Prerequisites
Before studying this topic, review:
- Vectors - Vector addition and components
- Newton’s Laws - Force and mass concepts
The Hook: Why Does Thor’s Mjolnir Spin Perfectly?
Ever wondered why Thor’s hammer Mjolnir spins so perfectly when he throws it? Or why a thrown stick wobbles in the air but rotates around one special point?
In the climactic battle scene of Thor: Ragnarok, when Thor hurls Mjolnir, it spins elegantly around a single point - its centre of mass. This isn’t movie magic; it’s fundamental physics!
Similarly, when cricketers throw the ball with a scrambled seam, it wobbles but still follows a parabolic path. That path is traced by one special point: the centre of mass.
The Question: Why does every object, no matter how irregularly shaped, have one special point that behaves as if all the mass is concentrated there?
The Core Concept
What is Centre of Mass?
The centre of mass (COM) is the point where the entire mass of a system can be considered to be concentrated for analyzing translational motion.
Centre of Mass is the “average position” of all the mass in a system.
Imagine you have to balance an irregular object on your finger - that balance point is near the centre of mass!
Interactive Demo: Visualize Centre of Mass
Explore how the centre of mass changes for different mass distributions.
Formula for Centre of Mass
For a system of particles:
$$\boxed{\vec{r}_{cm} = \frac{\sum_{i=1}^{n} m_i\vec{r}_i}{\sum_{i=1}^{n} m_i} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + \cdots + m_n\vec{r}_n}{m_1 + m_2 + \cdots + m_n}}$$In component form:
$$x_{cm} = \frac{\sum m_i x_i}{M}, \quad y_{cm} = \frac{\sum m_i y_i}{M}, \quad z_{cm} = \frac{\sum m_i z_i}{M}$$where $M = \sum m_i$ is the total mass.
For continuous bodies:
$$\boxed{\vec{r}_{cm} = \frac{1}{M}\int \vec{r}\,dm}$$Memory Tricks & Patterns
Mnemonic for Centre of Mass Formula
Memory Trick: “Children’s Marks Average” → Centre of Mass is Average position
Just like averaging test scores (multiply each score by weight, then divide by total weight), COM is the weighted average of positions!
$$\text{COM} = \frac{\text{(mass × position) for all parts}}{\text{total mass}}$$Pattern Recognition: Symmetry Shortcuts
If a body is symmetric about an axis, the COM lies on that axis.
If a body has two or more axes of symmetry, the COM is at their intersection.
Examples:
- Uniform rod: COM at geometric center
- Uniform disc/sphere: COM at geometric center
- Uniform square plate: COM at intersection of diagonals
- Hollow hemisphere: COM is NOT at geometric center (no full symmetry)
Centre of Mass for Common Systems
1. Two-Particle System
For two particles of masses $m_1$ and $m_2$ separated by distance $d$:
$$x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$Taking $m_1$ at origin ($x_1 = 0$) and $m_2$ at distance $d$ ($x_2 = d$):
$$\boxed{x_{cm} = \frac{m_2 d}{m_1 + m_2}}$$The COM divides the line joining them in the ratio $m_2 : m_1$ (inverse ratio of masses).
For two masses: COM is closer to the heavier mass.
If $m_1 = m_2$, then COM is exactly at the midpoint.
2. Three Particles (Triangle)
For three particles at vertices of a triangle:
$$x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}$$ $$y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3}$$Special case: If $m_1 = m_2 = m_3$, COM is at the centroid (intersection of medians).
3. Uniform Rod
For a uniform rod of length $L$:
- COM is at the geometric center (distance $L/2$ from either end)
4. Semicircular Ring
For a thin semicircular ring of radius $R$:
$$y_{cm} = \frac{2R}{\pi}$$(measured from center, along the axis of symmetry)
5. Semicircular Disc
For a uniform semicircular disc of radius $R$:
$$y_{cm} = \frac{4R}{3\pi}$$6. Hollow Hemisphere
For a thin hemispherical shell of radius $R$:
$$y_{cm} = \frac{R}{2}$$7. Solid Hemisphere
For a solid hemisphere of radius $R$:
$$y_{cm} = \frac{3R}{8}$$Quick Reference Table
| Object | Centre of Mass Location |
|---|---|
| Uniform rod (length $L$) | At center: $L/2$ from end |
| Semicircular ring (radius $R$) | $\frac{2R}{\pi}$ from center |
| Semicircular disc (radius $R$) | $\frac{4R}{3\pi}$ from center |
| Hollow hemisphere (radius $R$) | $\frac{R}{2}$ from base |
| Solid hemisphere (radius $R$) | $\frac{3R}{8}$ from base |
| Hollow cone (height $h$) | $\frac{h}{3}$ from base |
| Solid cone (height $h$) | $\frac{h}{4}$ from base |
Motion of Centre of Mass
Velocity of Centre of Mass
$$\boxed{\vec{v}_{cm} = \frac{\sum m_i \vec{v}_i}{M} = \frac{\vec{p}_{total}}{M}}$$Key insight: The velocity of COM is the total momentum divided by total mass.
Acceleration of Centre of Mass
$$\boxed{\vec{a}_{cm} = \frac{\sum m_i \vec{a}_i}{M} = \frac{\vec{F}_{ext}}{M}}$$Newton’s Second Law for a System:
$$\vec{F}_{ext} = M\vec{a}_{cm}$$Internal forces do NOT affect the motion of the centre of mass.
Only external forces can change the velocity of COM. This is why:
- A person standing on a frictionless floor cannot move by pushing themselves
- A bomb exploding in mid-air: COM continues on the same parabolic path
- A cat falling: cannot change COM trajectory by twisting in air
When to Use Centre of Mass
Use COM analysis when:
- Multiple objects are interacting (collisions, explosions)
- System of particles with internal forces
- Finding trajectory of complex objects (javelin throw, rotating stick)
- No external force → COM moves with constant velocity
- Need to simplify complex motion into translation of one point
Don’t use COM for:
- Rotational motion about the COM (use torque/angular momentum)
- Individual particle motion when forces are different on each
Common Mistakes to Avoid
Mistake: Thinking COM and centre of gravity are always the same.
Truth: They’re the same only when gravitational field is uniform. In non-uniform fields (like near a black hole), they differ.
For JEE: Treat them as the same unless specifically mentioned.
Mistake: Thinking COM must be inside the material of the object.
Truth: COM can be outside the body!
Examples:
- Hollow hemisphere: COM is in the hollow region
- Ring: COM is at the center (no material there)
- Boomerang/horseshoe: COM is in the gap
Mistake: Thinking a person can move their COM on frictionless surface by moving their arms.
Correct approach: Only external forces change COM motion. Internal forces (muscles) create equal and opposite reactions that cancel out.
Worked Examples
Level 1: Foundation (NCERT)
Two point masses 3 kg and 2 kg are placed at coordinates (1, 2) m and (3, 4) m respectively. Find the position of their centre of mass.
Solution:
$$x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{3(1) + 2(3)}{3 + 2} = \frac{3 + 6}{5} = \boxed{1.8 \text{ m}}$$ $$y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} = \frac{3(2) + 2(4)}{3 + 2} = \frac{6 + 8}{5} = \boxed{2.8 \text{ m}}$$Centre of mass: $(1.8, 2.8)$ m
A 2 kg mass moves with velocity 3 m/s and a 3 kg mass moves with velocity 2 m/s in the same direction. Find the velocity of their centre of mass.
Solution:
$$v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{2(3) + 3(2)}{2 + 3} = \frac{6 + 6}{5} = \boxed{2.4 \text{ m/s}}$$Level 2: JEE Main
Two uniform rods each of mass $m$ and length $L$ are joined to form an L-shape. Find the position of the centre of mass from the corner.
Solution:
Consider corner as origin. Each rod’s COM is at its center.
Horizontal rod: COM at $(L/2, 0)$ Vertical rod: COM at $(0, L/2)$
$$x_{cm} = \frac{m(L/2) + m(0)}{m + m} = \frac{L}{4}$$ $$y_{cm} = \frac{m(0) + m(L/2)}{m + m} = \frac{L}{4}$$COM position: $\boxed{\left(\frac{L}{4}, \frac{L}{4}\right)}$ from corner
Distance from corner: $\sqrt{(L/4)^2 + (L/4)^2} = \boxed{\frac{L}{2\sqrt{2}}}$
A projectile is fired with velocity 20 m/s at 60° to horizontal. At the highest point, it explodes into two equal fragments. One fragment returns to the launch point. Where does the other fragment land? (Take $g = 10$ m/s²)
Solution:
Key concept: COM continues on the original parabolic path.
Original range: $R = \frac{u^2 \sin 2\theta}{g} = \frac{20^2 \sin 120°}{10} = \frac{400 \times \sqrt{3}/2}{10} = 20\sqrt{3}$ m
At highest point, horizontal velocity = $u \cos\theta = 20 \cos 60° = 10$ m/s
After explosion:
- Fragment 1 returns to launch point (lands at $x = 0$)
- Fragment 2 lands at $x = x_2$
- COM must land at original range: $x_{cm} = 20\sqrt{3}$ m
The second fragment lands at twice the original range!
Level 3: JEE Advanced
A uniform chain of length $L$ and mass $M$ is placed on a smooth table with length $b$ hanging over the edge. The chain starts sliding. Find the velocity of the chain when it just leaves the table.
Solution:
Initial position: Length $b$ hanging, $(L-b)$ on table
Initial COM height from ground:
$$h_i = \frac{(L-b) \cdot h_{\text{table}} + b \cdot (h_{\text{table}} - b/2)}{L}$$Taking table height as reference (0):
$$h_i = \frac{0 + b(-b/2)}{L} = -\frac{b^2}{2L}$$Final position: Just leaving table - entire length $L$ hanging
Final COM height:
$$h_f = -\frac{L}{2}$$Using energy conservation:
$$Mg(h_i - h_f) = \frac{1}{2}Mv^2$$ $$g\left(-\frac{b^2}{2L} + \frac{L}{2}\right) = \frac{v^2}{2}$$ $$g\left(\frac{L}{2} - \frac{b^2}{2L}\right) = \frac{v^2}{2}$$ $$g\left(\frac{L^2 - b^2}{2L}\right) = \frac{v^2}{2}$$ $$\boxed{v = \sqrt{\frac{g(L^2 - b^2)}{L}}}$$Two particles of masses $m$ and $2m$ are initially at rest at distance $d$ apart. They move towards each other under mutual gravitational attraction. Where do they collide?
Solution:
Key concept: No external force → COM doesn’t move!
Initial COM position (taking $m$ at origin):
$$x_{cm} = \frac{m(0) + 2m(d)}{m + 2m} = \frac{2d}{3}$$Since COM doesn’t move, collision occurs at $\boxed{x = \frac{2d}{3}}$ from mass $m$.
Alternative insight: Distance traveled is inversely proportional to mass.
- Mass $m$ travels: $\frac{2d}{3}$
- Mass $2m$ travels: $\frac{d}{3}$
- Ratio: $2:1$ (inverse of mass ratio)
Real-World Applications
Rocket Propulsion: Fuel ejection doesn’t change rocket’s COM momentum in absence of external forces
High Jump: Athletes arch their backs - body passes over bar but COM can pass under it!
Stability: Lower COM → more stable (race cars, SUVs)
Spacecraft Orientation: Astronauts use conservation of angular momentum about COM
Ballet Pirouettes: Dancer spins around COM by redistributing mass
Quick Revision Box
| Situation | Formula/Approach | Key Point |
|---|---|---|
| Two particles | $x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$ | Closer to heavier mass |
| Symmetric body | Use symmetry | COM at geometric center |
| COM velocity | $\vec{v}_{cm} = \frac{\vec{p}_{total}}{M}$ | Total momentum / total mass |
| COM acceleration | $\vec{a}_{cm} = \frac{\vec{F}_{ext}}{M}$ | Only external forces matter |
| Explosion | COM continues original path | Internal forces cancel |
| No external force | $\vec{v}_{cm}$ = constant | Momentum conservation |
Practice Problems
Foundation Level
Three particles of masses 1 kg, 2 kg, and 3 kg are at positions (1, 2), (3, 1), and (2, 4) meters. Find the COM.
A uniform rod of length 2 m is cut into two pieces of lengths 0.5 m and 1.5 m. Find the shift in COM.
Find the COM of a uniform semicircular disc of radius 10 cm from its center.
JEE Main Level
A man of mass 60 kg stands on one end of a 140 kg boat of length 4 m. He walks to the other end. How far does the boat move? (Water is frictionless)
A shell explodes at the highest point of its trajectory into two equal fragments. One fragment has velocity 3 times the velocity at highest point in the forward direction. Find the velocity of the other fragment.
Four particles each of mass $m$ are placed at the corners of a square of side $a$. Find the COM when one corner mass is removed.
JEE Advanced Level
A uniform chain of mass $M$ and length $L$ is held vertically with its lower end just touching a horizontal table. It is released. What is the force on the table when a length $x$ has fallen?
Two particles of masses $m_1$ and $m_2$ start moving towards each other from rest due to mutual attraction. Find the position where they collide if initially separated by distance $d$.
A disc of radius $R$ has a circular hole of radius $R/2$ cut from it. The center of hole is at distance $R/2$ from center of disc. Find the COM of the remaining portion.
Connection to Other Topics
Prerequisites you should know:
- Vectors and Components - For 2D/3D COM calculations
- Newton’s Laws - For understanding forces on systems
What’s next:
- Moment of Inertia - Rotational analog of mass
- Linear Momentum - COM velocity relates to total momentum
- Collisions - COM analysis simplifies collision problems
Cross-chapter links:
- Projectile Motion - COM follows parabolic path
- Conservation of Energy - COM motion and energy
Teacher’s Summary
Centre of Mass is the weighted average position of all mass in a system - like averaging test scores!
Only external forces change COM motion - internal forces (explosions, muscle movements) don’t affect it
For symmetric bodies, use symmetry to instantly locate COM - saves 2-3 minutes in exam!
In JEE, look for:
- Explosion problems → COM continues original path
- Frictionless surface → COM can’t move if no external force
- Man-boat problems → use COM conservation
Memory aid: “COM behaves as if all mass is concentrated there” - makes complex systems simple!
“The centre of mass is nature’s way of simplifying chaos into elegant simplicity.”
Master this, and you’ve unlocked 3-4 guaranteed marks in JEE. More importantly, you’ll see why Thor’s hammer spins so beautifully - physics, not magic!