01
CONCEPT
Centre of Mass of Discrete Particles
In one line
The centre of mass is the point where the whole mass of a system can be taken to be concentrated and where all external forces can be taken to act.
The centre of mass is the point where the whole mass of a system can be taken to be concentrated and where all external forces can be taken to act.
SYSTEM
6.1.1 System and forces
- Rigid body: particles held so firmly that their relative positions do not change.
- System: a group of interacting particles that we choose to study.
- Internal forces: forces the particles of the system exert on each other. They occur in equal and opposite pairs.
- External forces: forces applied on the system from outside.
MEANING
6.1.2 What the centre of mass is
Position vector of the C.M. is the weighted mean of r₁, r₂, r₃
- Its position does not depend on the coordinate system chosen.
- It depends on the shape of the body and its mass distribution.
- It depends only on masses and positions, not on forces.
- It may lie outside the material: disc inside, ring outside.
- Rotation about an axis through the C.M. does not change its position.
- C.M. and centre of gravity coincide only when g is uniform over the body.
TWO PARTICLES
6.1.3 System of two particles
- The C.M. lies on the line joining the two particles, nearer the heavier one.
- Equal masses: it is at the mid-point.
- Distances from the C.M. are in the inverse ratio of masses. For separation d:
Heavier mass is nearer to C
SHIFTING
6.1.4 Moving a particle, C.M. fixed
Both move towards C (or both away from it)
- If m1 moves towards C, m2 must also move towards C.
- If the C.M. is allowed to move, its shift is:
MANY PARTICLES
6.1.5 System of n particles
- Choose the origin at one of the masses: its term drops out.
- Work out x and y separately.
- With the origin at the C.M.:
Three equal masses: C is at the centroid
- Identical spheres: replace each by a point mass at its centre.
- Three equal masses at the corners of a triangle: C.M. at the intersection of the medians.
- Equal masses in a straight line: C.M. at the mean of their positions.
02
CONCEPT
Continuous Bodies: Standard Results & Variable Density
In one line
For a continuous body, the sum over particles becomes an integral. For uniform symmetric bodies you can write the answer straight from the standard results.
For a continuous body, the sum over particles becomes an integral. For uniform symmetric bodies you can write the answer straight from the standard results.
FORMULA
6.2.1 Continuous mass distribution
- Split the body into small elements of mass dm and add them up.
- Uniform symmetric body: C.M. is at the geometric centre.
- Any axis or plane of symmetry contains the C.M., which saves one integral.
POSITIONS
6.2.2 Centre of mass of uniform bodies
| Body | Position of C.M. |
|---|---|
| Hollow or solid sphere | centre of the sphere |
| Circular ring or disc | centre |
| Rod | mid-point |
| Square, rectangle, parallelogram | intersection of diagonals |
| Triangular lamina | intersection of medians |
| Cube or cuboid | intersection of diagonals |
| Hollow or solid cylinder | mid-point of the axis |
| Solid cone or pyramid | on the axis, 3h4 from the vertex |
STANDARD
6.2.3 Results to learn by heart
C is marked in yellow
| Body | Distance of C.M. |
|---|---|
| Semicircular wire | 2Rπ from centre |
| Semicircular disc | 4R3π from centre |
| Quarter disc | 4R3π from each straight edge |
| Solid hemisphere | 3R8 from flat face |
| Hollow hemisphere | R2 from flat face |
| Solid cone | h4 from base |
| Triangular lamina | h3 from base |
- Always read which point the question measures from: vertex or base, centre or edge.
JEE · NEET
Corner
Rod with variable density
Mass element dm = λ dx at distance x
- Write λ as a function of x.
- Do both integrals over the whole length, from 0 to L.
- Divide. The same method works for any variable density.
Results worth remembering
| λ | xcm from x = 0 |
|---|---|
| constant | L2 |
| kx | 2L3 |
| kx² | 3L4 |
| kxⁿ | n + 1n + 2 L |
| a + bx | L(3a + 2bL)3(2a + bL) |
- As n grows, mass crowds at the far end and xcm moves towards L.
- C.M. always shifts towards the denser side.
Plate with variable density
- Density depends on x only ⇒ ycm stays at the middle, b2.
- Then xcm is found just as for a rod with the same variation.
03
CONCEPT
Composite & Cut-Out Bodies
In one line
Replace every piece by a point mass at its own centre of mass, then use the particle formula. A hole is simply a piece with negative mass.
Replace every piece by a point mass at its own centre of mass, then use the particle formula. A hole is simply a piece with negative mass.
COMPOSITE
6.3.1 Bodies made of simple pieces
- x1, x2 are the centres of the separate pieces.
- Uniform wire or rod: mass goes with length.
- Uniform plate: mass goes with area.
- Uniform solid: mass goes with volume.
Bent rod: each straight part acts from its mid-point
- Equal-mass rods: use equal weights even if the lengths differ.
- Bent wire of different lengths: use lengths as the weights.
Split an L-shaped plate into two rectangles
- Choose the split that gives simple rectangles.
- Use symmetry first: a symmetric plate fixes one coordinate at once.
- A force through the C.M. gives translation only, with no rotation.
CUT-OUT
6.3.2 Bodies with a hole
- M1: the whole body. M2: the part removed.
- The C.M. shifts away from the hole.
- Take the origin at the centre of the whole body: x1 = 0.
G moves to the side opposite the hole
- Hole of radius r in a disc of radius R, centres d apart:
| Hole touching the edge | Shift of C.M. |
|---|---|
| Disc R, hole R2 | R6 |
| Disc 2R, hole R | R3 |
| Sphere R, cavity R2 | R14 |
- Quarter of a uniform plate cut off: remaining mass is three quarters.
04
CONCEPT
Motion of Centre of Mass
In one line
The centre of mass moves like a single particle of mass M acted on by the total external force. Internal forces can never change its motion.
The centre of mass moves like a single particle of mass M acted on by the total external force. Internal forces can never change its motion.
VELOCITY
6.4.1 Velocity and acceleration of C.M.
- These are vector sums: keep the signs and directions.
- Opposite directions get opposite signs.
- If vcm and acm are parallel, the path of C.M. is a straight line.
- If they are not parallel and acm is constant, the path is a parabola.
NEWTON
6.4.2 Second law for a system
- Total momentum equals total mass times the velocity of C.M.
- Change in total momentum = Fext × time.
- Particles in free flight: acm = g, whatever their separate velocities.
- Two blocks over a pulley:
- Kinetic energy of a system:
ISOLATED
6.4.3 No external force
- An isolated system’s C.M. moves in a straight line with uniform velocity.
- If it starts at rest, the C.M. stays at rest for ever.
Mutual attraction: both move, C stays put
- Two particles attracting each other meet at the C.M.
- Speeds are in the inverse ratio of masses, and so are distances:
BOAT
6.4.4 Man on a boat or cart
No horizontal external force: C.M. does not shift
- Man of mass m walks a length L relative to a boat of mass M.
- Boat moves back while he goes forward:
- Measure from the ground: xman = L − xboat.
- Same result for a boy on a cart or a person walking on a plank.
EXPLOSION
6.4.5 Explosion in flight
Fragments spread out; C keeps the old parabola
- Explosion forces are internal, so the C.M. keeps the original path.
- The fragments themselves do not follow that path.
- C.M. lands at the original range R:
- A falling body that breaks: C.M. of the parts continues as before.
EXAM-DAY CARD
Just see this before the exam
rcm = Σ m rΣ m
particles
m₁d₁ = m₂d₂
C.M. fixed
rcm = ∫ r dm∫ dm
continuous
x = M₁x₁ − M₂x₂M₁ − M₂
hole
Fext = M acm
motion
xboat = mLm + M
boat
FORMULA SHEET
Whole chapter on one page
Two particles
m₁r₁ = m₂r₂
r₁ = m₂dm₁ + m₂
C.M. fixed: m₁d₁ = m₂d₂
Many particles
xcm = Σ m xΣ m
ycm = Σ m yΣ m
About C.M.: Σ m r = 0
Standard bodies
semicircular wire: 2Rπ
semicircular disc: 4R3π
hemisphere: 3R8; cone: h4
Variable density
xcm = ∫ x λ dx∫ λ dx
λ = kx: 2L3
λ = kxⁿ: n + 1n + 2 L
Cut-out
x = M₁x₁ − M₂x₂M₁ − M₂
disc: r²dR² − r²
sphere: r³dR³ − r³
Motion
vcm = Σ m vΣ m, acm = Σ m aΣ m
Fext = 0 ⇒ vcm constant
P = M vcm
TOP MISTAKES
Avoid these = easy marks
MISTAKE ✗
C.M. depends on the forces acting
Bent rod: taking equal masses for unequal lengths
Hole: adding the removed mass
After explosion, fragments keep the old path
CORRECT ✓
Only masses and positions matter
Mass goes with length, area or volume
Subtract: M₁x₁ − M₂x₂ over M₁ − M₂
Only the C.M. keeps the old path