Rigid Bodies and Centre of Mass | JEE Notes

A cricket ball can travel toward a batter while spinning. A door can turn without its hinge moving. A rolling wheel does both. These are not three separate branches of mechanics: every planar motion of a rigid body can be described by the motion of its centre of mass and a rotation about that centre.

What you must already be able to do

You need Newton’s laws, free-body diagrams, work–energy, linear momentum, basic differentiation and integration, and the meaning of a vector cross product. If a derivative looks unfamiliar, read $ω = d θ / d t$ as “angular position’s instantaneous rate of change.” If an integral looks unfamiliar, read $\int r^{2} d m$ as “add the contribution $r^{2} Δ m$ of every tiny mass element.” The calculus is bookkeeping for a physical idea.

A rigid body is an idealisation in which the distance between every pair of material points stays constant. The object may translate and rotate, but does not stretch or bend. A particle model discards orientation; a rigid-body model retains it. Use the latter when the location of a force, the distribution of mass, or the orientation of the body affects the answer.

For most of this chapter, the body moves in a plane and its rotation axis is perpendicular to that plane. Choose $+ x$ right, $+ y$ up and $+ z$ out of the page. Counterclockwise rotation is positive: $𝛚 = ω \hat{𝐳}$ . An object rotating clockwise therefore has $ω < 0$ , even though its angular speed $| ω |$ is positive. State this convention once and keep it.

The two vector operations you actually need

A dot product selects the component along a direction: $𝐀 \cdot 𝐁 = A B \cos ϕ$ . It produces a scalar and appears in work. A cross product measures a directed perpendicular effect: $| 𝐀 \times 𝐁 | = A B \sin ϕ$ . Its direction follows the right-hand rule. In a plane,

(𝐫 \times 𝐅)_{z} = x F_{y} - y F_{x} .

This one expression resolves many sign disputes. A rightward force applied above the origin produces a negative, clockwise moment because $- y F_{x} < 0$ . A force pointing directly toward the origin gives zero moment because its line of action passes through the origin. A longer position vector is not automatically a longer lever arm: the lever arm is the perpendicular distance to the force’s line of action.

Why the centre of mass separates the motion

For masses $m_{i}$ at positions $𝐫_{i}$ , define $M = \sum_{i} m_{i}$ and

𝐫_{C} = \frac{\sum_{i} m_{i} 𝐫_{i}}{M}, 𝐫_{i} = 𝐫_{C} + 𝛒_{i}, \sum_{i} m_{i} 𝛒_{i} = 𝟎 .

Differentiating the final identity gives $\sum_{i} m_{i} 𝐯_{i / C} = 𝟎$ . Thus the total momentum is $𝐏 = M 𝐯_{C}$ : internal rotation contributes no net linear momentum. Summing Newton’s second law over all particles cancels the internal action–reaction forces, leaving

𝐅_{e x t} = M 𝐚_{C} .

The centre of mass consequently moves as a particle acted on by the resultant external force. But this equation does not determine how the body turns. A second piece of information—the external moment—is needed for orientation. A force can change both the centre-of-mass motion and the rotation, so “translation equation” and “rotation equation” must describe the same physical body.

For planar rigid motion, $𝐯_{i / C} = 𝛚 \times 𝛒_{i}$ . Expanding the kinetic energy gives

K = \frac{12}{\sum_{i}} m_{i} | 𝐯_{C} + 𝐯_{i / C} |^{2} = \frac{12}{M} v_{C}^{2} + {\frac{12}{I}}_{C} ω^{2}, I_{C} = \sum_{i} m_{i} ρ_{i, ⊥}^{2} .

The cross term vanishes because $\sum_{i} m_{i} 𝐯_{i / C} = 0$ . This is why translational and rotational kinetic energies add without double counting when the rotation term is taken about the centre of mass. Do not replace $I_{C}$ by an arbitrary moment of inertia while retaining $\frac{12}{M} v_{C}^{2}$ .

The centre of mass need not lie inside the material

A ring’s centre of mass sits in empty space. It is still a valid reference point. For a continuous distribution, replace the discrete sum by $𝐫_{C} = M^{- 1} \int 𝐫 d m$ . Symmetry can save this integration: the centre lies on every symmetry plane of a uniform body, and at the intersection when these planes determine a unique point. Symmetry of the outline alone is insufficient if density varies asymmetrically.

“Centre of mass” and “centre of gravity” coincide in a uniform gravitational field. Under that approximation, the distributed gravitational forces have the same resultant and moment as a single force $M 𝐠$ applied at C. This is why a rigid body’s gravitational potential energy is $M g y_{C}$ , even while the body changes orientation. In a significantly nonuniform field, that shortcut requires reconsideration.

Worked example 01 · A nonuniform two-mass rotor

Problem. Particles of mass $m$ and $3 m$ are fixed to the ends of a massless rod of length $L$ . Their coordinates along the rod are $0$ and $L$ . The assembly’s centre moves with speed $V$ , and it rotates in its plane with angular speed $ω$ . Find its centre of mass and kinetic energy.

Step 1: locate the centre. $x_{C} = (m \cdot 0 + 3 m L) / (4 m) = 3 L / 4$ . The distances from the centre are $3 L / 4$ and $L / 4$ , respectively. The centre lies closer to the heavier particle.

Step 2: calculate rotational inertia.

I_{C} = m {(\frac{3 L}{4})}^{2} + 3 m {(\frac{L}{4})}^{2} = \frac{34}{m} L^{2} .

Step 3: add the two energies.

K = 2 m V^{2} + \frac{38}{m} L^{2} ω^{2} .

Checks. Both terms have dimensions $k g m^{2} s^{- 2}$ . Setting $ω = 0$ gives the energy of a translating mass $4 m$ ; setting $V = 0$ gives pure rotation about the centre. The geometric midpoint would give the wrong rotational inertia.

Worked example 02 · Zero resultant force can still rotate a body

Problem. Two forces act on a plate: $+ F \hat{𝐱}$ at $(0, a)$ , and $- F \hat{𝐱}$ at $(0, - a)$ . Find the resultant force and moment. Does shifting the reference origin change the answer?

Step 1: sum forces. $𝐅_{n e t} = 𝟎$ , so the centre of mass has zero acceleration from this pair.

Step 2: sum moments. The upper force contributes $- a F$ , and the lower force also contributes $- a F$ . Therefore $τ_{z} = - 2 a F$ : a clockwise couple.

Step 3: shift the origin. If the new origin is displaced by $𝐛$ , then $𝛕^{'} = \sum (𝐫_{i} - 𝐛) \times 𝐅_{i} = 𝛕 - 𝐛 \times \sum 𝐅_{i} = 𝛕$ . A pure couple has the same moment about every origin.

Checks. The moment has units $N m$ ; it vanishes as the force separation $2 a$ vanishes. Zero resultant force does not imply zero angular acceleration. Also, zero acceleration of the centre does not imply zero centre velocity.

Beyond JEE: when scalar rotation is not enough

The scope boundary: scalar rotation is not all of three-dimensional mechanics

For fixed-axis rotation, the angular-momentum component along that axis is $L_{a x i s} = I_{a x i s} ω$ . For a planar lamina rotating normally to its plane through its centre, the full rotational angular momentum is parallel to $𝛚$ . For a general three-dimensional body about an arbitrary axis, $𝐋$ need not be parallel to $𝛚$ ; the full relation uses an inertia tensor. Do not silently promote every scalar formula here to a universal vector identity.

Checkpoint · Can a body have zero total linear momentum and nonzero kinetic energy?

Yes. A body spinning about a stationary centre of mass has $𝐏 = M 𝐯_{C} = 0$ , but $K = {\frac{12}{I}}_{C} ω^{2} > 0$ . The particles’ momenta cancel as vectors; their positive kinetic energies do not.

Centre-of-mass toolkit

JEE Main lists centre of mass inside the rotation unit, and it is asked directly. Two methods cover nearly every question: the table below, and treating a hole as negative mass.

Uniform body	Centre	Measured from
Semicircular wire, radius $R$	$2 R / π$	Centre of curvature
Semicircular plate	$4 R / 3 π$	Straight edge
Circular arc of half-angle $α$	$R \sin α / α$	Centre of curvature
Solid hemisphere	$3 R / 8$	Flat face
Hollow hemisphere	$R / 2$	Open rim plane
Solid cone	$h / 4$	Base
Hollow cone (no base)	$h / 3$	Base plane
Triangular plate	$h / 3$	Base

Removal method. Treat a cut-out as negative mass at its own centre:

x_{C} = \frac{M_{1} x_{1} - m_{2} x_{2}}{M_{1} - m_{2}} .

The centre always moves away from the hole.

Motion of the centre. Internal forces cannot move it. In an explosion, or when people walk on a boat, ${\vec{v}}_{C}$ stays fixed unless an external force acts.

Rigid bodies: one translation, one rotation

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