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Returning to the particle on a beam system of Example 9.21, replace the particle with a flat disk that rolls on the beam without slipping.
a. How many degrees of freedom are there in this modified system?
b. Find the equations of motion of the disk and beam.
Example 9.21
Particle on a Beam
A classic example of a combination rigid-body–particle system is a point mass sliding on a pinned beam (or seesaw). This system is a common example that you might explore in a course on automatic control. As shown in Figure 9.18a, the system consists of a thin rigid beam, pinned to a support at its center of mass, and a particle, which slides on the beam without friction. Note that this problem only has two degrees of freedom and thus can be completely described using only two coordinates: θ, the angle of the beam from horizontal, and x, the distance of the particle from the center of the beam. We define an inertial frame and body frame both attached to the center of the beam O, with the body frame fixed to the beam (Figure 9.18b). The two frames are related by the transformation array
Figure 9.18c shows the free-body diagrams of the two components of the system. Note that there are two normal forces involved here: N, the force of the beam support on the beam, and NP , the force of the particle on the beam. Due to Newton’s third law, NP must appear with opposite signs in both free-body diagrams. We start with the particle P, whose kinematics are

Note that we have carried out the inertial derivatives using the body-frame coordinates.
Applying Newton’s second law and separating components in the er and eθ directions produces the following two equations for P:

We now have two equations of motion for the two degrees of freedom, but the problem is not yet solved, since NP is unknown. To find it, we must consider the rotational equation of motion of the beam. The beam is rotating about its center of mass O and the only force producing a moment about O is NP. All other forces in the free-body diagram pass through O and contribute no moment. Thus the rotational equation of motion is

where IO is the moment of inertia of the beam about O. From Eq. (9.60), we find

which is valid as long as x ≠ 0.
Substituting the expression for NP into Eq. (9.59) produces the equations of motion for this system:

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