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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