Big picture
A cricket bat can translate through space while also rotating about its centre. System mechanics separates the motion of a body's centre of mass from its rotation, making complex motion manageable. The same approach explains a balanced seesaw, a spinning wheel and a rolling cylinder.
A single force equation is insufficient when forces can turn a body. Translation depends on net force, while rotation depends on net torque about a specified axis or point. Always define the reference before calculating a moment, because a force can have zero torque about one point and nonzero torque about another.
Learning goals
Locate the centre of mass of discrete and continuous systems.
Apply momentum conservation and centre-of-mass motion.
Calculate torque and angular momentum using vector reasoning.
Use equilibrium conditions and moments of inertia.
Solve fixed-axis rotation and ideal rolling problems.
Core ideas
1. Centre of mass
The centre of mass, COM, has position R = Σmᵢrᵢ/Σmᵢ. In one dimension, xCOM = Σmᵢxᵢ/M; apply the same weighted average separately to y and z. For 2 kg at x = 0 and 3 kg at x = 5 m, xCOM = (2 × 0 + 3 × 5)/5 = 3 m. It lies nearer the larger mass, not necessarily at the geometric midpoint.
For continuous matter, replace sums by integrals, R = (1/M)∫r dm. Symmetry locates the COM of uniform rods, discs and spheres at their geometric centres. A ring's COM can lie in empty space. The centre of gravity is where the resultant gravitational force acts; it coincides with COM in a uniform gravitational field but need not in strongly varying fields.
2. Motion of a system
Total momentum is P = MVCOM. Differentiating gives Fexternal = MA COM, because internal forces cancel in pairs under the elementary system assumptions. Internal interactions can redistribute motion but cannot change total momentum without an external impulse.
If a stationary 5 kg system separates into 2 kg moving right at 6 m s⁻¹ and 3 kg moving with speed v, momentum conservation gives 2 × 6 + 3v = 0, so v = −4 m s⁻¹. Its COM remains at rest although both fragments move. An exploding projectile's COM follows the external-force trajectory even while fragments spread out, if air resistance is negligible.
3. Angular quantities and vector products
Angular displacement θ is measured in radians; angular velocity ω = dθ/dt and angular acceleration α = dω/dt. For rotation about a fixed axis, a point at perpendicular distance r has tangential speed v = rω, tangential acceleration aₜ = rα and radial acceleration aᵣ = rω². All points share angular velocity, but farther points have larger linear speed.
For constant α, ω = ω₀ + αt, θ = ω₀t + ½αt² and ω² = ω₀² + 2αθ. A wheel starting from rest with α = 3 rad s⁻² for 4 s reaches 12 rad s⁻¹ and turns through 24 rad. The vector product A × B has magnitude AB sinφ and direction given by the right-hand rule. Rotation-related vectors point along the axis, not along the circular path.
4. Torque and angular momentum
Torque about a point is τ = r × F, with magnitude rF sinφ, equivalently force times perpendicular lever arm. Its SI unit is N m. Although torque and energy share dimensions, torque is a vector moment, while energy is scalar; report torque in N m rather than casually treating it as a joule.
A 25 N force applied perpendicular to a 0.40 m handle produces 10 N m. The same force directed along the handle toward the pivot has zero torque. Particle angular momentum is L = r × p. About a fixed inertial origin, τexternal = dL/dt. For a particle moving radially relative to that origin, r and p are parallel, so angular momentum is zero even though linear momentum is nonzero.
5. Equilibrium and couples
A rigid body in static equilibrium requires ΣF = 0 and Στ = 0. Zero net force alone prevents translational acceleration but not angular acceleration. A couple consists of equal opposite forces on different parallel lines; net force is zero, but torque magnitude is Fd, where d is perpendicular separation. A steering wheel illustrates turning action with little net translation.
For a light horizontal plank supported at its ends 4 m apart, a 300 N load sits 1 m from the left end. Taking moments about the left gives Rright × 4 = 300 × 1, so Rright = 75 N. Vertical balance gives Rleft = 300 − 75 = 225 N. If plank weight is non-negligible, include its weight at its own COM. Choosing a pivot through an unknown support can eliminate that reaction's torque.
6. Moment of inertia and axis theorems
Moment of inertia, I = Σmᵢrᵢ² or ∫r²dm, measures rotational inertia about a stated axis. Its unit is kg m². It depends on mass distribution and axis, not angular speed. Radius of gyration, k, is defined by I = Mk².
Uniform body and axis | Moment of inertia |
|---|---|
Thin ring, central perpendicular axis | MR² |
Solid disc or cylinder, symmetry axis | MR²/2 |
Solid sphere, diameter | 2MR²/5 |
Thin spherical shell, diameter | 2MR²/3 |
Rod, centre perpendicular to length | ML²/12 |
Rod, end perpendicular to length | ML²/3 |
The parallel-axis theorem is I = ICOM + Md² for parallel axes separated by d. For a 3 kg rod of length 2 m, ICOM = 3 × 4/12 = 1 kg m²; about an end, I = 1 + 3 × 1² = 4 kg m². The perpendicular-axis theorem applies to a plane lamina: I𝓏 = Iₓ + Iᵧ for mutually perpendicular axes meeting in the plane. It cannot be applied to an arbitrary three-dimensional body.
7. Rotational dynamics and energy
For a rigid body rotating about a fixed suitable axis with constant I, τnet = Iα and Laxis = Iω. Rotational kinetic energy is Krot = ½Iω², work is ∫τ dθ and power is τω for aligned fixed-axis quantities. A body with I = 0.80 kg m² under 4.0 N m net torque has α = 4/0.80 = 5 rad s⁻².
For I = 0.50 kg m² and ω = 8 rad s⁻¹, Krot = 0.5 × 0.50 × 64 = 16 J. Torque is analogous to force, I to mass and angular acceleration to linear acceleration, but the analogy works only with clearly specified axes and constraints. In general three-dimensional rotation, L need not be parallel to ω; elementary formulas here concern the appropriate symmetry or fixed axis.
8. Angular momentum conservation
If external torque about a specified point or axis is zero, angular momentum is conserved. A skater pulling in the arms reduces I, so angular speed rises to keep Iω constant. If I changes from 4.0 to 2.5 kg m² with initial ω = 3 rad s⁻¹, final ω = 4 × 3/2.5 = 4.8 rad s⁻¹.
Rotational kinetic energy need not remain constant during this redistribution. Initial energy is ½ × 4 × 3² = 18 J and final energy ½ × 2.5 × 4.8² = 28.8 J. Muscular work supplies the 10.8 J increase. Conservation of angular momentum is therefore not equivalent to conservation of kinetic energy.
9. Rolling motion
For pure rolling on a stationary surface, vCOM = Rω. The contact point is instantaneously at rest relative to the ground, while the top point moves at 2vCOM. Total kinetic energy is K = ½MvCOM² + ½ICOMω². A 2 kg solid cylinder rolling at 3 m s⁻¹ has translational energy 9 J and rotational energy 4.5 J, totalling 13.5 J.
For ideal rolling down an incline, energy gives Mgh = ½Mv²[1 + ICOM/(MR²)]. A solid cylinder descending through h = 1.5 m with g = 10 reaches v² = 2 × 10 × 1.5/(1 + 1/2) = 20, so v = √20 ≈ 4.47 m s⁻¹. Static friction can enforce rolling without dissipating mechanical energy on a fixed ideal surface. Slipping changes this model and may involve kinetic-friction losses.
10. Demonstrations and diagrams
A balancing ruler demonstrates moments: shift a known mass until clockwise and anticlockwise moments balance. Mark the pivot, force directions and perpendicular arms rather than using distances measured along an arbitrary sloping line. A ring and solid disc of equal mass and radius can illustrate different rotational inertia because the ring places more mass farther from the axis.
In rolling diagrams, draw COM velocity, angular direction and contact point separately. In a centre-of-mass experiment, suspended irregular laminae hang with COM below the support; intersecting plumb lines locate it. These observations connect equations to measurable geometry.
A reliable approach to questions
Define the system, axis and reference frame.
Draw external forces and their perpendicular lever arms.
Decide whether to use force balance, torque balance, energy or momentum conservation.
Select a moment of inertia for the actual axis and apply a theorem if necessary.
For rolling, impose v = Rω only when slipping is absent.
Connections that matter
Linear mechanics supplies momentum, force and energy conservation.
Gravitation uses angular-momentum conservation for central forces.
Geometry supplies perpendicular distances and symmetry arguments.
Common mistakes
Using force-arm length instead of perpendicular distance gives wrong torque.
Applying equilibrium from net force alone ignores rotational acceleration.
Treating I as independent of axis overlooks mass distribution.
Equating conserved angular momentum with conserved kinetic energy overlooks internal work.
Revision checklist
I can locate a COM and explain its motion.
I can calculate torque and angular momentum about a stated origin.
I can solve rigid-body equilibrium problems.
I can apply axis theorems and rotational dynamics.
I can combine translation and rotation in pure rolling.