Question 11.12: A homogeneous thin body of mass M, shown in Fig. 11.8, is su...
A homogeneous thin body of mass M, shown in Fig. 11.8, is suspended in the vertical plane by a thin wire of mass m \ll M, length a, and with its ends hinged in smooth bearings. (i) Derive the equations for small vibrations of the body about its vertical equilibrium position \phi=\theta=0. Identify the inertia and stiffness matrices. (ii) Sketch the general formulation of the solution. Then let a=l,, and suppose that the body is a thin circular disk of radius R=\sqrt{2} l. Determine the normal mode frequencies P_m, the ratios of the amplitudese C_{l m}, and thus derive the solution of the coupled equations of motion . Identify the normal equations of motion and their normal mode solutions as functions of the generalized variables.
Solution of (i). The system is scleronomic with two degrees of freedom \theta and \phi defined in Fig. 11.8. Since the wire has negligible mass compared with M, its contribution to the total kinetic and potential energies is negligible. The total velocity of the center of mass G referred to the body frame \psi=\left\{G ; \mathbf{e}_r, \mathbf{e}_\phi\right\} is given by \mathbf{v}^*=\mathbf{v}_H + \omega \times \mathbf{l}=a \dot{\theta} \mathbf{t} + l \dot{\phi} \mathbf{e}_\phi, where t = \sin(\phi – \theta)e_r + \cos(\phi – \theta)e_q, The total kinetic energy of the system, by (10.101), is
K(\mathscr{B}, t)=\frac{1}{2} m \mathbf{v}^* \cdot \mathbf{v}^* + \frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{I}_C \boldsymbol{\omega} (10.101)
T=\frac{1}{2} M\left[a^2 \dot{\theta}^2 \sin ^2(\phi – \theta) + (a \dot{\theta} \cos (\phi – \theta) + l \dot{\phi})^2\right] + \frac{1}{2} I_G \dot{\phi}^2, (11.92a)
and the total gravitational potential energy is given by
V=M g[a(1 – \cos \theta) + l(1 – \cos \phi)]. (11.92b)
We note that no coordinates are absent. For small vibrations, only terms of second order in all of the small quantities \phi, \dot{\phi}, \theta, \dot{\theta} are retained in (11.92a) and (11.92b). Therefore , the first term in (11.92a) is of higher order and may be neglected ; and with \cos (\phi – \theta) \cong 1 – (\phi – \theta)^2 / 2, to terms of second order the total kinetic and potential energies of the system for small vibrations about the vertical equilibrium configuration are thus given by
\begin{aligned}T &=\frac{1}{2}\left(I_G + M l^2\right) \dot{\phi}^2 + \frac{1}{2} M\left(2 l a \dot{\phi} \dot{\theta} + a^2 \dot{\theta}^2\right), \\V &=\frac{1}{2} M g\left(l \phi^2 + a \theta^2\right)\end{aligned}. (11.92c)
The potential energy is a positive definite, homogeneous quadratic function of the generalized coordinates \theta and \phi. Hence, clearly, the equilibrium configuration \phi=\theta=0 is infinitesimally stable. Similarly, the total kinetic energy is a homogeneous quadratic function of the generalized velocities \dot{\theta} and \dot{\phi} With I_H=I_G + M l^2 in accordance with the parallel axis theorem in (11.92c), the symmetric inertia and stiffness matrices in (11.84) and (11.83) are thus identified as
T\left(\dot{q}_r\right)=\frac{1}{2} M_{k l} \dot{q}_k \dot{q}_l (11.84)
V\left(q_r\right)=\frac{1}{2} K_{k l} q_k q_l (11.83)
\left[M_{k l}\right]=\left[\begin{array}{ll}I_H & M l a \\M l a & M a^2\end{array}\right], \quad\left[K_{k l}\right]=\left[\begin{array}{ll}M g l & 0 \\0 & M g a\end{array}\right] . (11.92d)
We note in passing that the stiffness matrix is diagonal and all of its nontrivial components are positive; so, it is quite evident that K u \cdot u>0 holds for all u \neq 0.
The hinge constraints are workless, so the system is conservative with a Lagrangian function
L=\frac{1}{2} I_H \dot{\phi}^2 + \frac{1}{2} M\left(2 l a \dot{\phi} \dot{\theta} + a^2 \dot{\theta}^2\right) – \frac{1}{2} M g\left(l \phi^2 + a \theta^2\right), (11.92e)
and Lagrange’s equations (11.66) now yield the equations of small vibration,
\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_r}\right) – \frac{\partial L}{\partial q_r}=0, \quad r=1,2, \ldots, n (11.66)
\begin{gathered}I_H \ddot{\phi} + M a l \ddot{\theta} + M g l \phi=0, \\M l a \ddot{\phi} + M a^2 \ddot{\theta} + M g a \theta=0\end{gathered}. (11.92f)
Solution of (ii). The natural frequencies for the small vibrations are determined by (11.89). With (11.92d), we have
\operatorname{det}\left|K_{k l} – p^2 M_{k l}\right|=0 (11.89)
\operatorname{det}\left[\begin{array}{ll}M g l – p^2 I_H & -p^2 M l a \\-p^2 M l a & M g a – p^2 M a^2\end{array}\right]=0.
This reduces to a quadratic equation in p²:
\left(M g l – p^2 I_H\right)\left(M g a – p^2 M a^2\right) – p^4 M^2 l^2 a^2=0, (11.92g)
which yield s two eigenfrequencies p_m . Use of these in (11.90) provides two sets of equations for the ratios of the coefficients C_{l m}, m=1,2; namely,
\left(K_{k l}-p_m^2 M_{k l}\right) C_{l m}=0 (11.90)
\left[\begin{array}{ll}M g l – p_m^2 I_H & -p_m^2 M l a \\-p_m^2 M l a & M g a – p_m^2 M a^2\end{array}\right]\left[\begin{array}{l}C_{1 m} \\C_{2 m}\end{array}\right]=0. (11.92h)
Finally, use of these results in (11.91) yield s the solutions for q_1=\phi \text { and } q_2=\theta. We omit these general detail s and tum to a special case for illustration.
q_k=\sum_{m=1}^n C_{k m} \sin \left(p_m t + \phi_m\right), \quad k=1,2, \ldots, n (11.91)
Consider a circular disk of radius R=\sqrt{2} l and let a=l \text {. } Then I_G=M R^2 / 2=M l^2, and I_H=2 M l^2. The characteristic equation (11.92g) thus simplifies to p^4 – 3 p_0^2 p^2 + p_0^4=0 with positive roots
p_1=p_0 \sqrt{\frac{1}{2}(3 + \sqrt{5})}, \quad p_2=p_0 \sqrt{\frac{1}{2}(3 – \sqrt{5})} \text {, } (11.92i)
where p_0 \equiv \sqrt{g / l}, and ( 11.92h) becomes
\left[\begin{array}{ll}p_0^2 – 2 p_m^2 & -p_m^2 \\-p_m^2 & p_0^2 – p_m^2\end{array}\right]\left[\begin{array}{l}C_{1 m} \\C_{2 m}\end{array}\right]=0 . (11.92j)
For m = 1, 2 (no sum), we thus obtain the amplitude ratios
\frac{C_{21}}{C_{11}}=\frac{\left(p_0^2 – 2 p_1^2\right)}{p_1^2}, \quad \frac{C_{12}}{C_{22}}=\frac{p_2^2}{\left(p_0^2 – 2 p_2^2\right)}. (11.92k)
Substitution here of the characteristic roots (11.92i) yields
C_{21}=-\frac{1}{2}(1 + \sqrt{5}) C_{11}, \quad C_{12}=\frac{1}{2}(1 + \sqrt{5}) C_{22} . (11.92l)
Introducing these in (11.91) in which q_1=\phi and q_2=\theta, we obtain the general solution
\begin{aligned}\phi &=C_{11} \sin \left(p_1 t + \phi_1\right) + \frac{1}{2}(1 + \sqrt{5}) C_{22} \sin \left(p_2 t + \phi_2\right), \\\theta &=-\frac{1}{2}(1 + \sqrt{5}) C_{11} \sin \left(p_1 t + \phi_1\right) + C_{22} \sin \left(p_2 t + \phi_2\right)\end{aligned}. (11.92m)
The remaining constants C_{11}, C_{22}, {\phi}_1, and \phi_2 are determined upon specification of the initial data.
The normal mode motions, readily identified from (11.92m), are defined by
\xi_1=C_{11} \sin \left(p_1 t + \phi_1\right), \quad \xi_2=C_{22} \sin \left(p_2 t + \phi_2\right), (11.92n)
and upon solving (11.92m) for the \xi_k \mathrm{~s}, we obtain the normal coordinates in terms of the original generalized coordinates:
\begin{aligned}&\xi_1=-\frac{1}{10} \sqrt{5}(2 \theta + \phi(1 – \sqrt{5})), \\&\xi_2=-\frac{1}{20}(\sqrt{5} – 5)(2 \theta + \phi(1 + \sqrt{5}))\end{aligned}. (11.92o)
These normal coordinates uncouple the original equations of motion (11.92f) applied to the circular disk so that the normal mode equations of motion are
\ddot{\xi}_k + p_k^2 \xi_k=0, \quad k=1,2(\text { no sum }), (11.92p)
in which the normal mode frequencies p_k are given in (11.92i).