place H at origin, Cl on +x axis.)
\[X_{cm} = \frac{1(0) + 35.5(1.27)}{1 + 35.5} = \frac{45.085}{36.5} \approx 1.235\text{ Å}\]
The CoM is very close to the chlorine atom — only 1.27 − 1.235 = 0.035 Å from Cl, and ~1.235 Å from H. Heavier atoms dominate the CoM.
Exercise 6.2: Three-particle CoM
A child sits stationary at one end of a long trolley moving uniformly with speed v on a smooth track. Then she gets up and runs to the other end. How does the speed of the CoM of the (trolley + child) system change?
The track is smooth, so the only external force is gravity (vertical). The horizontal momentum of the (trolley + child) system is conserved. Hence the horizontal velocity of the CoM remains v throughout — even when the child is running and the trolley is reacting. The internal motion redistributes velocities between trolley and child but cannot change the CoM motion.
Exercise 6.3: Cross Product Verification
Show that \(\vec{a}\times\vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\) by direct calculation, taking \(\vec{a} = 2\hat{i} + \hat{j} - \hat{k}\), \(\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}\).
\[\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&-1\\1&2&3\end{vmatrix} = \hat{i}(3+2) - \hat{j}(6+1) + \hat{k}(4-1) = 5\hat{i}-7\hat{j}+3\hat{k}\]
Verify \(\vec{a}\cdot(\vec{a}\times\vec{b}) = 2(5)+1(-7)+(-1)(3) = 10-7-3=0\). ✓
Verify \(\vec{b}\cdot(\vec{a}\times\vec{b}) = 1(5)+2(-7)+3(3) = 5-14+9=0\). ✓
Both dot products are zero, confirming perpendicularity.
Verify \(\vec{b}\cdot(\vec{a}\times\vec{b}) = 1(5)+2(-7)+3(3) = 5-14+9=0\). ✓
Both dot products are zero, confirming perpendicularity.
Exercise 6.4: Triangle CoM
Three particles of masses 100 g, 150 g, and 200 g are placed at the vertices A(0,0), B(0.5, 0), C(0.25, 0.25√3) m of an equilateral triangle of side 0.5 m. Find the CoM.
Total mass = 0.45 kg.
\[X_{cm} = \frac{0.1(0)+0.15(0.5)+0.2(0.25)}{0.45} = \frac{0.075+0.05}{0.45} = \frac{0.125}{0.45} \approx 0.278\text{ m}\] \[Y_{cm} = \frac{0.1(0)+0.15(0)+0.2(0.25\sqrt 3)}{0.45} = \frac{0.0866}{0.45} \approx 0.193\text{ m}\] CoM at (0.278 m, 0.193 m).
\[X_{cm} = \frac{0.1(0)+0.15(0.5)+0.2(0.25)}{0.45} = \frac{0.075+0.05}{0.45} = \frac{0.125}{0.45} \approx 0.278\text{ m}\] \[Y_{cm} = \frac{0.1(0)+0.15(0)+0.2(0.25\sqrt 3)}{0.45} = \frac{0.0866}{0.45} \approx 0.193\text{ m}\] CoM at (0.278 m, 0.193 m).
Exercise 6.5: Angular Velocity & Linear Speed
A wheel of radius 0.5 m rotates at 1200 rpm. Find (a) angular speed in rad/s, (b) speed of a point on rim, (c) angular acceleration if it stops uniformly in 10 s.
(a) ω = 1200 × 2π/60 = 40π ≈ 125.66 rad/s.
(b) v = ωR = 125.66 × 0.5 ≈ 62.83 m/s.
(c) α = (0 − 125.66)/10 ≈ −12.57 rad/s² (deceleration).
(b) v = ωR = 125.66 × 0.5 ≈ 62.83 m/s.
(c) α = (0 − 125.66)/10 ≈ −12.57 rad/s² (deceleration).
Exercise 6.6: Solid Sphere Down Incline
A solid sphere of mass 1 kg radius 0.10 m rolls without slipping down an incline of 1.5 m height. Find KE_translational and KE_rotational at the bottom. (g = 10 m/s²)
Total energy released = mgh = 1 × 10 × 1.5 = 15 J.
For solid sphere: K_trans/K_total = (½Mv²)/(½Mv² + ½Iω²) = 1/(1 + I/MR²) = 1/(1 + 2/5) = 5/7.
K_trans = (5/7)(15) ≈ 10.71 J.
K_rot = (2/7)(15) ≈ 4.29 J.
For solid sphere: K_trans/K_total = (½Mv²)/(½Mv² + ½Iω²) = 1/(1 + I/MR²) = 1/(1 + 2/5) = 5/7.
K_trans = (5/7)(15) ≈ 10.71 J.
K_rot = (2/7)(15) ≈ 4.29 J.
Exercise 6.7: Conservation of L
A rod of length L is rotating freely about its centre with angular velocity ω₀. Two small beads slide outward to its ends. The moment of inertia changes from I₀ to 2I₀. Find the new ω.
No external torque. L conserved: I₀ω₀ = 2I₀ω ⇒ ω = ω₀/2. Rotation slows. KE drops to ½ original — energy is required to push beads outward against centripetal force; here centripetal demand is what slows them. Actually, beads sliding outward freely don't lose KE, so the algebra: K_f/K_i = (½)(2I₀)(ω₀/2)²/(½I₀ω₀²) = (2)(1/4) = 1/2 — half. The "missing" energy went into the kinetic energy of beads moving outward radially while their tangential component matched the new ω.
Exercise 6.8: Equilibrium of a See-Saw
Two children of masses 30 kg and 45 kg sit on a 4 m seesaw whose fulcrum is at the centre. The 30 kg child sits at one end. Where must the 45 kg child sit for balance?
Distances from fulcrum: 30 kg at 2 m. Let 45 kg at distance d on opposite side.
Torque balance: 30 × 2 = 45 × d ⇒ d = 60/45 ≈ 1.33 m from fulcrum.
The heavier child must sit closer to the fulcrum to balance the lighter child at the end.
Torque balance: 30 × 2 = 45 × d ⇒ d = 60/45 ≈ 1.33 m from fulcrum.
The heavier child must sit closer to the fulcrum to balance the lighter child at the end.
Exercise 6.9: Hollow vs Solid Sphere
A solid sphere and a thin spherical shell (each of same mass and radius) are released together at the top of an incline. Show that the solid sphere reaches the bottom first and find the ratio of their times.
For rolling, a = g sin θ /(1 + I/MR²).
Solid: I/MR² = 2/5 ⇒ a_s = (5/7)g sin θ.
Shell: I/MR² = 2/3 ⇒ a_h = (3/5)g sin θ.
Length L same; t = √(2L/a). Ratio: t_h/t_s = √(a_s/a_h) = √[(5/7)/(3/5)] = √(25/21) ≈ 1.091.
Hollow takes about 9% more time.
Solid: I/MR² = 2/5 ⇒ a_s = (5/7)g sin θ.
Shell: I/MR² = 2/3 ⇒ a_h = (3/5)g sin θ.
Length L same; t = √(2L/a). Ratio: t_h/t_s = √(a_s/a_h) = √[(5/7)/(3/5)] = √(25/21) ≈ 1.091.
Hollow takes about 9% more time.
🎯 Concept Check — Linear vs Rotational
L4 Analyse
Match the linear-rotational analogues for these quantities and laws.
- F = ma
- p = mv
- KE = ½mv²
- P = Fv
- W = Fd
Predict: What is the rotational analogue of each?
- τ = Iα (torque = moment of inertia × angular acceleration)
- L = Iω (angular momentum = MoI × angular velocity)
- K_rot = ½Iω² (rotational kinetic energy)
- P = τω (power as torque times angular velocity)
- W = τθ (work as torque times angular displacement)
This perfect parallel between translation and rotation is one of the most beautiful symmetries in classical mechanics.
🎯 Competency-Based Questions
A uniform thin rod of length 2 m and mass 4 kg is pivoted freely at one end and released from horizontal position. (g = 10 m/s²)
Q1. Find the moment of inertia about the pivot.L3 Apply
Answer: (b). I = ML²/3 = 4 × 4/3 = 16/3 kg·m².
Q2. Find the angular speed when rod swings down to vertical (using energy conservation).L4 Analyse
Answer: CoM falls by L/2 = 1 m. PE released = MgL/2 = 4 × 10 × 1 = 40 J. Equate to ½Iω² = ½(16/3)ω² = (8/3)ω². So ω² = 40 × 3/8 = 15 ⇒ ω ≈ 3.87 rad/s.
Q3. Find the linear speed of the free end at this instant.L3 Apply
Answer: v = ωL = 3.87 × 2 ≈ 7.75 m/s.
Q4. State whether TRUE or FALSE: "The angular acceleration of the rod is constant during the swing." L5 Evaluate
Answer: FALSE. The torque (τ = MgL/2 · cos θ) varies with the angle θ. So α = τ/I varies — maximum at horizontal, zero at vertical (since the gravitational torque vanishes when the rod is vertical).
Q5. HOT: If the rod were instead a uniform thin disc of same mass M = 4 kg, radius R = 1 m, pivoted at its rim and released from horizontal, find the angular speed at the lowest point. L6 Create
Answer: I_pivot = (1/2)MR² + MR² = (3/2)(4)(1) = 6 kg·m² (parallel-axis). CoM falls by R = 1 m. PE released = MgR = 40 J. ½Iω² = 40 ⇒ 3ω² = 40 ⇒ ω ≈ 3.65 rad/s.
🧠 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): The radius of gyration k of a body about an axis is given by I = Mk².
Reason (R): Radius of gyration represents the perpendicular distance at which the entire mass of the body could be concentrated for the same I.
Answer: (A). Both true; R explains A. Definition: k = √(I/M).
Assertion (A): A spinning top eventually falls because gravity exerts a torque.
Reason (R): Gravity always acts vertically downward and produces no torque about the spin axis.
Answer: (C). Assertion TRUE — eventually the top falls because friction at the contact point and air drag remove its angular momentum. Reason FALSE — gravity DOES produce a torque about the contact point (causing precession). The fall is mainly due to friction, not direct toppling by gravity.
Assertion (A): A wheel rolling without slipping has the contact point momentarily at rest.
Reason (R): Velocity of contact point = v_cm − Rω = 0 when v_cm = Rω.
Answer: (A). Both true; R explains A. The instantaneous axis of rotation is at the contact point.
Frequently Asked Questions - NCERT Exercises and Solutions: System of Particles and Rotational Motion
What are the key NCERT exercise types in Chapter 6 System of Particles and Rotational Motion?
NCERT Class 11 Physics Chapter 6 System of Particles and Rotational Motion exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question in the chapter, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in System of Particles and Rotational Motion?
For numerical problems in NCERT Class 11 Physics Chapter 6 System of Particles and Rotational Motion: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 6?
From NCERT Class 11 Physics Chapter 6 (System of Particles and Rotational Motion), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 11 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula (M^a L^b T^c). For Chapter 6 System of Particles and Rotational Motion problems, write the dimensions of each quantity, substitute, and simplify. If dimensions match the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 6 exercises?
Common mistakes in NCERT Class 11 Physics Chapter 6 System of Particles and Rotational Motion exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for vectors, (3) forgetting to consider all forces in free-body diagrams, (4) algebra errors during derivations, and (5) misreading the problem statement. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 11 Physics Chapter 6 System of Particles and Rotational Motion solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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