Noesis
1.5.1Rotational Mechanics

Rigid body — definition, degrees of freedom

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What is a rigid body?

WHY this definition? The whole point of "rigid" is no deformation. The only mathematical way to say "shape never changes" is to freeze all the inter-particle distances. If distances are locked, the object can still move (translate) and turn (rotate), but it can never squash, stretch, or twist out of shape.

Degrees of freedom (DOF)

WHAT does "independent" mean? A coordinate is independent if you cannot compute it from the others using the constraints. Constraints are equations that tie coordinates together (like the fixed-distance rule).

Deriving the DOF of a rigid body — from scratch

We will count, then subtract constraints. This is the master method: DOF=(total coordinates)−(independent constraint equations)\boxed{\text{DOF} = (\text{total coordinates}) - (\text{independent constraint equations})}

Step 1 — Single free particle

A point in 3D space needs (x,y,z)(x,y,z). So 3 DOF per free particle. Why? Three perpendicular directions; you need one number along each.

Step 2 — NN free particles (no constraints)

coordinates=3N\text{coordinates} = 3N Why? Each particle independently needs 3 numbers.

Step 3 — Lock them into a rigid body

Now impose: every pair-distance is fixed. The number of pairs among NN particles is (N2)=N(N−1)2\binom{N}{2} = \frac{N(N-1)}{2} Naively that's the number of constraints. But these constraints are NOT all independent. Once you've nailed down a few points, the rest are over-determined.

Step 4 — The smart trick: fix 3 points, the rest are automatic

To locate the whole rigid body, it's enough to fix the positions of 3 non-collinear points (think: pin three corners of a tabletop — the whole table is then frozen).

  • Point 1: free, 3 coordinates.
  • Point 2: constrained to fixed distance from Point 1 → loses 1 DOF → 2 coordinates (it lives on a sphere around Point 1).
  • Point 3: fixed distance from Point 1 and from Point 2 → loses 2 DOF → 1 coordinate (it lives on a circle).

DOF=3+2+1=6\text{DOF} = 3 + 2 + 1 = \boxed{6}

  • Every other particle (4,5,…,N4, 5, \dots, N) is fully determined by its fixed distances to these 3 points — 0 new DOF each.
Figure — Rigid body — definition, degrees of freedom

Special cases (always asked!)

System DOF Why
Single particle (3D) 3 (x,y,z)(x,y,z)
Rigid body in 3D 6 3 trans + 3 rot
Rigid body in 2D (plane) 3 (x,y)(x,y) + 1 rotation angle θ\theta
Rigid body fixed at a point 3 only rotation (3 rot)
Rigid body on a fixed axle (1 axis fixed) 1 only the spin angle θ\theta
Diatomic molecule (2 point atoms) 5 3 trans + 2 rot (spin about bond axis doesn't count for point masses)

Common mistakes

Flashcards

Define a rigid body.
A system of particles where the distance between every pair stays constant in time (no deformation).
Mathematical condition for rigidity?
∣r⃗i−r⃗j∣=|\vec r_i - \vec r_j| = constant for all pairs (i,j)(i,j).
What are degrees of freedom?
The number of independent coordinates needed to fully specify the configuration of a system.
Master formula for DOF?
DOF = (total coordinates) − (number of independent constraints).
DOF of one free particle in 3D?
3.
DOF of a rigid body free in 3D?
6 (3 translational + 3 rotational).
Why is the rigid-body DOF independent of N?
After fixing 3 non-collinear points (3+2+1=6), every other particle's position is forced by fixed distances, adding 0 DOF.
DOF of a rigid body confined to a plane?
3 — (x,y)(x,y) plus one rotation angle.
DOF of a rigid body fixed at one point?
3 (rotation only).
DOF of a rigid body with a fixed axis (e.g., wheel on axle)?
1 (the spin angle).
DOF of a diatomic (point-mass) molecule?
5 — 3 translation + 2 rotation (spin about bond axis counts 0).
For N rigid particles, how many independent constraints?
3N−63N - 6.
Recall Feynman: explain to a 12-year-old

Imagine a LEGO castle glued solid so no piece can move relative to another. To tell your friend exactly where the castle is and which way it's pointing, how many things do you say? First, where it sits — left/right, forward/back, up/down: 3 numbers. Then which way it faces — tilt it three ways like an airplane (nose up/down, roll, turn): 3 more. That's 6 numbers total, and it doesn't matter if the castle has 10 bricks or 10,000 — because they're all glued, once you place 6 numbers the whole thing is locked. That "6" is what we call the degrees of freedom.

Connections

  • Centre of Mass — the natural reference point for the 3 translational DOF.
  • Moment of Inertia — governs the 3 rotational DOF.
  • Rotational Kinematics — describes how the rotational DOF evolve in time.
  • Constraints and Lagrangian Mechanics — DOF = the count of generalised coordinates.
  • Equipartition Theorem — each DOF carries 12kBT\tfrac12 k_BT of energy (uses the 5 for diatomics).
  • Rolling Motion — a rolling-without-slipping constraint reduces DOF.

Concept Map

too hard to track

defined by

means

allows only

question asked

equals

acts as

subtracted in

give

feed into

smart trick

yields

Real object billions of particles

Rigid body idealization

All pair distances constant

No deformation

Translation and rotation

Degrees of freedom

Independent coordinates needed

Constraint equations

DOF = coords minus constraints

N free particles

3N coordinates

Fix 3 non-collinear points

6 DOF total

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, rigid body ka matlab simple hai: aisa object jismein koi bhi do particles ke beech ki distance kabhi change nahi hoti. Chahe kitna bhi force lagao, woh na khinchta hai na mudta hai. Real life mein perfectly rigid kuch nahi hota, par steel ki rod ya pahiya jaise cheezon ke liye hum maan lete hain ki deformation zero hai — yeh ek useful idealization hai jo physics ko aasaan bana deta hai.

Ab degrees of freedom (DOF) ka funda: yeh batata hai ki body ko poori tarah locate karne ke liye kitne independent numbers chahiye. Ek akela particle ko 3D mein rakhne ke liye (x,y,z)(x,y,z) — yaani 3 numbers. Par jab particles ek rigid body mein jud jaate hain, toh distances fix ho jaati hain, aur extra numbers ki zaroorat nahi padti.

Trick yeh hai: agar tum 3 non-collinear points ki position fix kar do, toh poora body lock ho jaata hai. Pehla point = 3 numbers, doosra (P1 se distance fixed) = 2 numbers, teesra (P1 aur P2 dono se distance fixed) = 1 number. Total 3+2+1=63+2+1 = 6. Isiliye 3D mein har rigid body ke 6 DOF hote hain — chahe usmein 4 particle ho ya 4 crore. In 6 mein se 3 translation (slide x,y,z) aur 3 rotation (spin x,y,z) hote hain.

Yeh important kyun hai? Kyunki agar koi constraint lag jaaye toh DOF kam ho jaate hain — jaise wheel on axle = sirf 1 DOF, ya body fixed at a point = 3 DOF. Exam mein yahi cases puchhe jaate hain, toh hamesha pehle dekho "kya kya fix hai", phir count karo. Yeh concept aage moment of inertia, rotational motion aur equipartition theorem mein bahut kaam aata hai.

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Connections

Centre of mass — definition for system of particlesPhysics · 1.4.9Moment of inertia I = Σmᵢrᵢ² — conceptPhysics · 1.5.5Equipartition theorem — ½k_BT per quadratic degree of freedomPhysics · 2.4.14Inertia tensor — principal axes, principal momentsPhysics · 2.1.22