1.5.1 · D5Rotational Mechanics
Question bank — Rigid body — definition, degrees of freedom
The recurring master formula you'll lean on:
True or false — justify
True or false: A rigid body with particles has degrees of freedom.
False. Free particles would give , but rigidity locks all inter-particle distances; after 3 non-collinear points are pinned (), every other particle is forced, so DOF saturates at .
True or false: Making a body "more rigid" changes its number of degrees of freedom.
False. Rigidity is all-or-nothing for the ideal model — once distances are frozen, the DOF is regardless of how stiff the material is. Stiffness is a material property, DOF is a counting result.
True or false: A rigid body confined to a 2D plane has DOF.
False. In a plane the centre needs only and there is just one rotation axis (out of the plane), giving DOF, not .
True or false: The number of independent constraints for a rigid body equals .
False. Every pair has a fixed-distance equation, but for those equations are redundant; the number of independent constraints is .
True or false: A diatomic molecule modelled as two point atoms has DOF.
False. Spinning point masses about their own connecting line moves nothing, so that rotational DOF is physically absent: DOF.
True or false: Fixing a single point of a rigid body removes exactly DOF.
True. Fixing one point kills all translations but leaves all rotations about that point, so remain.
True or false: A wheel on a fixed axle and a swinging door have the same number of DOF.
True. Both have a fixed axis, leaving only one spin/swing angle , so each has exactly DOF.
True or false: If you know the positions of any particles of a rigid body, you know the position of all of them.
True — provided the 3 are non-collinear. Their fixed distances to every other particle then pin each remaining particle uniquely; collinear points fail because they don't fix rotation about that line.
Spot the error
Spot the error: "A rigid body has DOF, so a -particle body has DOF."
The formula quoted is for constraints, not DOF. DOF is always ; counts the independent constraints ( here), and .
Spot the error: "To fix a rigid body I must fix points, and any points will do."
The points must be non-collinear. Three points on a straight line leave the body free to rotate about that line, so orientation isn't fully fixed.
Spot the error: "A rigid body always has DOF, period."
Only true when free in 3D. Constraints reduce it: fixed axis , fixed point , confined to a plane . Always count what is constrained first.
Spot the error: "Point 2 in the 3-point derivation contributes coordinates like Point 1."
Point 2 is constrained to a fixed distance from Point 1, so it lives on a sphere and contributes only independent coordinates ().
Spot the error: "A rod that bends slightly under load is not a rigid body, so we can't use rigid-body DOF for it."
The rigid body is an idealization; tiny deformations are dropped (the 80/20 move). For ordinary forces we still treat the rod as rigid and use DOF because the bending is negligible.
Spot the error: "The translational DOF describe how the body spins."
Translation describes where the body sits (sliding along without turning). Spinning is captured by the rotational DOF (roll, pitch, yaw).
Spot the error: "Since a diatomic loses one rotation, a triatomic bent molecule also has DOF."
A non-linear (bent) triatomic is not collinear, so rotation about every axis moves atoms; it keeps all rotations, giving DOF. Only linear arrangements of point masses lose one.
Why questions
Why do we subtract constraints instead of just counting the fixed-distance equations directly?
Because those pair-distance equations are not all independent; blindly counting over-counts. Subtracting the independent constraints () from correctly yields .
Why does the DOF count stop growing once you pass non-collinear points?
After such points fix both position and orientation, every further particle's location is determined by its fixed distances to them, adding new DOF.
Why does rotation about the bond axis not count as a DOF for a diatomic molecule?
The atoms are treated as point masses lying on that axis, so rotating about the line leaves every particle exactly where it was — no configuration change means no independent coordinate.
Why is a rigid body's DOF split as and never, say, ?
In 3D space there are exactly independent translation directions and exactly independent rotation axes; no fourth way to slide or turn exists, so the split is forced.
Why does confining a body to a plane leave only rotational DOF instead of ?
A rotation must keep the body in the plane, and only the axis perpendicular to the plane does that; rotating about an in-plane axis would tip the body out of the plane.
Why must the master formula use independent coordinates, not just any coordinates?
Dependent coordinates can be computed from others via constraints, so they carry no new information; only independent numbers genuinely pin down the configuration.
Edge cases
Edge case: How many DOF does a rigid body have if two of its points are fixed?
DOF. Fixing two points fixes an axis (the line through them), and the only remaining freedom is rotation about that axis — the spin angle .
Edge case: What is the DOF of a rigid body when three non-collinear points are fixed?
DOF. The body is completely frozen — position and orientation are both fully determined, so no independent coordinate remains.
Edge case: Does a single free point particle have rotational DOF?
No — a point has no orientation, so it has only translational DOF and rotational DOF.
Edge case: For which do the pair-distances happen to all be independent?
Exactly , where . For , , so some pair-constraints become redundant.
Edge case: A rigid body slides frictionlessly on a table but can also lift off — how many DOF?
. Being able to leave the surface means it is effectively free in 3D; the table only becomes a constraint if it forces the body to stay in contact (which would drop it to ).
Edge case: What happens to the DOF of a rigid rod (many particles on one line) free in 3D?
It has DOF, not — like a linear molecule, rotation about its own long axis moves no (point-like) particle, so one rotational DOF vanishes. A rod with real thickness recovers the full .
Recall One-line summary of the whole bank
Free particles give ; rigidity removes independent constraints, leaving for a body free in 3D — then every constraint (axis, point, plane, collinearity) chips that downward, and linear point-mass bodies quietly lose one rotation.