Chapter 4
Newton–Euler Formulations

The field of dynamics consists of the study of kinematics and kinetics. Chapter 2 of this book discusses the foundations of kinematics, and Chapter 3 presents specific formulations for kinematics of spatial robotic systems. The study of kinematics provides the language used to describe the geometry of motion. The field of kinetics studies the connection between the forces and moments that act on a mechanical system and its resulting motion. Two general approaches to the study of kinetics are covered in this chapter and the next. This chapter discusses the collection of methods known as Newton–Euler formulations of the dynamics of robotic systems. Chapter 5 presents an alternative approach, those methods based on techniques of analytical mechanics. Upon the completion of this chapter, the student should be able to

• Define and calculate the linear momentum of a rigid body.
• Define and calculate the angular momentum of a rigid body.
• Define and calculate the center of mass and inertia matrix of a rigid body.
• State and employ Euler's laws for the motion of rigid bodies in a robotic system.
• Employ recursive order formulations to study robotic system dynamics.

4.1 Linear Momentum of Rigid Bodies

Elementary principles that describe the dynamics of a point mass or particle define thelinear momentum as the product of the mass of the particle and its velocity. The linear momentum of a system of particles is the sum of the linear momenta of the individual particles in the system. As highlighted in the following definition, the linear momentum of a rigid body can be viewed as a limiting case of the definition for a system of particles. The linear momentum of a rigid body is the integral of the velocity over all the differential mass elements that make up the rigid body.

---

Definition 4.1

The linear momentum of a rigid body in the frame is defined to be

where is the velocity of the differential mass element in the frame , as shown in Figure 4.1.

---

Figure 4.1 Rigid body with differential mass element.

Definition 4.1 is given in terms of the velocity of each of the infinite collection of points that make up the rigid body under consideration. Thecenter of mass is introduced to obtain an expression for linear momentum that depends on the velocity of a single point. As a result, the center of mass is used to describe the motion of a mechanical system in terms of a finite set of variables.

---

Definition 4.2

The position in the frame of the center of mass of a rigid body is defined by the equation

where is the total mass of the rigid body and is the position vector of the differential mass in the frame .

---

The above definition of the center of mass can be used to derive an expression for the linear momentum of a rigid body in terms of the center of mass velocity. The linear momentum of a rigid body is the product of the rigid body center of mass velocity and the total rigid body mass.

---

Theorem 4.1

The linear momentum of a rigid body in the frame is given by

where is the mass of the rigid body and is the velocity of the center of mass in the frame.

---

---

By definition the linear momentum is given by

The center of mass velocity may be calculated by taking the time derivative of and utilizing Definition 4.2, such that

The equation above relies on the fact that in this case “the derivative of the integral” is equal to the “integral of the derivative,” which can be written as

This is a special case of the Leibnitz integral rule that assumes there is no flux of linear momentum across the domain of integration.

---

Applications to robotic systems often introduce numerous frames of reference to formulate the equations of motion. The next example uses the definition of the center of mass to calculate its location for a link in a typical robotic system.

Figure 4.2 SCARA robot.

Figure 4.3 SCARA robot link 2 inertia estimation.

---

Example 4.1

For the SCARA robot shown in Figure 4.2, calculate the location of the center of mass of the outer arm (link 2), shown in Figure 4.3a, relative to the frame. Construct this approximation by using the geometric primitive shown in Figure 4.3a constructed of two rectangular prisms with dimensions labeled. Also assume that the mass distribution within the link is uniform.

Solution

By definition, the location of the center of mass in the frame satisfies

where the vector connects the origin of the frame to the differential mass element . From introductory dynamics, it is known that when the body is comprised of discrete sub‐components this equation can be written as

where is the mass of sub‐component and is the location of the center of mass of body in the frame. The locations of the center of mass of each sub‐component are defined as

The final expression for the location of the center of mass of the composite rigid body is consequently

---

It will be shown later in this chapter that the initial step in deriving the equations of motion for a robot often requires the evaluation of the linear momentum of its links. The next examples illustrate how the calculation of the linear momentum for realistic three dimensional solids, such as the links of robotic systems, is simplified when the location and velocity of themass center are known.

---

Example 4.2

Calculate the linear momentum for the outer arm show in Figure 4.3a of the SCARA robot depicted in Figure 4.2 using Theorem 4.1 and the results of Example 4.1. Assume that the velocity of the point and angular velocity of the body are given as

Assume that at the moment under consideration, the frame and frames are aligned (i.e. is the identity matrix). Express the answer in terms of the basis for the frame.

Solution

From Theorem 4.1 the linear momentum is given by . The velocity of the center of mass can be calculated using the relative velocity Theorem 2.16 in Chapter 2 as

where in the instant shown. From Example 4.1, the components relative to the basis for of the vector are given by

The linear momentum in the frame expressed in terms of the basis for the frame is therefore given by

---

---

Example 4.3

A rectangular prism with uniform density and a body fixed frame moves relative to the frame as shown in Figure 4.4. Calculate the linear momentum of the body in the frame at the instant shown. At this instant assume that the velocity of point and the angular velocity are given by

Figure 4.4 Rectangular Prism.

Use Definition 4.1 to calculate the answer, and then check the result using Theorem 4.1.

Solution

As specified, the solution for this problem will first be generated by integrating over the domain of the rigid body. Then, the same solution will be calculated by considering the velocity of the center of mass and invoking Theorem 4.1. In addition the frame defined at point in the problem statement, an additional frame is defined parallel to with its origin at the center of mass . The velocity of a differential mass element can be written using Theorem 2.16 in Chapter 2 as

where the vector is given by

The linear momentum, by Definition 4.1, can be written as

The only term in the integrand that varies over the body is the vector . Thus, the terms and may be factored out of the integrand to obtain

From the definitions of the rigid body and the center of mass position, it is known that

Substituting these expressions into the formulation for results in

(4.1)

The location of the center of mass is . When , , and are substituted into Equation 4.1, the final expression is obtained as

Alternatively, Theorem 4.1 specifies that

(4.2)

The velocity of the center of mass may be derived using Theorem 2.16 as

where is the vector that connects the point to the center of mass. Substituting this into Equation 4.2 results in

(4.3)

which is identical to Equation 4.1, showing that both approaches yield the same answer.

---

4.2 Angular Momentum of Rigid Bodies

4.2.1 First Principles

Theangular momentum about a point of a single particle in motion is equal to the “moment of momentum.” That is, the angular momentum is the cross product of the position vector that connects the point to the particle and the linear momentum of the particle. The angular momentum of a system of particles is the sum of the angular momenta of the individual particles. As with the linear momentum of a rigid body, the definition of the angular momentum of a rigid body can be interpreted as a limiting case as the number of particles increases. The summation over all the particles in the system is replaced by an integral over all the differential mass elements that make up the rigid body.

---

Definition 4.3

The angular momentum of a rigid body in the frame about point is given by

where is the vector connecting the point to the differential mass and is the velocity of the differential mass in the frame .

---

There is no restriction on the point about which the angular momentum may be calculated in Definition 4.3. In practice, it is often convenient to calculate the angular momentum about an arbitrary point by relating it to the angular momentum about the center of mass. Theorem 4.2 describes this relationship.

---

Theorem 4.2

The angular momentum in the frame of a rigid body about the arbitrary point is given by

where is the angular momentum in the frame of the rigid body about its center of mass , is mass of the rigid body, is the vector connecting the point to the center of mass , and is the velocity of the center of mass in the frame .

---

---

The proof of Theorem 4.2 relies on a decomposition of the vector that locates the differential mass element into the sum of a vector from the point to the center of mass and the vector from the center of mass to the differential mass element . This decomposition is illustrated in Figure 4.1. Applying this decomposition to the definition of results in

Since the vector does not vary over the mass integral, it may be removed from the integral and

Based on prior analysis, the two integrals in this expression may be represented as

resulting in the desired final formulation of , such that

---

The next two examples show that the angular momentum of realistic three dimensional bodies, such as the rigid links of a robotic system, can be calculated directly from their definition. As when calculating linear momentum, the task of calculating the angular momentum can be facilitated by utilizing the knowledge of the position and velocity of the center of mass.

An important qualitative observation will also be made in these examples: the calculation of angular momentum for rigid bodies often introduces some common integrals over the body, namely thecross products of inertia and themoments of inertia. These integrals appear so frequently that they are discussed in detail in Sections 4.2.2 and 4.2.3 of this chapter. The cross products of inertia and moments of inertia also appear frequently in calculations of the kinetic energy of a robotic system; hence their important role in theanalytical mechanics formulations that are presented in Chapter 5.

---

Example 4.4

A rectangular prism with body fixed frame is shown in Figure 4.5. It rotates about the axis at the angular rate .

Figure 4.5 Rectangular prism.

Calculate the angular momentum in the frame about the point using Definition 4.3.

Solution

The velocity in of the differential mass element is given by

where is the position vector of the differential mass element. The vector is represented in terms of the basis for the frame as

where are the coordinates along the directions. The velocity of the differential mass of is consequently

The definition of angular momentum results in

(4.4)

where the fact that and do not vary with the integration over the body allowed those variables to be removed from the integrals. To evaluate the integrals, the single integration over the mass of the body can be re‐cast as a volumetric integral. Assuming is the uniform density of the solid, the differential mass may be defined as . along the three basis vectors , , and . Using this expansion, it can be seen that the integrals and are equal to zero. For the first,

A similar result holds for . The third integral may be evaluated as

(4.5)

where , based on the definitions of the prism density and the geometry in Figure 4.5. It is concluded that the angular momentum can be written in the form

where is the axis moment of inertia of the body with respect to the center of mass and is the axis moment of inertia of the body with respect to point .

It will be shown in the sections that follow that the quantities and are examples of the cross products of inertia of a rigid body that measure of the symmetry of a rigid body. The integral is an example of amoment of inertia of a rigid body. It measures the resistance of a rigid body to rotation.

---

---

Example 4.5

Calculate the angular momentum in of the rectangular prism studied in Example 4.4 using Theorem 4.2.

Solution

First, the angular momentum of the body is evaluated about its center of mass. For this example, the origin of the body fixed frame is fixed at the center of mass, as shown in Figure 4.6. The position of a differential mass with respect to this frame is defined as

where are the coordinates along the directions. By following the same essential steps as in Example 4.4, it can be shown that

Figure 4.6 Rectangular Prisms with Body Fixed Frame at Center of Mass.

A critical observation here is that the limits of integration in these three integrals differ from the corresponding expression in Equation 4.5. As a result,

The first two integrals, and , can be shown to be zero simply by virtue of the fact that

appears as a factor in both expressions. The last integral to be expanded is

The final expression for is therefore

Theorem 4.2 gives the following expression for the angular momentum about the point

For this example, and , resulting in

This is the same result as in Example 4.4. It will also be shown that the coefficient defined in the equation as

can also be obtained via the application of the parallel axis theorem for inertia discussed in Section 4.2.3.3.

---

Theorem 4.2 states a general relationship between the angular momentum about an arbitrary point and about the center of mass of a rigid body. The theorem was derived by decomposing the position vector into the sum . Another useful form for the angular momentum can be achieved by decomposing the velocity vector instead of the position vector. The next theorem makes use of this strategy. The resulting form for the angular momentum will be critical in the discussions of the inertia matrix in the following sections.

---

Theorem 4.3

The angular momentum of a rigid body in the frame about the point is given by

where is the total mass of the body, the vector connects the point to the center of mass , is the velocity of the point in the frame , is the angular velocity of the body frame in the frame , and is the vector from the point to the differential mass .

---

---

From the definition of the angular momentum for a rigid body,

Because the point and the differential mass are on the same body, the velocity may be defined with respect to the velocity of point using Theorem 2.16 in Chapter 2,

The velocity of the point does not vary over the integral, and the first term may be simplified by noting that

Combining these expressions results in

---

4.2.2 Angular Momentum and Inertia

The last section introduced the definition of angular momentum for rigid bodies and derived theorems that can be used for its calculation. As shown in Examples 4.4 and 4.5, the calculation of angular momentum can require the evaluation of integrals over the body that have the form , and . The integrals and are examples of the cross products of inertia, and the integral is an example of a moment of inertia of a rigid body. The cross products and moments of inertia are used to construct theinertia matrix for a rigid body. These definitions are important because most commonly encountered expressions for angular momentum in applications are cast in terms of theinertia matrix.

---

Definition 4.4

Let be a frame whose origin is located at the point . Suppose the position vector of a differential mass element in the rigid body is expressed in terms of the basis as

The inertial matrix of the rigid body about the point relative to the frame is given by

The diagonal terms are the moments of inertia, and the off‐diagonal entries are the products of inertia, about the point with respect to the basis.

---

The inertia matrix about the point relative to the frame utilizes the superscript to denote the basis vectors for which the matrix is defined. As in earlier chapters, the convention that a superscript denotes the coordinates of a vector relative to a specific choice of frame is used. For example, if is an arbitrary vector, denotes the components of the vector as expressed in terms of the basis. The inertia matrix about a point relative to a frame contains the components of theinertia tensor expressed in terms of thetensor basis . A vector is a first order tensor, and the inertia tensor is a second order tensor. Appendix A contains a detailed discussion of tensors for the interested reader.

The point used to define the angular momentum in Definition 4.3 can be any point in the mechanical system. The most general form for the angular momentum in terms of the inertia matrix that will be used is derived in the following theorem.

---

Theorem 4.4

Suppose that a body with fixed frame moves relative to the frame . The components of the angular momentum in relative to the basis of the frame are given by

(4.6)

where is the angular velocity of the frame in the frame, is the inertia matrix about the point relative to the frame , is the vector connecting the point to the center of mass , and is the velocity of the point in the frame .

---

This theorem will be proven while carrying out the proof of Theorem 4.5. Equation 4.6 is useful in many applications because of its generality. For example, this theorem will constitute the technical foundation for the most general variant ofEuler's second law of motion for rigid bodies presented in Section 4.3 of this chapter. This form of Euler's second law serves as the foundation for the numerically efficientrecursive order formulations of kinematics and dynamics of robotic systems.

Still, despite its generality, Theorem 4.4 is frequently modified so that it is easier to use in direct application to individual problems. It is most common to select the point in applications so that the term is identically equal to zero. This is most easily accomplished by choosing the point to be either fixed in the frame , or by selecting the point to be the center of mass of the rigid body.

---

Theorem 4.5

Suppose that a body with fixed frame moves relative to the frame , and that the point is either the center of mass of the rigid body or is a point of the rigid body that is fixed in the frame . Then the components of the angular momentum in relative to the basis for a frame are given by the matrix product

where is theangular velocity of the frame in the frame and is the inertia matrix about the point relative to the frame .

---

---

First, the conclusion reached in Theorem 4.5 will be expanded upon. Suppose that the angular velocity vector , position vector , and angular momentum vector are expressed in terms of the basis for the frame as

The matrix equation becomes

The proof of this theorem follows from Theorem 4.3 when it is noted that the point is by hypothesis either the center of mass or a point that is fixed in the ground frame. In the former case , and in the latter case . Theorem 4.3 implies that

Since each of the vectors is assumed to be given in terms of their components relative to the same basis , it can be directly computed that

This equation can be rewritten as

The steps above also prove Theorem 4.5, where the term is not equal to zero in general. In that case, it has been shown that .

---

Theorem 4.5 will be used in numerous applications, and it is vital to note the different roles of the frames , and in the theorem. The body fixed frame is , which is assumed to move relative to the frame . In the next section Euler's laws will be stated, which hold in aninertial frame. The inertial frame will be chosen to be . In other words, the frames and are connected to the physics of the problem at hand. The frame can be any frame whatsoever, and is used to provide a basis to conveniently express the components of the physical vectors and .

Figure 4.7 Satellite with body fixed frame.

---

Example 4.6

The inertia matrix of a satellite depicted in the configuration shown in Figure 4.7 about its own center of mass is given by

(4.7)

Suppose that the satellite is traveling in an orbit as described in Example 2.6 in Chapter 2. Figure 4.8 depicts the right ascension and inclination of the orbit. The inclination of the orbital plane is and the right ascension is . Assume that at the instant shown the distance from the center of the earth to the center of mass of the satellite is and that the velocity of the center of mass of the satellite is

(4.8)

where is the speed and measures the angle of rotation of the velocity vector with respect to the axis. The unit vector is along the direction of flight of the satellite and is depicted in Figure 4.9. Calculate the angular momentum in the frame of the satellite about the center of the earth. Express the answer in terms of the basis for the frame.

Figure 4.8 Orbital plane definition.

Figure 4.9 Frame definitions in the orbital plane.

Solution

Theorem 4.2 can be used to relate the angular momentum of the satellite about the center of the earth and about its own center of mass

(4.9)

The angular momentum of the vehicle about its own center of mass is calculated using Theorem 4.5 to be

The angular velocity of the satellite in the earth frame is

so that

(4.10)

The vector that connects the center of the earth to the mass center of the satellite is where

The second term in Equation 4.9 can be expanded as

(4.11)

The sum of terms in Equation 4.10 and 4.11 gives the desired result if these two terms are represented in a common basis. The relationship between the frame and is given by the rotation matrix

The angular momentum is

The expression for the components relative to the basis for the frame can be obtained by premultiplying this equation by the rotation matrix from Example 2.6 in Chapter 2,

As a result, the angular momentum of the satellite about the center of the earth is given by

where .

---

4.2.3 Calculation of the Inertia Matrix

The integrals that appear in the Definition 4.4 depend on the choice of the frame . With all the possible choices available for the frame in Theorem 4.5, it is important that efficient techniques are available for calculating inertia matrices relative to different coordinate frames. This section describes several theorems and techniques that facilitate the calculation of the inertia matrix in typical applications.

Sections 4.2.3.1 and 4.2.3.3 discuss theinertia rotation transformation law and theparallel axis theorem. The inertia rotation transformation law describes how to relate inertia matrices calculated with respect to two frames with the same origin but different orientations. The parallel axis theorem describes how to relate inertia matrices calculated with respect to two frames having parallel orientations, but with origins at the center of mass and an arbitrary point . By combining these two results, the inertia matrix relative to any arbitrary frame can be calculated if the inertia matrix relative to a given frame is known.

Section 4.2.3.2 summarizes the construction ofprincipal axes with respect to which the inertia matrix assumes a diagonal form. When possible, it is convenient to cast problems in terms of the principal axes, as it reduces the number of parameters in the equations of motion.

Section 4.2.3.4 details the role of symmetry in the calculation of the inertia matrix. It will be shown that by identifying coordinate planes of symmetry it is often possible to conclude that certainproducts of inertia in the inertia matrix are zero, without having to calculate them explicitly.

4.2.3.1 The Inertia Rotation Transformation Law

This section derives the equation used to relate rotation matrices that are defined from frames that have a common origin, but are rotated relative to one another. The equations derived in Theorem 4.6 are a special case of thetensor transformation laws familiar intensor analysis. A comprehensive treatment can be found in [7]. The theorem below summarizes the transformation laws that relate the inertia matrices associated with frames that are rotated relative to one another.

---

Theorem 4.6

Let and be theinertia matrices of a rigid body relative to the frames and , respectively, which have a common origin at . Theseinertia matrices satisfy

---

These expressions should be compared and contrasted to the trnasformation laws for the components of a vector ,

The coordinate change for a vector, which is a first order tensor, premultiplies the components by a rotation matrix. The coordinate transformation for the inertia matrix, which is a second order tensor, premultiplies and postmultiplies the components by a rotation matrix and its transpose, respectively. See Appendix A for a detailed discussion.

---

The inertia matrices satisfies the equation that relates the angular momentum and the angular velocity vectors in Theorem 4.5,

But the angular momentum and angular velocity are themselves vectors whose components transform via the application of the rotation matrices or , such that

If these are substituted into the equations above, the formulation results in

The second equation in the theorem follows from the first when the first is premultiplied by and postmultiplied by .

---

---

Example 4.7

The inertia matrix of the rigid body shown in Figure 4.10 relative to the body fixed frame that has its origin at the point is given by

The link is fixed with respect to the ground frame that shares an origin with frame at point . Frame is mapped into frame by a rotation about . Calculate the inertia matrix about the point relative to the frame shown in Figure 4.10.

Figure 4.10 Single link rotated with respect to ground frame.

Solution

The transformation law for the inertia matrix yields the following theorem that relates the inertia matrices and

The change of basis between the frames and can be derived by inspection of the geometry and is determined by the rotation matrices

Putting these results together yields the desired inertia matrix

---

---

Example 4.8

The inertia matrix of the satellite shown in Figure 4.11a about its center of mass relative to the frame is given by the sum

where are the moments of inertia of the central body about the system center of mass and are the moments of inertia of the solar arrays about the system center of mass.

Figure 4.11 Satellite with central body and solar panel fixed frames.

Use the inertia rotation transformation law to derive the system inertia matrix when the solar arrays are rotated by about the axis as depicted in Figure 4.11b.

Solution

Let the frame be a set of axes fixed relative to the solar arrays. The origin of the frame is located at the system center of mass, and the frame is parallel to the frame as shown in Figure 4.11a. In Figure 4.11b the solar arrays have been rotated by . The rotation matrix that relates the frame and the frame depicted in Figure 4.11b is given by

By the inertia rotation transformation law, the inertia matrix of the solar arrays about the system center of mass relative to the frame depicted in Figure 4.11a is given by

The inertia matrix for the satellite about its center of mass is consequently given by the sum

Note that the rotation transformation in this case agrees with intuition: the new inertia matrix is just obtained by permuting the entries of the original inertia matrix of the solar arrays.

---

4.2.3.2 Principal Axes of Inertia

Theorem 4.6 provides a direct way to calculate the inertia matrix relative to some rotated frame from the inertia matrix associated with a given frame, provided the two frames have a common origin. In general, the inertia matrix is a full matrix. There are certain choices of the frame that simplify the form of the inertia matrix. The definition of the principal axes, given below, describes a choice of coordinates that yields a diagonal inertia matrix.

---

Definition 4.5

The frame having origin at defines a set of principal axes at the point for a rigid body whenever the inertia matrix is diagonal,

The diagonal terms are theprincipal moments of inertia associated with the basis .

---

For rigid bodies with complex geometry, it may not be evident that there is any special structure of the inertia matrix. However, for any point in a rigid body there is always a set of axes at the point relative to which the inertia matrix is diagonal. That is, there exists a set of principal axes at every point in the rigid body. The following theorem summarizes this fact.

---

Theorem 4.7

Let be a point in a rigid body. There is a set ofprincipal axes fixed in the body that have their origin at point .

---

---

This theorem will be proven in detail because the procedure is constructive and provides a way of directly solving practical problems. Background material for this section can be found in Appendix A. Suppose an inertia matrix relative to the basis of the frame is given. First, theeigenvectors and eigenvalues are calculated from the solution of thealgebraic eigenvalue problem

From Theorem A.11 in Appendix A, since is real and symmetric, it is always possible to scale the eigenvectors so that themodal matrix has the properties that

From this, it is concluded that it is possible to choose the eigenvectors so that the matrix is orthogonal. In fact, the eigenvectors in the columns of may always be ordered and scaled so that they correspond to a right hand basis. Define the frame via the introduction of the rotation matrix . As a result,

The frame defines a set of principal axes at for the rigid body. The principal moments of inertia are the eigenvalues .

---

---

Example 4.9

Find a set of principal axes of the inertia matrix

Solution

The characteristic polynomial is

The eigenvalues, or principal moments of inertia, are the roots of the characteristic polynomial , , and . The principal axes are determined from the eigenvectors associated with each of these roots. For , the eigenvector must satisfy

From the first two equations, it must be the case that , while the third equation allows for to be arbitrary. A unit eigenvector associated with may then be defined as

Repeating this procedure for results in

The last equation yields , while the first two equations require that . As a result, a unit eigenvector associated with is

Repeating this procedure for results in

As before, the third equation specifies that and the first two equations require that . A unit eigenvector corresponding to is defined as

To obtain the desired rotation matrix, the unit eigenvectors must be assembled into a rotation matrix , taking care to make sure that det . This may be facilitated by multiplying the unit vector by .

It may be verified that this matrix satisfies the relationships

---

4.2.3.3 The Parallel Axis Theorem

Theorem 4.6 summarizes how the inertia relative to one set of axes may be used to calculate the inertia relative to a rotated basis. The two frames must have the same origin when applying Theorem 4.6. The parallel axis theorem, in contrast, relates the inertia matrix about a point in terms of the inertia matrix about the center of mass. In this case the frames are not rotated relative to each other; they are parallel.

---

Theorem 4.8

Let the position of a differential mass element in a rigid body be given by the coordinates relative to a basis whose origin is at , and by the coordinates relative to a parallel frame whose origin is at the center of mass. Suppose the two coordinates systems are related by the equations

The moments of inertia relative to these two frames satisfy

In these equations and are the moments of inertia relative to the frame that has its origin at point and point , respectively. The cross products of inertia relative to the two frames satisfy

In these equations and are the cross products of inertia relative to the frame that has its origin at points and point , respectively.

---

---

As all of the equations for the moments of inertia are similar, the equation is only derived for . By definition,

The first term on the right is equal by definition to the moment of inertia about the center of mass. The last two terms on the right are equal to zero by the definition of the center of mass, which is located at the origin of the coordinate system.

The analysis of the cross products of inertiais similar; only the equation for is derived. By definition,

The first term on the right is by definition thecross product of inertia . The last two terms are zero by the definition of the center of mass, which is located at the origin of the coordinate system.

---

The parallel axis theorem is lengthy, as written in Theorem 4.8. The following theorem expresses the parallel axis theorem in matrix form and can be useful in programming implementations.

---

Theorem 4.9

The parallel axis theorem in theorem 4.8 can be written in matrix form as

(4.12)

where is the mass of the rigid body, is the inertia matrix about the point relative to the frame , is the inertia matrix relative to a frame that is parallel to and that has its origin at the center of mass, is the skew operator, and .

---

---

The proof stems directly from the expansion of the term , which is

By inspection, Equation 4.12 is identical to the collection of scalar equations in the conclusion of Theorem 4.8.

---

---

Example 4.10

This problem calculates the inertia matrix for the horizontal arm of the cylindrical robot shown in Figure 4.12. Use the inertia rotation transformation law and the parallel axis theorem to calculate the inertia matrix for the horizontal arm with respect to the frame with origin at point from the inertia matrix

of the horizontal arm about its own center of mass at point . Note that is not parallel to for .

Figure 4.12 Cylindrical Robot.

Solution

First, a virtual frame will be defined at the horizontal link center of mass that is parallel with frame . The rotation matrix

maps the basis for the frame into the frame. The inertia matrix with respect to the frame is then

The inertia rotation transformation law yields the inertia that corresponds to the reordering of the principal moments of inertia . Let be the coordinates located at the center of mass along the directions, respectively. The coordinates are along the directions. In general,

(4.13)

The constants may be calculated by evaluating these equations at a particular point. For example, the origin of the frame has coordinates with respect to the frame, and it has coordinates with respect to the frame. When these values are substituted into Equation 4.13,

By the parallel axis theorem, the moments of inertia are

and the cross products of inertia are

Collecting these into results in

---

4.2.3.4 Symmetry and Inertia

The explicit calculation of the integrals that appear in the inertia matrix can be difficult to calculate analytically. In cases of complex geometry, the calculation of closed form expressions can be intractable. Numerical techniques to estimate the inertia matrix are now a standard feature of solid modeling and computer aided design software. The algorithms involved in such calculations are not complex, and require implementation of numerical quadrature formulae to estimate the integrals as a weighted sum of functional evaluations. The most difficult task in implementing this approximation procedure is the characterization of the geometry of the rigid body, but this is precisely the job at which solid modeling and computer aided design programs excel.

Still, even for the most complex geometric models, it is often possible to reduce the work required to evaluate the inertia matrix by utilizing the body's symmetry. The identification of a specific coordinate plane of symmetry implies that cross products of inertia involving the coordinate perpendicular to the plane of symmetry are equal to zero. Before stating this theorem, the mathematical definition of acoordinate plane of symmetry will be presented.

---

Definition 4.6

Suppose that are coordinates along the basis vectors of the frame . The plane of the frame is aplane of symmetry of a rigid body if for each differential mass element in the body located at there is a differential mass element located at . In other words,

for all in the body where the density of the rigid body is . Similar definitions of the 1–3 and 2–3 planes of symmetry hold.

---

This definition specifies that a plane of symmetry splits the body into two pieces that are reflections of each other across the plane of symmetry. The importance of identifying a plane of symmetry in applications is demonstrated in the following theorem.

---

Theorem 4.10

Suppose that the – plane of the frame is a plane of symmetry of the rigid body. Then the cross products of inertia

for and . If there exist at least twoplanes of symmetry relative to the frame , the frame defines a set of principal axes for the rigid body.

---

---

The proof of this theorem is built on a few fundamental facts from calculus.

---

---

Definition 4.7

A function is anodd function if and only if

for all .

---

Building on this definition for a single variable function, a multivariable function is anodd function in the argument if the function is a anodd function when the other variables are held fixed. In addition, there is a special property of odd functions related to integration.

---

Proposition 4.1

The integral of an odd functionover a symmetric domain is equal to zero.

---

---

The proof of this proposition will be presented in enough detail that the same sequence of arguments can be used to prove Theorem 4.8. Suppose that is an odd function. The integration over the symmetric domain will be performed by extending to be equal to zero for all that are not in the domain. The integral of over the domain is then

which proves the proposition.

---

The proof of Theorem 4.10 has the same structure. Suppose that the – plane of the frame is a plane of symmetry of the rigid body. By definition,

By using the fact that the density outside of the body, this integral can be written as

The integrand in is an odd function of that is integrated over a symmetric domain, and is equal to zero.□

---

Example 4.11

Suppose that link 1 of a PUMA robot is oriented relative to the frame as shown in Figure 4.13. Use symmetry arguments to deduce the form of its inertia matrix. Calculate the inertia matrix assuming that the base is the union of two cylindrical bodies having a common uniform density. Use symmetry arguments to confirm that the individual inertia matrices calculated as intermediate results have the correct form.

Figure 4.13 PUMA link 1 with single plane of symmetry.

Solution

From the figure it is apparent that the – coordinate plane of the frame is a plane of symmetry. Theorem 4.10 implies that all of the cross products of inertia are zero that involve the coordinate along the direction that is perpendicular to the plane of symmetry. Therefore,

which results in an inertia matrix of the form

The inertia for the composite body will be calculated with respect to the frame about the point by calculating the inertia of each cylinder and summing the result. First, the vertical cylinder will be considered. Fix a frame at the center of mass of the vertical cylinder, with each of its axes parallel to the corresponding axes . The inertia matrix of the vertical cylinder with respect to the frame located at its center of mass is given by

This inertia matrix must be diagonal since the – and – coordinate planes of the frame are planes of symmetry of the vertical cylinder. All the cross products of inertia with respect to coordinates perpendicular to these planes must vanish, which implies that all the cross products are zero.

Next, the parallel axis theorem is used to calculate the inertia of the vertical cylinder about the point with respect to the frame. Let be the coordinates along the axes, respectively, and let be the coordinates along the axes. The coordinates are related by the equations

The offsets may be calculated by evaluating the coordinates of some fixed point in the and frames. The origin of the frame has coordinates with respect to the frame, while its coordinates are relative to the frame. We substitute these values and obtain

The parallel axis theorem then yields the inertia of the vertical cylinder about the point relative to the frame as

(4.14)

Again, this inertia matrix has the anticipated form. The – and – coordinate planes of the frame are planes of symmetry of the vertical cylinder, and it follows that all the cross products of inertia are zero by Theorem 4.10.

The inertia matrix of the horizontal cylinder will now be calculated about the point with respect to the frame. Let the frame define a set of axes parallel to the frame, but whose origin is located at the center of mass of the horizontal cylinder. The inertia matrix of the horizontal cylinder about its center of mass with respect to the frame is given by

The axes are known to be principal axes of the cylinder since each of the coordinate planes of the frame are a plane of symmetry for the horizontal cylinder. As before, the parallel axis theorem is used to derive the inertia matrix of the horizontal cylinder about the the point with respect to the axes. Let now be the coordinates of a chosen point along the basis vectors, and let be the coordinates along the basis vectors. The location of point is relative to the frame and relative to the frame. When we introduce these two expressions in the change of coordinates, we find that

The parallel axis theorem guarantees that we have for the horizontal cylinder that

(4.15)

where . Only the – plane of the frame is a plane of symmetry for the the composite body. Theorem 4.10 asserts that the cross products and are equal to zero for the inertia matrix of this body about the point relative to the frame. The final solution for the inertia matrix of the composite body is obtained by summing the inertia matrices in Equations 4.14 and 4.15.

---

4.3 The Newton–Euler Equations

A first course in dynamics introduces Newton's laws of motion, which are applicable to bodies that are idealized as point masses.Newton's first law holds that in the absence of applied external forces, a mass at rest remains at rest, or if it is in motion it travels along a straight line with constant velocity.Newton's second law maintains that the sum of forces acting on a point mass is equal to the time rate of change of the linear momentum of the point mass. A critical feature of the Newton's laws of motion is that they are stated with respect to observations made in aninertial reference frame.

Robotic systems are most often comprised of collections of rigid bodies whose mass is distributed spatially. While it is sometimes possible to create reasonable approximations of robotic systems using lumped mass or point mass approximations, this is frequently not the case. The required generalization of Newton's laws of motion for rigid bodies having distributed mass are given byEuler's laws of motion.

---

Theorem 4.11 (Euler's first law)

The resultant force acting on a rigid body is equal to the time rate of change in an inertial frame of the linear momentum

---

---

Theorem 4.12 (Euler's second law)

Let be a point that is fixed in the inertial frame . The resultant moment about point acting on a rigid body is equal to the time rate of change in an inertial frame of the angular momentum about the point .

---

Theorems 4.11 and 4.12 can be used in applications, but it is often the case that alternative forms are sought. Euler's first law can also be stated as the fact that the resultant force acting on a rigid body is equal to the mass of the rigid body multiplied by the acceleration of the center of mass of the body in an inertial frame.

---

Theorem 4.13

The resultant force acting on a rigid body is equal to the mass multiplied by the acceleration in an inertial frame of the center of mass of the rigid body.

---

---

The time rate of change in the inertial frame of the linear momentum is

---

The form of Euler's second law stated in Theorem 4.12 requires that the point is fixed in the inertial frame . This restriction can be a serious drawback in many applications. For example, in space robotics it is not convenient to impose this condition. However, a suitable alternative form may be derived. Euler's second law can also be stated for the case when the point is selected to be the center of mass of the rigid body.

---

Theorem 4.14

The resultant moment acting on a rigid body about the center of mass is equal to the time rate of change in the inertial frame of the angular momentum about the center of mass.

---

---

Theorem 4.2 defines the relationship between the angular momentum about a point and the center of mass to be

Suppose that is some point fixed in the inertial frame . Substituting this expression into Euler's second law results in

The second term on the right is equal to zero, and Euler's first law used to substitute in the third term on the right to obtain

This equation can be rearranged as

---

The above proof illustrates that Theorems 4.12 and 4.14 are equivalent; either form could be selected as Euler's second law. As a final topic in this section, alternative laws are presented that are derived from Euler's second law that relate angular momentum and moments about an arbitrary point that is fixed on a rigid body.

---

Theorem 4.15

Let point be fixed on a rigid body, but otherwise arbitrary. Theresultant moment about of the external forces and moments acting on a rigid body and the angular momentum about of the rigid body satisfy the three following equivalent equations:

(4.16)

(4.17)

(4.18)

In these equations and are the angular momentum in the frame about the center of mass and point , respectively, is the vector connecting the point to the center of mass of the rigid body, is the inertia tensor relative to the point , is the angular velocity of the frame in the frame, is the acceleration of the point in the frame, is the acceleration of the center of mass in the frame, and is the velocity of the center of mass in the frame.

---

---

Euler's second law for the center of mass of the body states

(4.19)

The resultant moment about the point can be written in terms of the resultant moment about the center of mass as

(4.20)

where is the resultant force acting on the rigid body. The proof of Equation 4.16 utilizes Equations 4.19 and 4.20 to show that

For the proof of Equation 4.17, Theorem 4.2 states that

(4.21)

Substituting Equation 4.21 into Equation 4.19 yields

This equation can be rewritten using the identity

(4.22)

(4.23)

to show that

The mass multiplied by the acceleration of the center of mass is equal to the resultant applied force from Euler's first law, which shows that Equation 4.17 holds

(4.24)

The angular momentum about the point and the inertia matrix have been related in Theorem 4.4 via the identity

The proof of Equation 4.18 begins by substituting in the above expression for angular momentum into Equation 4.24, and expanding the velocity using the relative velocity Theorem 2.16 in terms of point , such that

Equation 4.23 may be used to replace the second term in the above equation and cancel out the final term, resulting in

The derivative in the second term of the above equation may be expanded and the third term expressed in terms of the velocity , such that

Combining the final two equations and applying the derivative results in the desired equation from Equation 4.18

---

4.4 Euler's Equation for a Rigid Body

Euler's first and second laws as stated in Theorems 4.11 and 4.12 can be used to derive the equations of motion for mechanical systems comprised of rigid bodies. When treating complex mechanical systems that consist of rigid bodies connected by ideal joints, the choice of kinematic variables and frames of reference can be quite complicated. The choice of kinematic variables is often tailored to the problem at hand to simplify calculations. The following theorem discusses one of the most common such applications of Euler's first and second laws,Euler's equations of rotational motion for a single rigid body. It is assumed in these equations that the body fixed frame defines a set of principal axes that have their origin at the center of mass.

---

Theorem 4.16

Suppose that a rigid body with body fixed frame moves in the inertial frame . Let the origin of the frame be located in the center of mass, and let the frame define a set of principal axes for the rigid body. Euler's equations for the rotational motion of the rigid body are given by

where the applied moment about the center of mass and the angular velocity are given by, respectively,

where are a basis for the frame and are the principal moments of inertia relative to the frame.

---

---

Let the principal moments of inertia with respect to the frame be , , and . The components in of the angular momentum in about the center of mass of the body is given by Theorem 4.5 as

(4.25)

where , and are the components of the angular velocity vector with respect to the body frame. Euler's second law in Theorem 4.14 can be written

(4.26)

using the derivative Theorem 2.12 in Chapter 2 above to obtain the expression in terms of the derivative . It is straightforward to expand Equation 4.26 using Equation 4.25, which is given in terms of the basis for the frame. The cross product term is

The final equations of motion are consequently

---

4.5 Equations of Motion for Mechanical Systems

Euler's first law in Theorem 4.11 and Euler's second law in Theorem 4.12 give a complete description of the dynamics of mechanical systems that consist of rigid bodies. This section discusses the use of these principles in realistic problems that arise when studying robotic systems.

4.5.1 The General Strategy

A general outline of the steps used to derive the equations of motion from Euler's laws for mechanical systems is given in Figure 4.14. The set of steps that appear in Figure 4.14 can be difficult to apply in practice, and experience is required to become adept at deriving the equations of motion for complex systems. The complexity of the resulting equations can vary dramatically based on the selection of the kinematic variables, or, depending on the choice of frames, in a particular problem.

1. 1. Derive the equations of motion for each body,
   1. 1.1 Create afree body diagram for body .
   2. 1.2 Formulate Euler's first law for body .
      1. 1.2.1 Calculate the acceleration of the center of mass of body .
      2. 1.2.2 Sum forces equal to the derivative of the linear momentum.
   3. 1.3 Formulate Euler's second law for body .
      1. 1.3.1 Calculate the derivative of the angular momentum.
      2. 1.3.2 Sum moments equal to derivative of the angular momentum.
2. 2. Assemble the collection of equations above into a governing set ofdifferential‐algebraic equations, orDAE's, having the form
   
3. 3. Eliminate algebraic unknowns and write the governing equations as a system ofordinary differential equations, or ODE's, having the form
   

Figure 4.14 Methodology to derive the equations of motion.

The successful application of the steps in Figure 4.14 requires numerous theorems and principles that have been introduced in Chapters 2, 3, and 4. The calculation of the acceleration of the center of mass builds on the foundations developed in Chapter 2 and on the specific formulations of kinematics for robotics summarized in Chapter 3. The determination of the derivative of the angular momentum in Steps 2.2.1 or 2.3.1 of Figure 4.14 requires a working knowledge of the use ofangular velocities introduced in Chapter 2 in Section 2.5.1 and the derivative theorem in Theorem 2.12. Even with a strong understanding of these definitions and underlying principles, practice will guide the careful selection of the kinematic variables and frames in a particular problem.

4.5.2 Free Body Diagrams

The creation of an accuratefree body diagram is essential in the sequence of steps in Figure 4.14. The free body diagram releases a rigid body from all external constraints on the body and represents the effect of these constraints by unknown reaction forces and moments that act between the bodies. The representation of constraint forces and moments that act between bodies in a mechanical system connected byideal joints can be treated systematically. The sum of thenumber of degrees of freedom of an ideal joint and the number of constraint reaction forces and moments that act between the bodies connected by that joint is always equal to six. In fact, much more can be said about the complementary nature of the kinematic constraints introduced by an ideal joint and the associated reaction forces and moments. This complementarity is the subject of the next theorem.

---

Theorem 4.17

The number of degrees of freedom of an ideal joint that connects two rigid bodies and the number of reaction forces and moments that act between these bodies at the joint satisfies the equation

If the ideal joint prevents relative translation along mutually orthogonal directions, there are in general reaction forces that act on the adjacent bodies along these directions. The reaction forces serve to constrain the relative translation. If the ideal joint prevents relative rotation between the two bodies about mutually orthogonal directions there are in general constraint reaction moments. The reaction moments that act along these directions serve to constrain the relative rotation. The internal reaction forces and moments acting on each body have the same magnitude, but opposite directions.

---

Before applying this theorem in a number of examples, it must be emphasized that the statement of the theorem concludes that there are in general reaction forces or moments. It can occur that the value of some of the components of these reaction forces or moments are equal to zero. Strictly speaking, it is perhaps more accurate to say that the joint in question can support reaction forces or moments, but that some of these reactions may be equal to zero. The next few examples will show how to create free body diagrams for some typical mechanical systems. Each example has been selected to emphasize the practical implications of Theorem 4.17 and the methodology outlined in Figure 4.14

---

Example 4.12

The base, vertical column and inner arm of a PUMA robot are depicted in Figure 4.15. In this example the ground frame is denoted as , the frame is fixed in the revolving vertical column, and the inner arm has body fixed frame . The revolute joint 1 defines a single axis rotation of the vertical column through the angle about the axis at point . The revolute joint 2 defines a single axis rotation of the the inner arm through the angle about the axis at point . Create free body diagrams for bodies 1 and 2 in this robotic system.

Figure 4.15 Base body and inner arm of a PUMA robot.

Solution

There is a revolute joint at point , and its joint variable is . There must be three orthogonal reaction forces that constrain the relative displacement between the base link 0 and link 1 at point . There must also be two orthogonal reaction moments that prevent rotation of link 1 relative to the – plane of the frame. These forces and moments that act on link 1 are represented in terms of the basis for the frame, so they have the form

In addition to the forces and moments due to the joint constraints, an actuator is included in the robotic system to move the joint through its degree of freedom. In a revolute joint, the actuator will supply a torque along the joint axis. For joint 1, this torque is applied about the axis, such that

The revolute joint at point allows only relative rotation between link 1 and link 2 about the axis. There must be three orthogonal reaction forces that prevent relative displacement between these two bodies, and there must be two orthogonal reaction moments that constrain rotation relative to the plane perpendicular to the axis. The constraint reaction forces and moments that act on link 1 at point in terms of the basis for the frame are

A second actuator also controls the relative rotation between the inner arm and vertical column by supplying a second torque along the revolute joint 2 axis. For joint 2, this torque is applied about the axis and is given by

By convention, free body diagrams are constructed from the terminal link to the base. Therefore, as shown in Figure 4.16b, the forces and moments for joint 2 are applied without modification to point on link 2. However, the loading acting on link 1 due to joint 2 must have equal magnitude but opposite direction in relation to the loading acting on link 2 due to joint 2, as shown in Figure 4.16b. Figure 4.16b also includes the loading on link 1 due to joint 1.

Figure 4.16 Link free body diagrams. (a) Link . (b) Link .

---

---

Example 4.13

Two solar arrays are to the base body of a satellite by two revolute joints, as shown in Figure 4.17. The frame is fixed in the base body of the satellite, and frames and are fixed in the solar arrays. Create free body diagrams of each body in this mechanical system.

Figure 4.17 Satellite with two solar arrays.

Solution

For this example, all of the reaction forces and moments are represented in terms of the basis for the frame that is fixed in the core body. The revolute joint between bodies and prevents all relative motion except rotation about the axis. The constraint forces and moments at this joint that act on the satellite body are shown in Figure 4.18b and are given by

Figure 4.18 Satellite free body diagrams. (a) Satellite body. (b) Solar panels.

The actuation torque acts on the satellite body and drives the relative motion about the degree of freedom axis . It is written

Similarly, the constraint forces and torques that act on the satellite body due to the solar array are expressed as

and the actuation moment that drives relative motion is

These reaction forces and moments, as well as the actuation torque, are also depicted in Figure 4.18b.

For the free body diagram of the satellite body shown in Figure 4.18a, the loading equal and opposite to the loading applied on each solar panel array is applied on the satellite body at the shared point between each pairs of bodies.

---

The free body diagrams developed in Examples 4.12 and 4.13 are prototypical of those derived in a host of mechanical systems encountered in applications. Additional examples are considered in the problems at the end of this chapter. The goal in creating free body diagrams is to enable the determination of the equations of motion from Euler's first and second laws. The following examples use the free body diagram to cast the equations of motion in simple case studies. In Example 4.14, the mass and inertia matrix that appear in Euler's laws are assumed to be negligible, and the force and moment summations take a particularly simple form. We use this analysis to confirm that the composite joint satisfies Theorem 4.17.

---

Example 4.14

The composite joint shown in Figure 4.19 is constructed from a rectangular bar, collar, and yoke that have body fixed frames , , and , respectively. Relative displacement is allowed between the rectangular bar and the collar along the axis, and relative rotation is allowed between the collar and yoke about the axis. Create free body diagrams for the rectangular bar, collar and yoke. If the collar shown in Figure 4.20b has negligible mass and inertia, determine equivalent free body diagrams for the loading transmission directly between the rectangular bar and the yoke.

Solution

The revolute joint between the collar and yoke permits only relative rotation about the axis, so there must be three perpendicular forces that restrain relative translation between these two bodies. There must also be two perpendicular moments acting between the two bodies that prevent relative rotation. The reaction forces and moments acting on the yoke are represented in terms of the basis for the frame as

Figure 4.19 Composite joint, prismatic and revolute.

Figure 4.20 Composite joint free body diagrams. (a) Yoke. (b) Collar.

Equal and opposite reaction forces and moments act on the collar due to this joint.

The prismatic joint between the collar and the rectangular bar permits relative displacement along the axis. There are two perpendicular forces that prevent relative displacement orthogonal to the axis, and there are three moments that act between the bodies to constrain relative rotation. The reaction forces and moments that act on the collar are denoted and , respectively, and are represented in terms of the basis for the frame as

Equal and opposite reaction forces and moments act on the rectangular bar due to this joint.

Figures 4.20a, 4.20b and 4.21 illustrate the three free body diagrams for the yoke, collar, and rectangular bar, respectively, which include the reaction forces and moments due to these two joints.

Figure 4.21 Composite joint rectangular bar free body diagram.

Next, the composite joint is analyzed assuming that the collar has negligible mass and inertia. Given that the collar has neither mass nor inertia, it is unnecessary to create a free body diagram for that component and to consider it as a purely geometric entity (instead of mechanical) that enforces the pair of joint constraints. The simplest approach for deriving the loading between these two bodies connected by a pair of ideal joints is to utilize a set of basis vectors that align with the two axes of the ideal joints. In this case, frame aligns with both the prismatic and revolute joints.

The prismatic joint does not permit a force along , and the revolute joint does not permit a moment about . The remaining four components of force and moment may be non‐zero. As a result, the internal forces and moments acting on the yoke may be defined as

As before, an equal and opposite reactive force and moment act on the rectangular bar.

Figures 4.22a and 4.22b illustrate the free body diagrams for the yoke and rectangular bar acting under the assumption that the dynamics of the collar may be ignored.

Figure 4.22 Composite joint free body diagrams assuming massless collar. (a) Yoke. (b) Rectangular bar.

---

---

Example 4.15

The composite joint shown in Figure 4.23 is constructed from a cylindrical shaft, collar, and yoke that have body fixed frames , , and , respectively. Relative displacement and rotation is allowed between the rectangular bar and the collar along/about the axis, and relative rotation is allowed between the collar and yoke about the axis. Create free body diagrams for the cylindrical rod, collar, and yoke. If the collar shown in Figure 4.24b has negligible mass and inertia, determine equivalent free body diagrams for the loading transmission directly between the cylindrical rod and the yoke.

Figure 4.23 Composite joint, cylindrical and Revolute.

Figure 4.24 Composite joint free body diagrams. (a) Yoke. (b) Collar.

Solution

As in Example 4.14, the revolute joint between the collar and yoke allows only relative rotation. There are consequently three perpendicular forces and two perpendicular moments acting between these two bodies. The reaction forces and moments acting on the yoke can be expressed in terms of the basis as

The cylindrical joint between the cylindrical shaft and collar permits rotation about and translation along the axis. It follows that there are two perpendicular reaction forces and two perpendicular reaction moments that act between these two bodies. The reaction forces and moments that act on the collar defined in terms of the basis for the frame are

Figures 4.24a, 4.24b and 4.25 illustrate the three free body diagrams for the yoke, collar, and cylindrical rod, respectively, that include the reaction forces and moments due to these two joints. As before, the revolute joint loading is applied directly to the yoke, and the equal and opposite loading is applied to the collar. Likewise, the cylindrical joint loading is applied directly to the collar, and the equal and opposite loading is applied to the cylindrical rod.

Next, the composite joint is analyzed assuming that the collar has negligible mass and inertia. As before, a set of basis vectors aligning with the to joint axes will be utilized; in this case, aligns with both the cylindrical and revolute joints. The cylindrical joint does not permit a force along or a moment about , and the revolute joint does not permit a moment about . The remaining three components of force and moment may be non‐zero. As a result, the internal forces and moments acting on the yoke may be defined as

Figure 4.25 Composite joint cylindrical rod free body diagram.

Figure 4.26 Composite joint free body diagrams assuming massless collar. (a) Yoke. (b) Cylindrical rod.

Figures 4.26a and 4.26b illustrate the free body diagrams for the yoke and cylindrical rod acting under the assumption that the dynamics of the collar may be ignored. As before, the composite joint loading is applied directly to the yoke, and the equal and opposite loading is applied to the cylindrical rod.

---

The last few examples have derived the appropriate free body diagrams that are associated with rigid bodies connected by ideal joints. Euler's first and second laws have been applied in the case where the mass and inertia matrix of one of the bodies are negligible; the resulting equations of motion resemble those obtained in elementary statics problems of engineering. The next two examples study robotic systems in three spatial dimensions. First, a single degree of freedom of motion is considered, followed by the generalization to the two degree of freedom case. The resulting equations of motion significantly increase in complexity as the number of degrees of freedom increase.

---

Example 4.16

This example studies the two link robot described in Example 4.12. In this example, revolute joint 1 between the base and vertical cylinder is fixed ( ), and revolute joint 2 between the vertical cylinder and inner arm is allowed to rotate through angle . Derive the equations of motion of the arm using Euler's first and second laws. The location of the mass centers in this system are depicted in Figure 4.27.

Figure 4.27 Links 0–2 of a PUMA robotic arm.

Solution

The free body diagram for the arm has already been constructed in Figure 4.16b. The acceleration of the center of mass of the arm is given by

where in this problem , , , and since is a constant and the vertical column is held fixed. Note that one does not simply differentiate in to conclude that the angular acceleration . The definition of the angular acceleration and the derivative Theorem 2.12 in Chapter 2 should always be used to obtain the correct expression. For example, in this problem,

However, in the next Example 4.17, it will be shown that it is not always the case that is “merely” a vector having magnitude equal to . Upon substitution, it can be seen that

Euler's first law yields

which can be rewritten as in terms of the components as

(4.27)

The – and – planes of the frame are planes of symmetry of the arm, so that the angular momentum in of the arm about its own center of mass is

Euler's second law can be written as

Since the cross product can be expanded as

it is possible to use the derivative Theorem 2.12 in Chapter 2 to obtain

(4.28)

The final system of governing equations for this problem include Equation 4.27 and 4.28. There is one set of three equations governing translation and one set of three equations governing rotation. The unknowns in these equations include and its derivatives, as well as the reaction forces and torques . Note that there are exactly six equations and six unknowns in the final form of the governing equations. The variable is known as adifferential unknown since derivatives of this variable appear in the governing equations. The variables are known asalgebraic unknowns because their derivatives do not appear in the governing equations.

---

---

Example 4.17

Derive the equations of motion for the two link robotic system shown in Figures 4.27, 4.16a, and 4.16b, and studied in Examples 4.12 and 4.16. In contrast to Example 4.16, both of the joint variables and are unknown in the discussion that follows.

Solution

First, the kinematics of the two links will be defined to facilitate calculation of the dynamic model during the link‐by‐link analysis. The rotation matrices between the three bodies are defined as

Next, the angular velocities and accelerations between the links will be defined. The angular velocity of link 1 may be defined with respect to frame as

The angular velocity of link 2 is determined by the addition Theorem 2.15 in Chapter 3, so that

The angular acceleration of these bodies may be found by taking the derivative of these angular velocities with respect to the frame. For link 1, this results in

However, for the angular acceleration of link 2, the derivative Theorem 2.12 of Chapter 2 must be used to reformulate the frame derivative in terms of frame ,

As noted in the previous example, the angular acceleration is not generally the time derivative of the joint angle coefficients (as was the case in link 1); the derivative theorem also accounts for the time rate of change of the basis vectors as well.

Next, the velocities points and will be calculated. Within link 1, the relative velocity Theorem 2.16 will be used to calculate the velocities of points and . (In the next step, the velocity of point will be used to calculate the velocity of point .) For point ,

Likewise, the velocity of point is

As points and are both fixed on body 2, the relative velocity theorem may also be used to determine the velocity of point from the velocity of point , such that

Next, this kinematic analysis will be used to construct the equations of motion for links 1 and 2, starting with link 2. Based on the free body diagram of link 2 in Figure 4.16b, Euler's first law for link 2 can be written as

where

and

Combining these expressions into a single equation results in

(4.29)

Despite it being link 2 being under consideration, because the simplest representation of is with respect to the frame, the equation of motion is analyzed with respect to this frame. When evaluating this vector equation, either these components will need to be mapped into the frame to match the force components, or the force components will need to be mapped into the frame.

Euler's second law applied to link 2 results in

The left hand side may be solved by utilizing the derivative theorem to formulate the derivative in terms of the frame, and utilizing the angular momentum representation ,

The angular velocity may be determined from the formulation of above using a change of basis,

By a similar transformation, may be determined to be

The load acting on link 2 may be evaluated by noting that , resulting in

Representing Euler's second law for link 2 with respect to the frame results in

or

(4.30)

Next, the equations of motion for link 1 will be calculated. Based on the free body diagram of link 1 in Figure 4.16a, Euler's first law for link 1 can be written as

where

and

Representing Euler's first law for link 1 with respect to the frame results in

(4.31)

Euler's second law applied to link 2 results in

As before, the left hand side may be solved by utilizing the derivative theorem to formulate the derivative in terms of the frame and utilizing the angular momentum representation in

The angular velocity and acceleration may be determined from the formulations of and above, with

In addition, for the inertia matrix, because there is only a single plane of symmetry, the and entries of the inertia matrix will be non‐zero. It follows that

The load acting on link 2 may be evaluated by representing it in terms of frame . Noting that and , it is written as

Combining the angular momentum derivative and loading into Euler's second law for link 1 results in

(4.32)

In summary, the complete set of governing equations for this example includes Equations 4.29, 4.30, 4.31, and 4.32. There is one set of three equations governing translation and one set of three equations governing rotation for each body. The unknowns in these equations include the geometric variables and their derivatives, and the forces and moments . Thus there are twelve unknowns and twelve equations in this problem. As in the last example, and are differential unknowns and the constraint forces and torques are algebraic unknowns.

---

The equations of motion for mechanical systems in realistic problems with three dimensional kinematics can be quite complicated in form. The next example studies another two link robotic system. Because of the symmetry of the bodies with respect to the selected frames of reference, and the restriction of motion so that only varies, the equations of motion take a familiar form in Example 4.18.

---

Example 4.18

The inner arm (link 1) of the SCARA robot shown in Figure 4.28 is locked in place relative to the base, and the outer arm (link 2) is articulated through the angle . Derive the equations of motion for the outer arm using Euler's first and second laws.

Figure 4.28 Links 0–2 of a SCARA robotic arm.

Solution

First, three frames , , and are defined at points , , and , respectively, as shown in Figure 4.28. Frame is the inertial frame, frame is fixed in the inner arm, and frame is fixed in the outer arm. The vertical column is denoted link 0, the inner arm is link 1 and the outer arm is link 2 in this example. The location of the center of mass of the outer arm is defined as

in terms of the body fixed basis. The inertia matrix with respect to a frame that has its origin at the center of mass and that is parallel to the frame is

Since the 1–3 plane of the frame is a plane of symmetry for the outer arm, it is known that . The acceleration of the center of mass of the outer arm in the inertial frame is given by

when is constant. A free body diagram of the outer arm is depicted in Figure 4.29. Euler's first law yields the components relative to the frame

(4.33)

Figure 4.29 Free body diagram of the outer arm.

Euler's second law is written , and the applied moment about the center of mass is given by

The angular momentum about the center of mass is

since when the inner arm is stationary. The derivative theorem from Chapter 2 is used to simplify the calculation in Euler's second law. The moment is

Combination of the moment equations above yields

(4.34)

In summary, a complete set of governing equations is obtained by collecting the six Equations 4.33 and 4.34 to solve for the six unknowns. These unknowns include the angle and its derivatives, and the reaction forces and moments and .

---

The equations of motion derived in Example 4.18 assumed a particularly simple form due to symmetry and the fact that there is but a single degree of freedom. In general, however, the equations of motion can be quite complex. In this next example it is shown that by allowing the joint variable in Example 4.18 to vary in time as an unknown, the resulting equations of motion for the system again become increasingly more complicated.

---

Example 4.19

Consider again the SCARA robot studied in Example 4.18. In this example, derive the equations of motion for the two degrees of freedom system, where and vary in time.

Solution

First, the outer arm, link 2, will be considered. Note the changes to the governing equations from Example 4.18 when both and vary with time. The angular velocity of the frame in the frame is now given by

The form of the angular acceleration follows from this definition and the application of the derivative theorem

The acceleration of the center of mass of the outer arm is now given by

The force summation for the free body diagram depicted in Figure 4.29 yields the equations

(4.35)

The angular momentum of the outer arm about its own center of mass becomes

The moment summation is now given by

(4.36)

Figure 4.30 Free body diagram of the inner arm.

The free body diagram of the inner arm is shown in Figure 4.30. The acceleration of the center of mass of the inner arm is given by

and the angular momentum about the center of mass is given by

The force summation for the free body diagram shown in Figure 4.30 yields

or

(4.37)

The applied moments about the center of mass can be computed from the expression

This expression may be simplified and written in terms of frame as

(4.38)

(4.39)

The moment summation for the free body diagram in Figure 4.30 is consequently

(4.40)

where is given in Equation 4.38.

The final system of governing equations for this example includes Equations 4.35, 4.36, 4.37, and 4.40. There is one set of three equations governing translation for each body, and there is one set of three equation governing rotation for each body. The differential unknowns in the governing equations include and their derivatives. The algebraic unknowns are the reaction forces and torques . There are a total of 12 unknowns and 12 equations.

---

4.6 Structure of Governing Equations: Newton–Euler Formulations

Examples 4.16, 4.17, 4.18, and 4.19 demonstrate that the system of equations obtained via the application of Euler's first and second law to robotics applications is unavoidably complicated. An essential step in the study of Newton–Euler formulations casts the derived equations into one of a few possible canonical mathematical forms. This step is important from a theoretical standpoint since in many cases analysts have developed a theory that can be consulted to study existence, uniqueness, or stability of solutions for general classes of problems. From a practical viewpoint, once the equations are written in a standard from, it is possible to employ common and well understood numerical techniques for their study. While there are many possible structural forms in which the governing equations can be cast, two standard forms will be considered in this book. The equations of motion will be written as a system ofdifferential algebraic equations or DAEs, or as a set ofordinary differential equations or ODEs.

4.6.1 Differential Algebraic Equations (DAEs)

There are two types of variables that appear in the governing equations in Newton–Euler formulations of dynamics, as shown in Examples 4.16, 4.17, 4.18, and 4.19. One set of variables are explicitly differentiated with respect to time, while others are not. The subset of variables that are differentiated with respect to time are referred to as differential unknowns, while those that are not differentiated are known as algebraic unknowns. A collection of equations that couples both types of variables constitutes a system of DAEs. The form of DAEs that will be used in this text have the structure

(4.41)

where is the collections of differential variables, is the collection of algebraic variables, the function and . There are a total of variables in the above equations. It is straightforward to express the equations derived via Newton–Euler methods in this form, as shown in the next examples.

---

Example 4.20

Write the governing equations in Example 4.16 as a set of DAEs that have the structure shown in Equation 4.41.

Solution

The unknown variables that appear in Equations 4.27 and 4.41 are , , , , , and . Define the differential variables and algebraic variables as, respectively,

The differential subset of equations becomes

(4.42)

The algebraic subset of equations is

(4.43)

With the functions and defined in Equations 4.42 and 4.43, the governing equations have the form of the DAEs in Equation 4.41.

---

---

Example 4.21

Write the governing equations in Example 4.18 as a set of DAEs that have the structure shown in Equation 4.34.

Solution

The unknown variables that appear in Equations 4.33 and 4.34 are , , , , , and . Define the differential and algebraic variables as, respectively,

The subset of differential equations becomes

The subset of algebraic equations is

With the functions and defined in Equations 4.42 and 4.43, the governing equations have the form of the DAEs in Equation 4.41.

---

4.6.2 Ordinary Differential Equations (ODEs)

In most applications of Newton–Euler formulations to realistic robotic systems, the resulting equations contain both differential and algebraic variables. As discussed in the last section, it is natural to study these systems as a collection of DAEs. Still, it is most common in the literature to eliminate the algebraic unknowns via algebraic manipulation. There are technical conditions, expressed in terms of the Jacobian of the algebraic constraint equations in Equation 4.41, that guarantee that it is possible to solve the last equations in 4.41 for the algebraic unknowns in terms of . The expressions for in terms of can then be substituted into the first line of Equation 4.41, thereby eliminating the algebraic unknowns. The interested reader should consult [27] or [2] for the complete description. When this procedure is feasible, the resulting form of the governing equations is a collection of ODEs of the form

(4.44)

where and . It must be emphasized that although the procedure discussed above to reformulate a set of DAEs in the form of Equation 4.41 into Equation 4.44 is easy to describe in principle, it can be difficult to achieve in practice. This is certainly the case for complex robotic systems. Despite the difficulty of the task, there are several reasons that motivate the attempt.

(1) Order reduction. For any system, the number of DAEs is always larger than the reduced number of ODEs. The difference in complexity can be substantial for robotic systems. In fact, most of the unknowns in a kinematic chain are associated with the algebraic variables. If it is desired to construct a feedback law that must be updated in real time, the cost in estimation of the solution of a set of DAEs can be prohibitive. For example, for an link kinematic chain having one degree of freedom at each joint, and .

(2) Theoretical foundations. The study of ODEs and their numerical approximation is much more mature than the corresponding state of affairs for DAEs. The study of DAEs is continuously evolving. Furthermore, the current theoretical framework for DAEs is more difficult to state and has more restrictions on its applicability. Broad statements about the existence, stability, convergence and stability of DAEs are less common. Again, references [2] and [27] can be consulted for advanced presentations.

(3) Control theoretic issues. It has already been noted in (1) above that the computational cost associated with the numerical solution of DAEs can be prohibitive in control applications. Just as importantly, nearly all of the supporting theory and algorithms for the control of robotic systems has been derived for systems of ordinary differential or discrete difference equations. This body of research is impressive and represents decades of research and development. The study of the control of DAEs does not have as mature a theoretical and computational infrastructure.

The next two examples illustrate how the algebraic unknowns can be eliminated for relatively simple problems.

---

Example 4.22

Obtain a governing ordinary differential equation from the DAEs in Example 4.16 by eliminating the algebraic unknowns.

Solution

The formulation of the DAEs for Example 4.16 presented in Example 4.20 will be used to obtain the desired ordinary differential equation form. The second differential equation in this example states

(4.45)

The desired ordinary differential equation is obtained by eliminating the algebraic variable from this equation. This can be done by considering the sole algebraic equation with as a factor,

Combining these two equations results in

This equation is the familiar equation in terms of the moment of inertia about the 2 axis through point . The moment of inertia about the 2 axis through point is given by the parallel axis theorem

and the final form of the governing equation is

---

---

Example 4.23

Obtain a governing ordinary differential equation from the DAEs in Example 4.18 by eliminating the algebraic unknowns.

Solution

Combining the equations from the second row in the Equation 4.33 and the third row of Equation 4.34 results in

for the rotational equation of motion about the vertical axis. This equation can be written in a simpler form by observing that the parallel axis theorem states that

where is the moment of inertia about the 3 axis passing through point . As a result,

is the moment equation about this axis.

---

4.7 Recursive Newton–Euler Formulations

4.7.1 Recursive Calculation of Forces and Moments

In Chapter 3 recursive algorithms for determining velocities and derivatives of velocities have been derived for the kinematic chain depicted in Figure 3.20 or Figure 3.21. The joints are numbered from to 1 starting at the joint nearest the base body and decreasing toward the tip. The recursive algorithms developed in Sections 3.4.1 and 3.4.3 start with the joint connected to the base body, and then solve for velocities or derivatives of velocities as we move from inboard to outboard joints.

This section will show that the calculation of joint forces and torques can be carried out starting at joint 1 toward the tip of the kinematic chain and proceeding inward toward the base. As in the study of velocities and accelerations, a recursive method is possible because of the special structures of the matrices that relates joint forces and torques, acceleration, and angular acceleration. The reason that the recursion in this section progresses from tip to base is because the coefficient matrix in Theorem 4.18 is exactly equal to the transpose of the coefficient matrix that appears in either Theorem 3.4 or Theorem 3.5. The following theorem summarizes this matrix equation.

---

Theorem 4.18

The forces and moments in for the kinematics chain depicted in Figure 3.20 and Figure 3.21 satisfy the equation

(4.46)

(4.47)

where is given in Equation 4.50, is defined in Equation 4.50, is defined in Equation 4.51, is given in Equation 4.56, and is given in Equation 4.57 for .

---

Figure 4.31 Joint loading convention.

4.1a

Recall that in this convention the links of the kinematic chain are numbered from the outer most body to the root or base body . The acceleration in the base frame of the center of mass of body is given by

where is the acceleration in the base frame of the origin of frame is the angular acceleration in the base frame of frame is the angular velocity in the base frame of frame , and is the vector from the origin of frame to the center of mass of link .

Applying Euler's first law to body , as shown in Figure 4.31, results in

(4.48)

For Euler's second law, the form given in Theorem 4.15 in which moments acting on body are summed about point ,

is used to obtain the equation

The derivative Theorem 2.12 is applied to obtain the familiar form

(4.49)

Equation 4.48 and Equation 4.49 give a complete description of the dynamics of body ; however, they must be expressed in terms of the vectors

(4.50)

and

(4.51)

The desired form can be achieved by recalling that the acceleration is defined as

(4.52)

and the angular acceleration is defined as

(4.53)

Substituting Equations 4.52 and 4.53 into 4.48 and 4.49 results in

and

This pair of equations can be written as a singlematrix equation

(4.54)

This equation may be written in the compact form

(4.55)

where the generalized mass or inertia matrix is defined as

(4.56)

and the inertial force vector is given as

(4.57)

Internal forces have equal magnitude and opposite direction, which implies that

It follows that and satisfy the equation . The rotation matrix appears since, by definition, the entries in are the components of the forces and moments in terms of the basis for the frame. The final form of the recursive equation for the forces is obtained by substituting this equation into Equation 4.55, which results in

(4.58)

Equation 4.46 is obtained when Equation 4.58 is expressed for each joint .

The coefficient matrix that appears in Equation 4.46

is the transpose of the coefficient matrix that appears in the recursive calculation of the velocity and acceleration in Equations (3.13) in Theorem 3.4 and (3.30) in Theorem 3.5. In both of those cases, the structure of the coefficient matrix enables a recursive procedure that starts with the last row, which corresponds to the root of the kinematic chain. In the current case, for solving the forces and moments, the structure of the lead coefficient matrix makes it possible to solve the first equation for the forces and moments in

Suppose that all of the velocities and angular velocities have been solved for using the recursive algorithm described in Section 3.4.1, and that all of the derivatives of velocities and angular velocities have been solved for using the recursive algorithm summarized in Section 3.4.3. When this is the case, it is possible to evaluate all of the terms on the right side of the above equation immediately and obtain . With this, the solution can proceed to the second row in Equation 4.46 and solve for as

Since all the terms required to evaluate the right side of this equation are known, may be calculated. This process is continued until all the joints have been processed. The following table summarizes this recursive algorithm for the calculation of the reaction forces and moments.

1. 1. Solve for the velocities and angular velocities using the recursive algorithm in Section 3.4.1.
2. 2. Solve for the derivatives of the velocities and angular velocities using the recursive algorithm in Section 3.4.3.
3. 3. Iterate from outboard joints to inboard joints.
   1. 3.1 Form the bias inertial force .
      
   2. 3.2 Form the transition operator
      
   3. 3.3 Form the generalized inertia and mass matrix
      
   4. 3.4 Calculate the forces and moments
      (4.59)

Figure 4.32 Recursive algorithm for calculating forces and moments.

---

Example 4.24

Use the recursive order algorithm to solve for the joint forces and torques for link 1 of the two link robotic arm presented in Example 3.6 and shown in Figure 4.33.

Figure 4.33 Two link robotic arm.

Solution

First, the joint forces and moments will be computed directly from first principles for body 1, and it will be shown that the same results are computed using the recursive algorithm. The free body diagrams for the two bodies that comprise the kinematic chain are shown in Figures 4.34a and 4.34b. Euler's first law for body 1 yields

Figure 4.34 Free body diagrams for a two link robotic arm. (a) Link . (b) Link .

The following quantities have been derived in Examples 3.6 and 3.7:

(4.60)

(4.61)

(4.62)

(4.63)

We transform the equations to obtain the force components relative to the frame that is fixed in body 1 using the rotation matrix in Example 3.6,

(4.64)

(4.65)

The summation of moments acting on body 1 about the point is given by

A change of basis for the point acceleration yields as

which in turn is used to compute the cross product

The final expression for moment acting on body 1 is

(4.66)

Equations 4.64 and 4.66 may be assembled to reach the final expression for the vector of joint forces and torques acting at point on body 1.

(4.67)

As previously stated, the goal is to verify that the recursive algorithm yields the same result as in 4.67, via an efficient, systematic, programmable procedure. The first row of the system equations for the forces and moments yields

For , the generalized mass matrix is found to be

since . The product is expanded directly as

(4.68)

The inertial force is defined to be

This complex expression can be evaluated quickly if the termwise expressions are computed and substituted, such that

This calculation employs the fact that , , and . The inertial force takes its final form upon substitution of these immediate results,

(4.69)

Finally, Equations 4.68 and Equation 4.69 are added. After cancelling like terms, the result

(4.70)

is the same expression as the one derived via direct application of first principles in Equation 4.67.

---

4.8 Recursive Derivation of the Equations of Motion

Sections , and 4.7.1 have shown that the forward kinematics problem for either velocities or accelerations, as well as the calculation of joint forces or torques, can be solved efficiently via recursive order algorithms. This section will show that the equations of motion for robotic systems that form a kinematic chain as depicted in Figure 3.20 and 3.21 can be written in terms of constituent matrices that were introduced in Sections , and 4.7.1. Reference [24] gives a comprehensive account of this formulation, or see [14] for a similar discussion. The velocities and angular velocities for the kinematic chain are collected in the vector

where point is attached to body on the outward side of joint . The terms are the components relative to the basis for the frame of the velocity vector . The terms are the components relative to the basis for the frame of the angular velocity vector . Equation (3.13) can be written in terms ofsystem vectors of unknowns as

(4.71)

where the assembled system vectors and are defined as

and the system matrices and are defined as

The matrix and its transpose will appear repeatedly in the derivatives that follow. A similar equation can be obtained from Theorem 3.5 for the derivatives of the velocities and angular velocities. The vectors that are used to formulate the kinematic problem for the accelerations are defined to be

The primary result in Section 3.4.3 as made explicit in Theorem 3.5 is that the derivatives of the velocity and angular velocities satisfy the system equation

(4.72)

where the system vectors and are defined as

Finally, recall the definition of the vector

which contains the components relative to frame of the force vectors and moments that are applied at point on body at the joint . Equation 4.48 in Theorem 4.18 holds that the joint forces and moments satisfy the system equation

(4.73)

where the system vectors and , and system matrix , are defined to be

The determination of the equations of motion for the kinematic chain depicted in Figures 3.20 and 3.21 is straightforward once Equations 4.71, 4.72, and 4.73 are derived. Equations 4.72 and 4.73 may be solved for and , respectively, to obtain

(4.74)

(4.75)

Equation 4.74 is then substituted into 4.75 to obtain an expression for the joint forces and torques

(4.76)

In this equation, the vector contains the joint forces and moments acting on each body in the kinematic chain.

The final equations of motion for the kinematic chain are obtained by noting that all of the forces and moments that act on the individual joints of the body can be decomposed into constraint forces and moments and actuation torques as

(4.77)

where is the actuation torque generated about the axis for . The system vector of actuation torques has been introduced into this equation as

By definition, the constraint forces and moments are orthogonal to the actuation torques. Since the direction of the actuation torque is given by , the condition that the constraint forces are perpendicular to the actuation forces and moments can be written as

Equation 4.77 can be premultiplied by to utilize this orthogonality condition and obtain the equation

(4.78)

This equation calculates the projection of all of the joint forces and moments along the direction of the actuation torques. Recall that each is a unit vector by definition, which guarantees that

with an appropriately dimensioned identity matrix. The final equations of motion for the robotic system can be obtained by combining Equations 4.76 and 4.78 to attain

4.9 Problems for Chapter 4, Newton–Euler Equations

4.9.1 Problems on Linear Momentum

Problems 4.1,4.2, and 4.3 refer to the robotic manipulator shown in Figures 4.35 and 4.36. This robot has revolute joints about the , , and axes. The joint variables , , and are measured about the , , and axes, respectively. In each case the angle is measured from the positive axis to the positive axis, for .

Figure 4.35 PUMA robot frame definitions.

Figure 4.36 PUMA robot joint and mass center offsets.

1. Problem 4.1 Compute the linear momentum in the 0 frame of link 1, the shoulder, for the robot shown in Figures 4.35 and 4.36. Express your answer in terms of the basis for the 1 frame and 0 frame.
2. Problem 4.2 Compute the linear momentum in the 0 frame of link 2, the inner arm, for the robot shown in Figures 4.35 and 4.36. Express your answer in terms of the basis for the 2, 1 and 0 frames.
3. Problem 4.3 Compute the linear momentum in the 0 frame of link 3, the outer arm, for the robot shown in Figures 4.35 and 4.36. Express your answer in terms of the basis for the 3, 2, 1 and the 0 frames.

   Problems 4.4,4.5, and 4.6 refer to the SCARA robot shown in Figures 4.37 and 4.38. This robotic manipulator has two revolute joints about the and directions and a prismatic joint along the direction. The joint variables and are measured about the and axes, respectively. In each case the angle is measured from the positive axis to the positive axis, for .

   

   Figure 4.37 SCARA robot frame definitions.

   

   Figure 4.38 SCARA robot joint and mass center offsets.

4. Problem 4.4 Compute the linear momentum in the 0 frame of link 1, the inner arm, for the robot shown in Figures 4.37 and 4.38. Express your answer in terms of the basis for the 1 frame and 0 frame.
5. Problem 4.5 Compute the linear momentum in the 0 frame of link 2, the outer arm, for the robot shown in Figures 4.37 and 4.38. Express your answer in terms of the basis for the 2, 1, and 0 frames.
6. Problem 4.6 Compute the linear momentum in the 0 frame of link 3, the tool carriage, for the robot shown in Figures 4.37 and 4.38. Express your answer in terms of the basis for the 3, 2, 1, and 0 frames.

   Problems 4.7,4.8, and 4.9 refer to the robotic manipulator shown in Figures 4.39 and 4.40. This robotic manipulator has a revolute joint along the direction, and it has a prismatic joints along the , , and directions. The tool carriage, which is not shown, has its center of mass located at point . The joint variable is measured about the axis. The angle is measured from the positive axis to the positive axis. The height from the point to the point is the joint variable measuring displacement along the axis, the length from the point to the point is the joint variable measuring displacement along the direction, and the distance from the point to the point is the joint variable along the axis.

   

   Figure 4.39 Cylindrical robot frame definitions.

   

   Figure 4.40 Cylindrical robot joint and mass center offsets.

7. Problem 4.7 Compute the linear momentum in the 0 frame of link 1, the vertical arm, for the robot shown in Figures 4.39 and 4.40. Express your answer in terms of the basis for the 1 and 0 frames.
8. Problem 4.8 Compute the linear momentum in the 0 frame of link 2, the horizontal arm, for the robot shown in Figures 4.39 and 4.40. Express your answer in terms of the basis for the 2, 1, and 0 frames.
9. Problem 4.9 Compute the linear momentum in the 0 frame of link 3, the tool carriage, for the robot shown in Figures 4.39 and 4.40. Express your answer in terms of the basis for the , and 0 frames.

4.9.2 Problems on the Center of Mass

1. Problem 4.10 Figure 4.41a depicts the terminal component in an industrial robot. Assume that the mass density is uniform and that the body can be approximated as the union of two cylinders having radii and lengths and . The location of the center of mass of each cylinder is shown in the figure. Find the mass center of this composite rigid body.

   

   Figure 4.41 Industrial robot end effector component. (a) Detailed design. (b) Geometric primative.

2. Problem 4.11 Figure 4.42a depicts the fixed base of a robotic assembly. Assume that the mass density is uniform and can be approximated as the union of a cylinder with radius and length and a rectangular prism of dimensions , as shown in 4.42b. The location of the center of mass of each cylinder is also shown in Figure 4.42b. Find the mass center of this composite rigid body relative to the frame.

   

   Figure 4.42 Industrial robot fixed base component. (a) Detailed design. (b) Geometric primative.

3. Problem 4.12 Figure 4.43a depicts the mounting bracket for an industrial robot. Assume that the mass density is uniform and can be approximated as the union of three rectangular prisms. The location of the center of mass of each is shown in Figures 4.43a and 4.43b. Find the mass center of this composite rigid body.

   

   Figure 4.43 Industrial robot mounting bracket component. (a) Detailed design. (b) Geometric primative.

   Problems 4.13 and 4.14 study a two link robot that consists of link 1 and 2, as shown in Figures 4.44a and 4.44b. The center of mass coordinates of link 1 with respect to frame 1 are , and the center of mass coordinates of link 2 with respect to frame 2 are .

   

   Figure 4.44 Links and of an industrial robot. (a) Horizontal configuration. (b) Vertical configuration.

4. Problem 4.13 The two link robot in Figure 4.44a is locked in the horizontal configuration shown. At the instant shown, assume that , or . Calculate the location of the center of mass of the robot in this configuration in terms of the frame 1.
5. Problem 4.14 The two link robot in Figure 4.44b is locked in the vertical configuration shown. As the instant shown, assume that , or at the instant shown. Calculate the location of the center of mass of the robot in this configuration in terms of frame 1.

4.9.3 Problems on the Inertia Matrix

1. Problem 4.15 Consider the rigid body studied in Problem 4.10. Assume the origin of frame (point ) is located along the axis of cylinder 2 and is parallel to the axis of cylinder 1. Use symmetry arguments to show that the form of the inertia matrix about the point relative to the frame is given by
   

   In other words, use symmetry arguments to show that the frame defines a set of principal axes for the composite body. Derive the entries of this matrix by using the parallel axis theorem for each of the components of this composite body.

2. Problem 4.16 Consider the rigid body studied in Problems 4.10 and 4.15. Calculate the inertia matrix relative to axes parallel to the frame but whose origin is at the center of mass of the composite body.
3. Problem 4.17 Consider the rigid body studied in Problem 4.12. Use symmetry arguments to show that the form of the inertia matrix about the point (the origin of frame ) relative to frame is given by
   

   In other words, use symmetry arguments to show that the frame defines a set of principal axes for the composite body. Derive the entries of this matrix by using the parallel axis theorem for each of the components of this composite body.

4. Problem 4.18 Consider the rigid body studied in Problems 4.12 and 4.17. Calculate the inertia matrix relative to axes parallel to the frame but whose origin is at the center of mass of the composite body.
5. Problem 4.19 Consider the rigid body studied in Problem 4.11. Use the parallel axis theorem to calculate the inertia matrix relative to frame at its origin.
6. Problem 4.20 Consider the rigid body studied in Problems 4.10,4.15, and 4.16. Calculate the inertia matrix about the point relative to the frame that is is obtained by rotating the frame by about the axis.
7. Problem 4.21 Consider the yoke studied in Problems 4.12,4.17, and 4.18. Calculate the inertia matrix about the point relative to the frame that is obtained by rotating the frame by about the axis.
8. Problem 4.22 Consider the two link robot shown in Figure 4.44a and studied in Problem 4.13. Calculate the inertia matrix with respect to frame 1 at its origin for the system in this configuration.
9. Problem 4.23 Consider the two link robot shown in Figure 4.44b and studied in Problem 4.14. Calculate the inertia matrix with respect to frame 1 at its origin for the system in this configuration.
10. Problem 4.24 Consider the satellite studied in Example 4.13 and shown in Figure 4.17. Let and be the joint variables that measure the rotation of solar arrays and about the axis. Suppose the principal moments of inertia of the satellite body about its own center of mass relative to the frame are given in the form
    

    and the principal moments of inertia of each solar array about their own centers of mass and respectively) relative to their own body fixed frame are given in the form

    

    Find an expression for the inertia matrix of the satellite relative to the frame for any joint angle and .

4.9.4 Problems on Angular Momentum

Problems 4.25,4.26, and 4.27 refer to the PUMA robot shown in Figures 4.35 and 4.36.

1. Problem 4.25 Compute the angular momentum in the 0 frame of link 1, the shoulder, shown in Figures 4.35 and 4.36 about its own center of mass. Express your answer in terms of a frame whose origin is at the center of mass and that is parallel to the 1 frame.
2. Problem 4.26 Compute the angular momentum in the 0 frame of link 2, the inner arm, depicted in Figures 4.35 and 4.36 about its own center of mass. Express your answer in terms of a frame whose origin is at the center of mass and that is parallel to the 2 frame.
3. Problem 4.27 Compute the angular momentum in the 0 frame of link 3, the outer arm, depicted in Figures 4.35 and 4.36 about its own center of mass. Express your answer in terms of a frame whose origin is at the center of mass and that is parallel to the 3 frame.

   Problems 4.28 and 4.29 refer to the robotic manipulator depicted in Figures 4.37 and 4.38.

4. Problem 4.28 Calculate the angular momentum of the 0 frame of link 1, the inner arm, shown in Figures 4.37 and 4.38 about its own center of mass. Express your answer in terms of a frame that has its origin at the center of mass and that is parallel to the 1 frame.
5. Problem 4.29 Calculate the angular momentum in the 0 frame of link 2, the outer arm, shown in Figures 4.37 and 4.38 about its own center of mass. Express your answer in terms of a frame that has its origin at the center of mass and that is parallel to the 2 frame.

   Problems 4.30,4.31, and 4.32 refer to the robotic manipulator depicted in Figures 4.39 and 4.40.

6. Problem 4.30 Compute the angular momentum in the 0 frame of link 1, the vertical arm, shown in Figures 4.39 and 4.40 about its own center of mass. Express your answer in terms of the basis for frames 1 and 0.
7. Problem 4.31 Compute theangular momentum in the 0 frame of link 2, the horizontal arm, shown in Figures 4.39 and 4.40 about its own center of mass. Express your answer in terms of the basis for frames 2, 1 and 0.
8. Problem 4.32 Compute theangular momentum in the 0 frame of link 3, the tool carriage, shown in Figures 4.39 and 4.40 about its own center of mass. Express your answer in terms of the basis for frames 4, 3, 2, 1 and 0.

4.9.5 Problems on the Newton–Euler Equations

1. Problem 4.33 Frame is fixed in the outer link and frame is fixed in the inner link of thespherical joint shown in Figure 4.45. What are the constraints on the displacement and rotation imposed by the spherical joint on the bodies having fixed frames and ? How many degrees of freedom does the spherical joint have? Draw consistent free body diagrams for the spherical joint.

   

   Figure 4.45 Spherical joint frames.

2. Problem 4.34 Theuniversal joint shown in Figure 4.46 is comprised of a yoke with body fixed frame , a yoke with body fixed frame , and the crossbar with body fixed frame . Draw a set of complete free body diagrams for three components that make up the universal joint. Then, assume the mass and inertia matrix of the cross bar are negligible and derive an equivalent set of free body diagrams that eliminate the crossbar. How many degrees of freedom does the universal joint have?

   

   Figure 4.46 Universal joint frames.

3. Problem 4.35 Use the results of Problem 4.3 to write Euler's first law for link 3, the outer arm, of the PUMA robot in Figure 4.35.
4. Problem 4.36 Use the results of Problem 4.6 to write Euler's first law for link 3, the tool carriage, for the SCARA robot in Figure 4.37.
5. Problem 4.37 Use the results of Problem 4.7 to write Euler's first law for link 1, the vertical arm, of the cylindrical robot in Figure 4.39.
6. Problem 4.38 Use the results of Problem 4.8 to write Euler's first law for link 2, the horizontal arm, of the cylindrical robot in Figure 4.39.
7. Problem 4.39 Use the results of Problem 4.9 to write Euler's first law for link 3, the tool carriage, of the cylindrical robot in Figure 4.39.
8. Problem 4.40 Use the results of Problem 4.27 to write Euler's second law for link 3, the outer arm, of the PUMA robot shown in Figure 4.35.
9. Problem 4.41 Use the results of Problem 4.30 to write Euler's second law for link 1, the vertical arm, of the cylindrical robot in Figure 4.39.
10. Problem 4.42 Use the results of Problem 4.31 to write Euler's second law for link 2, the horizontal arm, of the cylindrical robot in Figure 4.39.
11. Problem 4.43 Use the results of Problem (4.32) to write Euler's second law for link 3, the tool carriage, of the cylindrical robot in Figure 4.39.
12. Problem 4.44 Write Euler's second law for link 2, the inner arm, of the robot in Example 4.17, taking moments about point instead of the mass center of the arm. Note that the point is not fixed in the inertial frame in this problem.
13. Problem 4.45 Write Euler's second law for link 1, the shoulder, of the robot in Example 4.17, taking moments about the point . Note that the point is fixed in the inertial frame.