Energy and Momentum Methods for Rigid Bodies in Plane Motion

Newtons second law in determining kinetics relationships is not always
the most ef?cient, although it always works. As for particles, energy
and momentum methods are often useful to analyze
rigid bodies in plane motion.

Work of a Force on a Rigid Body

The work of a force acting on a rigid body moving from position 1 to 2 is

Work of a Moment

The work of a moment has a similar form, for angular positions ?,

In the common case where the moment vector M is perpendicular
to the plane of motion,
M ? d? = M d?.
It is important to note those forces that do no work:
1. Forces that act at ?xed points on the body do not do work.
For example, the reaction at a ?xed, frictionless pin does no work
on the body that rotates about that pin.
2. A force which is always perpendicular to the direction of the
motion does no work.
3. The weight of a body does no work when the bodys center
of gravity moves in a horizontal plane.
4. The friction force ^ at a point of contact on a body that rolls
without slipping does no work. This is because the point of contact
is the instantaneous center of zero velocity.

Kinetic Energy of a Rigid Body

The kinetic energy of a particle only consists of the energy associated
with its translational motion. The kinetic energy of a rigid body also
includes a term for the rotational energy of the body,

where C is the center of mass of the rigid body.
The kinetic energy of a rigid body rotating about an arbitrary
axis at point O is

Principle of Work and Energy

The principle of work and energy for a rigid body is the same
as used for particles with the addition of the rotational energy terms.

where  T₁  = initial kinetic energy of the body
T₂  = ?nal kinetic energy of the body
U₁₂  = work of all external forces and moments acting on the body
moving from position 1 to 2 This method is advantageous when
displacements and velocities are the desired quantities.
Conservation of Energy The conservation of energy in a conservative
rigid body system is

where  T  = kinetic energy
V  = total potential energy (gravitational and elastic)

Power

The net power supplied to or required of the system is

This can be calculated by taking time derivatives of the kinetic and
potential energy terms. Each term is considered positive when it
represents the power supplied to the system and negative when
power is taken from the system.

Impulse and Momentum of a Rigid Body

Impulse and momentum methods are particularly useful when time
and velocities are of interest. Figure 1.3.20 shows how rigid bodies
are to be considered for this kind of analysis. Notice that rotational
motion of the rigid body must be included in the modeling.

FIGURE 1.3.20 Impulse and momentum for rigid bodies.

The impulse of the external forces in the given interval is

where t is time, C is the center of mass, and ?F includes all
external forces.
The impulse of the external moments in the given interval is

For plane motion, if ?M is parallel to ?, the scalar expressions are

Impulse and Momentum of a System of Rigid Bodies

A system of rigid bodies can be analyzed using one of the two following
procedures, illustrated in Figure 1.3.21.
1. Apply the principle of impulse and momentum to each rigid
member separately. The mutual forces acting between members must be
included in the formulation of the solution.
2. Apply the principle of impulse and momentum to the entire system
of bodies, ignoring the mutual forces between members.

FIGURE 1.3.21 System of rigid bodies.

Conservation of Momentum

The principle of conservation of linear and angular momentum of
particles can be  extended to rigid bodies that have no external forces
or moments acting on them. The conservation of linear momentum
means that the center of mass C moves at a constant speed in
a constant direction,

Likewise, for conservation of angular momentum of rigid bodies,

For a system of rigid bodies, use the same ?xed reference point O for
all parts of the system. Thus, for plane motion,

There are two important points to remember when using these equations.
First, ?HC = 0 does not imply that ?HO = 0, or vice versa. Second,
conservation of momentum does not require the simultaneous
conservation of both angular and linear momenta (for example, there
may be an angular impulse while linear momentum is conserved).

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