Consider the planar pendulum shown in Figure 9.3, which consists of three slender rods. Find the linearized equations of motion of the system.

Question

Accepted Answer

Let us formulate the kinematics of the system. Note: Examples 9.2.1 and 9.2.2, however simple, were linearized after the equations of motion had been formulated. This can be a difficult task in multibody systems with more than three or four degrees of freedom. An alternative to this method is the use of Kane’s equation.

Let the joint angles in this example be given by the absolute angles measured from the vertical line.

The center mass velocities of [latex]G_1, G_2[/latex] and [latex]G_3[/latex] in R are:

[latex]\begin{aligned}&{ }^R \bar{v}^{G_1}=\frac{L}{2} \dot{ heta}_1 \bar{n}_1^1 (9.31) \&{ }^R \bar{v}^{G_2}=L \dot{ heta}_1 \bar{n}_2^1+\frac{L}{2} \dot{ heta}_2 n_2^2 (9.32) \&{ }^R \bar{v}^{G_3}=L \dot{ heta}_1 \bar{n}_2^1+L \dot{ heta}_2 \bar{n}_2^2+\frac{L}{2} \dot{ heta}_3 n_2^3 (9.33)\end{aligned}[/latex]

The angular velocities of bodies A, B and C in R are obtained as follows:

[latex]\begin{aligned}&{ }^R \bar{\omega}^A=\dot{ heta}_1 \bar{n}_3 (9.34) \&{ }^R \bar{\omega}^B=\left(\dot{ heta}_2 ight) \bar{n}_3 (9.35) \&{ }^R \bar{\omega}^C=\left(\dot{ heta}_3 ight) \bar{n}_3 (9.36)\end{aligned}[/latex]

Figure 9.3 shows the relationship between the Euler angles and the unit vectors of R.

The mass center acceleration of [latex]G_1, G_2[/latex] and [latex]G_3[/latex] corresponding to body A, B and C in R are:

[latex]\begin{aligned}{ }^R \bar{a}^{G_1} &=\frac{L}{2} \ddot{ heta}_1 \bar{n}_2^1-\frac{L}{2} \dot{ heta}_1^2 n_1^1 (9.37) \{ }^R \bar{a}^{G_2} &=L \ddot{ heta}_1 \bar{n}_2^1-L \dot{ heta}_1^2 \bar{n}_1^1+\frac{L}{2} \ddot{ heta}_2 \bar{n}_2^2-\frac{L}{2} \dot{ heta}_1^2 n_1^2 (9.38) \{ }^R \bar{a}^{G_3} &=L \ddot{ heta}_1 \bar{n}_2^1-L \dot{ heta}_1^2 \bar{n}_1^1+L \ddot{ heta}_2 \bar{n}_2^2-L \dot{ heta}_1^2 n_1^2+\frac{L}{2} \ddot{ heta}_3 \bar{n}_2^3-\frac{L}{2} \dot{ heta}_3^2 n_1^3 (9.39)\end{aligned}[/latex]

The way to proceed at this point is to express both the mass center velocities and accelerations with respect to the orthogonal unit vectors attached to R : namely, [latex]\bar{n}_1[/latex], [latex]\bar{n}_2[/latex] and [latex]\bar{n}_3[/latex]. In doing so, let us linearize those expressions as well. Let

[latex]\begin{aligned}&s heta \simeq heta \&c heta \simeq 1\end{aligned}[/latex]

Then

[latex]\begin{aligned}&s 1 \simeq heta_1 c 1 \simeq 1 \&s 2 \simeq heta_2 c 2 \simeq 1 \&s 3 \simeq heta_3 c 3 \simeq 1\end{aligned}[/latex]

The mass center velocities become

[latex]\begin{aligned}&{ }^R \bar{v}^{G_1}=\frac{L}{2} \dot{ heta}_1\left(- heta_1 \bar{n}_1+\bar{n}_2 ight) (9.40) \&{ }^R \bar{v}^{G_2}=-L\left( heta_1 \dot{ heta}_1+\frac{1}{2} heta_2 \dot{ heta}_2 ight) \bar{n}_1+L\left(\dot{ heta}_1+\frac{\dot{ heta}_2}{2} ight) \bar{n}_2 (9.41) \&{ }^R \bar{v}^{G_3}=-L\left( heta_1 \dot{ heta}_1+ heta_2 \dot{ heta}_2+\frac{1}{2} heta_3 \dot{ heta}_3 ight) \bar{n}_1+L\left(\dot{ heta}_1+\dot{ heta}_2+\frac{1}{2} \dot{ heta}_3 ight) \bar{n}_2 (9.42)\end{aligned}[/latex]

We can drop the terms that are doomed to be small, and the equations above reduce further to

[latex]\begin{aligned}{ }^R \bar{v}^{G_1} &=\frac{L}{2} \dot{ heta}_1 \bar{n}_2 (9.43) \{ }^R \bar{v}^{G_2} &=L\left(\dot{ heta}_1+\frac{\dot{ heta}_2}{2} ight) \bar{n}_2 (9.44)\{ }^R \bar{v}^{G_3}&=L\left(\dot{ heta}_1+\dot{ heta}_2+\frac{1}{2} \dot{ heta}_3 ight) \bar{n}_2 (9.45)\end{aligned}[/latex]

The linearized mass center accelerations are then found to be

[latex]\begin{aligned}{}^R \bar{a}^{G_1} &=\frac{L}{2} \ddot{ heta}_1 \bar{n}_2 (9.46) \{}^R \bar{a}^{G_2} &=L\left(\ddot{ heta}_1+\frac{\ddot{ heta}_2}{2} ight) \bar{n}_2 (9.47) \{ }^R \bar{a}^{G_3} &=L\left(\ddot{ heta}_1+\ddot{ heta}_2+\frac{1}{2} \dot{ heta}_3 ight) \bar{n}_2 (9.48)\end{aligned}[/latex]

These can, of course, be obtained by linearizing equations (9.37) to (9.39) directly. Since we are going to need the partial velocities for the formulation of the equations of motion, then we obtain the following terms

[latex]\frac{\partial \bar{v}_1^G}{\partial \dot{ heta}_1}=\frac{L}{2} \bar{n}_2, \quad \frac{\partial \bar{v}_2^G}{\partial \dot{ heta}_2}=0, \quad \frac{\partial \bar{v}_1^G}{\partial \dot{ heta}_3}=0[/latex] (9.49)

[latex]\frac{\partial \bar{v}_2^G}{\partial \dot{ heta}_1}=L \bar{n}_2, \quad \frac{\partial \bar{v}_2^G}{\partial \dot{ heta}_2}=\frac{L}{2} \bar{n}_2, \quad \frac{\partial \bar{v}_2^G}{\partial \dot{ heta}_3}=0[/latex] (9.50)

[latex]\frac{\partial \bar{v}^{G 3}}{\partial \dot{ heta}_1}=L \bar{n}_2, \quad \frac{\partial \bar{v}_3^G}{\partial \dot{ heta}_2}=L \bar{n}_2, \quad \frac{\partial \bar{v}_3^G}{\partial \dot{ heta}_3}=\frac{L}{2} \bar{n}_2[/latex] (9.51)

In a similar fashion we obtain the partial angular velocities.

Formulation of the Linearized Equations of Motion

Using Kane's equation, the generalized inertia and active forces need to be formulated. That is,

[latex]f_{\ell}^*+f_{\ell}=0(\ell=1,3)[/latex] (9.52)

where

[latex]f_{\ell}^*=\left(m_k \bar{a}^{G_k} ight) \cdot \frac{\partial \bar{v}^{G_k}}{\partial \dot{ heta}_{\ell}}+\left(\bar{T}_k^* ight) \cdot \frac{\partial \bar{\omega}^k}{\partial \dot{ heta}_{\ell}}[/latex] (9.53)

The value [latex]\bar{T}_k^*[/latex] represents the inertia torque of body k in R, and for each [latex]\ell[/latex] we have k that goes from 1 to 3 . Similarly, the generalized active forces are obtained making use of equation (9.53) by replacing the inertia and torque forces with the resulting external force and torque acting on each body. Since we only have the gravity force due to the weight, the generalized active forces contribution is simply obtained from

[latex]f \ell=\left(m_k g \bar{n}_1 ight) \cdot \frac{\partial \bar{v}^{G_1}}{\partial \dot{ heta}_{\ell}}[/latex] (9.54)

Substituting the linearized partial velocities and accelerations into the equations of motion, we obtain automatically the equations desired:

[latex]\begin{array}{r}\frac{7}{3} \ddot{ heta}_1+\frac{3}{2} \ddot{ heta}_2+\frac{1}{2} \ddot{ heta}_3+\frac{5}{2} \frac{g}{L} heta_1=0 (9.55) \\frac{3}{2} \ddot{ heta}_1+\frac{4}{3} \ddot{ heta}_2+\frac{1}{2} \ddot{ heta}_3+\frac{3}{2} \frac{g}{L} heta_2=0 (9.56) \\frac{1}{2} \ddot{ heta}_1+\frac{1}{2} \ddot{ heta}_2+\frac{1}{3} \ddot{ heta}_3+\frac{g}{L} heta_3=0 (9.57)\end{array}[/latex]

Question 9.2.3: Consider the planar pendulum shown in Figure 9.3, which cons...

Related Answered Questions

Use the subspace iteration method to solve for the eigenvalue problem given by Example 9.8.1. ...

Verified Answer:

Supposing that the mass matrix M = I , and the stiffness matrix is given by K = [3 -1 -1 -1 3 -1 -1 -1 3] Use the Jacobi method to find the eigenvalues and eigenvectors for such a system. ...

Verified Answer:

Suppose that K and M are given by K = [3 -1 -1 -1 3 -1 -1 -1 3], M = [ 1 0 0 0 1 0 0 0 1] Use the Rayleigh–Ritz method to solve for the eigenvalues and eigenvectors of the system. ...

Verified Answer:

Consider Example 6.12.3, dealing with the pantograph. Obtain the linearized equations of motion of the mechanical system (follow procedures outlined in the above example). ...

Verified Answer:

Consider the system shown in Figure 9.8, where the clamped–free slender bar of a parabolic side is assumed to oscillate longitudinally. Our task is to extract the eigenvalues and the corresponding mode shapes using the Rayleigh–Ritz method discussed above. ...

Verified Answer:

Find the eigenvalues and eigenvectors for the lumped mass system shown in Figure 9.7. ...

Verified Answer:

Referring to Example 3.9.3 obtain the linearized kinematical expressions needed for the equations of motion (see Figure 9.2). ...

Verified Answer: