Choose an axis, identify the constraint and let torque, energy and angular momentum play their distinct roles.
JEE Main · JEE Advanced · NEET UG85 min study guide9 indexed sections
BY THE END, YOU CAN
Use angular kinematics with tangential and normal acceleration.
Calculate torque, equilibrium and moments of inertia with correct axes.
Apply rotational work and angular-momentum conservation without assuming energy conservation.
Solve rolling and slipping problems with explicit friction-regime checks.
BEFORE YOU START
Newton’s laws, vector cross products and work–energy.
Constant-acceleration motion, simple integrals and rigid-body geometry.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. Angular variables describe the motion of a rigid shape
A rigid body keeps distances between its particles fixed. For rotation about a fixed axis, one angular coordinate θ describes its orientation. Angular displacement must be expressed in radians when using arc length s = rθ. Angular velocity is dθ/dt and angular acceleration dω/dt. Every point of the body shares the same angular velocity, but its linear speed depends on distance from the axis.
For fixed-axis motion, a point at perpendicular radius r has tangential speed r|ω|, tangential acceleration rα and inward normal acceleration rω². These acceleration components are perpendicular. Constant angular speed makes tangential acceleration zero but leaves the normal acceleration needed to turn the point’s velocity.
If angular acceleration is constant, integrate to get ω = ω₀ + αt, Δθ = ω₀t + αt²/2 and ω² = ω₀² + 2αΔθ. These have the same algebraic form as constant linear-acceleration equations, but angular displacement is not distance. A point’s travelled arc length requires the accumulated absolute angular motion if the direction reverses.
For general three-dimensional rotation, angular velocity and angular momentum are vectors and need not be parallel. Most elementary fixed-axis problems avoid that complexity by specifying a symmetry axis or constrained rotation. State the axis before using scalar formulas; changing the axis changes both the geometry and the moment of inertia.
ORIGINAL WORKED EXAMPLE
Spin-up with two kinds of acceleration
A disk starts from rest with constant angular acceleration 2 rad s⁻². At t = 3 s, find angular speed, angular displacement and acceleration components of a point 0.20 m from the axis.
Angular speed is αt = 6 rad s⁻¹ and displacement is αt²/2 = 9 rad.
Tangential acceleration is rα = 0.40 m s⁻².
Normal acceleration is rω² = 0.20 × 36 = 7.2 m s⁻², directed inward. The total is their vector sum, not arithmetic sum.
Answer 6 rad s⁻¹; 9 rad; tangential 0.40 m s⁻² and inward normal 7.2 m s⁻².
02 / UNDERSTAND & APPLY
2. Torque depends on both force and the chosen point
Torque about a point O is τO = r × F, where r runs from O to the force’s point of application. Magnitude is rF sinθ, or force times perpendicular distance from O to its line of action. A large force can have zero torque if its line of action passes through O. Moving a force along its own line does not change its torque about O.
For fixed-axis rotation with constant moment of inertia about that axis, the axis component of external torque satisfies Στ = Iα. Internal forces cancel in the appropriate rigid-body total under the usual model. The scalar form does not mean every applied force is tangential; take the relevant cross-product component.
A couple consists of equal and opposite forces with separated parallel lines of action. It has zero resultant force but nonzero torque independent of the choice of origin. A single force generally causes both centre-of-mass acceleration and rotation unless constraints provide additional forces. Translation and rotation are therefore separate equations, not competing explanations.
Rigid-body equilibrium requires both ΣF = 0 and Στ = 0. Choose a torque origin that eliminates unknown forces whose lines pass through it. A hinge reaction can have unknown components but zero moment about the hinge in the ideal point-hinge model. The choice simplifies algebra without removing the reaction from force balance.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Geometry uses a common length scale. Force-arrow length is symbolic. The line of action is extended backward to show the perpendicular foot.
ORIGINAL WORKED EXAMPLE
A horizontal supported beam
A uniform 4.0 m beam of mass 20 kg is hinged at its left end and held horizontal by a vertical cable at its right end. Use g = 10. Find cable tension and vertical hinge reaction.
The beam’s weight 200 N acts at its centre, 2.0 m from the hinge.
Taking moments about the hinge: T(4.0) − 200(2.0) = 0, giving T = 100 N.
Vertical force balance then gives hinge reaction +100 − 200 = 0, so hinge reaction is 100 N upward.
Answer Cable tension 100 N; vertical hinge reaction 100 N upward.
03 / UNDERSTAND & APPLY
3. Mass farther from the axis is harder to spin up
Moment of inertia about an axis is I = Σmiri² or ∫r⊥² dm for a continuous body. The squared perpendicular distance is why moving mass outward strongly increases rotational inertia. Two bodies with equal mass and outer radius can have different I because their mass distributions differ. Moment of inertia has units kg m² and is never a signed torque-like quantity.
For common uniform bodies about standard symmetry axes: a thin ring has MR²; a solid disk or cylinder has MR²/2; a solid sphere has 2MR²/5; a thin spherical shell has 2MR²/3. A slender rod about a perpendicular axis through its centre has ML²/12. These results name both shape and axis; “the moment of inertia of a rod” without an axis is incomplete.
The parallel-axis theorem is IO = ICM + Md², where the two axes are parallel and d is their perpendicular separation. The reference axis must pass through the centre of mass. For a planar lamina, the perpendicular-axis theorem gives Iz = Ix + Iy for mutually perpendicular axes through the same point with x and y lying in the lamina. It does not apply unchanged to a three-dimensional solid.
To derive a rod result, choose coordinate x along a uniform rod with dm = (M/L)dx. Integrating x² from −L/2 to L/2 yields ML²/12. Moving the axis to an end adds M(L/2)², giving ML²/3. Derivation makes the role of density and axis explicit and allows nonuniform rods to be handled by replacing dm appropriately.
ORIGINAL WORKED EXAMPLE
Shift an axis correctly
A uniform rod of mass 3 kg and length 2 m rotates about an axis perpendicular to it through one end. Find I and angular acceleration under 8 N m net torque.
Centre-axis inertia is ML²/12 = 3 × 4/12 = 1 kg m².
Parallel-axis shift by L/2 = 1 m adds Md² = 3 kg m², so Iend = 4 kg m².
Angular acceleration is τ/I = 8/4 = 2 rad s⁻².
Answer I = 4 kg m²; α = 2 rad s⁻².
04 / UNDERSTAND & APPLY
4. Rotational work and kinetic energy
For a rigid body rotating about a fixed axis, each mass element has speed rω. Adding its kinetic energy gives Krot = Iω²/2. The work of the axis torque is W = ∫τ dθ, and instantaneous power is τω when torque and angular velocity refer to the same fixed-axis component. For general vectors, power is τ·ω.
A constant torque through angle Δθ does work τΔθ, but if torque changes with angle, integrate the actual function. Net work changes total kinetic energy. If a body both translates and rotates about its centre of mass, K = MvCM²/2 + ICMω²/2. These two terms count different parts of the particle velocities and must both be included.
A body rotating about a fixed point can alternatively use I about that point and K = IOω²/2. Do not add a separate centre-of-mass kinetic term to that expression unless the decomposition is being changed consistently; otherwise translation is counted twice. The parallel-axis theorem connects the two valid descriptions.
Conservative gravitational potential energy can convert into both translational and rotational kinetic energy. Constraints such as ideal rolling determine their ratio. A friction force may be essential for the constraint while doing zero work at an instantaneously stationary contact point in the ideal no-slip rigid-surface model.
ORIGINAL WORKED EXAMPLE
Work during angular acceleration
A wheel with I = 2 kg m² speeds from 3 to 7 rad s⁻¹. Find net work. If a constant net torque acts through 10 rad, find its magnitude.
Kinetic-energy change is (1/2)I(ωf² − ωi²) = 1 × (49 − 9) = 40 J.
For constant torque, W = τΔθ, so τ = 40/10 = 4 N m.
The angular kinematic relation also gives α = (49 − 9)/(2 × 10) = 2 rad s⁻², consistent with Iα = 4 N m.
Answer 40 J; torque 4 N m.
05 / UNDERSTAND & APPLY
5. Conserve angular momentum only about a justified origin
A particle’s angular momentum about O is LO = r × p. Its time derivative equals external torque about O when O is fixed in an inertial frame, with suitable extensions for other origins. For a rigid body about a fixed principal axis, the relevant component is L = Iω. In more general three-dimensional rotation, L and ω need not be parallel.
If net external torque about the chosen axis is zero, that angular-momentum component is conserved. Pulling mass inward reduces I and increases ω so Iω stays fixed, but rotational kinetic energy can increase because internal muscular or mechanical work is done. Conservation of angular momentum does not automatically imply conservation of kinetic energy.
In a brief collision with a hinged body, the hinge can exert a large impulse, so total linear momentum of the body-plus-projectile system may not be conserved. Its torque impulse about the hinge can nevertheless be zero because the hinge force acts at that point. Angular momentum about the hinge can then be the useful conserved quantity during the collision, while mechanical energy may be lost.
For a force-free particle travelling along a straight line, angular momentum about an origin off the line can be nonzero and constant: magnitude is p times the perpendicular impact parameter. Angular momentum is not restricted to visibly circular trajectories. Origin dependence is physical bookkeeping, not an inconsistency.
ORIGINAL WORKED EXAMPLE
A sticking collision at a rod end
A 1 kg lump moving perpendicular to a stationary 2 m uniform rod strikes and sticks to its free end at 6 m s⁻¹. The rod has mass 3 kg and pivots frictionlessly at its other end. Find angular speed immediately after impact.
During the short collision, neglect gravity’s impulse and take angular momentum about the fixed hinge, whose impulsive reaction has zero moment there.
Initial angular momentum is mvL = 1 × 6 × 2 = 12 kg m² s⁻¹.
Rod inertia about the hinge is ML²/3 = 4 kg m². The attached lump adds mL² = 4 kg m², giving total 8.
Thus ω = 12/8 = 1.5 rad s⁻¹. Kinetic energy need not be conserved because the lump sticks.
Answer 1.5 rad s⁻¹.
06 / UNDERSTAND & APPLY
6. No slipping links translation and rotation
For pure rolling on a stationary flat surface, the contact point is instantaneously at rest relative to the surface. The kinematic constraint is vCM = Rω in magnitude, and for compatible acceleration components aCM = Rα. The top point moves at 2vCM relative to ground and the centre at vCM; the contact point’s zero instantaneous velocity does not make its acceleration generally zero.
For a rigid body rolling without slipping down an incline under gravity, let ICM = βMR². Translation along the slope gives Mg sinθ − f = Ma. Torque about the centre gives fR = ICMα, and the rolling constraint gives α = a/R. Eliminating f yields a = g sinθ/(1 + β). Static friction points uphill in this unpowered descent to supply the torque.
The required friction is f = βMg sinθ/(1 + β). No-slip motion is possible only if this is at most μₛMg cosθ, giving μₛ ≥ β tanθ/(1 + β). If the available friction is insufficient, the body slips and the rolling constraint cannot be imposed. A perfectly smooth slope generally does not spin up a body released from rest into pure rolling.
For release from rest, energy conservation over a drop h gives Mgh = Mv²/2 + ICMv²/(2R²), yielding v² = 2gh/(1 + β). A solid sphere accelerates faster than a disk, and a disk faster than a thin hoop under the same ideal conditions because less of the energy is tied to rotation for smaller β. Static friction direction in other rolling problems can differ; do not promote the downhill result into a universal rule.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Velocity arrows use a common scale of 2:1. They are not force arrows. Zero contact velocity does not imply zero contact acceleration.
ORIGINAL WORKED EXAMPLE
A rolling disk on an incline
A uniform solid disk rolls without slipping down a 30° incline. Use g = 10. Find acceleration and minimum static-friction coefficient needed for this motion.
For a disk β = 1/2. Acceleration is g sin30°/(1 + 1/2) = 5/(1.5) = 10/3 m s⁻².
Thus μₛ must be at least approximately 0.192. Friction acts uphill in this gravity-driven descent.
Answer a = 10/3 m s⁻²; μₛ ≥ 0.192.
07 / UNDERSTAND & APPLY
7. Separate a rolling constraint from a hoped-for outcome
A wheel initially sliding without the matching spin is not in pure rolling. Kinetic friction changes its centre-of-mass velocity and angular velocity until the contact’s relative sliding speed may reach zero. Only then test whether static friction can maintain rolling. Imposing v = Rω from the start erases the very transient the problem asks you to analyse.
For a uniform disk sliding on a horizontal rough surface with initial speed v₀ > 0, zero spin and kinetic-friction coefficient μₖ > 0, friction reduces v and spins the disk. Taking magnitudes during this regime, v = v₀ − μₖgt and ω = 2μₖgt/R. The no-slip condition is reached at t = v₀/(3μₖg), with v = 2v₀/3. Some initial kinetic energy has become heat, so mechanical-energy conservation would be wrong during the slip.
For compound rigid bodies, moments of inertia add about the same axis. Remove a hole by subtracting the missing part’s inertia about that axis, using the parallel-axis theorem when its centre is displaced. Do not subtract only a central formula if the hole’s axis is offset. Mass density and the remaining centre of mass may also change.
When a rotating body interacts with a string or pulley, no slip at the contact supplies a tangential acceleration relation. A massive pulley needs a net torque and hence different tensions on its two sides. Treating the pulley as massive in its kinetic energy but massless in the force equations is an inconsistent hybrid model.
08 / UNDERSTAND & APPLY
8. Original practice: axis, constraint and conservation law
Use ideal rigid bodies and the specified axes. Where g is needed, take 10 m s⁻². Identify whether energy, angular momentum or both are conserved before calculating.
ORIGINAL PRACTICE · 1
A wheel turns through 10 rad in 2 s at constant angular speed. A point lies 0.30 m from its axis. Find speed.
Reveal answer & reasoning
1.5 m s⁻¹.
ω = 10/2 = 5 rad s⁻¹ and v = rω.
ORIGINAL PRACTICE · 2
A constant 6 N force acts with perpendicular lever arm 0.40 m. Find torque magnitude.
Reveal answer & reasoning
2.4 N m.
Torque is force times perpendicular lever arm.
ORIGINAL PRACTICE · 3
A force’s line of action passes through the chosen pivot. What is its torque about that pivot?
Reveal answer & reasoning
Zero.
Perpendicular lever arm is zero, even if force magnitude is large.
ORIGINAL PRACTICE · 4
A thin hoop and solid disk have equal M and R about their central axes. Compare I.
Reveal answer & reasoning
Hoop inertia is twice disk inertia.
Ihoop = MR² and Idisk = MR²/2.
ORIGINAL PRACTICE · 5
A rod’s centre-axis inertia is ML²/12. Find its parallel end-axis inertia.
Reveal answer & reasoning
ML²/3.
Add M(L/2)² using the parallel-axis theorem.
ORIGINAL PRACTICE · 6
With zero external torque, a skater halves I. How do ω and rotational energy change?
Reveal answer & reasoning
ω doubles; rotational energy doubles.
L = Iω remains fixed and K = L²/(2I). Internal work supplies the energy increase.
ORIGINAL PRACTICE · 7
A particle moves in a straight line with momentum p at perpendicular distance b from an origin. Find angular-momentum magnitude.
Reveal answer & reasoning
pb.
Magnitude of r×p equals momentum times perpendicular distance to the line.
ORIGINAL PRACTICE · 8
A hoop rolls without slipping at centre speed v. Find total kinetic energy.
Reveal answer & reasoning
Mv².
Translation contributes Mv²/2; rotation contributes (MR²)(v/R)²/2, equal to the first term.
ORIGINAL PRACTICE · 9
A rolling wheel’s centre moves at 3 m s⁻¹ over stationary ground. Find top-point and contact-point speeds.
Reveal answer & reasoning
6 m s⁻¹ and zero.
Add or subtract rotational velocity relative to the centre.
ORIGINAL PRACTICE · 10
A solid sphere rolls down an incline without slip. Express acceleration in terms of g sinθ.
Reveal answer & reasoning
(5/7)g sinθ.
β = 2/5, so a = g sinθ/(1+2/5).
ORIGINAL PRACTICE · 11
Can angular momentum about a hinge be conserved during a sticking impact while linear momentum is not?
Reveal answer & reasoning
Yes.
The hinge can deliver a linear impulse but zero torque impulse about itself; sticking can dissipate mechanical energy.
ORIGINAL PRACTICE · 12
A uniform disk initially slides at v₀ without spin on a rough horizontal surface. What speed does it have when pure rolling first begins in the stated kinetic-friction model?
Reveal answer & reasoning
2v₀/3.
Friction gives v=v₀−μgt and Rω=2μgt; equating them gives t=v₀/(3μg).
09 / UNDERSTAND & APPLY
9. A rotational solution must name its axis and regime
Start with an axis, a force diagram and a kinematic constraint only if the physical contact supports it. Use moment of inertia about that same axis. For impacts, examine external torque impulse rather than assuming both linear and angular momentum conservation. For rolling, verify that the friction required by the no-slip solution is available.
Check units and limiting cases: I is kg m², torque is N m, angular momentum is kg m² s⁻¹ and energy is joules. A force through the pivot should give zero torque. Moving mass outward should increase inertia. A body with smaller β should reach a greater speed after the same no-slip gravitational descent. These qualitative tests expose an incorrectly chosen axis or double-counted kinetic energy.
All points share ω, not the same linear speed.
Torque depends on force line and chosen origin.
Moment of inertia requires both mass distribution and axis.
Conserved angular momentum does not imply conserved kinetic energy.
Pure rolling is a contact constraint, not a default assumption.
Translation plus centre-of-mass rotation is a complete energy decomposition.
KEEP THE BIG IDEAS
Quick revision
Rotational variables only become definite after an axis is specified.
Mass distribution controls rotational inertia.
Torque changes angular momentum; work changes kinetic energy.
Rolling couples translation and rotation only when slipping is absent.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.