## Rigid-Body Dynamics

Below are selected topics from *rigid-body dynamics*, a subtopic
of *classical mechanics* involving the use of Newton's laws of
motion to solve for the motion of *rigid bodies* moving in 1D,
2D, or 3D space.^{B.11} We may think
of a rigid body as a *distributed mass*, that is, a mass that has
length, area, and/or volume rather than occupying only a single point
in space. Rigid body models have application in stiff strings
(modeling them as disks of mass interconnect by ideal springs), rigid
bridges, resonator braces, and so on.

We have already used Newton's to formulate mathematical dynamic
models for the ideal *point-mass* (§B.1.1), *spring*
(§B.1.3), and a simple mass-spring system (§B.1.4).
Since many physical systems can be modeled as assemblies of masses and
(normally damped) springs, we are pretty far along already. However,
when the springs interconnecting our point-masses are very stiff, we
may approximate them as rigid to simplify our simulations. Thus,
rigid bodies can be considered mass-spring systems in which the
springs are so stiff that they can be treated as rigid massless rods
(infinite spring-constants , in the notation of §B.1.3).

So, what is new about *distributed* masses, as opposed to the
point-masses considered previously? As we will see, the main new
ingredient is *rotational dynamics*. The *total momentum*
of a rigid body (distributed mass) moving through space will be
described as the *sum* of the *linear momentum* of its
*center of mass* (§B.4.1 below) plus the *angular
momentum* about its center of mass (§B.4.13 below).

### Center of Mass

The *center of mass* (or *centroid*) of a rigid body is
found by averaging the spatial points of the body
weighted by the mass of those points:^{B.12}

*mass-weighted average location*of the object. For a continuous mass distribution totaling up to , we can write

*mass density*at the point . The total mass is

A nice property of the center of mass is that gravity acts on a
far-away object as if all its mass were concentrated at its center of
mass. For this reason, the center of mass is often called the
*center of gravity*.

#### Linear Momentum of the Center of Mass

Consider a system of point-masses , each traveling with vector velocity , and not necessarily rigidly attached to each other. Then the total momentum of the system is

Thus, the *momentum*
of any collection of masses
(including rigid bodies) equals the total mass times the
*velocity of the center-of-mass*.

####

Whoops, No Angular Momentum!

The previous result might be surprising since we said at the outset
that we were going to *decompose* the total momentum into a sum
of linear plus angular momentum. Instead, we found that the total
momentum is simply that of the center of mass, which means any angular
momentum that might have been present just went away. (The center of
mass is just a point that cannot rotate in a measurable way.) Angular
momentum does not contribute to linear momentum, so it provides three
new ``degrees of freedom'' (three new energy storage dimensions, in 3D
space) that are ``missed'' when considering only linear momentum.

To obtain the desired decomposition of momentum into linear plus
angular momentum, we will choose a fixed reference point in space
(usually the center of mass) and then, with respect to that reference
point, decompose an arbitrary mass-particle travel direction into the
sum of two mutually orthogonal vector components: one will be the
vector component pointing *radially* with respect to the fixed
point (for the ``linear momentum'' component), and the other will be
the vector component pointing *tangentially* with respect to the
fixed point (for the ``angular momentum''), as shown in
Fig.B.3. When the reference point is the center of mass, the
resultant radial force component gives us the force on the center of
mass, which creates *linear momentum*, while the net tangential
component (times distance from the center-of-mass) give us a resultant
*torque* about the reference point, which creates *angular
momentum*. As we saw above, because the tangential force component
does not contribute to linear momentum, we can simply sum the external
force vectors and get the same result as summing their radial
components. These topics will be discussed further below, after some
elementary preliminaries.

### Translational Kinetic Energy

The *translational kinetic energy* of a collection of masses
is given by

More generally, the *total energy* of a collection of masses
(including distributed and/or rigidly interconnected point-masses) can
be expressed as the *sum* of the *translational* and
*rotational* kinetic energies [270, p.
98].

### Rotational Kinetic Energy

The *rotational kinetic energy* of a rigid assembly of masses (or
mass distribution) is the sum of the rotational kinetic energies of
the component masses. Therefore, consider a point-mass
rotating^{B.13} in a circular orbit
of radius and angular velocity (radians per second), as
shown in Fig.B.4. To make it a closed system, we can
imagine an effectively infinite mass at the origin
. Then the
speed of the mass along the circle is , and its kinetic
energy is
. Since this is what we want
for the *rotational* kinetic energy of the system, it is
convenient to define it in terms of angular velocity in
radians per second. Thus, we write

where

is called the

*mass moment of inertia*.

### Mass Moment of Inertia

The *mass moment of inertia* (or simply *moment of
inertia*), plays the role of *mass* in rotational dynamics, as
we saw in
Eq.(B.7) above.

The mass moment of inertia of a rigid body, relative to a given axis of rotation, is given by a weighted sum over its mass, with each mass-point weighted by the square of its distance from the rotation axis. Compare this with the center of mass (§B.4.1) in which each mass-point is weighted by its vector location in space (and divided by the total mass).

Equation (B.8) above gives the moment of inertia for a single point-mass
rotating a distance from the axis to be . Therefore,
for a rigid collection of point-masses ,
,^{B.14} the
moment of inertia about a given axis of rotation is obtained by adding
the component moments of inertia:

where is the distance from the axis of rotation to the th mass.

For a continuous mass distribution, the moment of inertia is given by integrating the contribution of each differential mass element:

(B.10) |

where is the distance from the axis of rotation to the mass element . In terms of the

*density*of a continuous mass distribution, we can write

#### Circular Disk Rotating in Its Own Plane

For example, the moment of inertia for a uniform circular disk of total mass and radius , rotating in its own plane about a rotation axis piercing its center, is given by

#### Circular Disk Rotating About Its Diameter

The moment of inertia for the same circular disk rotating about an axis in the plane of the disk, passing through its center, is given by

### Perpendicular Axis Theorem

In general, for any 2D distribution of mass, the moment of inertia about an axis orthogonal to the plane of the mass equals the sum of the moments of inertia about any two mutually orthogonal axes in the plane of the mass intersecting the first axis. To see this, consider an arbitrary mass element having rectilinear coordinates in the plane of the mass. (All three coordinate axes intersect at a point in the mass-distribution plane.) Then its moment of inertia about the axis orthogonal to the mass plane is while its moment of inertia about coordinate axes within the mass-plane are respectively and . This, the perpendicular axis theorem is an immediate consequence of the Pythagorean theorem for right triangles.

### Parallel Axis Theorem

Let denote the moment of inertia for a rotation axis passing
through the center of mass, and let denote
the moment of inertia for a rotation axis parallel to the first
but a distance away from it. Then the *parallel axis theorem*
says that

*point mass*rotating a distance from the center of rotation.

### Stretch Rule

Note that the moment of inertia does not change when masses are moved
along a vector parallel to the axis of rotation (see, *e.g.*,
Eq.(B.9)). Thus, any rigid body may be ``stretched'' or
``squeezed'' parallel to the rotation axis without changing its moment
of inertia. This is known as the *stretch rule*, and it can be
used to simplify geometry when finding the moment of inertia.

For example, we saw in §B.4.4 that the moment of inertia
of a point-mass a distance from the axis of rotation is given
by . By the stretch rule, the same applies to an ideal
*rod* of mass parallel to and distance from the axis of
rotation.

Note that mass can be also be ``stretched'' along the circle of
rotation without changing the moment of inertia for the mass about
that axis. Thus, the point mass can be stretched out to form a
mass *ring* at radius about the axis of rotation without
changing its moment of inertia about that axis. Similarly, the ideal
rod of the previous paragraph can be stretched tangentially to form a
*cylinder* of radius and mass , with its axis of symmetry
coincident with the axis of rotation. In all of these examples, the
moment of inertia is about the axis of rotation.

### Area Moment of Inertia

The *area moment of inertia* is the second moment of an area
around a given axis:

Comparing with the definition of *mass moment of inertia* in
§B.4.4 above, we see that mass is replaced by area in
the area moment of inertia.

In a planar mass distribution with total mass uniformly
distributed over an area (*i.e.*, a constant mass density of
), the mass moment of inertia is given by the area
moment of inertia times mass-density :

### Radius of Gyration

For a planar distribution of mass rotating about some axis in the
plane of the mass, the *radius of gyration* is the distance from
the axis that all mass can be concentrated to obtain the same mass
moment of inertia. Thus, the radius of gyration is the ``equivalent
distance'' of the mass from the axis of rotation. In this context,
*gyration* can be defined as *rotation* of a planar region
about some axis lying in the plane.

For a bar cross-section with area , the radius of gyration is given by

where is the area moment of inertia (§B.4.8) of the cross-section about a given axis of rotation lying in the plane of the cross-section (usually passing through its centroid):

#### Rectangular Cross-Section

For a rectangular cross-section of height and width , area , the area moment of inertia about the horizontal midline is given by

The radius of gyration can be thought of as the ``effective radius'' of the mass distribution with respect to its inertial response to rotation (``gyration'') about the chosen axis.

Most cross-sectional shapes (*e.g.*, rectangular), have at least two
radii of gyration. A circular cross-section has only one, and its
radius of gyration is equal to half its radius, as shown in the next
section.

#### Circular Cross-Section

For a circular cross-section of radius , Eq.(B.11) tells us that the squared radius of gyration about any line passing through the center of the cross-section is given by

Using the elementrary trig identity , we readily derive

For a circular *tube* in which the mass of the cross-section lies
within a circular *annulus* having inner radius and outer
radius , the radius of gyration is given by

### Two Masses Connected by a Rod

As an introduction to the decomposition of rigid-body motion into
*translational* and *rotational* components, consider the
simple system shown in Fig.B.5. The excitation force
density^{B.15} can be
applied anywhere between and along the connecting rod.
We will deliver a vertical impulse of momentum to the mass on the
right, and show, among other observations, that the total kinetic
energy is split equally into (1) the *rotational* kinetic energy
about the center of mass, and (2) the *translational* kinetic
energy of the total mass, treated as being located at the center of
mass. This is accomplished by defining a *new frame of
reference* (*i.e.*, a moving coordinate system) that has its origin at
the center of mass.

First, note that the driving-point impedance (§7.1)
``seen'' by the driving force varies as a function of .
At , The excitation sees a ``point mass'' , and no
rotation is excited by the force (by symmetry). At , on the
other hand, the excitation
only sees mass at time
0, because the vertical motion of either point-mass initially only
rotates the other point-mass via the massless connecting rod. Thus,
an observation we can make right away is that the *driving point
impedance seen by depends on the striking point and,
away from , it depends on time as well*.

To avoid dealing with a time-varying driving-point impedance, we will
use an impulsive force input at time . Since momentum is the
time-integral of force (
), our
excitation will be a *unit momentum* transferred to the two-mass
system at time 0.

#### Striking the Rod in the Middle

First, consider . That is, we apply an upward unit-force impulse at time 0 in the middle of the rod. The total momentum delivered in the neighborhood of and is obtained by integrating the applied force density with respect to time and position:

*velocity of the center of mass*, again obvious by symmetry. Continuing to refer to Fig.B.5, we have

The kinetic energy of the system after time zero is

#### Striking One of the Masses

Now let . That is, we apply an impulse of vertical momentum to the mass on the right at time 0.

In this case, the unit of vertical momentum is transferred entirely to the mass on the right, so that

Note that the velocity of the center-of-mass is the
*same* as it was when we hit the midpoint of the rod. This is an
important general equivalence: The sum of all external force vectors
acting on a rigid body can be applied as a single resultant force
vector to the total mass concentrated at the center of mass to find
the linear (translational) motion produced. (Recall from §B.4.1
that such a sum is the same as the sum of all radially acting external
force components, since the tangential components contribute only to
rotation and not to translation.)

All of the kinetic energy is in the mass on the right just after time zero:

However, after time zero, things get more complicated, because the mass on the left gets dragged into a rotation about the center of mass.

To simplify ongoing analysis, we can define a *body-fixed frame
of reference*^{B.16} having its origin at the center of mass. Let
denote a velocity in this frame. Since the velocity of the center of
mass is
, we can convert any velocity in the
body-fixed frame to a velocity in the original frame by adding
to it, *viz.*,

In the body-fixed frame, all kinetic energy is *rotational* about
the origin. Recall (Eq.(B.9)) that the moment of inertia for this
system, with respect to the center of mass at , is

*rotational kinetic energy*(§B.4.3) is found to be

*half*of the kinetic energy we computed in the original ``space-fixed'' frame (Eq.(B.13) above). The other half is in the

*translational kinetic energy*not seen in the body-fixed frame. As we saw in §B.4.2 above, we can easily calculate the translational kinetic energy as that of the total mass traveling at the center-of-mass velocity :

In summary, we defined a moving body-fixed frame having its origin at the center-of-mass, and the total kinetic energy was computed to be

It is important to note that, after time zero, both the linear
momentum of the center-of-mass (
), and the angular momentum in the body-fixed frame
(
) remain
*constant* over time.^{B.17} In the original space-fixed
frame, on the other hand, there is a complex transfer of momentum back
and forth between the masses after time zero.

Similarly, the translational kinetic energy of the total mass, treated as being concentrated at its center-of-mass, and the rotational kinetic energy in the body-fixed frame, are both constant after time zero, while in the space-fixed frame, kinetic energy transfers back and forth between the two masses. At all times, however, the total kinetic energy is the same in both formulations.

### Angular Velocity Vector

When working with rotations, it is convenient to define the
*angular-velocity vector* as a vector
pointing
along the *axis of rotation*. There are two directions we could
choose from, so we pick the one corresponding to the *right-hand
rule*, *i.e.*, when the fingers of the right hand curl in the direction
of the rotation, the thumb points in the direction of the angular
velocity vector.^{B.18} The
*length*
should obviously equal the angular
velocity . It is convenient also to work with a unit-length
variant
.

As introduced in Eq.(B.8) above, the mass moment of inertia is
given by where is the distance from the (instantaneous)
axis of rotation to the mass located at
. In
terms of the angular-velocity vector
, we can write this as
(see Fig.B.6)

where

Using the *vector cross product* (defined in the next section),
we will show (in §B.4.17) that can be written more succinctly as

### Vector Cross Product

The *vector cross product* (or simply *vector product*, as
opposed to the *scalar product* (which is also called the
*dot product*, or *inner product*)) is commonly used in
*vector calculus*--a basic mathematical toolset used in
mechanics [270,258],
acoustics [349], electromagnetism [356], quantum
mechanics, and more. It can be defined symbolically in the form of
a *matrix determinant*:^{B.19}

where denote the

*unit vectors*in . The cross-product is a vector in 3D that is orthogonal to the plane spanned by and , and is oriented positively according to the

*right-hand rule*.

^{B.20}

The second and third lines of Eq.(B.15) make it clear that . This is one example of a host of identities that one learns in vector calculus and its applications.

#### Cross-Product Magnitude

It is a straightforward exercise to show that the cross-product
magnitude is equal to the product of the vector lengths times the sine
of the angle between them:^{B.21}

where

*vector cosine*of the angle between two vectors is given by their inner product divided by the product of their norms [451].)

To derive Eq.(B.16), let's begin with the cross-product in matrix form as using the first matrix form in the third line of the cross-product definition in Eq.(B.15) above. Then

where
denotes the identity matrix in
,
denotes the orthogonal-projection matrix onto
[451],
denotes the projection matrix onto
the orthogonal *complement* of
,
denotes the component of
orthogonal to
, and we used the fact that orthogonal projection matrices
are *idempotent* (*i.e.*,
) and
*symmetric* (when real, as we have here) when we replaced
by
above. Finally,
note that the length of
is
, where is the angle
between the 1D subspaces spanned by
and
in the plane
including both vectors. Thus,

*orthogonal projection*of onto ( ) or vice versa ( ). In this geometric picture it is clear that the cross-product magnitude is maximized when the vectors are orthogonal, and it is zero when the vectors are collinear. It is ``length times orthogonal length.''

The direction of the cross-product vector is then taken to be orthogonal to both and according to the right-hand rule. This orthogonality can be checked by verifying that . The right-hand-rule parity can be checked by rotating the space so that and in which case . Thus, the cross product points ``up'' relative to the plane for and ``down'' for .

#### Mass Moment of Inertia as a Cross Product

In Eq.(B.14) above, the mass moment of inertia was expressed in terms of orthogonal projection as , where . In terms of the vector cross product, we can now express it as

#### Tangential Velocity as a Cross Product

Referring again to Fig.B.4, we can write the
*tangential velocity vector*
as a vector cross product of
the angular-velocity vector
(§B.4.11) and the position
vector
:

To see this, let's first check its direction and then its magnitude. By the right-hand rule, points up out of the page in Fig.B.4. Crossing that with , again by the right-hand rule, produces a tangential velocity vector pointing as shown in the figure. So, the direction is correct. Now, the magnitude: Since and are mutually orthogonal, the angle between them is , so that, by Eq.(B.16),

### Angular Momentum

The *angular momentum* of a mass rotating in a circle of
radius with angular velocity (rad/s), is defined by

*mass moment of inertia*of the rigid body (§B.4.4).

#### Relation of Angular to Linear Momentum

Recall (§B.3) that the momentum of a mass traveling with velocity in a straight line is given by

*tangential speed*of the mass along the circle of radius is given by

Thus, the angular momentum is times the linear momentum .

Linear momentum can be viewed as a renormalized special case of angular momentum in which the radius of rotation goes to infinity.

### Angular Momentum Vector

Like linear momentum, angular momentum is fundamentally a vector in . The definition of the previous section suffices when the direction does not change, in which case we can focus only on its magnitude .

More generally, let
denote the 3-space coordinates
of a point-mass , and let
denote its velocity
in . Then the *instantaneous angular momentum vector*
of the mass relative to the origin (not necessarily rotating about a
fixed axis) is given by

where denotes the

*vector cross product*, discussed in §B.4.12 above. The identity was discussed at Eq.(B.17).

For the special case in which
is *orthogonal* to
, as in Fig.B.4, we have that
points, by the right-hand rule, in the direction of the angular
velocity vector
(up out of the page), which is
satisfying. Furthermore, its magnitude is given by

In the more general case of an arbitrary mass velocity vector
, we know from §B.4.12 that the magnitude of
equals the product of the distance from the axis
of rotation to the mass, *i.e.*,
, times the length of
the component of
that is orthogonal to
, *i.e.*,
, as needed.

It can be shown that vector angular momentum, as defined, is
*conserved*.^{B.22} For
example, in an orbit, such as that of the moon around the earth, or
that of Halley's comet around the sun, the orbiting object speeds up
as it comes closer to the object it is orbiting. (See Kepler's laws
of planetary motion.) Similarly, a spinning ice-skater spins faster
when pulling in arms to reduce the moment of inertia about the spin
axis. The conservation of angular momentum can be shown to result
from the principle of least action and the isotrophy of space
[270, p. 18].

#### Angular Momentum Vector in Matrix Form

The two cross-products in Eq.(B.19) can be written out with the help
of the vector analysis identity^{B.23}

where

The matrix is the Cartesian representation of the

*mass moment of inertia tensor*, which will be explored further in §B.4.15 below.

The vector angular momentum of a rigid body is obtained by summing the angular momentum of its constituent mass particles. Thus,

In summary, the angular momentum vector is given by the mass moment of inertia tensor times the angular-velocity vector representing the axis of rotation.

Note that the angular momentum vector
does *not* in general
point in the same direction as the angular-velocity vector
. We
saw above that it does in the special case of a point mass traveling
orthogonal to its position vector. In general,
and
point
in the same direction whenever
is an *eigenvector* of
, as will be discussed further below (§B.4.16). In this
case, the rigid body is said to be *dynamically balanced*.^{B.24}

### Mass Moment of Inertia Tensor

As derived in the previous section, the *moment of inertia
tensor*, in 3D Cartesian coordinates, is a three-by-three matrix
that can be multiplied by any angular-velocity vector to
produce the corresponding angular momentum vector for either a point
mass or a rigid mass distribution. Note that the origin of the
angular-velocity vector
is always fixed at
in the space
(typically located at the center of mass). Therefore, the moment of
inertia tensor
is defined relative to that origin.

The moment of inertia tensor can similarly be used to compute the
*mass moment of inertia* for any normalized angular velocity
vector
as

Since rotational energy is defined as (see Eq.(B.7)), multiplying Eq.(B.22) by gives the following expression for the rotational kinetic energy in terms of the moment of inertia tensor:

We can show Eq.(B.22) starting from Eq.(B.14). For a point-mass located at , we have

where again denotes the three-by-three identity matrix, and

which agrees with Eq.(B.20). Thus we have derived the moment of inertia in terms of the moment of inertia tensor and the normalized angular velocity for a point-mass at .

For a collection of masses located at , we simply sum over their masses to add up the moments of inertia:

#### Simple Example

Consider a mass at . Then the mass moment of inertia tensor is

*should*be zero about that axis. On the other hand, if we look at , we get

#### Example with Coupled Rotations

Now let the mass be located at so that

We expect to yield zero for the moment of inertia, and sure enough . Similarly, the vector angular momentum is zero, since .

For , the result is

#### Off-Diagonal Terms in Moment of Inertia Tensor

This all makes sense, but what about those off-diagonal terms in ? Consider the vector angular momentum (§B.4.14):

*coupling*of rotation about with rotation about . That is, there is a component of moment-of-inertia that is contributed (or subtracted, as we saw above for ) when

*both*and are nonzero. These cross-terms can be eliminated by

*diagonalizing*the matrix [449],

^{B.25}as discussed further in the next section.

### Principal Axes of Rotation

A *principal axis of rotation* (or *principal direction*) is
an eigenvector of the *mass moment of inertia tensor* (introduced
in the previous section) defined relative to some point (typically the
center of mass). The corresponding eigenvalues are called the
*principal moments of inertia*.
Because the moment of inertia tensor is defined relative to the point
in the space, the principal axes all pass through that point
(usually the center of mass).

As derived above (§B.4.14), the angular momentum vector is given by the moment of inertia tensor times the angular-velocity vector:

*principal moment of inertia*. If we set the rigid body assocated with rotating about the axis , then is the mass moment of inertia of the body for that rotation. As will become clear below, there are always three mutually orthogonal principal axes of rotation, and three corresponding principal moments of inertia (in 3D space, of course).

#### Positive Definiteness of the Moment of Inertia Tensor

From the form of the moment of inertia tensor introduced in Eq.(B.24)

*symmetric*. Moreover, for any normalized angular-velocity vector we have

since
is unit length, and projecting it onto any other vector
can only shorten it or leave it unchanged. That is,
, with equality occurring for
for any nonzero
. Zooming out,
*of course* we expect any moment of inertia for a positive
mass to be nonnegative. Thus,
is *symmetric
nonnegative definite*. If furthermore
and
are not
collinear, *i.e.*, if there is any nonzero angle between them, then
is *positive definite* (and ). As is well known in
linear algebra [329], real, symmetric, positive-definite
matrices have *orthogonal eigenvectors* and *real, positive
eigenvalues*. In this context, the orthogonal eigenvectors are
called the *principal axes of rotation*. Each corresponding
eigenvalue is the moment of inertia about that principal axis--the
corresponding principal moment of inertia. When angular velocity
vectors
are expressed as a linear combination of the principal
axes, there are no cross-terms in the moment of inertia tensor--no
so-called *products of inertia*.

The three principal axes are *unique* when the eigenvalues of
(principal moments of inertia) are *distinct*. They are
not unique when there are repeated eigenvalues, as in the example
above of a disk rotating about any of its diameters
(§B.4.4). In that example, one principal
axis, the one corresponding to eigenvalue , was
(*i.e.*,
orthogonal to the disk and passing through its center), while any two
orthogonal diameters in the plane of the disk may be chosen as the
other two principal axes (corresponding to the repeated eigenvalue
).

Symmetry of the rigid body about any axis
(passing through the
origin) means that
is a principal direction. Such a symmetric
body may be constructed, for example, as a *solid of
revolution*.^{B.26}In rotational dynamics, this case is known as the *symmetric top*
[270]. Note that the center of mass will lie
somewhere along an axis of symmetry. The other two principal axes can
be arbitrarily chosen as a mutually orthogonal pair in the (circular)
plane orthogonal to the
axis, intersecting at the
axis. Because of the circular symmetry about
, the two
principal moments of inertia in that plane are equal. Thus the moment
of inertia tensor can be diagonalized to look like

###

Rotational Kinetic Energy Revisited

If a point-mass is located at
and is rotating about an
axis-of-rotation
with angular velocity , then the
distance from the rotation axis to the mass is
,
or, in terms of the vector cross product,
. The tangential velocity of the mass is
then , so that the kinetic energy can be expressed as
(*cf.* Eq.(B.23))

where

In a collection of masses having velocities , we of course sum the individual kinetic energies to get the total kinetic energy.

Finally, we may also write the rotational kinetic energy as half the
*inner product* of the angular-velocity vector and the
angular-momentum vector:^{B.27}

### Torque

When twisting things, the rotational force we apply about the center
is called a *torque* (or *moment*, or *moment of
force*). Informally, we think of the torque as the *tangential
applied force* times the *moment arm* (length of the
*lever arm*)

as depicted in Fig.B.7. The moment arm is the distance from the applied force to the point being twisted. For example, in the case of a wrench turning a bolt, is the force applied at the end of the wrench by one's hand, orthogonal to the wrench, while the moment arm is the length of the wrench. Doubling the length of the wrench doubles the torque. This is an example of

*leverage*. When is increased, a given twisting angle is spread out over a larger arc length , thereby reducing the tangential force required to assert a given torque .

For more general applied forces
, we may compute the
tangential component
by *projecting*
onto the
tangent direction. More precisely, the *torque* about the
origin
applied at a point
may be defined by

where is the applied force (at ) and denotes the cross product, introduced above in §B.4.12.

Note that the torque vector is orthogonal to both the lever arm and the tangential-force direction. It thus points in the direction of the angular velocity vector (along the axis of rotation).

The torque magnitude is

### Newton's Second Law for Rotations

The rotational version of Newton's law is

where denotes the

*angular acceleration*. As in the previous section, is

*torque*(tangential force times a moment arm ), and is the

*mass moment of inertia*. Thus, the net applied torque equals the time derivative of angular momentum , just as force equals the time-derivative of linear momentum :

To show that Eq.(B.28) results from Newton's second law ,
consider again a mass rotating at a distance from an axis
of rotation, as in §B.4.3 above, and
let denote a *tangential* force on the mass, and
the corresponding tangential acceleration. Then we have, by Newton's
second law,

*torque*. Thus, we have derived

In summary, force equals the time-derivative of linear momentum, and torque equals the time-derivative of angular momentum. By Newton's laws, the time-derivative of linear momentum is mass times acceleration, and the time-derivative of angular momentum is the mass moment of inertia times angular acceleration:

### Equations of Motion for Rigid Bodies

We are now ready to write down the general equations of motion for rigid bodies in terms of for the center of mass and for the rotation of the body about its center of mass.

As discussed above, it is useful to decompose the motion of a rigid body into

- (1)
- the
*linear*velocity of its center of mass, and - (2)
- its
*angular*velocity about its center of mass.

The linear motion is governed by Newton's second law , where is the total mass, is the velocity of the center-of-mass, and is the sum of all external forces on the rigid body. (Equivalently, is the sum of the radial force components pointing toward or away from the center of mass.) Since this is so straightforward, essentially no harder than dealing with a point mass, we will not consider it further.

The angular motion is governed the *rotational* version of
Newton's second law introduced in §B.4.19:

where is the vector torque defined in Eq.(B.27), is the angular momentum, is the mass moment of inertia tensor, and is the angular velocity of the rigid body about its center of mass. Note that if the center of mass is moving, we are in a moving coordinate system moving with the center of mass (see next section). We may call the

*intrinsic momentum*of the rigid body,

*i.e.*, that in a coordinate system moving with the center of the mass. We will translate this to the non-moving coordinate system in §B.4.20 below.

The driving torque
is given by the *resultant moment* of
the external forces, using Eq.(B.27) for each external force to
obtain its contribution to the total moment. In other words, the
external moments (tangential forces times moment arms) sum up for the
net torque just like the radial force components summed to produce the
net driving force on the center of mass.

#### Body-Fixed and Space-Fixed Frames of Reference

Rotation is always about some (instantaneous) axis of rotation that is
free to change over time. It is convenient to express rotations in a
coordinate system having its origin (
) located at the
center-of-mass of the rigid body (§B.4.1), and its coordinate axes
aligned along the principal directions for the body (§B.4.16).
This *body-fixed frame* then moves within a stationary
*space-fixed frame* (or ``star frame'').

In Eq.(B.29) above, we wrote down Newton's second law for angular
motion in the *body-fixed frame*, *i.e.*, the coordinate system
having its origin at the center of mass. Furthermore, it is simplest
(
is diagonal) when its axes lie along principal directions
(§B.4.16).

As an example of a local body-fixed coordinate system, consider a
spinning top. In the body-fixed frame, the ``vertical'' axis
coincides with the top's axis of rotation (spin). As the top loses
rotational kinetic energy due to friction, the top's rotation-axis
*precesses* around a circle, as observed in the space-fixed
frame. The other two body-fixed axes can be chosen as any two
mutually orthogonal axes intersecting each other (and the spin axis)
at the center of mass, and lying in the plane orthogonal to the spin
axis. The space-fixed frame is of course that of the outside
observer's inertial frame^{B.28}in which the top is spinning.

#### Angular Motion in the Space-Fixed Frame

Let's now consider angular motion in the presence of linear motion of the center of mass. In general, we have [270]

#### Euler's Equations for Rotations in the Body-Fixed Frame

Suppose now that the body-fixed frame is rotating in the space-fixed frame with angular velocity . Then the total torque on the rigid body becomes [270]

Similarly, the total external forces on the center of mass become

*cf.*Eq.(B.15))

Substituting this result into Eq.(B.30), we obtain the following equations of angular motion for an object rotating in the body-fixed frame defined by its three principal axes of rotation:

These are call *Euler's
equations:*^{B.29}Since these equations are in the body-fixed frame, is the mass
moment of inertia about principal axis , and is the
angular velocity about principal axis .

#### Examples

For a uniform sphere, the cross-terms disappear and the moments of inertia are all the same, leaving , for . Since any three orthogonal vectors can serve as eigenvectors of the moment of inertia tensor, we have that, for a uniform sphere, any three orthogonal axes can be chosen as principal axes.

For a cylinder that is not spinning about its axis, we similarly
obtain two uncoupled equations
, for , given
(no spin). Note, however, that if we replace the
circular cross-section of the cylinder by an *ellipse*, then
and there is a coupling term that drives
(unless happens to cancel it).

**Next Section:**

Properties of Elastic Solids

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Momentum