Conservation of mechanical energy is assumed if kinetic energy
(T) and potential energy (V) change back and forth in a conservative
system (the dissipation of energy is considered negligible). Equation
1.3.22 formalizes such a situation, where position 1 is the initial state
and position 2 is the ?nal state. The reference (datum) should be chosen to
reduce the number of terms in the equation.

T₁ + V₁ = T₂ + V₂

Linear and Angular Momentum Methods

The concept of linear momentum is useful in engineering when the
accelerations of particles are not known but the velocities are.
The linear momentum is derived from Newtons second law,

G  = mv

The time rate of change of linear momentum is equal to force.
When mv is constant, the conservation of momentum equation results,

The method of angular momentum is based on the momentum of a
particle about a ?xed point, using the vector product in the general case
(Figure 1.3.8).

H_o = r x mv

FIGURE 1.3.8 De?nition of angular momentum for a particle.

The angular momentum equation can be solved using a scalar
method if the motion of the particle remains in a plane,

If the particle does not remain in a plane, then the general space motion
equations apply. They are derived from the cross-product r ? mv,

Time Rate of Change of Angular Momentum

In general, a force acting on a particle changes its angular momentum:
the time rate of change of angular momentum of a particle is equal to the
sum of the moments of the forces acting on the particle.

A special case is when the sum of the moments about point O is zero.
This is the conservation of angular momentum. In this case
(motion under a central force), if the distance r increases, the velocity
must decrease, and vice versa.

Impulse and Momentum

Impulse and momentum are important in considering the motion of
particles in impact. The linear impulse and momentum equation is

Conservation of Total Momentum of Particles

Conservation of total momentum occurs when the initial
momentum of n particles is equal to the ?nal momentum of those same
n particles,

When considering the response of two deformable bodies to direct
central impact, the coef?cient of restitution is used. This coef?cient e
relates the initial velocities of the particles to the ?nal velocities,

For real materials, 0 < e < 1. If both bodies are perfectly elastic, e = 1,
and if either body is perfectly plastic, e = 0.

Kinetics of Systems of Particles

There are three distinct types of systems of particles: discrete particles,
continuous particles in ?uids, and continuous particles in rigid or deformable
bodies. This section considers methods for discrete particles that have
relevance to the mechanics of solids. Methods involving particles in
rigid bodies will be discussed in later sections.

Newtons Second Law Applied to a System of Particles

Newtons second law can be extended to systems of particles,

Motion of the Center of Mass

The center of mass of a system of particles moves under the action of
internal and external forces as if the total mass of the system and all the
external forces were at the center of mass. Equation 1.3.32
de?nes the position, velocity, and acceleration of the center of mass of
a system of particles.

Work and Energy Methods for a System of Particles

Gravitational Potential Energy. The gravitational potential energy of
a system of particles is the sum of the potential energies of the individual
particles of the system.

where g = acceleration of gravity
y_C = vertical position of center of mass with respect to a reference level
Kinetic Energy. The kinetic energy of a system of particles is the
sum of the kinetic energies of the individual particles of the system with
respect to a ?xed reference frame,

A translating reference frame located at the mass center C of a
system of particles can be used advantageously, with

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The work and energy equation for a system of particles is similar to the
equation stated for a single particle.

Momentum Methods for a System of Particles

Moments of Forces on a System of Particles. The moments of
external forces on a system of particles about a point O are given by

Linear and Angular Momenta of a System of Particles. The resultant
of the external forces on a system of particles equals the time rate of change
of linear momentum of that system.

The angular momentum equation for a system of particles about a
?xed point O is

The last equation means that the resultant of the moments of the external
forces on a system of particles equals the time rate of change of
angular momentum of that system.

Angular Momentum about the Center of Mass

The above equations work well for reference frames that are stationary,
but sometimes a special approach may be useful, noting that the
angular momentum of a system of particles about its center of mass C
is the same whether it is observed from a ?xed frame at point O or from the
centroidal frame which may be translating but not rotating. In this case

Conservation of Momentum

The Moments of Forces on a System of Particles equations for
a system of particles is analogous to that for a single particle.

Impulse and Momentum of a System of Particles

The linear impulse momentum for a system of particles is

The angular impulse momentum for a system of particles is

Kinematics of Rigid Bodies

Rigid body kinematics is used when the methods of particle kinematics
are inadequate to solve a problem. A rigid body is de?ned as one in which
the particles are rigidly connected. This assumption allows for some
similarities to particle kinematics.  There are two kinds of rigid body
motion,  translation and rotation. These motions may occur separately
or in combination.

Translation

Figure 1.3.9 models the translational motion of a rigid body.

FIGURE 1.3.9 Translational motion of a rigid body.

These equations represent an important  fact: when a rigid body is in
translation, the motion of a single point completely speci?es the motion
of the whole body.

Rotation about a Fixed Axis

Figure 1.3.10 models a point P in a rigid body rotating about a ?xed axis
with an angular velocity ?. The velocity v of point P is determined assuming
that the magnitude of r is constant,

v =  ? ? r

FIGURE 1.3.10 Rigid body rotating about a ?xed axis.

The acceleration a of point P is determined conveniently by using normal
and tangential components,

Note that the angular acceleration ? and angular velocity ? are valid for
any line perpendicular to the axis of rotation of the rigid body at
a given instant.

Kinematics Equations for Rigid Bodies Rotating in a Plane

For rotational motion with or without a ?xed axis, if displacement is
measured by an angle ?,

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General plane motion of a rigid body is de?ned by simultaneous
translation and rotation in a plane. Figure 1.3.11 illustrates how the velocity
of a point A can be determined using Equation 1.3.46, which is based on
relative motion of particles.

FIGURE 1.3.11 Analysis of velocities in general plane motion.

There are ?ve important points to remember when solving general plane
motion problems, including those of interconnected rigid bodies.
1. The angular velocity of a rigid body in plane motion is independent
of the reference point.
2. The common point of two or more pin-jointed members must have
the same absolute velocity
even though the individual members may have different angular velocities.
3. The points of contact in members that are in temporary contact may
or may not have the same absolute velocity. If there is sliding between the
members, the points in contact have different absolute velocities.
The absolute velocities of the contacting particles are always the same
if there is no sliding.
4. If the angular velocity of a member is not known, but some points
of the member move along de?ned paths (i.e., the end points of a piston rod),
these paths de?ne the directions of the velocity vectors and are useful
in the solution.
5. The geometric center of a wheel rolling on a ?at surface moves in
rectilinear motion. If there is no slipping at the point of contact, the linear
distance the center point travels is equal to that portion of the rim
circumference that has rolled along the ?at surface.

Instantaneous Center of Rotation

The method of instantaneous center of rotation is a geometric
method of determining the angular velocity when two velocity vectors
are known for a given rigid body. Figure 1.3.12 illustrates the method. This
procedure can also be used to determine velocities that are parallel to
one of the given velocities, by similar triangles.

FIGURE 1.3.12 Schematic for instantaneous center of rotation.

Velocities vA and vB are given; thus the body is rotating about point I
at that instant. Point I has zero velocity at that instant, but generally has an
acceleration. This method does not work for the determination
of angular accelerations.

Acceleration in General Plane Motion

Figure 1.3.13 illustrates a method of determining accelerations
of points of a rigid body. This is similar to (but more dif?cult than)
the procedure of determining velocities.

FIGURE 1.3.13 Accelerations in general plane motion.

There are six key points to consider when solving this kind of
a problem.
1. The angular  velocity and acceleration of a rigid body in plane
motion are independent of the reference point.
2. The common points of pin-jointed members must have the same
absolute acceleration even though the individual members may have
different angular velocities and angular accelerations.
3. The points of contact in members that are in temporary contact may
or may not have the same absolute acceleration. Even when there is no
sliding between the members, only the tangential accelerations of the
points in contact are the same, while the normal accelerations are
frequently different in magnitude and direction.
4. The instantaneous center of zero velocity in general has an
acceleration and should not be used as a reference point for
accelerations unless its acceleration is known and included
in the analysis.
5. If the angular acceleration of a member is not known, but some
points of the member move along de?ned paths, the geometric
constraints of motion de?ne the directions of normal and tangential
acceleration vectors and are useful in the solution.
6. The geometric center of a wheel rolling on a ?at surface moves
in rectilinear motion. If there is no slipping at the point of contact,
the linear acceleration of the center point is parallel to the ?at
surface and equal to r? for a wheel of radius r and
angular acceleration ?.

General Motion of a Rigid Body

Figure 1.3.14 illustrates the complex general motion (three-dimensional)
of a rigid body. It is important to note that here the angular velocity
and angular acceleration vectors are not necessarily in the same
direction as they are in general plane motion. Equations 1.3.48 give the
velocity and acceleration of a point on the rigid body.
These equations are the same as those presented for plane motion.

FIGURE 1.3.14 General motion of a rigid body.

The most dif?cult part of solving a general motion problem is
determining the angular acceleration vector. There are three cases for
the determination of the angular acceleration.
1. The direction of ? is constant. This is plane motion and
? = ? ? can be used in scalar solutions of problems.
2. The magnitude of ? is constant but its direction changes.
An example of this is a wheel which travels at a constant speed on
a curved path.
3. Both the magnitude and direction of ? change. This is space motion
since all or some points of the rigid body have three-dimensional paths.
An example of this is a wheel which accelerates on a curved path.

A useful expression can be obtained from item 2 and Figure 1.3.15.
The rigid body is ?xed at point O and ? has a constant magnitude.
Let ? rotate about the Y axis with angular velocity ?. The angular
acceleration is determined from Equation 1.3.49.

FIGURE 1.3.15 Rigid body ?xed at point O.

For space motion it is essential to combine the results of items 1
and 2, which provide components of ? for the change in magnitude
and the change in direction. The following  example illustrates the
procedure.

Example 9

The rotor shaft of an alternator in a car is in the horizontal plane.
It rotates at a constant angular speed of 1500 rpm while the car
travels at v = 60 ft/sec on a horizontal road of 400 ft radius
(Figure 1.3.16). Determine the angular acceleration of the rotor
shaft if v increases at the rate of 8 ft/sec2.

FIGURE 1.3.16 Schematic of shafts motion.

Solution. There are two components of ?. One is the change in the direction
of the rotor shafts ?x, and the other is the change in magnitude from the
acceleration of the car.
1. Component from the change in direction. Determine ?c of the car.

2. Component from the acceleration of the car. Use Equation 1.3.9:

The angular acceleration of the rotor shaft is

This problem could also be solved using the method in the next section.

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