Drag the masses along the rod. The rod itself is treated as weightless
Controls
Scene
Mass to move
Readings
- Total mass
- 6.0kg
- Centre of mass, x
- 59.17cm
- Support (pivot) position
- 50.0cm
- Net torque about the support
- 5.39N·m
How to use this simulation
- In the “Masses on a rod” scene, drag masses A, B and C with the mouse or a finger; the yellow diamond is the centre of mass, and its position is in the readings.
- Slide the support and watch which way the rod tips, then switch on “Move the support under the centre of mass” to level it.
- In the “Masses on a plate” scene follow the dashed lines to see the centre of mass move in both x and y.
- In the “Toppling block or bus” scene change the width, height and bottom ballast, then tilt the base slowly and watch the body go over at the critical angle.
The ruler-on-two-fingers trick
Rest a ruler loosely on your two index fingers, one near each end. Now slide your fingers slowly together. Strange thing: they never move at the same time. One slides, then the other, taking turns, and in the end they meet exactly in the middle. That meeting point is the ruler’s centre of mass.
Stick a lump of modelling clay on one end and try again. The fingers now meet nearer the clay. The centre of mass has moved towards the heavy end. With one tiny experiment you have caught its real character: it is the average position, weighted by mass.
Acrobats on a rope, a load of bamboo on a rickshaw, luggage piled on a bus roof, a news story about a bus rolling over on a bend: all of them hang on the same idea. In this lesson we build the formula, play with it in the simulation, and then work some numbers.
From zero: what the centre of mass actually is
A body is made of billions of particles, and Earth pulls on every one of them separately. To keep our sums manageable we would love to replace all those pulls with a single arrow. There is a special point where, if you pretend all the mass sits there, the body’s overall motion from place to place comes out right. That point is the centre of mass.
Think of a spinning hammer tossed across a room. The hammer tumbles, and the motion looks messy. Yet its centre of mass traces a perfectly smooth parabola, exactly as a small pebble would. However the body twists, the centre of mass moves like a single point mass under the same outside forces.
The centre of mass does not have to be inside the body. A ring has it in the empty middle. A semicircular wire has it off the wire, on the inner side of the curve. We will even locate the centre of mass of the Earth and Moon together, and the result is surprising.
Centre of mass and centre of gravity are different ideas. The centre of mass depends only on how mass is spread out. The centre of gravity is the point where the total weight acts. When g is the same everywhere, as it is for everything near the ground, the two points coincide. They would separate only for something so large that g changes across it. In school problems, treat them as one.
Words you must know
These terms come up again and again in this chapter. Each has a plain meaning and a familiar example.
| Term | Plain meaning | Familiar example |
|---|---|---|
| Centre of mass | The balance point of the mass distribution | The spot where a pencil balances on a finger |
| Centre of gravity | The point where the whole weight acts | The point under which a hanging body’s string stops |
| Point mass | Mass imagined at a single point, size ignored | A small metal bead glued to a rod |
| Torque (moment) | Force times perpendicular distance; turning power | A push on a door handle |
| Support (pivot) | The point where the body is propped up | The knife-edge in the middle of a balance |
| Equilibrium | Net force and net torque both zero | A balance hanging perfectly still |
| Stable equilibrium | Returns by itself after a small push | A marble in the bottom of a bowl |
| Unstable equilibrium | Runs further away after a small push | A marble on top of an upturned bowl |
| Neutral equilibrium | Stays wherever it is left | A marble on a flat table |
The formula: a weighted average of positions
In an ordinary average every number counts equally. In the centre-of-mass average each position counts as much as the mass sitting there. Heavy masses pull harder on the average, so the centre of mass sits closer to the heavy side.
x_cm = (m₁x₁ + m₂x₂ + … + mₙxₙ) / (m₁ + m₂ + … + mₙ)along a line
x_cm = Σmx / Σm, y_cm = Σmy / Σmin a plane, two numbers
x_cm = (1/M) ∫ x dmfor a continuous body
Two particles: the easiest case
Take 4 kg and 1 kg, 100 cm apart. Put the origin at the 4 kg mass: x_cm = (4×0 + 1×100) / (4 + 1) = 20 cm. The centre of mass lies on the line joining them, much closer to the heavy one.
The ratio of the two distances is the inverse of the ratio of the masses: distance from the heavy mass ÷ distance from the light mass = light mass ÷ heavy mass. This is just the law of the balance.
Many particles and the simulation’s defaults
At the start A = 2 kg at 20 cm, B = 1 kg at 60 cm and C = 3 kg at 85 cm. Total mass 6 kg. The moments add up: 2×20 + 1×60 + 3×85 = 40 + 60 + 255 = 355. So x_cm = 355 ÷ 6 = 59.17 cm.
A lovely check: multiply each mass by its distance from the centre of mass and add. You must get zero: Σm(x − x_cm) = 0. The “mass × distance” on one side of the centre of mass exactly cancels the other side.
Bodies with a known shape
A uniform body has its centre of mass at its geometric centre. For shapes where the centre is not obvious, here are the standard results, evaluated for R = 10 cm and H = 12 cm.
| Body | Measured from | Formula | Example |
|---|---|---|---|
| Semicircular wire | Midpoint of the diameter | 2R/π | 6.37 cm |
| Semicircular disc | Midpoint of the diameter | 4R/3π | 4.24 cm |
| Solid hemisphere | Centre of the flat face | 3R/8 | 3.75 cm |
| Hollow hemisphere | Centre of the flat face | R/2 | 5.00 cm |
| Triangular plate | The base, along the height | H/3 | 4.00 cm |
| Solid cone | The base, along the height | H/4 | 3.00 cm |
Where the formula comes from
Memorising the formula is easy, but why this formula? Two short arguments settle it: one from the balance, one from Newton’s laws.
From balancing moments
Put masses m₁, m₂, … on a light rod and prop the rod at a point P. Each weight mg pulls down at a distance (x − P) from the prop. The rod balances if the clockwise and anticlockwise moments about P are equal.
That means Σ m g (x − P) = 0. Cancel g and write Σ m x − P Σ m = 0, so P = Σ m x / Σ m. The prop position that balances the rod is exactly the centre of mass.
Σ m g (x − P) = 0 ⇒ P = Σmx / Σm
From Newton: it moves like one point mass
Write Newton’s second law for each particle and add them: Σ mᵢaᵢ = Σ Fᵢ. On the right the internal forces cancel in pairs, because by Newton’s third law A pulls B exactly as hard as B pulls A the other way. Only the external forces survive.
On the left, Σ mᵢaᵢ = M a_cm, which you get by differentiating the centre-of-mass definition twice. So M a_cm = F_ext. That is why, after an explosion with no external force, the fragments may fly anywhere but their centre of mass carries on exactly as before.
Σ mᵢaᵢ = Σ Fᵢadd every particle’s equation
M a_cm = F_extinternal forces drop out
Equilibrium and stability
A body is in equilibrium when the net force and the net torque on it are both zero. But not all equilibria are alike. Nudge the body a little and see what happens; that sorts equilibrium into three kinds.
The most useful rule for exams: if a nudge raises the centre of mass, the equilibrium is stable; if it lowers it, unstable; if the height stays the same, neutral. Raising the centre of mass costs work, so the body wants to roll back.
tan φ_c = (b/2) / hcritical angle for toppling
φ < φ_c ⇒ stable, φ > φ_c ⇒ topplestilt φ against the critical angle
When does a bus topple
A bus stands on the rectangle between its wheels, its base. Drop a vertical from the centre of mass. If it lands inside the base, the weight turns the bus back down. If it lands outside, the weight itself turns the bus over.
With base width b and centre-of-mass height h, the critical tilt satisfies tan φ_c = (b/2)/h. A wider base or a lower centre of mass raises φ_c, so the bus can lean further before going over.
Why racing cars are wide and low
A racing car has a wide track and a centre of mass very near the road. Both raise φ_c. A double-decker with luggage tied to the roof does the opposite: the centre of mass climbs, and the safe angle shrinks.
| Kind | Centre of mass when nudged | What happens | Example |
|---|---|---|---|
| Stable | Rises | Returns | Marble in a bowl, a hanging swing |
| Unstable | Falls | Runs away | A pencil standing on its point |
| Neutral | Stays at the same height | Stays put | A ball on a flat floor |
Try it in the simulation
Before each experiment, guess what will happen, then move the sliders or the masses.
- In the rod scene put the support at 50 cm. Which way does the rod tip? Why is the torque reading 5.39 N·m positive?
- Switch on “Move the support under the centre of mass”. The torque reads zero and the rod levels. Now drag mass B to the right. What changes?
- Raise mass C from 3 to 6 kg. Which way did the centre of mass move, and by how much?
- Make all three masses equal. Does the centre of mass land on the plain average of the three x values?
- In the plate scene drag one mass into a corner. Do x_cm and y_cm both change?
- In the bus scene switch on “Tilt slowly”. What happens as the tilt touches the critical angle? Now add ballast at the bottom and tilt again.
- Set the width to 0.4 m and the height to 3 m, a tall thin box. How much smaller is the critical angle?
Solved problems
For each problem draw a picture first, choose an origin, and then substitute. The numbers match the simulation.
Problem 1: the default rod
A = 2 kg at 20 cm, B = 1 kg at 60 cm, C = 3 kg at 85 cm on a light rod. Find the centre of mass and the net torque about a support at 50 cm.
Step 1: x_cm = 355 ÷ 6 = 59.17 cm. Step 2: torque about 50 cm = g × Σ m(x − 50)/100 = 5.39 N·m. Positive means the right side goes down, since the centre of mass is 9.17 cm to the right of the support.
Problem 2: the same masses on a plate
The masses sit at (20, 15), (60, 20) and (85, 50) cm. x_cm = 59.17 cm and y_cm = (2×15 + 1×20 + 3×50) ÷ 6 = 33.33 cm. Treat the two directions separately, then put them together.
Problem 3: Earth and Moon
Earth 5.97×10²⁴ kg, Moon 7.35×10²² kg, centres 384,400 km apart. Distance of the centre of mass from Earth’s centre = (Moon’s mass × separation) ÷ total mass = 4,675 km.
Earth’s radius is 6,371 km, so the centre of mass lies inside Earth, a little below the surface. It is more accurate to say that both bodies orbit this point than that the Moon orbits Earth.
Problem 4: a person walking on a boat
A 60 kg person walks 4 m along a 120 kg boat in still water. There is no external horizontal force, so the centre of mass of person and boat stays where it is.
The boat moves 60×4 ÷ (60 + 120) = 1.33 m the other way. Relative to the shore the person moves 2.67 m. The heavier boat moves less.
Problem 5: a disc with a hole
A uniform disc of radius 30 cm has a circle of radius 15 cm cut out so that the hole touches the disc’s centre. How far does the centre of mass shift?
Mass goes with area, and the hole is 0.25 of the disc. The centre of mass moves away from the hole by (hole fraction × 15) ÷ (1 − hole fraction) = 5 cm. The trick is to treat the cut-out as a negative mass.
Problem 6: how far can the bus lean
A bus has base width 2.4 m and its centre of mass 1.5 m up. Critical angle = tan⁻¹((2.4/2)/1.5) = 38.7°.
Put heavy batteries on the floor so the centre of mass is 0.75 m and the angle becomes 58.0°. Pile luggage on the roof so it is 2.0 m and it drops to 31.0°. Roof luggage eats much of the safe lean.
Mistakes everyone makes
Most lost marks in centre-of-mass problems come from a handful of slips.
- Assuming the centre of mass is always the geometric centre. With uneven mass it moves, and for a hollow body it can lie outside the body.
- Writing “centre of mass” and “centre of gravity” as synonyms. They are different ideas that coincide only in uniform g.
- Dividing by the number of particles instead of the total mass in the x_cm formula.
- Thinking the centre of mass changes when you move the origin. The point is the body’s own; only its coordinate changes.
- Mixing up left and right distances, and so their signs, when taking moments about a support.
- Dropping the vertical for a toppling body from the middle of the bus instead of from the centre of mass.
- Believing stable equilibrium means “does not move”. It means “returns after a nudge”, and an unstable equilibrium can sit still for a while.
Centre of mass in real life
A tightrope walker carries a long pole. The drooping pole drags the combined centre of mass down towards, or below, the rope, so a small wobble gives a restoring torque instead of a fall.
A high jumper arches over the bar so that the body’s centre of mass can pass under the bar while the body passes over it. That is a centre of mass outside the body, put to work.
The Leaning Tower of Pisa still stands because the vertical through its centre of mass falls inside its base. Engineers stabilised the foundation to stop the lean from growing.
In boats, cranes, trucks and rickshaws, designers keep the centre of mass low. This is also why a loaded truck has heavy cargo at the bottom and light cargo on top.
Exam corner: NCERT and CBSE style
In Class 11 this chapter appears as short-answer and numerical questions. The pattern below mirrors the board paper.
Sample question and answer
Question: A light rod 0.8 m long carries 3 kg at one end and 5 kg at the other. (a) Define the centre of mass. (b) Where is the centre of mass measured from the 3 kg end? (c) Explain why a support placed there keeps the rod horizontal.
Answer: (b) x = 5×0.8 ÷ (3 + 5) = 0.5 m, which is 50 cm. (c) About the centre of mass the total torque Σ m g (x − x_cm) is zero, so there is no turning effect.
How to present it
For two-mark conceptual questions give the reason in one or two sentences, not the formula. For numericals write the formula, substitute with units, and state the answer with its unit.
Quick revision
This list is enough the night before an exam.
- Centre of mass = mass-weighted average position: x_cm = Σmx/Σm.
- In a plane, y_cm = Σmy/Σm; a uniform body has it at its geometric centre.
- The prop point that balances a rod is the centre of mass; Σm(x − x_cm) = 0.
- M a_cm = F_ext; internal forces cannot change the centre of mass’s motion.
- The centre of mass shifts toward the heavier part, staying close to the larger mass.
- Equilibria come in three kinds: nudge raises the CM → stable, lowers it → unstable, same height → neutral.
- A body topples when the vertical through its centre of mass leaves the base: tan φ_c = (b/2)/h.
- A wide base and a low centre of mass give more stability.
Frequently asked questions
What is the centre of mass?
The centre of mass is the point where the whole mass of a body or system can be taken as concentrated, so that its overall motion under external forces is that of a single particle. For point masses x_cm = Σmx/Σm. In a uniform body it lies at the geometric centre.
What is the difference between centre of mass and centre of gravity?
The centre of mass depends only on how mass is distributed. The centre of gravity is the point where the total weight acts. When g is uniform across the body they coincide; for extremely large bodies, where g varies, they differ slightly.
Can the centre of mass lie outside the body?
Yes. A ring has it in the empty centre, and a semicircular wire has it off the wire, inside the curve. For the Earth-Moon system the centre of mass does lie inside Earth, but it is not at Earth’s centre.
Why do buses topple over?
A bus topples when, as it tilts, the vertical through its centre of mass leaves the base between its wheels. Then its weight turns it further over. The critical angle is tan φ_c = (b/2)/h, so a narrow base or a high centre of mass makes toppling easier.
How do stable, unstable and neutral equilibrium differ?
After a small push, a body in stable equilibrium returns, and its centre of mass rises. In unstable equilibrium it moves further away, and its centre of mass falls. In neutral equilibrium it stays in its new position and the centre of mass stays at the same height.
How do I find the centre of mass of masses on a rod?
Use x_cm = (m₁x₁ + m₂x₂ + …) ÷ (m₁ + m₂ + …). For 2 kg at 20 cm, 1 kg at 60 cm and 3 kg at 85 cm the answer is 59.17 cm.
What happens to the centre of mass in an explosion?
With no external force it keeps moving exactly as before. The fragments may fly in any directions, but their mass-weighted average position follows the original path, because the internal forces cancel in pairs.
What path does the centre of mass of a thrown, spinning object follow?
A smooth parabola, since gravity is the only external force. However the object tumbles, its centre of mass moves like a projectile point mass.
Keep studying this topic
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Enzyme Action
Biology
The Nerve Impulse and the Action Potential
Biology
The Nephron and Urine Formation
Biology
Meiosis Cell Division
Biology
Digestive System
Biology
Breathing Mechanism: Inhalation and Exhalation
Biology
Protein synthesis
Biology
Blood circulation: double circulation and cardiac output
Biology
Cell structure and organelles: plant cell vs animal cell
Biology
Reflex action and the reflex arc
Biology
The human eye: parts, image formation and accommodation
Biology
Muscle contraction and the sliding-filament model
Biology
Transport in plants: xylem, phloem and transpiration
Biology
Food chain and food web: energy flow and the 10 % rule
Biology
Population growth: exponential and logistic curves
Biology
Natural selection: peppered moths and the sooty tree
Biology
Immune response: primary and secondary antibody response
Biology
Pollination and fertilisation
Mathematics
Pythagorean Theorem
Mathematics
Trigonometric Ratios and the Unit Circle
Mathematics
Quadratic functions and their graphs
Mathematics
Circle theorems
Mathematics
Heights and distances: elevation and depression angles
Mathematics
Probability: experimental vs theoretical probability
Mathematics
Arithmetic and Geometric Sequences
Mathematics
Mean, Median and Mode
Mathematics
Sets and Venn Diagrams
Mathematics
Linear equations: graphs, slope and simultaneous pairs
Mathematics
Graphs of sin, cos and tan
Mathematics
Completing the Square and the Quadratic Formula
Mathematics
Area of a Circle: Proof, Formula and Solved Problems
Mathematics
Volume and Surface Area of Solids: Cube, Cuboid, Cylinder, Cone, Sphere
Mathematics
Similar Triangles: Scale Factor, AA, SSS, SAS and Shadow Heights
Mathematics
Coordinate Geometry: Distance, Midpoint, Slope, Section Formula and Locus
Mathematics
Conic Sections: Circle, Ellipse, Parabola and Hyperbola
Mathematics
Function Transformations: Shifts, Stretches and Reflections
Mathematics
Derivative as the Slope of a Tangent: Secant to Tangent
Mathematics
Riemann Sums and Integration: Area Under a Curve
Mathematics
Limits and the ε–δ Definition: A Visual Guide
Mathematics
3D Vectors: Sum, Dot Product and Cross Product
Mathematics
Matrix Transformations: Rotation, Reflection, Shear and the Determinant
Mathematics
Complex Numbers: Sum, Product, Powers and Roots on the Argand Plane
Mathematics