Lecture11(1)

Lecture 11 Dynamics of Rigid Bodies

We now consider the motion of rigid bodies, extended objects whose sizes and shapes are fixed. The assumption of rigidity is equivalent to assuming that the elastic moduli (resistance to elastic deformations) are infinite. The motion of rigid bodies can be decomposed into that of their centers of mass and rotations about this point, which are characterized by 3 angles in 3 dimensions (such as the Euler angles).

Inertia tensor

We distinguish between the position of particles in a fixed (inertial) coordinate system and a system rotating about a fixed point in the rigid body (the body or rotating system). If r is the position of particle with respect to the fixed point P, the velocity of the particle in the fixed coordinate system is:

where V is the velocity of the fixed point and is the rotation vector about an axis parallel to that passes through P. The kinetic energy can be expressed in terms of this velocity:

The position of the center of mass is:

where M is the total mass. If P is chosen to be the center of mass, the middle term in the above expression vanishes and the kinetic energy can be decomposed into translation and rotational components:

Using the vector identity:

the rotational kinetic energy can be rewritten as:

We define the summation over as the inertia tensor:

yielding:

In the limit that is an eigenvector of I, this reduces to our traditional expression for the kinetic energy in terms of the momenta of inertia and angular rotation. The inertia tensor is symmetric; its diagonal elements Iii are called the moments of inertia while the negative of its off-diagonal elements -Iij are called the products of inertia. For a continuous body, the inertia tensor is given by:

Example: inertia tensor of a uniform cube about a corner

If we have a cube of density , side b, and mass M =b3, the above equation yields:

This gives us the desired inertia tensor.

Angular momentum

The angular momentum about a point P in the body coordinate system is:

where in the last equation we have made use of the vector identity:

Two convenient choices for this point are a point that is fixed in the inertial frame or the center of mass. If we expand the above expression for L in index notation, we can show:

This shows that the angular momentum L and rotation will not be parallel unless is an eigenvector of I. We will explore this possibility in later examples. This definition of L also shows that:

Example: frequency of pendulum with 2 masses

Let's consider a pendulum of length b with a bob of mass m1 suspended at its end and a second mass m2suspended halfway down its length. What is the frequency of small oscillations?

Let's use Lagrangian methods. To construct the Lagrangian, we first must find the kinetic and potential energies. To find the kinetic energy, we need to find the inertia tensor. We define this tensor about the pivot point P of the pendulum, using a Cartesian basis with the x direction to the right and the y direction being vertical. From the expression above, this tensor is:

If rotations are restricted to the z direction:

the kinetic energy is:

The potential energy is:

The Lagrange equation is therefore:

For small oscillations sin , the frequency of oscillations is therefore:

Principal axes of inertia

Expressions involving the inertia tensor are greatly simplified when a fixed point P and axes are chosen such that the inertia tensor is diagonalized. These axes are known as the principal axes of inertia. If the inertia tensor is diagonal:

implying that a rotation about one of the principal axes will lead to L || . Finding the principal axes of inertia is thus equivalent to finding the eigenvectors of I. The three eigenvalues I1, I2, and I3 corresponding to these eigenvectors are known as the principal moments of inertia. Since the inertia tensor is Hermitian, it has three orthogonal eigenvectors with real eigenvalues. An inertia tensor with three degenerate eigenvalues (I1 = I2 = I3) is known as a spherical top, if two eigenvalues are degenerate (I1 = I2 I3) it is called a symmetric top, and if all three eigenvalues are distinct (I1 I2 I3) it is called an asymmetric top. Symmetric tops with (I1 < I3) are prolate (like a football) while those with (I1 > I3) are oblate (like a frisbee).

Example: find the principal moments of inertia and principle axes of a cube about its corner

Earlier in this lecture, we calculated the inertia tensor of a cube about one corner and found:

The principal moments of inertia are the eigenvalues of this tensor, which can be found solving the equation:

We see that the cube is an oblate symmetric top. The minor axis is in the direction of the eigenvector corresponding to the eigenvalue = Mb2/6:

i.e. along the cube's diagonal, while the other two principal axes of inertia span the plane perpendicular to this vector, for example:

In this basis, the inertia tensor becomes:

Steiner's parallel-axis theorem:

If we change the fixed point in the body system from r to R = r + a without rotating the axes, the inertia tensor becomes:

However, if the inertia tensor I was defined about the body's center of mass, the 3rd term on the right-hand side vanishes. This implies that if we know the inertia tensor J about any point, we can find the inertia tensor I about the center of mass:

where a is the vector pointing from the center of mass to the point about which J is defined.

Example: find the inertia tensor about a cube's center of mass

Earlier, we found the inertia tensor J about a corner of the cube. The vector from the center of mass to the corner is:

Inserting these expressions into the right-hand side of Steiner's parallel-axis theorem yields:

We see that a cube is a spherical top about its center of mass; the principal moments of inertia are Mb2/6 about any axis and any choice of orthogonal axes can serve as principal axes. The inertia tensor about the center of mass contains insufficient information to distinguish spheres from cubes.

Transformation properties of the inertia tensor:

We have shown that the inertia tensor relates the angular momentum and angular velocity vectors:

Requiring L and are to transform as vectors under rotations fixes the transformation properties of the

inertia tensor. If vectors transform under rotations like:

then:

If we matrix multiply both sides of the above equation by on the left:

where in the 2nd equation we have made use of the fact that rotation matrices are unitary:

and we have defined:

In much simpler matrix notation:

implying:

This is how rank-2 tensors transform, so we have shown that the inertia tensor is in fact a tensor. This equation should look familiar from your linear algebra class. Since I is a real, symmetric matrix, we can always find a matrix t of orthonormal eigenvectors that that diagonalizes I'. The diagonal elements of the matrix I' are the principal moments of inertia of the inertia tensor I.

Example: diagonalizing the inertia tensor for a cube with origin at its corner

We showed in a previous example that the inertia tensor of a cube of uniform density with origin at its corner has three orthogonal principal axes, with the non-degenerate axis along the cube's diagonal. We can rotate explicitly into a basis with x1 along this diagonal. Rotating about an axis perpendicular to the plane spanned by the x1 and x'1 axes is accomplished by the matrix:

We can place the x'1 axis along the diagonal by first rotating by 45 degrees about the x1 axis, then rotating by cos-1 (2/3)1/2 about the new x'2 axis. These rotations are accomplished by the combined rotation matrix:

When we use this matrix to calculate I' according to the above transformation formula, we indeed find:

Eulerian angles

In the general case, the axes in the inertial coordinate system x' will not be aligned with those in the body coordinate system x. In three dimensions, we need three angles to fully specify a rotation to transform from the inertial to body coordinate systems. Although there is no unique choice of these three angles, the Eulerian angles first defined by Leonhard Euler in 1776 are often a convenient choice.

The first of the three rotations is by an angle about the x1' axis:

The second rotation is by an angle about the new x1'' axis, called the line of nodes:

The third and final rotation is by an angle about the new x3''' axis:

The application in succession of these three rotations yields a total rotation given by the matrix:

Because infinitesimal rotations commute, angular velocities can be expressed as vectors including rotations described by the Eulerian angles:

It will be useful to express this rotation in the x (body) coordinate system:

Euler's equation for a rigid body

Let us begin by considering force-free motion, in which case we can boost to a frame in which the center of mass is at rest (no translation kinetic energy). In this case:

where the body axes have been chosen to be the principal axes. We can choose the three Eulerian angles as our generalized coordinates, in which case the Lagrange equation associated with is:

Using the above expressions for and T, we find:

However, our decision to label a particular principal axis x3 was entirely arbitrary, so we can permute this equation to find similar equations for the remaining two components of the angular velocity:

These are the Euler equations for force free motion of an asymmetric top. If the rigid body experiences a torque N in the inertial (fixed) frame, its angular momentum will evolve according to the equation:

The component of this equation in the x3 direction is:

We can again permute the indices to yield the remaining two equations for the evolution of the angular velocity due to an external torque:

Example: barbell

Consider a barbell consisting of two masses m/2 connected by a rigid massless rod of length 2b. What is the inertia tensor of this body, what is its angular momentum if it is rotating about a fixed axis through its center of mass inclined at an angle with respect to its length, and what torque is required to maintain this rotation?

The principal axes of this body are clearly along the rod and in the plane perpendicular to the rod. If

we choose the x3 direction to be along the rod and the x2 direction to be in the plane spanned by the rod and , we find:

The angular velocity is:

The angular momentum is therefore:

According the the Euler equations for an asymmetric top, the torque required to maintain this angular momentum is:

Force-free motion of a symmetric top

A symmetric top has I1 = I2; according to the above Euler equations 3 is constant and:

If we differentiate the first equation with respect to time and substitute the second equation, we find:

We should all know the solution to this equation by now:

We see that the vector precesses in a cone about the principal axis e3 with nondegenerate eigenvalue with opening angle

and precession frequency . This precession cone is called the body cone. Because this is force-free motion, both the angular momentum L and rotational kinetic energy

are constants, implying that in the inertial frame the angular velocity precesses about L on what is called the space cone. Since

we know that L, , and e3 all lie in a plane. The motion in the inertial frame can therefore be described as the space and body cones rolling on one another.

If we define the x3' axis to point in the direction of L and the x2 axis to lie in the plane spanned by L and , we have in the body frame:

If is the angle between and the x3 axis, then

The angular momentum can also be written as L = I , in components:

These relation imply:

If our symmetric top is prolate (I1 = I2 > I3), this implies that < and and have opposite signs. Conversely, an oblate top (I1 = I2 < I3) has > and and have the same signs:

Motion of a symmetric top (I1 = I2 I3) with one point fixed

Expressed in terms of the Euler angles , , and , the Lagrangian for this system is

We can define conjugate momenta

Since L is independent of and , both of these momenta are conserved. They are in fact the angular momenta about the x3' and x3 axes respectively; conservation of these angular momenta is consistent with the torque r x (-mg) being directed along the line of nodes. Since the system is conservative, the energy

is conserved, as is the quantity

where we have expressed d/dt in terms of the momenta in the last equation. We can define an effective potential

in terms of which

This first-order ordinary differential equation can be solved to yield

from which differential equations for (t) and (t) can be obtained from the expression for the conserved conjugate momenta. We plot the effective potential below:

The functional form of the effective potential implies that it will have a minimum at some value 0 only if

If at d/dt = 0 at = 0, the top will precess in a cone with fixed opening angle about x3'. In the more general case, will undergo nutation between the turning points. The precession d/dt may change sign during the motion depending on the values of p and p.

Stability of rigid body rotation

Let us consider a fully asymmetric top (I1 < I2 < I3) that is rotating about an axis that is nearly equal to the principal axis e1:

where ,

axis with intermediate principal moment of inertia are unstable. If two of the principal moments are equal, the frequencies about those axes vanish and perturbations are linearly unstable. Perturbations about the nondegenerate axis are stable.