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.

Question

Figure 1.14 depicts a rigid rod of length l, mass m, and mass moment of inertia about its mass center, [latex]I_c[/latex]. 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.

Accepted Answer

For a given angular orientation θ, the free body diagram of the rod
is shown in Fig. 1.15. In this figure, [latex]R_x[/latex] and [latex]R_y[/latex] 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 [latex]R_x[/latex] and [latex]R_y[/latex] 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

[latex]M_a = M_{eff}[/latex] (1.53)

[latex]M_a = T − mg \frac{1}{2} \sin θ − (c\dot{x}_a + kx_a)l \cos θ [/latex]

where [latex]x_a[/latex] and [latex]\dot{x}_a [/latex] are the displacement and velocity of point a expressed in terms of the angle θ as

[latex]x_a = l \sin θ, \dot{x}_a = l \dot{θ} \cos θ[/latex]

Substituting these two equations into the expression for [latex]M_a[/latex], one obtains

[latex]M_{a} = T − mg\frac{l}{2} \sin θ − (c \dot{θ} \cos θ + k sin θ)l^{2} \cos θ[/latex]

The moments of the effective forces and moments about O are

[latex]M_{eff} = I_{c} \ddot{θ} + m\ddot{x}c\frac{l}{2} \cos θ + m \ddot{y_{c}}\frac{l}{2} \sin θ[/latex]

where the coordinates of the center of mass [latex]x_{c}[/latex] and [latex]y_{c}[/latex] are given by

[latex] x_{c} =\frac{l}{2} sin θ, y_{c} = −\frac{l}{2}cos θ [/latex]

The velocity and acceleration of the center of mass can be obtained by direct differentation of the above equations, that is,

[latex]\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[/latex]

[latex]\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[/latex]

Substituting the expressions of the accelerations into the [latex]M_{eff}[/latex] moment equation, yields

[latex]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[/latex]

[latex]+ m [\frac{1}{2} \ddot{θ} \sin θ + \frac{1}{2} \dot{θ}^2 \cos θ] \frac{1}{2} \sin θ[/latex]

[latex]= I_c \ddot{θ} + m (\frac{1}{2})^2 \ddot{θ} [\cos^2 θ + \sin^2 θ][/latex]

Using the trigonometric identity, cos²θ + sin² θ = 1, the effective moment [latex]M_{eff}[/latex] reduces to

[latex]M_{eff} = I_{c} \ddot{θ} + \frac{ml^{2}}{4} \ddot{θ} = ( I_{c} +\frac{ml^{2}}{4} ) \ddot{θ} = I_O \ddot{θ} [/latex]

where [latex]I_O[/latex] is the mass moment of inertia of the rod about point O and is defined as

[latex] I_O = I_c + \frac{ml^{2}}{4}[/latex]

Using Eq. 1.53, one has [latex] M_{a} = M_{eff}[/latex], that is,

[latex] T − mg \frac{l}{2} \sin θ − (c \dot{θ} \cos θ + k \sin θ)l^{2} cos θ = I_{O}\ddot{ θ}[/latex]

which can be written as

[latex] I_O \ddot{θ} + cl^2 \dot{θ} \cos^2 θ + kl^2 \sin θ \cos θ + mg \frac{l}{2} \sin θ = T [/latex]

Question 1.3: Figure 1.14 depicts a rigid rod of length l, mass m, and mas...

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