This MCQ module is based on: Centre of Mass
TOPIC 24 OF 33
Centre of Mass
🎓 Class 11
Physics
CBSE
Theory
Ch 6 – System of Particles and Rotational Motion
⏱ ~14 min
🌐 Language: [gtranslate]
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Centre of Mass
6.1 Introduction
So far we have treated bodies as point particles. But real objects — wheels, cricket bats, planets, top — have extent. They can translate AND rotate. To handle them we need new tools: the centre of mass (CoM), rigid body, moment of inertia, torque, angular momentum.
6.2 Centre of Mass
6.2.1 Two-particle system
For two particles of masses \(m_1, m_2\) at positions \(x_1, x_2\), the centre of mass is at:
\[X_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\]
It lies on the line joining the particles, dividing it in the inverse ratio of masses (closer to the heavier mass).
6.2.2 N-particle system
Generalising to \(n\) particles in 3D:
\[\vec{R}_{\text{cm}} = \frac{1}{M}\sum_{i=1}^n m_i \vec{r}_i, \quad M = \sum m_i\]
For a continuous body (rigid solid):
\[\vec{R}_{\text{cm}} = \frac{1}{M}\int \vec{r}\,dm\]
| Body | Where is CoM? |
|---|---|
| Uniform rod (length L) | Mid-point |
| Uniform circular ring | Geometric centre (NOT on the ring itself) |
| Uniform circular disc | Centre |
| Uniform solid sphere | Centre |
| Triangular plate (uniform) | Intersection of medians (centroid) |
| L-shaped bracket | Outside the body — depends on dimensions |
6.3 Motion of Centre of Mass
Differentiate the CoM definition with respect to time:
\[M \vec{V}_{\text{cm}} = \sum m_i \vec{v}_i = \vec{P}_{\text{total}}\]
So the total linear momentum equals total mass times velocity of CoM. Differentiating again:
\[M \vec{a}_{\text{cm}} = \sum m_i \vec{a}_i = \vec{F}_{\text{external}}\]
(Internal forces cancel by Newton's third law.) Hence:
Theorem of CoM motion: The CoM of a system moves as if all the mass were concentrated there and all external forces acted on it. This justifies treating extended bodies as point particles for translational motion.
Spectacular consequence: A grenade thrown into the air follows a parabola because of gravity. When it explodes mid-flight, the FRAGMENTS scatter, but the CoM still follows the original parabolic trajectory — the explosion is internal!
🎯 Interactive Simulation: Centre of Mass Locator
Vary the masses and positions of two particles. See how the CoM shifts toward the heavier mass.
Xcm = 8.00 m
Example 6.1: Three-particle CoM
Three particles of masses 1, 2, 3 kg are placed at the corners A(0,0), B(1,0), C(0,1) m of a right triangle. Find the centre of mass.
\[X_{cm} = \frac{1(0)+2(1)+3(0)}{1+2+3} = \frac{2}{6} = 0.333\text{ m}\]
\[Y_{cm} = \frac{1(0)+2(0)+3(1)}{6} = \frac{3}{6} = 0.500\text{ m}\]
CoM at (0.333 m, 0.500 m) — closer to the heaviest (3 kg) particle on the y-axis.
Example 6.2: CoM of an L-shaped Plate
A uniform L-shaped plate is made of two rectangles each of side 2L × 1L, joined to form an L. Find its CoM relative to the corner of the L.
Let mass per area = σ. Take the corner at origin. Vertical strip (2L × 1L) has area 2L², mass 2σL², CoM at (0.5L, L). Horizontal strip (1L × 2L) has area 2L², mass 2σL², CoM at (L, 0.5L). Total mass 4σL².
\[X_{cm} = \frac{2σL² × 0.5L + 2σL² × L}{4σL²} = \frac{1.5L}{2} = 0.75L\] \[Y_{cm} = \frac{2σL² × L + 2σL² × 0.5L}{4σL²} = 0.75L\] CoM at (0.75L, 0.75L) — outside the inner corner but inside the L overall.
\[X_{cm} = \frac{2σL² × 0.5L + 2σL² × L}{4σL²} = \frac{1.5L}{2} = 0.75L\] \[Y_{cm} = \frac{2σL² × L + 2σL² × 0.5L}{4σL²} = 0.75L\] CoM at (0.75L, 0.75L) — outside the inner corner but inside the L overall.
Example 6.3: Exploding Shell
A shell of mass 4 kg moving horizontally at 20 m/s explodes into two pieces of masses 1 kg and 3 kg. The 1 kg piece flies straight up at 30 m/s after explosion. Find the velocity of the 3 kg piece.
Explosion conserves momentum (internal force).
Initial: \(\vec{P} = (4)(20, 0) = (80, 0)\) kg·m/s.
After: \(\vec{P}_1 = (1)(0, 30) = (0, 30)\). Need \(\vec{P}_2\) such that total = (80, 0).
\(\vec{P}_2 = (80, -30)\) ⇒ \(\vec{v}_2 = (80, -30)/3 = (26.67, -10)\) m/s.
Magnitude: \(|\vec{v}_2| = \sqrt{26.67^2 + 100} \approx 28.49\) m/s.
Direction: \(\tan^{-1}(10/26.67) = 20.6°\) below horizontal.
Initial: \(\vec{P} = (4)(20, 0) = (80, 0)\) kg·m/s.
After: \(\vec{P}_1 = (1)(0, 30) = (0, 30)\). Need \(\vec{P}_2\) such that total = (80, 0).
\(\vec{P}_2 = (80, -30)\) ⇒ \(\vec{v}_2 = (80, -30)/3 = (26.67, -10)\) m/s.
Magnitude: \(|\vec{v}_2| = \sqrt{26.67^2 + 100} \approx 28.49\) m/s.
Direction: \(\tan^{-1}(10/26.67) = 20.6°\) below horizontal.
⚖ Activity 6.1 — Find the CoM of an Irregular Shape
L3 Apply
Materials: Cardboard cut in any irregular shape, pin, plumb bob (string with a weight), pencil.
Procedure:
- Pierce the cardboard near one edge with a pin and let it hang freely.
- Hang a plumb bob from the same pin. Mark the vertical line on the cardboard.
- Repeat from a second pin location far away — draw another vertical line.
- The intersection is the CoM.
Predict: Will the CoM always lie within the cardboard? Why or why not?
Conclusion: CoM is the unique point where any vertical line through the suspended body must pass. For convex shapes, CoM is inside; for concave shapes (e.g., a horseshoe or a ring), CoM may lie outside the actual material — yet the body still pivots about that imaginary point under gravity.
🎯 Competency-Based Questions
A 60 kg girl stands at one end of a 120 kg boat 6 m long, initially at rest on still water. She walks to the other end. Neglect water friction.
Q1. While she walks, the boat moves backward. What stays fixed?L2 Understand
Answer: (c). No external horizontal force ⇒ CoM stays fixed by Newton's first law applied to the system.
Q2. Find the displacement of the boat when the girl walks 6 m relative to the boat.L4 Analyse
Answer: Let boat move backward by Δ. Girl moves forward (6 − Δ) relative to ground. CoM fixed: 60(6 − Δ) − 120Δ = 0 ⇒ 360 = 60Δ + 120Δ = 180Δ ⇒ Δ = 2 m. The boat slides backward 2 m, the girl moves forward 4 m relative to the water.
Q3. The CoM of a uniform circular ring lies at: L1 Remember
Answer: (b). By symmetry, but interestingly NOT on the ring itself — there is no matter at the CoM!
Q4. State whether TRUE or FALSE: "External forces alone determine the motion of the CoM." L5 Evaluate
Answer: TRUE. Internal forces between particles come in action-reaction pairs and cancel pairwise. Hence M·a_CoM = F_external. This is one of the most powerful theorems in mechanics.
Q5. HOT: Two astronauts (60 kg, 80 kg) are at rest 10 m apart in deep space. They pull each other via a rope. Where do they meet? L6 Create
Answer: CoM is fixed (no external force). Place coordinates so 60 kg at 0, 80 kg at 10. CoM = (60×0 + 80×10)/140 = 800/140 ≈ 5.71 m. They MUST meet at the CoM. The 60 kg astronaut moves 5.71 m; the 80 kg astronaut moves 4.29 m. The lighter astronaut moves a longer distance.
🧠 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): An exploding projectile's CoM continues on its parabolic path.
Reason (R): Explosion forces are internal and do not change CoM motion.
Answer: (A). Both true; R explains A. Only gravity (external) acts on the system, so CoM still has constant horizontal velocity and falls with g.
Assertion (A): The CoM of a body is always located within its volume.
Reason (R): Mass is distributed continuously inside the body.
Answer: (D). Assertion FALSE — CoM of a ring is at the centre of the empty hole. Reason TRUE for solid bodies but does not imply CoM is inside.
Assertion (A): If two equal masses are at the ends of a uniform rod, the CoM is at the midpoint of the rod.
Reason (R): The CoM divides the joining line in inverse ratio of the masses.
Answer: (A). Both true; R explains A. Equal masses → 1:1 inverse ratio → midpoint.
Frequently Asked Questions - Centre of Mass
What is the main concept covered in Centre of Mass?
In NCERT Class 11 Physics Chapter 6 (System of Particles and Rotational Motion), "Centre of Mass" 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 Centre of Mass useful in real-life applications?
Real-life applications of Centre of Mass 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 Centre of Mass?
Key formulas in Centre of Mass (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. Centre of Mass 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 Centre of Mass?
CBSE board questions from Centre of Mass 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 Centre of Mass 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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