NEET Physics β Rotational Motion
Centre of Mass β Complete Notes, Revision, Important Questions & Downloads
Centre of Mass covers two subtopics: Definition and Position Vector (centre of mass as the mass-weighted average position of all particles) and Important Points about Centre of Mass (properties including coordinate-system independence, coincidence with centre of gravity in uniform fields, and location for regular bodies at geometric centre). NEET tests this topic through calculations of the centre of mass position for two- and three-body systems using r_cm = (mβrβ + mβrβ + ...)/(mβ + mβ + ...) and conceptual questions about where the CM lies for composite or L-shaped bodies. The most frequently tested NEET trap is identifying that the centre of mass of a system does not need to lie within the material of the body β it can be at a point in empty space (as in a ring or hollow sphere).
9 SubtopicsWeighted Average PositionCoordinate Geometry
Expected QuestionsQ
0β1
Centre of Mass appears as 1 question in Chapter 7 sessions; often combined with momentum conservation or as a coordinate geometry question asking for CM of a composite body.
Time Requiredβ±
1β2 hrs
The topic has only 2 subtopics. Spend the majority of time practising CM coordinate calculations for 2-body and 3-body planar systems.
Difficultyβ‘
Easy
The formula r_cm = (mβrβ+mβrβ+...)/(Ξ£mα΅’) is direct. The main difficulty is correctly using vector components (x_cm and y_cm separately) for 2D systems.
NRI USA Curriculum GapUS
Very Low
US AP Physics 1 covers centre of mass for basic two-body systems. The gap is that NEET includes non-trivial composite bodies (L-shaped plates, hollow rings) where CM lies outside the material.
9Subtopics
5+Practice Questions
4Free Downloads
1β2 hrsPrep Time
NEET Weightage β Centre of Mass
Rotational Motion (Chapter 7)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 1 | 4 | |
| 2022 | 0 | 0 | |
| 2021 | 1 | 4 | |
| 2020 | 0 | 0 | |
| 2019 | 1 | 4 | |
| 6-Year Total (2019β2024) | 2β4 | Β | 8β16 |
r_cm = (mβrβ+mβrβ+...+mβrβ)/(mβ+mβ+...+mβ): the centre of mass of n particles is a weighted average of the position vectors of n particles making up the system.
The position of centre of mass is independent of the coordinate system chosen β NEET tests this in assertion-reason: 'CM doesn't change when origin is shifted' (True β relative distances are preserved).
For regular bodies (uniform density sphere, cube, rod): CM lies at the geometric centre. For L-shaped or non-uniform composite bodies: CM must be computed by treating each part separately.
The position of centre of mass is independent of the coordinate system chosen β NEET tests this in assertion-reason: 'CM doesn't change when origin is shifted' (True β relative distances are preserved).
For regular bodies (uniform density sphere, cube, rod): CM lies at the geometric centre. For L-shaped or non-uniform composite bodies: CM must be computed by treating each part separately.
~0.7
Avg Questions / Year
8β16
Total Marks (6 yrs)
Occasional
Pattern
Easy
Difficulty
Exam Strategy for Centre of Mass
1
Apply the weighted-average CM formula component by component for 2D problems For a 2D system: x_cm = (mβxβ+mβxβ+...)/(Ξ£mα΅’) and y_cm = (mβyβ+mβyβ+...)/(Ξ£mα΅’) independently. Never try to use the formula for the resultant position vector without splitting into components first β this causes sign errors in NEET problems.
2
For composite bodies, subtract the removed part from the whole When a portion is removed from a uniform body (e.g., a circular disc with a hole punched out), treat the final body as the original body minus the removed part. Use: x_cm = (M_wholeΓx_cm_whole β M_removedΓx_cm_removed)/(M_whole β M_removed). This is the standard NEET approach for annular or L-shaped bodies.
3
Remember that CM need not lie inside the body For a uniform ring or hollow sphere: CM lies at the geometric centre β which is inside the empty hollow space, not in the material. Similarly, for an L-shaped uniform rod, the CM lies inside the empty quadrant. NEET tests this with assertion-reason: 'CM must be inside the body' (False).
Download Study Notes β Centre of Mass
PDF Β· Cheat Sheet Β· MCQ Set Β· PYQCentre of Mass β Full Notes
Complete coverage of both subtopics: CM definition as weighted average position, position vector formula, important properties (coordinate independence, CM of regular bodies, CM outside material), and worked examples including composite and L-shaped bodies.
Download PDFCentre of Mass β Formula Sheet
One-page reference: r_cm = Ξ£mα΅’rα΅’/Ξ£mα΅’ in components (x_cm and y_cm), CM of regular shapes table, and composite body subtraction method.
Download PDFCentre of Mass β MCQ Practice
15 NEET-style MCQs: CM of two- and three-body systems, CM of regular bodies (rod, disc, sphere, ring), CM of composite L-shaped bodies, and assertion-reason on CM properties.
Download PDFCentre of Mass β NEET-Style PYQ Practice
NEET-style questions on CM position for two-particle systems, CM of composite bodies (disc with hole, uniform ring), and CM motion under external forces.
Download PDFSubtopics in Centre of Mass
2-Column TableRapid Revision β Centre of Mass
Concept β Trap β Example1) Definition and Position Vector
Weighted Average PositionCentre of mass of a system (body) is a point that moves as though all the mass were concentrated there and all external forces were applied there. Position vector: r_cm = (mβrβ+mβrβ+...+mβrβ)/(mβ+mβ+...+mβ). In components: x_cm = Ξ£mα΅’xα΅’/Ξ£mα΅’; y_cm = Ξ£mα΅’yα΅’/Ξ£mα΅’.
- the centre of mass of n particles is a weighted average of the position vectors of n particles making up the system β each position vector is weighted by the particle's mass.
- For a continuous body: x_cm = β«x dm/M where M is total mass. This reduces to the geometric centre for uniform symmetric bodies.
- Common NEET trap: computing CM for a two-body problem using average position (unweighted) β only works when masses are equal. Always weight by mass.
Example (NEET-style)Two masses: mβ=3 kg at xβ=0, mβ=1 kg at xβ=4 m. x_cm = (3Γ0+1Γ4)/(3+1) = 4/4 = 1 m. The CM is at 1 m from the origin, closer to the heavier mass.
2) Important Points about Centre of Mass
Properties and Special CasesThe position of centre of mass is independent of the coordinate system chosen. CM of regular uniform bodies (sphere, cube, rod, disc) lies at the geometric centre. CM of a ring or hollow body lies at the geometric centre β which may be inside the empty space (not inside material).
- In a uniform gravitational field: centre of mass coincides with centre of gravity (the point where g can be considered to act for torque calculations).
- For a system of two particles, CM lies on the line joining them, at distances inversely proportional to mass: CM divides the line in ratio mβ:mβ from mβ.
- CM of a uniform ring of radius R: at the geometric centre (x=0, y=0) β inside the hollow, not inside the material. This is the most common NEET assertion-reason question for this subtopic.
Example (NEET-style)Uniform rod of length L: CM at L/2 from either end. Uniform disc of radius R: CM at centre. Uniform ring of radius R: CM at centre (inside the hollow). Equilateral triangle plate: CM at the centroid (intersection of medians, at height h/3 from base).
US Curriculum Gaps β Centre of Mass
NRI students from US high schools may find these gaps when preparing for NEET Centre of Mass problems.CM of Composite and Hollow Bodies (AP Physics 1 β Unit 5: Momentum)
AP Physics 1 covers CM for simple point-mass systems but does not typically require CM calculations for composite bodies (L-shaped plates) or hollow bodies (rings, cylindrical shells) where CM lies outside the material.
- AP Physics 1 treats two-point-mass CM as the standard problem; hollow body CM (at geometric centre of the hollow) is not a regular AP Physics 1 problem type.
- NEET tests CM of a disc with a circular hole (CM subtraction method); a ring (CM at centre of hollow space); and L-shaped uniform rods.
- Practise: disc of mass M, radius R, with a hole of mass m and radius r at distance d from centre β CM = (MΓ0 β mΓd)/(Mβm).
Coordinate Independence of CM (AP Physics 1 β Conceptual Gap)
AP Physics 1 does not explicitly test the assertion that the CM position is the same regardless of which coordinate origin is chosen. NEET tests this in assertion-reason format.
- The CM position relative to other bodies in the system does not change when the coordinate origin is shifted β only the numerical coordinates change.
- This is because the CM is defined by a ratio of positions; shifting origin shifts all positions by the same amount, and the net offset cancels when taking weighted averages.
- Practise: compute CM of two masses (3 kg at x=2, 1 kg at x=6) in two different coordinate systems; verify the relative position of CM is unchanged.
NEET-Style Practice Questions β Centre of Mass
5 NEET-style practice questions1Centre of mass of a system (body) is a point that moves as though all the mass were concentrated there. For a uniform ring of radius R, the position of the centre of mass is:NEET-style
On the circumference
At the geometric centre (inside the hollow)
At a distance R/2 from the geometric centre
Depends on which section is heavier
For a uniform ring (constant linear mass density), by symmetry the CM lies at the geometric centre β the centre of the ring, which is inside the hollow space (not in the material of the ring itself). This is a classic NEET concept: the CM does not have to lie within the material of the body. Option (a) places CM on the ring material β incorrect, by symmetry the CM is at the centre. Option (c) R/2 would be the CM of a semicircular arc, not a full ring. Option (d) is incorrect β a uniform ring has uniform density so all parts contribute equally; the result is symmetric. Correct: (b).
2Two particles of masses 4 kg and 2 kg are placed at positions x = 2 m and x = 8 m respectively on the x-axis. The position of their centre of mass is:NEET-style
5 m
4 m
6 m
3 m
x_cm = (mβxβ + mβxβ)/(mβ+mβ) = (4Γ2 + 2Γ8)/(4+2) = (8+16)/6 = 24/6 = 4 m. The CM lies at x = 4 m, which is closer to the 4 kg mass (at x=2 m) than to the 2 kg mass (at x=8 m), as expected since the heavier mass dominates the weighted average. Option (a) 5 m is the simple geometric midpoint (unweighted average of 2 and 8), neglecting mass differences. Option (c) 6 m would require the lighter mass to dominate, which is not physical. Option (d) 3 m is closer to the larger mass but not the correct weighted average. Correct: (b) 4 m.
3Three particles of masses 1 kg, 2 kg, and 3 kg are located at positions (1,1), (2,β1), and (β1,0) metres respectively. The x-coordinate of their centre of mass is:NEET-style
0.5 m
β0.17 m
0.17 m
1 m
x_cm = (mβxβ + mβxβ + mβxβ)/(mβ+mβ+mβ) = (1Γ1 + 2Γ2 + 3Γ(β1))/(1+2+3) = (1+4β3)/6 = 2/6 = 1/3 β 0.17 m. Steps: mβxβ = 1Γ1 = 1, mβxβ = 2Γ2 = 4, mβxβ = 3Γ(β1) = β3. Sum = 1+4β3 = 2. Total mass = 6 kg. x_cm = 2/6 = 1/3 m β 0.17 m. Option (a) 0.5 m = simple average of x-coordinates (1+2β1)/3 without mass weighting. Option (b) β0.17 m has wrong sign. Option (d) 1 m is one of the x-coordinates, not the CM. Correct: (c) 0.17 m.
4The position of centre of mass is independent of the coordinate system chosen. A system of two particles has its CM at x = 6 m in frame A. In frame B (origin shifted 2 m to the right), the position of the CM is:NEET-style
6 m
4 m
8 m
3 m
Shifting origin 2 m to the right (frame B origin is at x=2 m of frame A) means all positions in frame B = positions in frame A β 2. CM in frame B = 6 β 2 = 4 m. The CM is the same physical point; its coordinate changes when the origin changes, but its position relative to other bodies in the system is unchanged. Option (a) 6 m would mean coordinates are the same in both frames β false when origin shifts. Option (c) 8 m = 6+2 corresponds to shifting origin to the left (frame B origin at x=β2 of frame A). Option (d) 3 m = 6/2 is incorrect (dividing by 2 has no physical basis). Correct: (b) 4 m.
5A uniform square plate of side 2a has a square hole of side a cut from one corner. The CM of the remaining plate is shifted from the centre of the original plate by:NEET-style
a/6 towards the corner without hole
a/3 towards the corner with hole
a/2 towards the hole
No shift β CM remains at centre
Original plate: mass M, side 2a, CM at origin (0,0). Removed square: side a, corner at (a,a), mass M/4 (area ratio = aΒ²/(4aΒ²) = 1/4), CM at (a + a/2, a + a/2) β wait, if hole is cut from one corner, say corner at (a,a), the removed square occupies from (a/2Γ2 = a) to (a + a = 2a). Wait: let centre at origin, original square from (βa,βa) to (+a,+a). Hole of side a at corner (+a,+a): occupies from (0,0) to (a,a). Removed mass = M/4, CM = (a/2, a/2). Remaining mass = 3M/4. x_cm = [MΓ0 β (M/4)(a/2)] / (3M/4) = [βMa/8]/(3M/4) = βa/6. The CM shifts by a/6 from origin towards the corner opposite the hole (left-bottom direction, away from the removed section). Option (b) towards the corner with hole is the wrong direction. Option (c) a/2 overstates the shift. Option (d) is wrong β removing mass from one corner must shift CM away from that corner. Correct: (a).
Practice Questions β Centre of Mass
Click "Reveal Answer" after attempting1A 3 kg and 6 kg mass are 9 m apart. Where is their CM from the 3 kg mass?
3 m
6 m
4.5 m
9 m
π Reveal Answer
Option (b) 6 m. x_cm = (3Γ0 + 6Γ9)/(3+6) = 54/9 = 6 m from the 3 kg mass. The CM divides the line in the inverse ratio of masses: 6 kg:3 kg = 2:1 from 3 kg end β 9Γ2/3 = 6 m from the 3 kg mass.
2For which of the following bodies does the CM NOT lie within the material of the body?
Solid sphere
Solid cube
Hollow ring
Solid cylinder
π Reveal Answer
Option (c) Hollow ring. The CM of a hollow ring lies at the geometric centre β which is inside the hollow space, not in the ring material. All solid bodies (sphere, cube, cylinder) have CM that lies within their material.
3Two particles of masses 2 kg and 3 kg are at (3,2) and (β2, β1) m. Find the x-coordinate of their CM.
0 m
0.2 m
β0.2 m
1 m
π Reveal Answer
Option (c) β0.2 m. x_cm = (2Γ3 + 3Γ(β2))/(2+3) = (6β6)/5 = 0/5 = 0 m. Wait: 2Γ3=6, 3Γ(β2)=β6, sum=0. x_cm = 0/5 = 0 m. Select option (a) 0 m.
4A uniform L-shaped body has two arms: arm 1 (mass 2 kg, CM at x=1 m, y=0) and arm 2 (mass 1 kg, CM at x=0, y=1 m). Find the y-coordinate of the CM.
1/3 m
2/3 m
1/2 m
0 m
π Reveal Answer
Option (a) 1/3 m. y_cm = (mβyβ + mβyβ)/(mβ+mβ) = (2Γ0 + 1Γ1)/(2+1) = 1/3 m.
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Frequently Asked Questions β Centre of Mass
Notes Β· Downloads Β· Revision Β· Important QuestionsWhat is the definition of centre of mass?
Centre of mass of a system is a point that moves as though all the mass of the system were concentrated there and all external forces were applied there. Its position is the mass-weighted average of all particle positions: r_cm = (mβrβ+mβrβ+...+mβrβ)/(mβ+mβ+...+mβ). For a continuous body: r_cm = β«r dm / M.
Why is the position of centre of mass independent of the coordinate system?
The centre of mass is a physical point in space β it has a definite location. When you change the coordinate origin, all position vectors shift by the same amount, but the mass-weighted average shifts by the same amount as well. The CM's position relative to other bodies in the system remains unchanged. The numerical coordinates of CM change, but the physical point does not move.
Can the centre of mass lie outside the body?
Yes. The CM is a mathematical point, not necessarily within the material. For a uniform ring, the CM is at the geometric centre β inside the empty space, not in the ring material. For an L-shaped plate, the CM may lie inside the empty quadrant. The condition for this is any shape with a cavity or concavity where the geometric centre falls outside the material.
How do you find the CM of a composite body?
Method 1 (direct): treat each part as a point mass at its own CM; apply x_cm = Ξ£mα΅’xα΅’/Ξ£mα΅’. Method 2 (subtraction): for a body with a removed portion (e.g., disc with hole): x_cm = (M_whole Γ x_cm_whole β M_removed Γ x_cm_removed) / (M_whole β M_removed). In both methods, find the CM of each simple part individually (at its geometric centre for uniform shapes) before combining.
Where is the CM of a uniform ring, disc, sphere, and rod?
Uniform ring: at geometric centre (inside hollow space). Uniform disc: at geometric centre (at its flat face centre). Uniform solid sphere: at geometric centre. Uniform rod: at midpoint (L/2 from either end). Uniform thin triangular plate: at centroid (intersection of medians, at h/3 from base and 2h/3 from apex).
How does the CM move under external forces?
The acceleration of the CM: F_external = Ma_cm, where M is the total mass and F_external is the net external force. Internal forces do not alter the CM's motion β only external forces do. This is why, in an explosion or collision with no external forces, the CM continues at constant velocity (Newton's First Law applied to the CM).
For two particles, how does the CM divide the line joining them?
The CM divides the line segment joining the two particles in the ratio mβ:mβ from the particle mβ. That is, the distance of CM from mβ is d Γ mβ/(mβ+mβ) and from mβ is d Γ mβ/(mβ+mβ), where d is the total separation. The CM is always closer to the heavier particle.
What is the difference between centre of mass and centre of gravity?
Centre of mass is the mass-weighted average position of all particles. Centre of gravity is the point at which the body's weight can be assumed to act for the purpose of torque calculation. In a uniform gravitational field (g = constant throughout the body), CM and CG coincide exactly. In a non-uniform gravitational field (very large bodies near a massive object), they are slightly different β CG is biased toward the region of stronger gravitational pull.
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