22-Mec-A2 Kinematics and Dynamics of Machines · December 2014

Question 5 of 6: Spring–dashpot rigid-bar pendulum — damped free vibration

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams, December 2014 — 07-Mec-A2, Kinematics and Dynamics of Machines. 3 hours, OPEN BOOK. Six questions of 20 marks each; candidates answer any FIVE. All SIX are solved in full below (Part A — mechanisms and machine dynamics, Q1–Q4; Part B — mechanical vibration, Q5–Q6).

Reference texts: R. L. Norton, Design of Machinery, 5th ed. (instant centres and velocity analysis, four-bar position/velocity/acceleration, planetary gear kinematics, engine balancing); S. S. Rao, Mechanical Vibrations, 6th ed. (single- and multi-degree-of-freedom free and forced response, modal analysis); J. E. Shigley & C. R. Mischke, Mechanical Engineering Design (epicyclic train-value cross-check).

Question 5: Spring–dashpot rigid-bar pendulum — damped free vibration (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Bar hinged at the top (O), hanging vertically, mass $m_c$ at the free (bottom) end. Both springs attach at $a_k=L/3$ from O; both dashpots at $a_c=2L/3$ from O. Values:

$k_1=k_2$	$c_1=c_2$	$m_b$	$m_c$	$L$	$g$
100 N/m	5 N·s/m	1 kg	0.5 kg	1.2 m	9.81 m/s²

Find. (a) the small-angle EOM; (b) $\omega_d$; (c) $\theta(t)$ for $\theta(0)=0,\ \dot\theta(0)=2\ \text{rad/s}$.

Figure 5.1 — hanging rigid bar hinged at O; springs $k_1,k_2$ at $L/3$ and dashpots $c_1,c_2$ at $2L/3$ act horizontally; concentrated mass $m_c$ at the bottom.

Approach. Take the angle $\theta$ from vertical as the single coordinate; sum moments about O (inertia, spring, dashpot and gravity), linearise for small $\theta$, then read off the natural frequency, damping ratio and the underdamped free response.

Inertia about O. $$I_O=\tfrac13 m_bL^2+m_cL^2=\tfrac13(1)(1.2)^2+0.5(1.2)^2=0.48+0.72=1.20\ \text{kg}\cdot\text{m}^2.$$
Generalised stiffness. A horizontal spring at height $a$ deflects $a\theta$ and returns a moment $k\,a^2\theta$; gravity on a hanging pendulum adds a restoring moment $(m_b g\tfrac L2+m_c g L)\theta$. Hence $$k_\theta=(k_1+k_2)a_k^2+\Bigl(m_bg\tfrac L2+m_cgL\Bigr)=200(0.4)^2+(5.886+5.886)=32+11.77=43.77\ \tfrac{\text{N}\cdot\text{m}}{\text{rad}}.$$
Generalised damping. $$c_\theta=(c_1+c_2)a_c^2=10(0.8)^2=6.40\ \text{N}\cdot\text{m}\cdot\text{s/rad}.$$
(a) Equation of motion. Summing moments about O, $$1.20\,\ddot\theta+6.40\,\dot\theta+43.77\,\theta=0.$$
(b) Frequencies and damping. $$\omega_n=\sqrt{\frac{k_\theta}{I_O}}=\sqrt{\frac{43.77}{1.20}}=6.04\ \text{rad/s},\qquad \zeta=\frac{c_\theta}{2\sqrt{k_\theta I_O}}=\frac{6.40}{2\sqrt{43.77\cdot1.20}}=0.442.$$ Underdamped, so $\omega_d=\omega_n\sqrt{1-\zeta^2}=5.42\ \text{rad/s}$ ($f_d=0.863\ \text{Hz}$).
(c) Free response. With $\theta_0=0$, $\dot\theta_0=2\ \text{rad/s}$ and $\zeta\omega_n=c_\theta/2I_O=2.667\ \text{s}^{-1}$, the underdamped solution $\theta(t)=e^{-\zeta\omega_nt}\!\left[\theta_0\cos\omega_dt+\dfrac{\dot\theta_0+\zeta\omega_n\theta_0}{\omega_d}\sin\omega_dt\right]$ reduces to $$\theta(t)=0.369\,e^{-2.667t}\sin(5.42\,t)\ \text{rad}.$$ The amplitude envelope $0.369\ \text{rad}$ ($\approx21^{\circ}$) decays as $e^{-2.667t}$.

Quantity	Value
$I_O$	1.20 kg·m²
$k_\theta$ (springs 32 + gravity 11.77)	43.77 N·m/rad
$c_\theta$	6.40 N·m·s/rad
$\omega_n$ / $\zeta$	6.04 rad/s / 0.442
Damped natural frequency $\omega_d$	5.42 rad/s (0.863 Hz)
Free vibration $\theta(t)$	$0.369\,e^{-2.667t}\sin(5.42t)$ rad

Check. The bar hangs (hinge at top), so gravity is a restoring moment and raises the stiffness; an inverted (bottom-hinged) bar would subtract the same 11.77 N·m/rad and lower $\omega_n$. Springs are at $L/3$ and dashpots at $2L/3$ per the figure — the lever arms enter squared, so the dashpot arm dominates the damping.

Topic. Single-DOF damped free vibration of a rigid-body (rotational) pendulum with gravity, springs and dashpots.

Key relations. $I_O\ddot\theta+c_\theta\dot\theta+k_\theta\theta=0$ with $k_\theta=\sum k_ia_i^2+k_{grav}$, $c_\theta=\sum c_ia_i^2$; $\omega_n=\sqrt{k_\theta/I_O}$, $\zeta=c_\theta/2\sqrt{k_\theta I_O}$, $\omega_d=\omega_n\sqrt{1-\zeta^2}$.

Why this works. For small angles every element’s displacement is proportional to its lever arm $a_i$, so its restoring/dissipative moment scales as $a_i^2$ — reducing the distributed system to one equivalent rotational spring–mass–dashpot. Gravity contributes a linear restoring moment because $\sin\theta\approx\theta$.

Common pitfall. Using $\tfrac1{12}m_bL^2$ (centroidal) instead of $\tfrac13m_bL^2$ (about the end O), and forgetting to square the lever arms or to include the gravity stiffness with the correct sign. (Rao, Mechanical Vibrations, ch. 2.)

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