System of Particles and Rotational Motion: NEET notes

1. Rigid bodies and their motion

NCERT §6.1

• A particle is an idealised point mass with no size. Real bodies are extended, and an extended body is best seen as a system of particles.
• An ideal rigid body keeps a perfectly definite shape: the distance between every pair of its particles stays fixed. No real body is truly rigid, but wheels, tops, steel beams, molecules and planets can be treated as rigid when their bending, twisting and vibration are negligible.
• In pure translation every particle of the body has the same velocity at each instant, like a block sliding down an incline without turning.
• A cylinder rolling down the same incline is translating plus rotating: its particles have different velocities at the same instant, and if it rolls without slipping the point in contact with the incline is momentarily at rest.
• Fixed along a line, a body has only rotation open to it. Each of its particles then traces a circle whose plane is at right angles to the axis and whose centre lies on it; particles on the axis stay still. A ceiling fan and a potter's wheel turn this way, and so do the giant wheel and merry-go-round at a fair.
• In some rotations only one point is fixed and the axis moves: a spinning top's axis sweeps out a cone about the vertical through its tip (precession), and an oscillating table fan's axis swings about its pivot.
• Summary: a body that is not pivoted either translates or translates and rotates; a pivoted or fixed body rotates. This chapter treats rotation about a fixed axis only.

2. Centre of mass

NCERT §6.2

• For two particles on the x-axis, the centre of mass is at X = (m₁x₁ + m₂x₂)/(m₁ + m₂), the mass-weighted mean of their positions. For equal masses it lies midway.
• In three dimensions, with total mass M = Σmᵢ: X = Σmᵢxᵢ/M, Y = Σmᵢyᵢ/M and Z = Σmᵢzᵢ/M; compactly, R = Σmᵢrᵢ/M.
• Three equal masses: the centre of mass is at the centroid of their triangle. If the centre of mass is taken as the origin, Σmᵢrᵢ = 0.
• For a continuous body the sums become integrals: R = (1/M)∫r dm. By reflection symmetry, uniform rods, rings, discs, spheres and cubes have their centre of mass at their geometric centre.
• Masses 100 g, 150 g and 200 g at the corners (0, 0), (0.5, 0) and (0.25, 0.25√3) m of an equilateral triangle of side 0.5 m have their centre of mass at (5/18 m, 1/(3√3) m), not at the centroid, because the masses are unequal.
• A uniform triangular plate can be cut into thin strips parallel to one side; each strip balances at its midpoint, so the centre of mass lies on every median, at the centroid.
• A uniform 3 kg L-shaped plate made of three 1 m squares: treat each square as 1 kg at its centre, (½, ½), (3/2, ½), (½, 3/2) m. The centre of mass is at (5/6 m, 5/6 m), on the plate's line of symmetry.
• The centre of mass need not lie inside the material of the body; for a ring it is at the empty centre.

3. Motion of the centre of mass

NCERT §6.3, §6.4

• Differentiating MR = Σmᵢrᵢ gives MV = Σmᵢvᵢ and MA = Σmᵢaᵢ, for masses that do not change.
• Internal forces come in equal and opposite pairs (third law) and cancel in the sum, so MA = F_ext: the centre of mass moves as if the whole mass were there and all the external forces acted there.
• No knowledge of the internal forces is needed to find the motion of the centre of mass, whatever the body is doing inside, rotating or breaking up.
• A shell that explodes in mid-flight: the explosion is internal, gravity is unchanged, so the centre of mass of the fragments keeps to the original parabola.
• The total linear momentum of a system is P = Σmᵢvᵢ = MV, and dP/dt = F_ext: Newton's second law for a system.
• If F_ext = 0, P is constant (conservation of linear momentum) and the centre of mass moves with constant velocity, even though the particles may follow complicated paths. The vector law is three scalar laws: Pₓ, P_y and P_z are each constant.
• A radium nucleus decaying into radon and an alpha particle: the products move so that their centre of mass continues along the path of the radium nucleus. In the frame where the centre of mass is at rest, they fly apart back to back.
• A binary star with no external force: the centre of mass moves in a straight line at constant speed, and in the centre-of-mass frame the two stars circle it, always on opposite sides.

4. Vector product of two vectors

NCERT §6.5

• The vector (cross) product c = a × b has magnitude ab sin θ, where θ is the smaller angle between a and b, and points perpendicular to the plane of a and b.
• Direction by the right-hand rule: curl the fingers of the right hand from a to b through the smaller angle; the thumb gives c. A right-handed screw turned from a to b advances along c.
• The cross product is not commutative: b × a = −(a × b), same size, opposite direction. The scalar product is commutative.
• Under reflection every component changes sign, so a → −a and b → −b, but a × b is unchanged.
• Both products are distributive: a × (b + c) = a × b + a × c. Also a × a = 0.
• Unit vectors: i × i = j × j = k × k = 0; i × j = k, j × k = i, k × i = j, and the products in the reverse order are negative.
• In components, a × b = (a_y b_z − a_z b_y) i + (a_z b_x − a_x b_z) j + (a_x b_y − a_y b_x) k, which is the determinant with rows (i, j, k), (aₓ, a_y, a_z), (bₓ, b_y, b_z).
• For a = 3i − 4j + 5k and b = −2i + j − 3k: a · b = −25 and a × b = 7i − j − 5k, so b × a = −7i + j + 5k.

5. Angular velocity and angular acceleration

NCERT §6.6, §6.6.1

• Let a particle of the spinning body turn by Δθ in a time Δt. The limit of Δθ/Δt as Δt shrinks to zero is the angular velocity, ω = dθ/dt.
• Every particle of the rigid body has the same ω at a given instant, so ω is the angular velocity of the whole body. Pure rotation means every part has the same angular velocity, just as pure translation means every part has the same velocity.
• The speed of a particle at perpendicular distance r from the axis is v = ωr; particles on the axis (r = 0) are at rest.
• Angular velocity is a vector along the axis, in the direction a right-handed screw advances when turned with the body. Reversing the sense of rotation reverses ω.
• In vector form v = ω × r, with r measured from an origin on the axis; v is tangent to the particle's circle and has magnitude ωr⊥. The same relation holds for rotation about a fixed point, such as a top.
• For rotation about a fixed axis the direction of ω does not change; only its magnitude can. In more general rotation both can change.
• Angular acceleration is α = dω/dt. For a fixed axis it is along the axis, and the vector equation becomes the scalar α = dω/dt.

6. Torque and angular momentum

NCERT §6.7

• Torque (moment of force) is the rotational analogue of force: τ = r × F, where r is the position of the point of application from the origin. A door turns most easily when pushed at right angles at its outer edge; a push on the hinge line does nothing.
• Magnitude τ = rF sin θ = r⊥F = rF⊥, where r⊥ is the perpendicular distance of the line of action from the origin (the moment arm) and F⊥ the component of F perpendicular to r.
• Torque is zero if F = 0, r = 0, or the line of action passes through the origin (θ = 0° or 180°). Its SI unit is N m and dimensions M L² T⁻², the same as work, but torque is a vector and work a scalar.
• Angular momentum of a particle about a point: l = r × p, magnitude l = rp sin θ = r⊥p.
• dl/dt = τ: the rate of change of angular momentum equals the torque, the rotational form of F = dp/dt.
• For a system, L = Σrᵢ × pᵢ and dL/dt = τ_ext. Internal torques cancel provided the forces between particles are equal, opposite and along the line joining them.
• If τ_ext = 0, L is constant: conservation of angular momentum, three scalar laws for Lₓ, L_y, L_z.
• A particle moving with constant velocity has constant angular momentum about any point: r sin θ is the fixed distance of its straight path from the point, and the direction of l does not change.
• A fast-spinning bicycle rim held by one string at one end of its axle does not fall; its angular momentum precesses about the string, turning about an axis perpendicular to both L and the torque.

7. Equilibrium of a rigid body

NCERT §6.8

• A rigid body is in mechanical equilibrium when neither its linear momentum nor its angular momentum changes: it has no linear and no angular acceleration.
• Translational equilibrium: ΣF = 0. Rotational equilibrium: Στ = 0. Each is three scalar equations, six conditions in all.
• If all the forces lie in one plane, three conditions are enough: force components add to zero along two perpendicular directions in the plane, and torques about an axis perpendicular to the plane add to zero.
• If ΣF = 0, the total torque is the same about every point, so the origin for torques can be chosen freely.
• Partial equilibrium is possible. Two equal parallel forces in the same direction at the ends of a light rod: zero net torque about the centre but a net force. Equal and opposite forces at the two ends: zero net force but a net torque.
• A couple is a pair of equal and opposite forces with different lines of action. It produces rotation without translation. Turning a bottle lid and a compass needle in the earth's magnetic field (not pointing north–south) are examples.
• The moment of a couple is AB × F, where AB joins the points of application; it does not depend on the point about which moments are taken.
• A particle can only be in translational equilibrium; the forces on it act at one point, so they are concurrent.

8. Principle of moments and centre of gravity

NCERT §6.8.1, §6.8.2

• An ideal lever is a light rod pivoted at the fulcrum, like a see-saw or the beam of a balance. With load F₁ at distance d₁ and effort F₂ at d₂ from the fulcrum, the reaction there is R = F₁ + F₂.
• Principle of moments: d₁F₁ = d₂F₂, load arm × load = effort arm × effort. Anticlockwise moments are usually taken positive. It also holds when the parallel forces are not perpendicular to the lever.
• Mechanical advantage M.A. = F₁/F₂ = d₂/d₁. When the effort arm is longer than the load arm, M.A. > 1 and a small effort lifts a large load.
• The centre of gravity is the point about which the total gravitational torque on the body is zero; a cardboard balances horizontally on a pencil tip placed there.
• In uniform gravity Σmᵢrᵢ × g = 0 gives Σmᵢrᵢ = 0, so the centre of gravity coincides with the centre of mass. For a body so large that g varies across it, they differ; the centre of mass depends only on how the mass is distributed.
• A plate hung freely from a point settles with its centre of gravity on the vertical through that point. Two or three such verticals from different points cross at the centre of gravity.
• A 70 cm, 4.00 kg uniform bar on knife-edges 10 cm from each end carries 6.00 kg at 30 cm from one end. With g = 9.8 m/s², forces give R₁ + R₂ = 98.00 N and moments about the centre give R₁ − R₂ = 11.76 N: R₁ = 54.88 N and R₂ = 43.12 N, about 55 N and 43 N.
• A 3 m, 20 kg ladder rests against a frictionless wall with its foot 1 m out. Moments about the foot give the wall's push F₁ = W/(4√2) = 34.6 N; the floor supplies N = 196.0 N up and friction 34.6 N, a resultant of 199.0 N at about 80° to the horizontal.

9. Moment of inertia

NCERT §6.9

• Each particle of a body rotating with ω has speed ωrᵢ, so the kinetic energy is K = ½(Σmᵢrᵢ²)ω² = ½Iω², where I = Σmᵢrᵢ² is the moment of inertia about the axis and rᵢ is the perpendicular distance from the axis.
• Comparing ½Iω² with ½mv² shows that I plays the role of mass in rotation: it measures a body's resistance to a change in its rotation (rotational inertia).
• A thin ring of mass M and radius R about the axis through its centre, perpendicular to its plane: I = MR², since all its mass is at R. Two masses M/2 at the ends of a light rod of length l, about a perpendicular axis through the middle: I = Ml²/4.
• Standard results: ring about a diameter MR²/2; thin rod about a perpendicular axis at the midpoint ML²/12; disc about the perpendicular central axis MR²/2 and about a diameter MR²/4; hollow cylinder about its axis MR²; solid cylinder about its axis MR²/2; solid sphere about a diameter 2MR²/5.
• Unlike mass, I is not fixed for a body: it depends on the mass, the shape and size, how the mass is spread about the axis, and the position and orientation of the axis.
• Radius of gyration k: I = Mk². It is the distance from the axis at which a point mass equal to the whole mass would have the same I. Rod about its middle: k = L/√12; disc about a diameter: k = R/2.
• Dimensions of I are M L², SI unit kg m².
• A flywheel is a disc of large moment of inertia on an engine's shaft. It resists sudden changes in speed, so a vehicle's speed changes gradually and the ride is smooth.

10. Kinematics of rotation

NCERT §6.10

• With the axis fixed, a single angle θ describes the whole body (one degree of freedom). It is read for any one particle, from a fixed reference line in that particle's plane of motion.
• θ, ω = dθ/dt and α = dω/dt correspond to x, v and a in linear motion.
• For constant α: ω = ω₀ + αt, θ = θ₀ + ω₀t + ½αt², and ω² = ω₀² + 2α(θ − θ₀), where θ₀ and ω₀ are the values at t = 0.
• ω = ω₀ + αt follows by integrating dω/dt = α with ω = ω₀ at t = 0; integrating once more gives the θ equation.
• Convert rpm to rad/s by multiplying by 2π/60. 1200 rpm = 40π rad/s and 3120 rpm = 104π rad/s.
• A motor wheel going from 1200 rpm to 3120 rpm in 16 s at constant α: α = (104π − 40π)/16 = 4π rad/s². It turns through θ = 40π × 16 + ½ × 4π × 16² = 1152π rad, that is 576 revolutions.

11. Dynamics of rotation

NCERT §6.11

• For a fixed axis only the torque components along the axis matter; the bearings supply the constraint forces that cancel the perpendicular components. So only forces in planes perpendicular to the axis, and only the parts of position vectors perpendicular to the axis, need be considered.
• A force F₁ on a particle at distance r₁ from the axis does work dW = F₁r₁ sin α₁ dθ = τ₁ dθ as the body turns through dθ. For all the forces together, dW = τ dθ.
• Power delivered by a torque: P = τω, like P = Fv.
• For a rigid body there is no internal motion, so this work all goes into kinetic energy: τω = d(½Iω²)/dt = Iωα, which gives τ = Iα, Newton's second law for rotation about a fixed axis.
• The angular acceleration is directly proportional to the torque and inversely proportional to the moment of inertia.
• Analogues: x ↔ θ, v ↔ ω, a ↔ α, M ↔ I, F = Ma ↔ τ = Iα, dW = F ds ↔ dW = τ dθ, ½Mv² ↔ ½Iω², Fv ↔ τω, p = Mv ↔ L = Iω.
• A 20 kg flywheel of radius 20 cm (I = MR²/2 = 0.4 kg m²) pulled by a steady 25 N on a cord wound on its rim: τ = 5.0 N m and α = 12.5 s⁻². When 2 m of cord unwinds, the pull does 50 J of work; θ = 2/0.2 = 10 rad, ω² = 2 × 12.5 × 10 = 250 (rad/s)², so K = ½ × 0.4 × 250 = 50 J, equal to the work since the bearings are frictionless.

12. Angular momentum and its conservation

NCERT §6.12, §6.12.1

• Take one particle at distance r⊥ from the fixed axis: the part of its angular momentum along the axis is mr⊥²ω. Adding over the body gives L_z = Iω.
• For a particle, l need not be along the axis: l and ω are not necessarily parallel, unlike p and v, which always are.
• If the axis is a symmetry axis of the body, the perpendicular parts of the particles' angular momenta cancel in pairs, and L = Iω exactly along the axis.
• For a fixed axis, dL_z/dt = τ along the axis, while the component of L perpendicular to the axis stays constant. With I constant this gives τ = Iα again.
• Conservation: if the external torque about the axis is zero, Iω = constant. If I decreases, ω increases in the same ratio, and the other way round.
• A person on a frictionless swivel chair spinning with arms folded slows down on stretching the arms out (I increases) and speeds up again on pulling them in.
• Acrobats, divers, skaters and dancers doing a pirouette on the toes of one foot use this: pulling the arms and legs in reduces I and raises the spin rate.
• Kinetic energy ½Iω² = L²/2I is not conserved when I changes: pulling the arms in raises it, and the extra energy comes from the work done by the person's muscles.

Must-know facts

1. Rigid body: all interparticle distances fixed. Pure translation: same velocity for every particle; fixed-axis rotation: same ω for every particle.
2. X = Σmᵢxᵢ / Σmᵢ; for two equal masses the centre of mass is midway, for three equal masses at the centroid.
3. L-shaped 3 kg plate of three 1 m squares: centre of mass at (5/6 m, 5/6 m).
4. MA = F_ext; internal forces cannot move the centre of mass.
5. An exploding projectile's centre of mass continues on the original parabola.
6. P = MV; if F_ext = 0, P and V_cm are constant.
7. |a × b| = ab sin θ; a × b = −b × a; i × j = k, j × k = i, k × i = j.
8. v = ωr, or v = ω × r; ω points along the axis by the right-hand rule.
9. τ = r × F, τ = rF sin θ = r⊥F; unit N m, dimensions M L² T⁻², a vector unlike work.
10. l = r × p; dL/dt = τ_ext; L constant if τ_ext = 0.
11. Equilibrium: ΣF = 0 and Στ = 0; a couple gives zero net force but non-zero torque, the same about every point.
12. Lever: d₁F₁ = d₂F₂; M.A. = d₂/d₁.
13. Centre of gravity = centre of mass only when g is uniform over the body.
14. I = Σmr², K = ½Iω², I = Mk².
15. Ring MR²; disc MR²/2; rod about its middle ML²/12; solid sphere 2MR²/5; hollow cylinder MR²; disc about a diameter MR²/4.
16. ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ; rpm × 2π/60 = rad/s.
17. τ = Iα, W = τθ for constant torque, P = τω.
18. L = Iω for a symmetric body about its axis; Iω constant with no external torque; K = L²/2I rises when I falls.

Common traps

Formulas

Key terms

Rigid body

A body whose particles all keep fixed distances from one another.

Pure translation

Motion in which every particle of the body has the same velocity at each instant.

Axis of rotation

The line, held fixed, around which each particle of a turning body traces a circle.

Precession

The slow turning of a spinning body's axis about another direction, as a top's axis sweeps a cone.

Centre of mass

The mass-weighted mean position of a system, which moves as if all the mass and external force were there.

Vector product

A vector of size ab sin θ perpendicular to a and b, directed by the right-hand rule.

Torque

The turning effect of a force about a point, r × F.

Angular momentum

The moment of linear momentum about a point, r × p.

Couple

Two equal and opposite forces along different lines, which turn a body without moving it along.

Mechanical advantage

The ratio of load to effort for a lever, equal to effort arm over load arm.

Centre of gravity

The point about which the gravitational torques on a body add to zero.

Moment of inertia

Σmr² about an axis: the rotational counterpart of mass.

Radius of gyration

The distance k at which the whole mass, as a point, would have the same moment of inertia.

Flywheel

A heavy wheel with a large moment of inertia that smooths out changes in an engine's speed.