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Centre of Mass Formulas for JEE 2027
Centre of Mass formulas for JEE 2027 help you determine the position, velocity and acceleration of the centre of mass of a system. This formula sheet covers particle systems, continuous mass distributions, rigid bodies, composite objects, momentum and collisions. It brings together important formulas, their conditions and worked examples for JEE Main and Advanced preparation.
Use this sheet while solving JEE Centre of Mass Questions, practising JEE chapter-wise PYQs and reviewing mock test. Focus on choosing a suitable origin, using symmetry and distinguishing internal forces from external forces. Revisit the quick revision section after practice to strengthen your understanding and correct mistakes involving units, coordinates and directions.
Notation
| Symbol | Stands for | Units |
|---|---|---|
| $$m_i$$ | Mass of the i-th particle | kg |
| $$M$$ | Total mass of the system | kg |
| $$\vec r_i,\,x_i,y_i,z_i$$ | Position vector / Cartesian coordinates of $$m_i$$ from chosen origin | m |
| $$\vec R,\,X,Y,Z$$ | COM position vector / coordinates | m |
| $$\rho(x,y,z)$$ | Volume mass density | kg m-3 |
| $$\sigma(x,y)$$ | Surface mass density | kg m-2 |
| $$\lambda(x)$$ | Linear mass density | kg m-1 |
| $$V,A,L$$ | Volume, area, length of elemental piece | m3, m2, m |
| $$\vec v_i,\vec V$$ | Particle and COM velocities | m s-1 |
| $$\vec a_i,\vec A$$ | Particle and COM accelerations | m s-2 |
| $$\vec p_i, \vec P$$ | Individual and total linear momentum | N s |
| $$\vec F_{\text{ext}}$$ | Net external force on system | N |
| $$\vec J$$ | Impulse on system | N s |
| $$t$$ | Time | s |
| $$\dot m$$ | Rate of mass change (dm/dt) | kg s-1 |
| $$\vec u_{\text{rel}}$$ | Velocity of ejected/added mass relative to COM | m s-1 |
| $$\mu$$ | Reduced mass $$m_1m_2/(m_1+m_2)$$ | kg |
| $$E_{\text{CM}},E_{\text{lab}}$$ | Kinetic energy in COM frame / Lab frame | J |
Discrete Particle System Formulas
This is the foundation: a handful of point masses at known coordinates. Nearly every JEE problem reduces the given shape or body to such a system when symmetry allows. Master this before touching integration.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$M=\sum_i m_i$$ | Total mass of system | Always |
| $$\vec R=\dfrac{\sum_i m_i\vec r_i}{\sum_i m_i}$$ | Vector position of COM | Origin fixed in inertial frame |
| $$X=\dfrac{\sum_i m_ix_i}{M}$$, $$Y$$, $$Z$$ similar | Component form – faster in cartesian questions | — |
| $$\vec P=\sum_i m_i\vec v_i = M\vec V$$ | Total momentum equals mass times COM velocity | — |
| $$\vec A=\dfrac{\vec F_{\text{ext}}}{M}$$ | Acceleration of COM | No hidden mass loss |
Worked Example
Three masses $$2\,\text{kg},\;3\,\text{kg},\;5\,\text{kg}$$ occupy $$(1,0,0)\,\text{m},\;(0,2,0)\,\text{m},\;(0,0,3)\,\text{m}$$ respectively. Find the COM.
- Total mass $$M=2+3+5=10\,\text{kg}$$.
- Compute $$X=\dfrac{2(1)+3(0)+5(0)}{10}=0.2\,\text{m}$$.
- Compute $$Y=\dfrac{2(0)+3(2)+5(0)}{10}=0.6\,\text{m}$$.
- Compute $$Z=\dfrac{2(0)+3(0)+5(3)}{10}=1.5\,\text{m}$$.
Answer: $$\vec R=(0.2\,\hat i+0.6\,\hat j+1.5\,\hat k)\,\text{m}$$.
Shortcuts and Special Cases
- Two-mass line: $$X=\dfrac{m_2d}{m_1+m_2}$$ from $$m_1$$ toward $$m_2$$ (takes 2 s to recall).
- Equal masses: COM is simple arithmetic mean of coordinates.
- Symmetric placement about origin: numerator zero ⇒ COM at origin; saves total two-thirds of calculation.
Watch Out
- Forgetting to divide by total mass when shifting from vector to component form.
- Using internal forces in $$\vec A=\vec F/M$$; only external forces count.
- Mixing cm with m in coordinates: unit inconsistency kills answer.
- Writing $$\sum m_i r_i$$ as $$\sum m_ir_i^2$$ by rote error—dimension check avoids it.
Continuous Mass Distribution Formulas
When the body cannot be split into a few lumps, you integrate densities. Many Advanced problems pick a rod, plate or lamina with varying density to test set-up skill. You now extend the discrete sum to an integral.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$M=\int dm$$ | Total mass by integration | Density defined |
| $$dm=\rho\,dV$$ | Elemental mass (3-D) | Uniform density within element |
| $$dm=\sigma\,dA$$ | Elemental mass (2-D lamina) | Thin plate |
| $$dm=\lambda\,dL$$ | Elemental mass (1-D rod/wire) | Thin wire |
| $$\vec R=\dfrac{1}{M}\int \vec r\,dm$$ | Vector COM for continuous body | — |
| $$X=\dfrac{1}{M}\int x\,dm$$ and likewise for $$Y,Z$$ | Component form | — |
Worked Example
A uniform semicircular wire of radius $$R$$ lies in the $$xy$$-plane, centre at origin, spanning $$\theta=0$$ to $$\pi$$. Find the x-coordinate of its COM.
- Linear density $$\lambda=M/(\pi R)$$ (not needed explicitly as it cancels).
- Element $$dm=\lambda R\,d\theta$$ with $$x=R\cos\theta$$.
- $$X=\dfrac{\int_{0}^{\pi} x\,dm}{M}=\dfrac{\lambda R\int_{0}^{\pi}R\cos\theta\,d\theta}{\lambda\pi R}=\dfrac{R^2[ \sin\theta ]_{0}^{\pi}}{\pi R}=0$$.
Answer: COM lies on the y-axis, $$X=0$$.
Shortcuts and Special Cases
- If density is uniform and geometry symmetric about an axis, COM lies on that axis – no integral required.
- For a uniform thin semicircular arc, $$Y=\dfrac{2R}{\pi}$$ – worth committing to memory.
- For a right triangle lamina, COM from base and side is at one-third distances – classical result used in 4-mark tasks.
Watch Out
- Forgetting the Jacobian: polar $$dA=r\,dr\,d\theta$$, cylindrical $$dV=r\,dr\,d\theta\,dz$$.
- Pasting volume densities into area integrals; keep eye on units.
- Missing proportional cancelation of uniform density – write it then strike it, don’t skip mentally.
- Setting wrong limits when angle starts at $$-\pi/2$$ or $$\theta=-\pi$$.
COM of Common Rigid Bodies & Composite Objects Formulas
Most paper-setters avoid raw integrals and expect you to quote ready-made results for spheres, cylinders, cones and use subtraction/addition of masses. This sub-topic bridges continuous formulae to plug-and-play answers.
Formulas
| Formula / Result | What it gives you | Valid when |
|---|---|---|
| Uniform cone: $$Z=\dfrac{h}{4}$$ from base | Axial COM position | Right circular cone |
| Uniform hemisphere (solid): $$Z=\dfrac{3R}{8}$$ from flat face | Axial COM | Solid, not shell |
| Uniform thin semicircular arc: $$Y=\dfrac{2R}{\pi}$$ | Perpendicular distance from diameter | Wire |
| Uniform quarter-circle plate: $$X=Y=\dfrac{4R}{3\pi}$$ | Coordinates from right-angle corner | Thin lamina |
| Composite: $$\vec R=\dfrac{\sum \vec R_j M_j}{\sum M_j}$$ | Combine known sub-COMs | Sub-bodies non-overlapping |
Worked Example
A solid hemisphere of radius $$R$$ is glued to the base of a solid cylinder of height $$h$$ and same radius. Find the COM measured from the flat end of the cylinder.
- Masses: $$M_1=\rho \pi R^2 h$$ (cylinder), $$M_2=\dfrac{2}{3}\rho \pi R^3$$ (hemisphere).
- COM positions: $$Z_1=\dfrac{h}{2}$$ (cylinder), $$Z_2= -\dfrac{3R}{8}$$ (hemisphere, measured downward from join).
- Using composite formula $$Z=\dfrac{M_1Z_1+M_2Z_2}{M_1+M_2}$$.
- Substitute and simplify: $$Z=\dfrac{\pi R^2 \rho [h(h/2)]+(2/3)\pi R^3\rho(-3R/8)}{\pi R^2\rho[h+2R/3]}=\dfrac{h^2/2 - R^2/4}{h+2R/3}$$.
Answer: $$Z=\dfrac{h^2/2 - R^2/4}{h+2R/3}$$ above the join (positive into the cylinder).
Shortcuts and Special Cases
| Body | Location to quote | Exam use |
|---|---|---|
| Uniform rod | Mid-point | Appears in 1-mark Mains |
| Ring / thin circular disc | Geometric centre | Used in rolling questions |
| Right triangle lamina | Intersection of medians (1/3 from each side) | Subtracted area method |
| Hollow hemisphere | $$Z=\dfrac{R}{2}$$ from the centre of flat face | Often paired with solid counterpart |
Watch Out
- Quoting solid cone $$h/4$$ result for hollow cone – wrong; hollow is $$h/3$$.
- Mixing sign convention while attaching bodies above/below reference.
- For composite method, forget to subtract mass when cavity removed.
- Writing hemisphere mass as $$\tfrac{2}{3}\pi R^3\rho$$ but forgetting to multiply by density when numbers provided separately.
Velocity & Acceleration of COM; System Dynamics Formulas
This piece connects kinematics to dynamics. JEE often hides a variable mass or explosion somewhere; knowing how velocities add and momenta cancel yields the answer in two lines.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$\vec V=\dfrac{\sum m_i\vec v_i}{M}$$ | COM velocity | Discrete system |
| $$\vec P=M\vec V$$ | Expresses total momentum through COM motion | — |
| $$\vec A=\dfrac{\vec F_{\text{ext}}}{M}$$ | COM acceleration equals external force per unit mass | No mass loss/gain |
| Impulse: $$\vec J=\int \vec F_{\text{ext}}dt=M\Delta\vec V$$ | Velocity jump after finite force | Inertial frame |
| Explosion (no ext force): $$M\vec V_{\text{before}}= \sum m_i\vec v_{i,\text{after}}$$ | Momentum conservation centred on COM | Impulse internal only |
Worked Example
A 4 kg projectile moving at $$300\,\text{m s}^{-1}$$ explodes into two pieces of masses 1 kg and 3 kg. The lighter piece moves at $$500\,\text{m s}^{-1}$$ perpendicular to the original direction. Find the speed of the heavier piece immediately after explosion.
- Initial momentum $$\vec P_i=4\times300=1200\,\hat i$$ N s.
- After explosion: $$\vec P_f=1(500\,\hat j)+3\vec v_h$$.
- Set $$\vec P_i=\vec P_f$$ (no external force). Hence $$1200\,\hat i =500\,\hat j+3\vec v_h$$.
- Components: $$3v_{hx}=1200\Rightarrow v_{hx}=400$$, $$3v_{hy}=-500\Rightarrow v_{hy}=-166.7$$.
- Speed $$v_h=\sqrt{400^2+166.7^2}\approx 433\,\text{m s}^{-1}$$.
Answer: $$v_h\approx 4.3\times10^{2}\,\text{m s}^{-1}$$.
Shortcuts and Special Cases
- If no external force, COM continues on same trajectory regardless of internal explosions – time saver for trajectory tracing problems.
- Equal masses after explosion form right triangle momenta ⇒ speeds combine vectorially like sides.
- For symmetrical breakup, heavier half almost keeps original velocity – a quick sanity check.
Watch Out
- Adding vectorially: never equate magnitudes instead of components.
- Using internal forces in $$\vec F_{\text{ext}}$$ – they cancel.
- Missing factor of 2 when impulse given as average force × time.
- Treating gravitational force as negligible for long explosions – only instant impulses allow that.
External Forces, Impulse and Work in COM Frame Formulas
After you know positions and basic velocities, this layer tells what forces do to the system energy wise. JEE Advanced blends it with block-pulley and spring sets.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Work-energy: $$E_{\text{lab}}=E_{\text{CM}}+\dfrac{P^2}{2M}$$ | Kinetic energy split | Any isolated system |
| Power of external forces $$=\vec F_{\text{ext}}\cdot\vec V$$ | Rate of change of COM kinetic energy | — |
| Angular momentum about COM $$\vec L_{\text{int}}=\sum (\vec r_i-\vec R)\times m_i\vec v_i^{\prime}$$ | Internal angular momentum | Rigid or deformable |
| If $$\vec F_{\text{ext}}=0$$ ⇒ $$\vec V=\text{constant},\,E_{\text{lab}}=\text{const}$$ | Conservation statements | — |
Worked Example
Two skaters, 50 kg and 70 kg, stand at rest on frictionless ice 10 m apart. They pull on a rope; when they meet, how far has the heavier skater moved?
- No external horizontal force ⇒ COM position unchanged.
- Initial COM measured from lighter skater: $$X_0=\dfrac{70\times10}{120}=5.83\,\text{m}$$.
- When they meet, both occupy same spot at some $$X_f=5.83\,\text{m}$$ from original light skater’s position.
- Therefore lighter skater travels $$5.83\,\text{m}$$, heavier skater $$10-5.83=4.17\,\text{m}$$.
Answer: Heavier skater moves $$4.2\,\text{m}$$ toward the lighter one.
Shortcuts and Special Cases
- In rope-pull problems, distance moved ratio equals inverse mass ratio.
- If floor suddenly removed under two stacked masses, vertical COM falls with $$g$$ – quick drop-time value.
- Kinetic split: internal vs COM motion question? Compute $$E_{\text{CM}}$$ using reduced masses to save lines.
Watch Out
- Choosing wrong origin when declaring COM fixed; origin must be inertial.
- Assuming work by internal forces zero – true only for rigid bodies with fixed separations.
- Forgetting vertical forces are external in horizontal COM problems.
- Mixing energy frames: always subtract translational energy first.
Variable Mass Systems & Rocket Equation Formulas
Rockets, leaky sand carts and rain on trucks all sit here. Mains usually gives a one-step momentum balance; Advanced may demand differential form.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| General: $$\dfrac{d}{dt}(M\vec V)=\vec F_{\text{ext}}+\vec u_{\text{rel}}\dot m$$ | Momentum balance for variable mass | Mass leaves/enters at rate $$\dot m$$ |
| Tsiolkovsky: $$\Delta V=v_e\ln\dfrac{M_0}{M}$$ | Rocket speed gain in space | $$\vec F_{\text{ext}}=0$$, constant exhaust speed $$v_e$$ |
| Mass accretion (rain): $$M\vec A=\vec F_{\text{ext}}- \vec u_{\text{rel}}\dot m$$ | Acceleration of cart gaining mass | Sign of $$\dot m$$ positive for gain |
| Ejection from rest: $$M\vec V= -\vec u_{\text{rel}}\Delta m$$ | Instant recoil velocity | Neglect external forces in instant |
Worked Example
A 1000 kg rocket in deep space ejects fuel at 20 kg s-1 with exhaust speed $$v_e=500\,\text{m s}^{-1}$$ relative to the rocket. Find its acceleration when its mass is 800 kg.
- Use $$A=\dfrac{u_{\text{rel}}\dot m}{M}$$; here $$u_{\text{rel}}=v_e=500$$ (opposite to motion), $$\dot m=-20$$ kg s-1 (mass decreases).
- Magnitude $$A=\dfrac{500(20)}{800}=12.5\,\text{m s}^{-2}$$.
Answer: Acceleration $$12.5\,\text{m s}^{-2}$$ opposite to exhaust jet.
Shortcuts and Special Cases
- If exhaust velocity given relative to rocket, sign is opposite direction of thrust by default.
- For sand leaking vertically down at same speed as cart, no horizontal momentum change – acceleration zero.
- When $$\vec F_{\text{ext}}=Mg$$ in vertical launch, replace by $$-Mg$$ in equation; quickly produce net thrust.
Watch Out
- Confusing $$\dot m$$ sign: take positive when mass is entering.
- Plugging weight into deep-space rocket; $$g$$ only on Earth.
- Using Newton’s second law $$F=ma$$ blindly; variable mass needs the generalized form.
- Applying Tsiolkovsky with exhaust speed varying – restrict to constant $$v_e$$ questions.
Collisions in the COM Frame Formulas
Elastic and inelastic collision problems simplify dramatically in the COM frame. JEE Advanced expects you to switch frames without fuss and back-transform the answers.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| COM velocity: $$\vec V=\dfrac{m_1\vec u_1+m_2\vec u_2}{m_1+m_2}$$ | Frame shift speed | Binary collision |
| Velocities in COM: $$\vec u_1^{\prime}=\vec u_1-\vec V,\;\vec u_2^{\prime}=\vec u_2-\vec V$$ | Pre-collision velocities viewed in COM | — |
| For perfectly elastic: $$\vec v_1^{\prime}=-\vec u_1^{\prime},\;\vec v_2^{\prime}=-\vec u_2^{\prime}$$ | Speeds unchanged, directions reversed | 1-D or general |
| Coefficient of restitution $$e=-\dfrac{v_2^{\prime}-v_1^{\prime}}{u_2^{\prime}-u_1^{\prime}}$$ | Same in any inertial frame | 1-D |
| Lab speeds after 1-D elastic: $$v_1=\dfrac{m_1-m_2}{m_1+m_2}u_1+ \dfrac{2m_2}{m_1+m_2}u_2$$ | Direct result without frame shift | Second mass initially at $$u_2$$ |
Worked Example
A 2 kg mass moving at 10 m s-1 collides elastically with a 3 kg mass at rest. Find final speeds.
- COM velocity $$V=\dfrac{2(10)+3(0)}{5}=4\,\text{m s}^{-1}$$.
- COM frame speeds: $$u_1^{\prime}=10-4=6$$, $$u_2^{\prime}=0-4=-4$$.
- Post-collision: reverse sign ⇒ $$v_1^{\prime}=-6$$, $$v_2^{\prime}=4$$.
- Transform back: $$v_1=v_1^{\prime}+4=-2$$ m s-1, $$v_2=4+4=8$$ m s-1.
Answer: First mass rebounds at 2 m s-1 opposite, second moves forward at 8 m s-1.
Shortcuts and Special Cases
- In perfectly inelastic collision, final speed in lab equals COM speed directly.
- Equal masses swap velocities in elastic 1-D – know by heart.
- If one mass >> other, heavy one hardly changes speed; treat it as wall for estimation.
Watch Out
- Mixing signs when back-transforming to lab frame.
- Using 1-D formulas in oblique collision questions – switch to vectors.
- Assuming kinetic energy conserved when restitution given <1.
- Forgetting the COM speed is constant across collision even with external forces if duration negligible.
Two-Body Problem and Reduced Mass Formulas
This final block underpins gravitational or electrostatic orbit questions but lands inside COM chapter whenever examiner asks for energy in mutual frame. It is shorter yet powerful.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Reduced mass $$\mu=\dfrac{m_1m_2}{m_1+m_2}$$ | Effective single-body mass | Central forces only |
| Relative coordinate $$\vec r=\vec r_1-\vec r_2$$ | Separation vector | — |
| Kinetic energy $$=\dfrac{1}{2}\mu \dot r^2+\dfrac{L^2}{2\mu r^2}$$ | In COM frame with angular momentum $$L$$ | Cylindrical coords |
| Angular momentum $$L=\mu r^2\dot \theta$$ | Conserved under central force | — |
| Total energy lab $$=E_{\text{CM}}+\dfrac{1}{2}MV^2$$ | Splits like earlier but $$E_{\text{CM}}$$ uses $$\mu$$ | — |
Worked Example
Two satellites, 800 kg and 1200 kg, orbit their mutual COM under negligible Earth influence at 2 km separation with relative speed 1 m s-1. Find kinetic energy of the system in the lab frame where COM at rest.
- Reduced mass $$\mu=\dfrac{800\times1200}{2000}=480\,\text{kg}$$.
- Kinetic energy in COM frame $$E_{\text{CM}}=\tfrac12\mu v_{\text{rel}}^2=\tfrac12(480)(1)^2=240\,\text{J}$$.
- COM at rest ⇒ lab energy same.
Answer: System kinetic energy $$240\,\text{J}$$.
Shortcuts and Special Cases
- If COM at rest, drop the $$\tfrac12MV^2$$ term entirely.
- In gravitational two-body with one mass >> other (planet-satellite), $$\mu\approx m_{\text{sat}}$$ – so usual single-body equations suffice.
- Binding energy uses $$\mu$$ in place of smaller mass – save recalc later.
Watch Out
- Forgetting to convert separation km to m when computing energies.
- Plugging total mass instead of reduced mass in kinetic term.
- Assuming COM at rest when external forces present – verify first.
- Mixing this $$\mu$$ with friction coefficient symbol in other chapters.
Commonly Confused Centre of Mass Formulas for JEE
| This formula | Confused with | How to choose correctly |
|---|---|---|
| $$\vec R=\dfrac{\sum m_i\vec r_i}{M}$$ | Electric dipole moment $$\vec p=\sum q_i\vec r_i$$ | Check numerator: mass vs charge; units kg m vs C m |
| $$\vec A=\dfrac{\vec F_{\text{ext}}}{M}$$ | Particle acceleration $$\vec a=\dfrac{\vec F}{m}$$ | System question? count many masses ⇒ use capital symbols |
| Rocket $$\dfrac{d}{dt}(MV)=F_{\text{ext}}+u\dot m$$ | Standard $$F=ma$$ | Mass changing? If yes, include second term |
| Reduced mass $$\mu$$ | Chemical μ (coefficient of viscosity) | Physics JEE uses kg; check context central forces |
| Elastic collision lab speed formula | Inelastic relative speed with restitution | Elastic only when KE conserved or e=1 stated |
Common Mistakes to Avoid When Using Centre of Mass Formulas for JEE
- Carrying internal forces into net external force calculation.
- Dropping the total mass denominator when fatigue hits during late-night JEE Study material binge.
- Integrating density over wrong limits, especially forgetting the extra $$r$$ in polar area elements.
- Using positive $$\dot m$$ for mass ejection in rocket equation – reverses thrust direction.
- Assuming COM fixed in Earth frame when external friction acts.
- Plugging numbers before cancelling density – floods calculator with decimals and hides algebraic insight.
- Missing vector nature of momentum: equating magnitudes instead of components in explosion/collision problems.
Quick Revision of Centre of Mass Formulas for JEE 2027
- $$\vec R=\dfrac{\sum m_i\vec r_i}{M}$$ – discrete COM, five-second recall.
- Uniform cone COM at $$h/4$$ above base – quoted directly in many PYQs.
- Acceleration of COM $$\vec A=\vec F_{\text{ext}}/M$$ – decide if motion follows projectile path.
- Tsiolkovsky rocket $$\Delta V=v_e\ln(M_0/M)$$ – appears in integer answer questions.
- Elastic 1-D equal masses swap speeds – kill time in Advanced Section 2.
- Semicircular arc COM distance $$2R/\pi$$ from centre – saves an integral mid-paper.
- Reduced mass $$\mu=m_1m_2/(m_1+m_2)$$ – central force shortcut.
- Impulse relation $$\vec J=M\Delta\vec V$$ – rope pulls and ice skaters.
Centre of Mass Formulas for JEE 2027: Conclusion
Centre of Mass formulas for JEE 2027 help you solve problems involving particle systems, continuous mass distributions, composite bodies and system motion. Understanding symmetry, choosing a clear reference point and identifying external forces are essential for applying these formulas correctly. Use the worked examples to connect centre of mass concepts with momentum, impulse and collisions.
Strengthen your preparation by solving JEE Centre of Mass Questions and JEE chapter-wise PYQs after each revision session. Keep this JEE Main formula sheet alongside your JEE study material, and use a JEE Main mock test to practise under time limits. Review mistakes involving coordinates, mass distribution and vector directions, then reattempt those questions to improve accuracy and confidence.
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