Systems of Particles and Rotational Motion

This chapter takes you from a "point particle" view of mechanics to a "real object with size" view, in a fixed sequence of ideas that each build on the last.

• It starts by asking what kinds of motion a rigid body (fixed shape, unchanging inter-particle distances) can have: pure translation, pure rotation about a fixed axis, or a combination such as rolling.
• It defines the centre of mass for discrete particles and continuous bodies, and shows that the centre of mass moves exactly like a single particle under the total external force alone — internal forces cancel completely, by Newton's third law.
• It introduces the vector (cross) product, the mathematical tool needed to write angular velocity, torque, and angular momentum precisely, and uses it to connect linear velocity $\vec{v}$ of any point on a spinning body to the body's single angular velocity $\vec{\omega}$ via $\vec{v}=\vec{\omega}\times\vec{r}$.
• It defines torque and angular momentum, first for a single particle and then for a system, and proves the rotational analogue of Newton's second law, $d\vec{L}/dt = \vec{\tau}_{\text{ext}}$, along with the conservation law that follows when the external torque vanishes.
• It states the two independent conditions for the mechanical equilibrium of a rigid body (zero net force and zero net torque), introduces the principle of moments and the lever, and distinguishes centre of gravity from centre of mass.
• It defines moment of inertia, tabulates it for standard shapes (rings, discs, rods, cylinders, spheres), and introduces radius of gyration.
• It closes with the full kinematics and dynamics of rotation about a fixed axis — the rotational analogues of every linear-motion equation you already know — culminating in $\tau = I\alpha$, rotational kinetic energy $K=\tfrac{1}{2}I\omega^2$, and the conservation of angular momentum $L=I\omega=\text{constant}$ that explains skaters, divers, and rotating platforms.

Rigid bodies: translation, rotation, and everything in between

A rigid body is an idealisation: a body whose shape is perfectly definite and unchanging, so the distance between any two of its particles never changes no matter what forces act on it. No real object is truly rigid — a steel beam bends a little — but for wheels, tops, beams, planets, and molecules the deformation is negligible enough to ignore. A free rigid body can undergo pure translation (every particle has the same velocity at any instant, as when a block slides down a frictionless incline) or a combination of translation and rotation (as when a cylinder rolls down the same incline — its point of contact is momentarily at rest while the top moves at twice the centre's speed). A body constrained to a fixed line can only rotate about that fixed axis: every particle moves in a circle lying in a plane perpendicular to the axis, centred on the axis. This chapter mostly restricts itself to rotation about a fixed axis, the simplest and commonest case (ceiling fans, potter's wheels, flywheels), rather than the more general motion of a top whose axis itself precesses.

The centre of mass

For two particles of mass $m_1$ and $m_2$ on a line at positions $x_1$ and $x_2$, the centre of mass (CM) is the mass-weighted average position $X = (m_1x_1+m_2x_2)/(m_1+m_2)$; for equal masses it sits exactly midway. Generalising to $n$ particles distributed anywhere in space, and writing everything as vectors, gives the compact definition every later section of this chapter builds on. Three or more non-collinear particles give a genuinely two- or three-dimensional CM, and a continuous body (a rod, a lamina, a sphere) is treated as an enormous number of infinitesimal mass elements $dm$, turning the sum into an integral.

where: $\vec{r}_i$ is the position vector of the $i$-th particle of mass $m_i$, $M=\sum m_i$ is the total mass, and the second form is the continuous limit, with $M=\int dm$. The physical reading is that $\vec{R}$ is not the geometric centre unless the mass is symmetrically distributed about it — a heavier particle pulls the CM toward itself, exactly as a heavier weight on a see-saw shifts the balance point. By reflection symmetry, the CM of any homogeneous ring, rod, disc, or sphere coincides with its geometric centre, because every mass element at $+x$ is matched by an equal one at $-x$. For an irregular shape (an L-shaped lamina, say), the trick is to split it into simple pieces whose individual centres of mass are known, treat each piece's entire mass as concentrated at its own CM, and then find the CM of those few point masses.

Newton's second law for a system: motion of the centre of mass

Differentiating $\vec{R}$ twice with respect to time turns the centre-of-mass definition into a dynamical statement about how the whole system moves, and it is one of the most useful shortcuts in mechanics. However complicated the internal motions of a system are — particles colliding, an object exploding into fragments, a gymnast twisting mid-air — the centre of mass itself obeys a strikingly simple law, proved in full in the first derivation below.

where: $\vec{A}=d\vec{V}/dt$ is the acceleration of the centre of mass and $\vec{F}_{\text{ext}}$ is the vector sum of only the external forces on the system — every internal force, by Newton's third law, cancels in equal-and-opposite pairs and contributes nothing. The physical reading: the centre of mass moves as if the entire mass of the system were concentrated there and every external force acted there directly. A shell that explodes mid-flight has its fragments scatter unpredictably, but their combined centre of mass sails on along the same parabola gravity alone would have produced, because the explosive forces are entirely internal. The same idea underlies why, in a system with no external force at all (a radioactive nucleus decaying, or the two stars of a binary system orbiting each other), the centre of mass either stays at rest or moves in a straight line at constant velocity, no matter how complicated the individual trajectories look in the ordinary frame of reference. Multiplying $\vec{V}=d\vec{R}/dt$ by the total mass gives the total linear momentum of the system, $\vec{P}=M\vec{V}=\sum m_i\vec{v}_i$, and Eq. (2) can then be rewritten as $d\vec{P}/dt=\vec{F}_{\text{ext}}$ — Newton's second law extended from a single particle to an entire system of particles.

where: the conservation statement is the direct system-level generalisation of conservation of momentum for two colliding particles, which you met in earlier chapters as a special case. It holds separately along each of the $x$, $y$, $z$ directions.

The vector product and angular velocity

Rotational quantities like torque and angular momentum need a way to combine two vectors — a position and a force, or a position and a momentum — into a third vector genuinely perpendicular to both originals. That tool is the vector (cross) product. Given two vectors $\vec{a}$ and $\vec{b}$ with an angle $\theta$ between them, $\vec{c}=\vec{a}\times\vec{b}$ has a magnitude built from $\sin\theta$ (vanishing when the vectors are parallel) and a direction fixed by a right-hand rule: curl the fingers from $\vec{a}$ toward $\vec{b}$, and the thumb points along $\vec{c}$.

where: $a$ and $b$ are the magnitudes of $\vec{a}$ and $\vec{b}$, and the determinant form is the practical way to compute components once $\vec{a}$ and $\vec{b}$ are written in $\hat{i},\hat{j},\hat{k}$ form; it follows from the elementary results $\hat{i}\times\hat{i}=0$, $\hat{i}\times\hat{j}=\hat{k}$, $\hat{j}\times\hat{k}=\hat{i}$, $\hat{k}\times\hat{i}=\hat{j}$ and their sign-reversed opposites. Unlike the scalar (dot) product, the vector product is not commutative: $\vec{a}\times\vec{b}=-\vec{b}\times\vec{a}$, because reversing the order reverses the sense of the right-hand curl even though the magnitude stays the same. For a rigid body spinning about a fixed axis, every particle traces its own circle with its own radius, but the whole body shares one angular velocity $\omega=d\theta/dt$ — the same for every particle, which is exactly what "rigid" and "one axis" together guarantee. Treated as a vector, $\vec{\omega}$ points along the axis of rotation (by the right-hand screw rule again), and the linear velocity of any particle follows from a single clean vector equation, proved in the second derivation below.

where: $\vec{r}$ is the particle's position vector measured from any origin on the axis. The physical reading is that this one equation automatically reproduces the familiar scalar fact $v=\omega r_\perp$ (with $r_\perp$ the perpendicular distance from the axis) and points the velocity in the correct tangential direction, without having to argue separately about magnitude and direction for every particle. The time derivative of $\omega$ is the angular acceleration $\alpha=d\omega/dt$, the direct rotational analogue of linear acceleration.

Torque and angular momentum

Force alone does not tell you whether a door will swing open: push on the hinge line and nothing rotates, however hard you push, while the same force applied far from the hinge, perpendicular to the door, swings it open easily. The quantity that actually governs rotation is torque (moment of force), $\vec{\tau}=\vec{r}\times\vec{F}$, and its translational-motion partner is angular momentum, $\vec{l}=\vec{r}\times\vec{p}$ — both position-dependent vector products, both measured about a chosen origin.

where: $\vec{r}$ locates the point of application of the force (for torque) or the particle (for angular momentum) relative to the chosen origin, $\vec{F}$ is the force and $\vec{p}=m\vec{v}$ the linear momentum. Both vanish if $\vec{r}=0$, the relevant vector is zero, or the line of action passes straight through the origin — a force pointed at or away from the pivot produces no rotation. For a system of $n$ particles the total angular momentum is $\vec{L}=\sum_i \vec{l}_i$, and differentiating $\vec{L}$ with Newton's third law eliminating internal torques gives (derived fully below):

where: only the external torque changes the system's total angular momentum — this is the conservation law behind skaters, divers, and rotating platforms, explored fully in the last physical-picture subsection and the third derivation below.

Equilibrium of a rigid body

A rigid body is in mechanical equilibrium only when neither its linear momentum nor its angular momentum is changing — which means both of the following independent conditions must hold simultaneously.

where: the first condition alone (net force zero) guarantees the centre of mass does not accelerate, but says nothing about spin — a couple (two equal, opposite, parallel forces with different lines of action, like fingers twisting a bottle cap) has zero net force yet produces pure rotation, precisely because it violates the second condition. A body can be in translational equilibrium without rotational equilibrium, or vice versa; true mechanical equilibrium needs both. For a light rod pivoted at a fulcrum, such as a see-saw, rotational equilibrium about the fulcrum reduces to the principle of moments.

where: $F_1$ (the load) acts at distance $d_1$ (the load arm) and $F_2$ (the effort) acts at distance $d_2$ (the effort arm) from the fulcrum, both measured perpendicular to the line of action. The ratio $F_1/F_2 = d_2/d_1$ is the lever's mechanical advantage: a long effort arm lets a small effort lift a large load. The centre of gravity (CG) of a body is the single point where the total gravitational torque on the body is zero — the point at which the body balances on a pencil tip. Because $g$ is essentially the same at every point of an everyday-sized object, the CG coincides with the centre of mass; the two concepts separate only for bodies so large that $g$ genuinely varies across them (a skyscraper or a mountain, in principle).

Moment of inertia

Just as mass measures how hard it is to change an object's linear velocity, moment of inertia measures how hard it is to change its angular velocity about a given axis — the rotational stand-in for mass, derived directly from summing the kinetic energy of every particle of a spinning rigid body (full derivation below).

where: $r_i$ is the perpendicular distance of the $i$-th particle from the chosen axis, and the sum runs over every particle of the body. Moment of inertia is not a fixed property of a body the way mass is: it depends on how mass is distributed relative to the specific axis, so the same disc has a different $I$ about its central perpendicular axis than about a diameter. The radius of gyration $k$, defined by $I=Mk^2$, is the distance from the axis at which the whole mass, concentrated as a point, would give the same $I$.

A large moment of inertia resists sudden changes in angular speed, which is exactly why engines are fitted with a heavy flywheel — its large $I$ smooths out jerky torque fluctuations into a steady rotation.

Kinematics and dynamics of rotation about a fixed axis

Because rotation about a fixed axis needs only one number ($\theta$) to describe the body's orientation at any instant, every kinematic equation for linear motion under constant acceleration has an exact rotational twin, obtained by the substitutions $x\to\theta$, $v\to\omega$, $a\to\alpha$.

where: $\theta_0$ and $\omega_0$ are the angular displacement and angular velocity at $t=0$, and $\alpha$ is assumed constant, exactly as $a$ was assumed constant in the linear versions of these equations. The dynamical partner of these kinematic equations is the rotational form of Newton's second law, $\tau=I\alpha$, derived below from the work-energy theorem, alongside rotational power $P=\tau\omega$ (the analogue of $P=Fv$) and work done by a constant torque, $W=\tau\,\Delta\theta$.

Angular momentum of a rotating body and its conservation

For a rigid body rotating about a fixed axis that is also an axis of symmetry, the general vector relation $\vec{L}=\sum\vec{r}_i\times\vec{p}_i$ collapses to a single clean scalar-times-direction statement, because the perpendicular components of angular momentum from symmetric pairs of particles cancel exactly.

where: $I$ is the moment of inertia about the fixed axis and $\omega$ the angular speed; $L$ and $\vec{\omega}$ are not generally parallel for a lopsided body spinning off its symmetry axis, which is why this simple form is restricted to symmetric bodies. The conservation statement is the single most exam-tested consequence of this chapter: if the net external torque about the axis is zero, then any change in $I$ must be compensated by an opposite change in $\omega$ so that their product stays fixed.

Board level - Centre of mass of three particles at a triangle's vertices

Three particles of mass $100\ \text{g}$, $150\ \text{g}$ and $200\ \text{g}$ sit at the vertices $O=(0,0)$, $A=(0.5,0)$ and $B=(0.25,0.25\sqrt3)$ metres of an equilateral triangle of side $0.5\ \text{m}$. Find the centre of mass.

Step 1. Total mass $M=100+150+200=450\ \text{g}$. $X$-coordinate: $X=\dfrac{100(0)+150(0.5)+200(0.25)}{450}=\dfrac{125}{450}=\dfrac{5}{18}\ \text{m}\approx0.278\ \text{m}$.

Step 2. $Y$-coordinate: $Y=\dfrac{100(0)+150(0)+200(0.25\sqrt3)}{450}=\dfrac{50\sqrt3}{450}=\dfrac{\sqrt3}{9}\ \text{m}\approx0.192\ \text{m}$.

Step 3. Compare with the geometric centroid, the unweighted vertex average: $\left(\tfrac{0+0.5+0.25}{3},\tfrac{0+0+0.25\sqrt3}{3}\right)=(0.25,\,0.144)\ \text{m}$. The CM sits higher in $y$ because $B$, the heaviest vertex, pulls the mass-weighted average toward itself. Answer: the centre of mass is at $\left(\dfrac{5}{18},\dfrac{\sqrt3}{9}\right)\ \text{m}\approx(0.278\ \text{m},\,0.192\ \text{m})$, not the geometric centroid.

Board level - Torque, angular acceleration, and work on a flywheel

A cord of negligible mass is wound round the rim of a flywheel of mass $20\ \text{kg}$ and radius $0.20\ \text{m}$, mounted on a frictionless horizontal axle. A steady pull of $25\ \text{N}$ is applied to the cord. Find (a) the angular acceleration, (b) the work done when $2\ \text{m}$ of cord is unwound, and (c) the kinetic energy gained, starting from rest.

Step 1. Torque from the tension: $\tau=FR=25\times0.20=5.0\ \text{N}\,\cdot\,\text{m}$. Moment of inertia of a solid cylinder about its axis: $I=\dfrac{MR^2}{2}=\dfrac{20\times(0.20)^2}{2}=0.4\ \text{kg}\,\cdot\,\text{m}^2$. So $\alpha=\tau/I=5.0/0.4=12.5\ \text{rad/s}^2$.

Step 2. Work done by the pull: $W=F\times d=25\times2=50\ \text{J}$, since the cord does not slip on the rim.

Step 3. Angular displacement for $2\ \text{m}$ of unwound cord: $\theta=d/R=2/0.20=10\ \text{rad}$. With $\omega_0=0$: $\omega^2=2\alpha\theta=2\times12.5\times10=250\ (\text{rad/s})^2$. Kinetic energy gained: $K=\tfrac{1}{2}I\omega^2=\tfrac{1}{2}(0.4)(250)=50\ \text{J}$. Answer: $\alpha=12.5\ \text{rad/s}^2$, $W=50\ \text{J}$, $K=50\ \text{J}$ — equal to the work done, exactly as the rotational work-energy theorem requires on a frictionless axle.

Competitive level - Ladder against a frictionless wall (JEE/NEET pattern)

A uniform $3\ \text{m}$ ladder of mass $20\ \text{kg}$ leans against a frictionless wall, its foot on rough ground $1\ \text{m}$ from the wall's base. Take $g=9.8\ \text{m/s}^2$. Find the normal reaction of the floor, the friction force at the floor, and the magnitude and direction of their resultant.

Step 1. Geometry: the height reached is $\sqrt{3^2-1^2}=\sqrt8=2\sqrt2\ \text{m}$. The weight $W=20\times9.8=196\ \text{N}$ acts at the midpoint, so its horizontal distance from the foot is $0.5\ \text{m}$. The frictionless wall can only push horizontally ($F_1$); the floor supplies a vertical normal reaction $N$ and horizontal friction $F$ toward the wall.

Step 2. Vertical equilibrium: $N=W=196\ \text{N}$. Horizontal equilibrium: $F=F_1$.

Step 3. Rotational equilibrium about the foot: the moment of $F_1$ (at height $2\sqrt2$) balances the moment of $W$ (at horizontal distance $0.5\ \text{m}$): $2\sqrt2\,F_1=0.5\,W$, so $F_1=\dfrac{W}{4\sqrt2}=\dfrac{196}{4\sqrt2}\approx34.6\ \text{N}=F$.

Step 4. Resultant floor reaction: $F_2=\sqrt{F^2+N^2}=\sqrt{34.6^2+196^2}\approx199.0\ \text{N}$, at an angle $\alpha=\tan^{-1}(N/F)=\tan^{-1}(196/34.6)\approx80°$ from the horizontal. Answer: $N=196\ \text{N}$, $F\approx34.6\ \text{N}$, resultant $\approx199\ \text{N}$ at about $80°$ to the horizontal. Timing note: about two minutes. The whole problem is two force balances plus one torque balance taken about the foot — choosing the foot as the pivot eliminates both unknowns $N$ and $F$ from the moment equation in one stroke, leaving a single equation for $F_1$. Always pick the pivot that kills the most unknowns first.

Competitive level - Angular momentum conservation on a rotating turntable (JEE/NEET pattern)

A child stands at the centre of a frictionless turntable rotating at $\omega_1=40\ \text{rev/min}$ with arms outstretched. By folding his arms he reduces his moment of inertia to $\tfrac{2}{5}$ of its initial value. Find his new angular speed, and the ratio of his final to initial rotational kinetic energy.

Step 1. No external torque acts about the vertical axis (frictionless turntable, gravity/normal force along the axis), so angular momentum is conserved: $I_1\omega_1=I_2\omega_2$, with $I_2=\tfrac{2}{5}I_1$.

Step 2. Solve for $\omega_2$: $\omega_2=\omega_1\dfrac{I_1}{I_2}=\omega_1\times\dfrac{5}{2}=40\times2.5=100\ \text{rev/min}$.

Step 3. Compare kinetic energies: $K_1=\tfrac{1}{2}I_1\omega_1^2$ and $K_2=\tfrac{1}{2}I_2\omega_2^2=\tfrac{1}{2}\left(\tfrac{2}{5}I_1\right)(2.5\,\omega_1)^2=\tfrac{1}{2}I_1\omega_1^2\times\left(\tfrac{2}{5}\times6.25\right)=2.5\,K_1$. Answer: $\omega_2=100\ \text{rev/min}$, and $K_2/K_1=2.5$ — the rotational kinetic energy increases by a factor of $2.5$, even though angular momentum stayed exactly constant. Timing note: under 90 seconds once you recognise the setup. The trap is assuming kinetic energy is conserved along with angular momentum — it is not: pulling the arms in requires the child's muscles to do internal work against the effective outward pull of rotation, and that work is exactly what raises the kinetic energy. Angular momentum conservation follows from zero external torque; energy conservation is a completely separate question.

• Confusing centre of mass with the geometric centre or centroid. They coincide only for a uniform, symmetric mass distribution; unequal masses (as in the triangle example above) pull the CM toward the heavier side.
• Forgetting that internal-force cancellation needs Newton's third law and the assumption that the force between any two particles lies along the line joining them. Without the second assumption, $\vec{\tau}_{\text{int}}=0$ does not follow, even though $\vec{F}_{\text{int}}=0$ still does.
• Treating $\omega$ and $\vec\tau$ as scalars and losing sign/direction, or using $r$ instead of the perpendicular distance $r_\perp$ (or $r\sin\theta$) when computing torque or angular momentum — only the component perpendicular to $\vec r$ contributes.
• Assuming zero net force implies zero net torque, or vice versa. A couple is the standard counter-example: zero net force, non-zero net torque. Equilibrium genuinely needs both conditions checked separately.
• Applying $L=I\omega$ to a body spinning off its own axis of symmetry. The simple scalar relation only holds when the rotation axis is a symmetry axis of the body; otherwise $\vec L$ and $\vec\omega$ are not even parallel.
• Forgetting moment of inertia is axis-dependent, and mixing up radius of gyration $k$ with the body's physical radius $R$. "The moment of inertia of a disc" is meaningless without stating the axis — $MR^2/2$ about the central perpendicular axis versus $MR^2/4$ about a diameter for the very same disc; $k=R$ only for a thin ring.

• locate the centre of mass of a system of discrete particles, or of a simple continuous body, using symmetry arguments or the mass-weighted average formula
• explain why the centre of mass of a system obeys $M\vec A=\vec F_{\text{ext}}$ regardless of how complicated the internal motion is, and apply this to projectile explosions or systems with no external force
• compute a vector (cross) product using the determinant rule, and find its direction using the right-hand rule
• connect the linear velocity of any point of a rotating rigid body to the body's angular velocity via $\vec v=\vec\omega\times\vec r$
• define torque as $\vec r\times\vec F$ and angular momentum as $\vec r\times\vec p$ for a single particle, and state how they differ from force and linear momentum
• test whether an extended body is in full mechanical equilibrium by checking both $\sum\vec F=0$ and $\sum\vec\tau=0$, and use the principle of moments to solve lever, beam, and ladder problems
• distinguish centre of mass from centre of gravity, and state the one condition under which they coincide
• state or compute the moment of inertia of a ring, disc, rod, cylinder, or sphere about a stated axis, and use the radius of gyration $I=Mk^2$
• use $\tau=I\alpha$ and $K=\tfrac12I\omega^2$ as the rotational analogues of $F=ma$ and $\tfrac12mv^2$ in dynamics and work-energy problems
• apply conservation of angular momentum, $I\omega=\text{constant}$ when $\tau_{\text{ext}}=0$, to explain and calculate outcomes for skaters, divers, and rotating platforms

Ask the tutor: if you are stuck on which pivot point to choose in an equilibrium problem, or want to see why a diver's angular speed changes mid-air even though nothing is pushing them, ask the tutor to walk through it with you.