Rigid body — definition, degrees of freedom: Worked examples

The scenario matrix

Every rigid-body DOF question is one (or a blend) of the cells below. Each worked example is tagged with the cell it covers.

Cell	Case class	What is special	Example
A	Free body, 3D	nothing constrained → the pure 6	Ex 1
B	Confined to a plane (2D)	one translation + two rotations killed	Ex 2
C	Fixed point (pivot)	all translation killed, all rotation kept	Ex 3
D	Fixed axis (hinge/axle)	only one spin angle survives	Ex 4
E	Degenerate geometry: collinear points	the "3 non-collinear points" rule breaks	Ex 5
F	Zero/limit case: single particle & two-particle "body"	smallest $N$ ; a rotation vanishes	Ex 6
G	Constraint-counting by subtraction	use DOF to find hidden constraint number	Ex 7
H	Real-world word problem	translate English into constraints	Ex 8
I	Exam twist: composite / linked bodies	add DOF, subtract linking constraints	Ex 9

Figures accompany the geometric cells (A, C, D, E).

Ex 1 — Cell A: a free brick tumbling in space

Worked example A brick thrown across a room

Statement: A solid brick is thrown so it both flies and tumbles freely. How many independent numbers pin it down at any instant?

Forecast: Guess before reading. (Trans + rot in 3D...)

Locate one reference point of the brick. That needs $(x, y, z)$ → 3 numbers. Why this step? A rigid body is frozen in shape, so once one point and the orientation are known, every other point follows. Start by placing that one point.
Orient the brick about that point. In 3D there are three independent turning axes (roll, pitch, yaw) → 3 more numbers. Why this step? After fixing a point the brick can still spin; three axes exhaust every possible spin in 3D. Look at the three cyan rotation arcs in the figure.
Add them: $3 + 3 = 6$ .

Answer: 6 DOF.

Verify: Cross-check with the "3 non-collinear points" method from the parent: $3 + 2 + 1 = 6$ . ✔ Both routes agree, which is the whole reason the number is trustworthy.

Ex 2 — Cell B: a coin sliding and spinning flat on a table

Worked example A coin confined to a tabletop

Statement: A coin lies flat and may slide anywhere on the table and spin about its centre, but never lifts off. DOF?

Forecast: One less freedom in each family than the free case?

Position of the centre. It lives in the 2D plane of the table → $(x, y)$ → 2 numbers. Why this step? Confinement to the plane is a constraint: the height $z$ is fixed, killing one translational DOF (from 3 down to 2).
Orientation. In a plane there is only one independent rotation axis — the vertical, straight out of the table. So one angle $θ$ → 1 number. Why this step? Tipping the coin about any in-plane axis would lift part of it off the table, which the confinement forbids. Two of the three 3D rotations are killed; one survives.
Add: $2 + 1 = 3$ .

Answer: 3 DOF.

Verify: Start from the free 6 and subtract what the plane removes: it kills 1 translation ( $z$ ) and 2 rotations (the two tipping axes): $6 - 1 - 2 = 3$ . ✔

Ex 3 — Cell C: a rigid body pinned at one point

Worked example A spinning top pinned at its tip

Statement: A rigid top touches the floor at a single fixed point (its tip does not slide). How many DOF?

Forecast: Translation gone, rotation intact?

Kill translation. The fixed point cannot move, so $(x, y, z)$ of the reference point are all pinned → 0 translational DOF. Why this step? A fixed point is three constraints at once (one per coordinate). $3 - 3 = 0$ .
Keep rotation. The body can still swivel about that point in every direction → 3 rotational DOF. Why this step? Pinning one point stops sliding but not turning; all three spin axes through that point remain free. See the three amber rotation arcs about the pinned tip in the figure.
Add: $0 + 3 = 3$ .

Answer: 3 DOF.

Verify: From the free 6, a fixed point removes exactly the 3 translations: $6 - 3 = 3$ . ✔ (Matches the parent's "fixed at a point → 3" row.)

Ex 4 — Cell D: a door on its hinges (fixed axis)

Worked example A swinging door

Statement: A door hangs on a fixed vertical hinge line. DOF?

Forecast: Almost everything is nailed down here...

The hinge line is a fixed axis — a whole line of the body is frozen, not just a point. Why this step? Fixing an axis is stronger than fixing a point: it stops all 3 translations and stops the 2 rotations whose axes are not the hinge.
Count what survives. Only rotation about the hinge axis itself is allowed → one swing angle $θ$ → 1 number. Why this step? You can neither slide the door nor tip it; the single red arc in the figure is the only motion left.
Result: $DOF = 1$ .

Answer: 1 DOF.

Verify: From the free 6, a fixed axis removes 3 translations + 2 rotations $= 5$ : $6 - 5 = 1$ . ✔ Same as a wheel on an axle. This is exactly the single-angle world where all of rotational kinematics ( $θ, ω, α$ ) lives.

Ex 5 — Cell E: the degenerate case — three collinear points

Worked example Why "3 points" secretly means "3

non-collinear points" Statement: Someone tries to fix a rigid rod by nailing three points that all lie on the same straight line. Does this freeze the rod? What DOF remain?

Forecast: Careful — does a line of nails stop rolling?

Fix point 1: 0 DOF left for it (it's pinned) — this pins the position. Why this step? One fixed point removes all 3 translations, exactly as in Ex 3.
Fix a second point on the line: it removes 2 of the 3 remaining rotations, leaving spin about the line through the two points. Why this step? Two fixed points define an axis; only rotation about that axis survives. So far this is identical to the fixed-axis door.
Try to fix a third point that is also on that same line. It adds no new information: it already lies on the fixed axis, so its position is automatically forced by the first two. It removes 0 further DOF. Why this step? The "3 points lock a body" rule needs the three points not on one line, so that the third point can pin down the leftover spin. Collinear points cannot, because they are all invariant under that spin. Look at the figure: rotating about the dashed axis leaves all three amber dots exactly where they were.
Remaining DOF: the spin about the line $= 1$ .

Answer: 1 DOF survives — the rod can still roll about its own long axis. This is the degenerate trap.

Verify: Two distinct points give a fixed axis (Ex 4 logic): $6 - 5 = 1$ ; the collinear third point adds $0$ : $1 - 0 = 1$ . ✔

Ex 6 — Cell F: the limit cases — one particle, and a two-atom molecule

Worked example Smallest systems:

$N = 1$ and a diatomic molecule Statement: (a) A single point particle in 3D. (b) A diatomic molecule modelled as two point masses joined by a rigid bond. DOF of each?

Forecast: (a) obvious. (b) is famously not 6 — guess why.

(a) Single particle.

Place it: $(x, y, z)$ → 3. There is nothing to orient (a point has no "facing"). Why this step? Orientation only means something for a body with extent; a mathematical point is the same from every angle, so 0 rotational DOF.
Answer (a): 3 DOF.

(b) Diatomic (two point masses, rigid bond).

Translation of the pair: locate the whole molecule's reference point → 3.
Rotation: the bond line can tumble in 3D — but rotation about the bond axis itself moves nothing, because both masses are points sitting on that axis. Why this step? A point mass on the rotation axis does not move when you spin about that axis, so that "rotation" is physically undetectable and does not count as a DOF. Only the two tumbling rotations (axes perpendicular to the bond) are real → 2.
Add: $3 + 2 = 5$ .
Answer (b): 5 DOF.

Verify: (a) $3 = 3$ . ✔ (b) Full rigid body would be 6; kill the one useless bond-axis spin: $6 - 1 = 5$ . ✔ This $5$ is exactly the number the Equipartition Theorem feeds into the specific heat of a diatomic gas. Contrast with the full rigid body's 6.

Ex 7 — Cell G: back out the constraint count

Worked example How many independent constraints hold a 5-particle rigid body together?

Statement: Five point masses are locked into one rigid body. Find the number of independent rigidity constraints.

Forecast: Not $(2 5) = 10$ — that overcounts. Guess the real number.

Total free coordinates: $3 N = 3 \times 5 = 15$ . Why this step? Before locking, each of the 5 particles needs 3 numbers.
DOF of any 3D rigid body is 6 (from Ex 1). Why this step? Rigidity always saturates at 6 regardless of $N$ .
Independent constraints $= coords - DOF = 15 - 6 = 9$ . Why this step? Rearranging the master formula: constraints $=$ coords $-$ DOF.
Answer: 9 independent constraints (even though there are $(2 5) = 10$ pair-distances — one is redundant).

Verify: General rule $3 N - 6$ with $N = 5$ : $3 (5) - 6 = 9$ . ✔ And $(2 5) = 10 > 9$ , so exactly $10 - 9 = 1$ pair-distance is redundant, matching the parent's warning about over-counting pairs.

Ex 8 — Cell H: real-world word problem

Worked example A robot forearm gripping a fixed workbench

Statement: A rigid robot forearm is bolted to a fixed shoulder joint (a ball-and-socket that allows all rotations but no sliding), and its far end is clamped so it cannot rotate about its own length. Give the DOF.

Forecast: Ball-and-socket = fixed point; then one more clamp removes something.

The bolted shoulder is a fixed point → removes all 3 translations, leaves 3 rotations (this is the Ex 3 pivot situation) → 3 so far. Why this step? "Bolted, cannot slide" is precisely a fixed point.
The clamp forbids rotation about the arm's own long axis → removes 1 of those 3 rotations. Why this step? One extra independent constraint removes exactly one DOF; here it kills the self-spin, leaving only the two "point-in-a-direction" swings.
Add: $3 - 1 = 2$ .

Answer: 2 DOF — the forearm can point its tip anywhere on a spherical patch (two swing angles), but cannot slide or self-twist.

Verify: From free 6: fixed point $- 3$ , self-spin clamp $- 1$ → $6 - 3 - 1 = 2$ . ✔ (Sanity of the physical picture: two angles = latitude + longitude of the pointing direction, which is indeed a 2-parameter direction — consistent with pinning a point and one axis.)

Ex 9 — Cell I: exam twist — two rods joined by a hinge

Worked example Composite body: two rigid rods linked by a pin

Statement: Two separate rigid rods float freely in 3D but are joined at one end by a pin (they share that single point, yet can still angle freely about it). Total DOF of the combined system?

Forecast: $6 + 6$ minus the linking constraint — but how much does a shared point remove?

DOF if the rods were independent: each free rigid rod has 6, so $6 + 6 = 12$ . Why this step? Two disconnected rigid bodies simply add their DOF (independent configurations).
Impose the pin: the two rods must share one point in space. That is 3 constraint equations (their $x$ , $y$ , $z$ at the joint must match). Why this step? Forcing two points to coincide ties down 3 coordinates; a pin allows any relative rotation, so it constrains only the shared position, not orientation.
Subtract: $12 - 3 = 9$ .

Answer: 9 DOF.

Verify: Independent check — describe the linked system directly: rod 1 free (6), rod 2 must keep its pinned end glued to rod 1's pinned end (that fixes rod 2's position, 3 translations gone, leaving 3 rotations): $6 + 3 = 9$ . ✔ Both counts agree. Handling such joint constraints systematically is exactly what Constraints and Lagrangian Mechanics is built for.

Recall One-line answer bank (guess before revealing)

Free rigid body in 3D? ::: 6 Rigid body confined to a plane? ::: 3 Rigid body pinned at one point? ::: 3 Rigid body on a fixed axis (door/wheel)? ::: 1 Three collinear fixed points leave? ::: 1 (spin about the line) Single point particle? ::: 3 Diatomic point-mass molecule? ::: 5 Independent constraints for N=5 rigid particles? ::: 9 (from $3 N - 6$ ) Robot arm: fixed point + no self-spin? ::: 2 Two free rods joined by a pin? ::: 9

Related building blocks: Centre of Mass (the natural reference point for the 3 translational DOF), Moment of Inertia and Rotational Kinematics (what fills the rotational DOF with dynamics), and Rolling Motion (a body with a rolling constraint that trims DOF further).