Rigid body dynamics

Rigid body dynamics studies translation and rotation of bodies whose internal distances are fixed.

Two equal uniform rods of length $2l$ and mass $M$, one pivoted to the floor and the other endpoint sliding on it, have arch coordinates $B=(2l\cos \theta ,2l\sin \theta )$ and $C=(4l\cos \theta ,0)$. The kinetic energy decomposition about the center of mass gives $T=\frac{4}{3}M{l}^2(1+3{\sin }^2\theta ){\dot{\theta }}^2$, while gravitational potential energy is $2Mgl\sin \theta$. Conservation of energy from rest at angle $\alpha$ gives the displayed equation and hinge impact speed $\sqrt{6gl\sin \alpha }$, provided contact is maintained. The endpoint's normal reaction becomes negative during this constrained fall if $\sin \alpha >35/54$; a unilateral floor then permits detachment and the constrained formula no longer describes the entire motion.

In solid-body rotation the angular velocity is independent of distance from the rotation axis, so tangential speed is proportional to cylindrical radius. A galactic region with this galaxy rotation curve has radial epicyclic frequency $\kappa =2\Omega$.

A symmetric top is a rigid body with two equal principal moments of inertia, conventionally $I_1=I_2$. Its third principal axis is its symmetry axis.

A heavy symmetric top is a symmetric top with one point fixed in a uniform gravitational field. Its vertical and body-axis rotational symmetries give two cyclic Euler angles.

A heavy symmetric top precesses uniformly when its inclination is constant and its azimuthal speed is constant. For a nonvertical inclination, the condition is $I_1\cos \theta {\Omega }^2-I_3{\omega }_3\Omega +Mgl=0$. The real-root condition depends on whether the body axis points above or below the horizontal.

Precession is the rotation of a body's spin axis about another direction, such as the vertical or the conserved angular-momentum vector.

Angular velocity is the vector whose direction gives the instantaneous rotation axis and whose magnitude gives the angular speed.

Angular acceleration is the time derivative of angular velocity, $\mathbf{\alpha }=d\mathbf{\omega }/dt$. For rotation about a fixed axis, its signed scalar component obeys $\tau =I\alpha$ when the axial moment of inertia is constant and $\tau$ is the total applied torque about that axis.

Angular speed is the nonnegative magnitude of angular velocity: $\omega =|\mathbf{\omega }|$. It measures the rate of rotation about an instantaneous axis without specifying its direction. For uniform circular motion of radius $R$, the ordinary speed is $v=R\omega$.

A fixed space frame is an inertial orthonormal basis whose axes do not rotate with the body.

A principal body frame is an orthonormal basis fixed in the body and aligned with the principal axes of its inertia tensor. In this frame the inertia tensor is diagonal with principal moments $I_1,I_2,I_3$.

A principal axis is an eigenvector direction of the inertia tensor. Choosing three orthogonal principal axes diagonalizes the inertia tensor.

If $e_i$ is fixed in a rigid body with angular velocity $\omega$, then its derivative in the fixed space frame is

If a vector $k$ is fixed in space while a body frame rotates with angular velocity $\omega$, its body-frame components satisfy

In principal-axis body coordinates, torque-free angular velocity satisfies

The Euler equations for a torque-free rigid body conserve rotational kinetic energy and squared angular momentum,Their common level sets are the intersection of two ellipsoids in angular-velocity space.

Eliminating ${\omega }_1^2$ and ${\omega }_2^2$ between the conserved energy and squared angular momentum givesThe right-hand side is a quartic polynomial in ${\omega }_3$; its real turning points delimit the nutation of the body-frame angular velocity.

For principal moments $I_1=I_2=I$ and symmetry-axis unit vector $e_3$, angular momentum and angular velocity obeyThus $L$, $\omega$, and the symmetry axis are coplanar at every instant.

For $I_1<I_2<I_3$, steady torque-free rotation about the first or third principal axis is stable, while rotation about the intermediate principal axis is unstable.

Linearize the Euler equations for a torque-free rigid body about rotation with angular speed $\Omega$ along the third principal axis. The transverse perturbations obeyand henceCyclic permutation gives the corresponding equations about the other two axes. For $I_1<I_2<I_3$, the squared frequency is positive about axes one and three, whereas the intermediate-axis equation has a positive exponential-growth coefficient.

On the invariant level $L^2=2E{I}_2$, the torque-free rigid-body orbit is a separatrix joining the two intermediate-axis rotations. Withone orientation has ${\omega }_2(t)=\mu \tanh (\lambda (t-t_0))$; the other components are constant multiples of $\mathrm{sech}(\lambda (t-t_0))$.

Moment of inertia is the mass-weighted squared distance from a rotation axis and determines rotational kinetic energy.

A uniform solid sphere of mass $M$ and radius $a$ has the indicated moment of inertia about any diameter. Integrating the squared perpendicular distance against uniform mass density gives $I=8\pi \rho a^5/15=2M{a}^2/5$. It is different from the value for a thin spherical shell and determines the rotational contribution to kinetic energy during rolling without slipping.

The mass second-moment tensor about a centre of mass is $I_{jk}=\int y_j{y}_{k}dm$. It measures the shape of a mass distribution and appears in tidal torque theory. It differs from the mechanical inertia tensor, which is ${\mathcal{I}}_{jk}=\mathrm{tr}(I){\delta }_{jk}-I_{jk}$. For a uniform ellipsoid of mass $M$ and semiaxes $a_1,a_2,a_3$, its principal values are $M{a}_j^2/5$.

The inertia tensor about a point isIt maps angular velocity to angular momentum.

The principal moments of inertia are the eigenvalues of the inertia tensor, measured along its principal axes.

For torque-free rigid-body motion, the kinetic-energy ellipsoid fixed in the body rolls without slipping on the invariable plane perpendicular to the conserved angular momentum. Its point of contact is the angular-velocity vector.

The moment about a parallel axis equals the centre-of-mass moment plus mass times the squared axis separation.

For a planar lamina in the $xy$-plane, the moments about three mutually perpendicular axes through one point satisfyThis follows pointwise from $x^2+y^2=y^2+x^2$ in the defining mass integrals.

For a uniformly filled, negligibly light disc of fixed radius $R$, remaining mass $m(t)$, angular velocity $\omega (t)$, and tangential exhaust speed $u$ relative to its rim, angular-momentum balance givesAt zero relative exhaust speed, $\omega /m$ is constant.

For a uniform solid cone of height $l$, base radius $R$, and mass $M$, whose symmetry axis points from vertex to base, the centre of mass lies at $3l/4$ from the vertex. The principal moments at the vertex are

Consider the uniform solid of revolution about the $x_3$-axis whose boundary isFor total mass $M$, the center of mass lies a distance $3h/5$ from the pointed origin, and the principal moments about that origin are

Euler angles describe the orientation of a rigid body through three successive angular rotations. Their convention specifies the axes and order of the rotations. For a symmetric top, a common convention takes $\theta$ as the inclination, $\varphi$ as precession and $\psi$ as spin about the body axis.

With the convention in which $\theta$ is inclination, $\varphi$ is precession, and $\psi$ is body-axis spin, an axisymmetric body's kinetic energy is

For a fixed point a distance $a$ from the centre of mass and inclination measured from the upward vertical,The spin and precession angles are cyclic.

A Lagrange top is an axisymmetric rigid body with one point fixed on its symmetry axis, moving in a uniform gravitational field with its center of mass on that axis.

For a Lagrange top whose inclination $\theta$ is measured from the upward vertical, steady precession at angular speed $\Omega$ obeysThus steady precession at a prescribed $0<\theta <\pi /2$ exists precisely when

With transverse and axial moments $I,J$, the Hamiltonian is

Fixing the cyclic momenta $p_{\psi }=\lambda$ and $p_{\varphi }=\mu$ leaves the canonical pair $(\theta ,p_{\theta })$ and a one-degree-of-freedom Hamiltonian with effective potential

At the upright spinning state, $p_{\varphi }=p_{\psi }=\lambda$. The reduced potential is locally minimized at zero inclination whenso sufficiently large axial angular momentum stabilizes the gravitationally inverted top.

Rolling without slipping imposes that the contact point is instantaneously at rest, giving v=R omega.