Question 1.3: Figure 1.14 depicts a rigid rod of length l, mass m, and mas...
Figure 1.14 depicts a rigid rod of length l, mass m, and mass moment of inertia about its mass center, I_c. One end of the rod is connected to the ground by a pin joint, and the other end is connected to the ground by spring and damper.
The spring stiffness is k and the viscous damping coefficient is c. If T is an externally applied moment that acts on the rod, derive the system differential equation of motion.
For a given angular orientation θ, the free body diagram of the rod
is shown in Fig. 1.15. In this figure, R_x and R_y denote the reaction forces of the pin joint. The pendulum system shown in Fig. 1.14 has only one degree of freedom which is described by the angular rotation θ, and therefore, the dynamics of this system can be described by only one differential equation.
In order to avoid the unknown reaction forces R_x and R_y we use Eq. 1.53 and take the moment about point O. From the free body diagram shown in Fig. 1.15, the moment of the applied forces about O is given by
M_a = M_{eff} (1.53)
M_a = T − mg \frac{1}{2} \sin θ − (c\dot{x}_a + kx_a)l \cos θwhere x_a and \dot{x}_a are the displacement and velocity of point a expressed in terms of the angle θ as
x_a = l \sin θ, \dot{x}_a = l \dot{θ} \cos θSubstituting these two equations into the expression for M_a, one obtains
M_{a} = T − mg\frac{l}{2} \sin θ − (c \dot{θ} \cos θ + k sin θ)l^{2} \cos θThe moments of the effective forces and moments about O are
M_{eff} = I_{c} \ddot{θ} + m\ddot{x}c\frac{l}{2} \cos θ + m \ddot{y_{c}}\frac{l}{2} \sin θwhere the coordinates of the center of mass x_{c} and y_{c} are given by
x_{c} =\frac{l}{2} sin θ, y_{c} = −\frac{l}{2}cos θThe velocity and acceleration of the center of mass can be obtained by direct differentation of the above equations, that is,
\dot{x}_{c} = \frac{l}{2} \dot{θ} \cos θ, \ddot{x_{c}} = \frac{l}{2}\ddot{θ} \cos θ − \frac{l}{2} \dot{θ^{2}} \sin θ\dot{y}_{c} = \frac{l}{2} \dot{θ} \sin θ, \ddot{y_{c}} = \frac{l}{2}\ddot{θ} \sin θ + \frac{l}{2} \dot{θ^{2}} cos θ
Substituting the expressions of the accelerations into the M_{eff} moment equation, yields
M_{eff} = I_c \ddot{θ} + m [\frac{1}{2} \ddot{θ} \cos θ – \frac{1}{2} \dot{θ}^2 \sin θ] \frac{1}{2} \cos θ
+ m [\frac{1}{2} \ddot{θ} \sin θ + \frac{1}{2} \dot{θ}^2 \cos θ] \frac{1}{2} \sin θ
= I_c \ddot{θ} + m (\frac{1}{2})^2 \ddot{θ} [\cos^2 θ + \sin^2 θ]
Using the trigonometric identity, cos²θ + sin² θ = 1, the effective moment M_{eff} reduces to
M_{eff} = I_{c} \ddot{θ} + \frac{ml^{2}}{4} \ddot{θ} = ( I_{c} +\frac{ml^{2}}{4} ) \ddot{θ} = I_O \ddot{θ}where I_O is the mass moment of inertia of the rod about point O and is defined as
I_O = I_c + \frac{ml^{2}}{4}Using Eq. 1.53, one has M_{a} = M_{eff}, that is,
T − mg \frac{l}{2} \sin θ − (c \dot{θ} \cos θ + k \sin θ)l^{2} cos θ = I_{O}\ddot{ θ}which can be written as
I_O \ddot{θ} + cl^2 \dot{θ} \cos^2 θ + kl^2 \sin θ \cos θ + mg \frac{l}{2} \sin θ = T