Rigid body — definition, degrees of freedom: Exercises

Level 1 — Recognition

Goal: recall the standard DOF numbers and where they come from.

L1.1

Recall Solution

A point in space is pinned down by three numbers: how far along $x$ , along $y$ , along $z$ . Nothing ties them together, so all three are independent. $DOF = 3$

L1.2

Recall Solution

$DOF = 6 = translation 3 + rotation 3$ Three numbers say where the body is (slide along $x, y, z$ ); three numbers say which way it faces (turn about $x, y, z$ ).

L1.3

Recall Solution

Fixing the axle kills all 3 translations and 2 of the 3 rotations, leaving only the spin angle $θ$ . $DOF = 1$

Level 2 — Application

Goal: apply "coordinates minus constraints" to concrete objects.

Step 1 — locate a point. The coin's centre lives in a 2D plane, needing $(x, y)$ → 2 numbers. Step 2 — orient it. In a plane there is only one rotation axis (straight up out of the table), so one angle $θ$ . $DOF = 2 + 1 = 3$ Look at figure s01: the two arrows are the slide directions, the curved arrow is the single spin.

Figure — Rigid body — definition, degrees of freedom

L2.2

Recall Solution

Step 1 — what's fixed? The hinge line is a fixed axis: translation gone, and 2 of 3 rotations gone. Step 2 — what's left? Just the swing angle $θ$ . $DOF = 1$ Same physics as the wheel on the axle (L1.3).

L2.3

Recall Solution

Fixing one point removes all 3 translations (the pinned point can't move). But the body can still rotate freely about that point in all 3 rotational ways. $DOF = 0_{trans} + 3_{rot} = 3$

Level 3 — Analysis

Goal: count constraints, and see when pair-distances over-count.

L3.1

Recall Solution

(a) Each particle needs 3 numbers: $3 N = 3 \times 4 = 12$ . (b) A 3D rigid body has 6 DOF, so $constraints = coords - DOF = 12 - 6 = 6.$ (c) Number of distinct pairs among 4 particles: $(2 4) = \frac{4 \cdot 3}{2} = 6.$ For $N = 4$ these 6 pair-distances are exactly independent — the count matches. (For $N > 4$ they would not all be independent.)

L3.2

Recall Solution

Total pair equations: $(2 8) = \frac{8 \cdot 7}{2} = 28.$ True independent constraints: $3 N - 6 = 3 \cdot 8 - 6 = 18.$ Redundant (over-counted) equations: $28 - 18 = 10.$ This is why we fix 3 points (giving $3 + 2 + 1 = 6$ DOF) instead of blindly counting all pairs.

L3.3

Recall Solution

Point 1: free → 3 numbers. Point 2: fixed distance to Point 1 → lives on a sphere → 2 numbers. Point 3: fixed distance to Points 1 and 2 → lives on a circle → 1 number. Running total: $3 + 2 + 1 = 6$ . Every further particle ( $4, 5, \dots, N$ ) has fixed distances to these 3 non-collinear anchors, which pin it uniquely. Each adds 0 DOF. Figure s02 shows the sphere → circle → point shrinking of freedom. $DOF = 6 for all N \geq 3.$

Level 4 — Synthesis

Goal: combine ideas — molecules, thermal DOF, and mixed constraints.

L4.1

Recall Solution

Translation: the molecule as a whole slides in 3D → 3. Rotation: it can tumble about two axes perpendicular to the bond → 2. The missing one: rotation about the bond axis itself spins two point masses that lie on that axis — nothing physically moves. That rotational DOF does not exist. $DOF = 3 + 2 = 5.$

L4.2

Recall Solution

A rigid diatomic has $f = 5$ active (quadratic) DOF: 3 translational + 2 rotational. Energy per molecule: $⟨ E ⟩ = f \cdot \frac{1}{2} k_{B} T = \frac{5}{2} k_{B} T .$ Molar internal energy (multiply by Avogadro's number, $N_{A} k_{B} = R$ ): $U = \frac{5}{2} R T .$ Molar heat capacity at constant volume: $C_{V} = \frac{d U}{d T} = \frac{5}{2} R .$ Neighbouring idea: Equipartition Theorem.

L4.3

Recall Solution

Before rolling constraint: a rigid body in a plane has 3 DOF — position $x$ along the ground, height $y$ , and spin angle $θ$ . Constraint 1 — stays on the ground: $y$ is fixed at the radius → removes 1 → leaves $x, θ$ (2 DOF). Constraint 2 — no slipping: the contact point is instantaneously at rest, tying $x$ to $θ$ by $x = R θ$ → removes 1 more. $DOF = 3 - 1 - 1 = 1.$ One number (say $θ$ , or equivalently $x$ ) tells you everything. See Rolling Motion.

Level 5 — Mastery

Goal: derive DOF for a genuinely new system from first principles.

L5.1

Recall Solution

Translation: to place a reference point in $d$ -space you need $d$ numbers. Rotation: a rotation acts in a plane, and a plane is chosen by picking 2 of the $d$ axes. The number of independent rotation planes is $(2 d) = \frac{d ( d - 1 )}{2} .$ So $DOF (d) = d + \frac{d ( d - 1 )}{2} = \frac{d ( d + 1 )}{2} .$ Check $d = 2$ : $\frac{2 \cdot 3}{2} = 3$ ✓ (matches the coin, L2.1). Check $d = 3$ : $\frac{3 \cdot 4}{2} = 6$ ✓ (the familiar answer).

L5.2

Recall Solution

Method 1 — physical angles. Rod A can only swing about its ground pin → 1 angle $θ_{A}$ . Rod B swings about the hinge at A's tip → 1 more angle $θ_{B}$ . $DOF = 2.$ Method 2 — bookkeeping. Each rigid rod in a plane has 3 DOF, so two free rods = $6$ .

Ground pin fixes one end of A in the plane: 2 constraints.
Hinge ties A's tip and B's end to the same point: 2 constraints. $DOF = 6 - 2 - 2 = 2. ✓$ Both roads give 2. See Constraints and Lagrangian Mechanics for the general recipe.

L5.3

Recall Solution

Start: free rigid body = 6 DOF (3 trans + 3 rot). Constraint: one marked point must lie on a line. A point on a line has only 1 free coordinate instead of 3, so the translation part drops from 3 to 1 → removes 2. Rotation: untouched — the body still faces any way → keeps 3. $DOF = (3 - 2) + 3 = 1 + 3 = 4.$

Recall One-line summary of the whole page

DOF = coordinates − independent constraints. Free rigid body: $6$ in 3D, $3$ in 2D, $\frac{d ( d + 1 )}{2}$ in $d$ -D. Each fixed point/axis/plane/wire subtracts exactly the freedoms it kills — count the removed ones.