Nonholonomic mechanics · Free motion

Chaplygin Sleigh

Can a sleigh on a knife edge slow its turning without losing any energy? Yes. The blade only forbids sideways motion at its contact, so its push acts across the blade's motion and does no work. That push shifts energy from turning into forward motion; the total stays fixed. The sleigh turns ever more slowly and settles into a nearly straight glide, mass center first. With the mass center over the blade, the push exerts no turning torque and the sleigh circles forever.

Launch a three-dimensional sleigh that rides on a knife edge. Move its center of mass, launch it backward or put the mass right over the blade. Compare the curved path the blade allows with the path of a freely sliding center of mass.

Blade forward speed $0$

Yaw rate $0$

Mechanical energy $0$

Maximum sampled relative energy drift $0$

Signed lateral reaction $0$

Blade lateral velocity $0$

Geometry & initial state

Change the constraint and see how the motion changes

Signed blade-to-mass offset

a

0.50\,\mathrm m

Radius of gyration about mass center

k

0.40\,\mathrm m

Initial blade forward speed

u_0

-1.00\,\mathrm{m/s}

Initial yaw rate

\omega_0

1.50\,\mathrm{rad/s}

Integration: standard or fine

b_{\rm fine}

0.00\,

Initial heading

\theta_0

20.00\,{}^\circ

The blade can move along the body axis but not sideways. The gold sphere marks the centre of mass; it is not an extra point mass. The total mass is one kilogram, and the radius of gyration sets the moment of inertia. The chassis and supports are schematic.

Cyan traces the centre of mass and gold traces where the blade touches the ground. Dashed grey shows the centre-of-mass path if the blade were removed at release, starting with the same velocity. That reference belongs to a different experiment.

The motion is integrated over eight seconds with fourth-order Runge–Kutta and cubic Hermite interpolation, using 4096 steps for standard and 8192 for fine. The drift shown checks the stored step states; it is not a rigorous error bound. Conserved energy alone does not prove the trajectory is accurate.

The sleigh is an ideal rigid body moving in a plane, with one knife edge that blocks sideways motion. Drag, friction loss at the blade, lifting off and tipping are not modeled. Mass and inertia are set directly; the chassis drawing is only schematic.

Start with the essentials

Focus question: How can rotation fade away without losing mechanical energy?
One-sentence intuition: The sideways constraint moves energy back and forth between translation and rotation. When the mass center is offset, the stable direction of travel has the blade trailing behind the mass.

Core mathematical model

Knife velocity and rigid-body dynamics

\dot x=u\cos\theta,\quad\dot y=u\sin\theta,\quad\dot\theta=\omega,\qquad\dot u=a\omega^2,\quad\dot\omega=-\frac{a u\omega}{k^2+a^2}

The coordinates locate the knife contact, not the mass center. A positive offset points forward along the blade, and the moment of inertia about the mass center equals the mass times the radius of gyration squared.

Energy and the blade's sideways force

E=\frac m2\left[u^2+(k^2+a^2)\omega^2\right],\quad N=\frac{m k^2 u\omega}{k^2+a^2},\quad P_N=Nv_{P,\perp}=0

The mass center moves sideways even though the blade does not. The reaction can be nonzero and still do no work.

Common difficulties

The contact limits velocity, not position

Typical misconceptionReturning a control to its start must undo the motion, and slower turning must mean lost energy.

Better mental modelWatch the total energy stay fixed while its two parts trade off. The constraint does no work.

Run the experiment

01
Experiment 1
Release it backward and watch the forward speed change sign as the energy stays constant.
What to observe: Compare the path with the numerical readouts, which update together.
02
Experiment 2
Choose Mass over blade, then reverse the offset. Compare the circular and straightening paths.
What to observe: Compare the path with the numerical readouts, which update together.
03
Experiment 3
Compare standard and fine integration at the largest yaw rate.
What to observe: Compare the path with the numerical readouts, which update together.

Primary theory: Simoes & Colombo · rolling disk · Chang & Perlmutter · Chaplygin sleigh