Centre of mass for JEE Main

The centre of mass is the point that moves as if all the mass of a system were concentrated there and all the external forces acted on it. That one idea turns messy problems into easy ones. A shell explodes in mid-air, a person walks along a boat, two skaters pull on a rope: in each case the centre of mass tells you the answer before you touch the details.

In the syllabus it sits at the start of the rotational motion unit, as the centre of mass of a two-particle system and of a rigid body. It is also the quickest route through many momentum questions.

Systems of particles

For particles of masses mᵢ at positions rᵢ:

r_cm = Σmᵢrᵢ / Σmᵢ, applied separately to x, y and z.

For two particles the centre of mass lies on the line joining them, closer to the heavier one. The distances from the two masses are in the inverse ratio of the masses: r₁/r₂ = m₂/m₁.

Continuous bodies

Replace the sum with an integral: x_cm = ∫x dm / ∫dm. Choose a small element, write dm in terms of a length, area or volume, and integrate.

Derivation 1: a rod with non-uniform density. A rod of length L has linear density λ = kx, measured from one end.
x_cm = ∫kx² dx / ∫kx dx from 0 to L = (L³/3)/(L²/2) = 2L/3. The centre of mass moves towards the denser end, as it should.

Derivation 2: a semicircular ring. Take a ring of mass M and radius R with its diameter along the x-axis. An element at angle θ has dm = (M/π) dθ and height y = R sin θ.
y_cm = (1/M) ∫ R sin θ × (M/π) dθ from 0 to π = (R/π) × [−cos θ] from 0 to π = 2R/π.

Standard results

Notice the pattern. Hollow objects keep their mass further out, so their centre of mass is further from the flat face or the base than for the solid version.

Bodies with cavities

Treat the removed part as a negative mass placed where the hole is. The full body (with its centre of mass known) plus the negative piece gives the remaining body.

For uniform plates, mass is proportional to area; for uniform solids, to volume. So a disc of radius R/2 has one-quarter the mass of a disc of radius R, and a sphere of radius R/2 has one-eighth the mass of a sphere of radius R.

Motion of the centre of mass

Differentiate the definition:

• v_cm = Σmᵢvᵢ / M, so total momentum = M × v_cm.
• M a_cm = F_external. Internal forces, however large, cannot change the motion of the centre of mass.

Three consequences you will use often:

• If the net external force is zero and the system starts at rest, the centre of mass stays where it is. This solves person-on-boat and particles-attracting-each-other problems.
• If a projectile explodes in mid-air, the centre of mass of the fragments continues along the original parabola until a fragment hits the ground.
• The acceleration of the centre of mass depends only on the total external force, not on which body it is applied to.

Worked problems

Problem 1: an L-shaped frame. Two identical uniform rods, each of length L, are joined at one end to form an L. Find the centre of mass from the corner.

Put the corner at the origin, one rod along x and one along y. Each rod's centre is at its midpoint: (L/2, 0) and (0, L/2). With equal masses, the centre of mass is at (L/4, L/4), a distance L/(2√2) from the corner along the bisector.

Problem 2 (numerical answer): person on a boat. A 60 kg person stands at one end of a 140 kg boat that is 4 m long, at rest on still water. The person walks to the other end. How far, in metres, does the boat move? Ignore water resistance.

Let the boat move d backwards. The person then moves (4 − d) forwards relative to the shore.
Centre of mass fixed: 60(4 − d) = 140d, so 240 = 200d and d = 1.2 m.
Check: the person moves 2.8 m; 60 × 2.8 = 168 = 140 × 1.2.

Problem 3: exploding projectile. A projectile would land 120 m from the launch point. At the highest point it explodes into two equal pieces. One piece falls vertically, starting from rest. Where does the other land?

At the top both pieces have zero vertical velocity, so they reach the ground together. Their centre of mass lands where the projectile would have: at 120 m. The first piece lands at 60 m (below the highest point).
(60 + x)/2 = 120, so x = 180 m from the launch point.

Problem 4: a plate with a hole. From a uniform square plate of side 2a, a square of side a is cut from one corner. Find the shift of the centre of mass.

Take the origin at the centre of the full plate. Masses are in proportion to area: full plate 4m, removed piece m, centred at (a/2, a/2).
x_cm = (4m × 0 − m × a/2)/(4m − m) = −a/6, and similarly y_cm = −a/6.
The centre of mass shifts a√2/6 along the diagonal, away from the hole.

Practice set

1. Find the centre of mass of 1 kg at (0, 0), 1 kg at (2, 0) and 2 kg at (0, 2), in metres.
2. Particles of 2 kg and 3 kg are 1 m apart. How far is the centre of mass from the 2 kg particle?
3. A rod of length L has linear density proportional to the distance from one end. Where is its centre of mass?
4. A uniform solid sphere of radius R has a spherical cavity of radius R/2 whose centre is R/2 from the sphere's centre. Find the centre of mass of the remaining body.
5. A 2 kg block moves east at 6 m/s and a 4 kg block moves west at 3 m/s. Find the velocity of their centre of mass.
6. Particles of 1 kg and 2 kg start at rest and attract each other, with no external force. When the 1 kg particle has moved 2 m, how far has the 2 kg particle moved?
7. Blocks of 1 kg and 3 kg rest on a smooth floor. A force of 8 N acts on the 1 kg block only. Find the acceleration of the centre of mass.
8. Find the distance of the centre of mass of a thin semicircular ring of radius 11 cm from its centre (take π = 22/7).

Answers

1. x = (0 + 2 + 0)/4 = 0.5; y = (0 + 0 + 4)/4 = 1, so (0.5 m, 1 m).
2. From the 2 kg particle: 3 × 1/5 = 0.6 m.
3. 2L/3 from the lighter end, as derived above.
4. Masses 8 : 1 by volume. x = −(1 × R/2)/(8 − 1) = R/14 from the centre, on the side away from the cavity.
5. (12 − 12)/6 = 0. The centre of mass is at rest.
6. The centre of mass stays fixed: 1 × 2 = 2 × d, so 1 m towards the other particle.
7. a_cm = 8/(1 + 3) = 2 m/s², whichever block the force acts on.
8. 2R/π = 22 × 7/22 = 7 cm.

What to do next

• Derive the semicircular ring and non-uniform rod results by integration once.
• Learn the standard results table, including which face each distance is measured from.
• Solve 15 cavity problems and 10 "centre of mass stays fixed" problems.
• Revise momentum and collisions, then move on to rotational dynamics.