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22-Mec-A2 Kinematics and Dynamics of Machines · May 2016

Question 6 of 6: Inverted pendulum with a torsional spring

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

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Paper format: EGBC National Exam, 22-Mec-A2 “Kinematics and Dynamics of Machines”, May 2016 (source code 07-Mec-A2). Open-book, 3 hours; six questions of equal value (20 marks each), candidates answer any five. Part A — mechanisms and machine dynamics (Q1–Q4); Part B — mechanical vibration (Q5–Q6). All six are solved in full below.

Reference texts: R. L. Norton, Design of Machinery, 5th ed. (mechanical advantage & virtual work, cam-motion synthesis, four-bar force balancing, epicyclic trains); S. S. Rao, Mechanical Vibrations, 6th ed. (base excitation / transmissibility, single-DOF free vibration); J. E. Shigley, Mechanical Engineering Design (planet spacing & gear geometry cross-check).

Question 6: Inverted pendulum with a torsional spring (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. $m=4$ kg, $I_O=0.02$ kg·m2 (about the pivot $O$), mass-centre offset $L=0.12$ m, torsional-spring stiffness $k_t=6.4$ N·m/rad, $g=9.81$ m/s2.

Find. (i) natural period about a stable equilibrium; (ii) free response given $\dot\theta(0)=1.2$ rad/s at that equilibrium.

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Figure 6.1 — Inverted pendulum: gravity acts to topple the mass (destabilizing), while the torsional spring at the pivot restores it. Upright is a stable equilibrium provided $k_t>mgL$.

Approach. Take $\theta$ from the upright. Moment balance about $O$ gives the small-angle equation; the effective stiffness is the spring stiffness minus the gravitational “negative stiffness.” The period follows from $\omega_n$, and the given initial angular velocity produces a pure-sine free response.

  1. Equation of motion (linearized about upright). Gravity gives a destabilizing moment $mgL\sin\theta$, the spring a restoring moment $-k_t\theta$: $$I_O\ddot\theta+(k_t-mgL)\theta=0,\qquad mgL=4(9.81)(0.12)=4.709\ \text{N}\!\cdot\!\text{m}.$$ Since $k_t=6.4>mgL$, the effective stiffness $k_{\mathrm{eff}}=6.4-4.709=1.691\ \text{N}\!\cdot\!\text{m/rad}>0$ — the upright position is stable.
  2. Natural frequency and period. $$\omega_n=\sqrt{\frac{k_{\mathrm{eff}}}{I_O}}=\sqrt{\frac{1.691}{0.02}}=9.196\ \text{rad/s},\quad \boxed{T=\frac{2\pi}{\omega_n}=0.683\ \text{s}}.$$
  3. Free response from the equilibrium. With $\theta(0)=0$ and $\dot\theta(0)=1.2$ rad/s, the solution is $\theta(t)=(\dot\theta_0/\omega_n)\sin\omega_n t$: $$\boxed{\theta(t)=0.1305\,\sin(9.196\,t)\ \text{rad}}\quad(\text{amplitude }7.48^\circ,\ \text{small-angle valid}).$$
QuantityValue
Gravitational moment coefficient $mgL$4.709 N·m/rad
Effective stiffness $k_{\mathrm{eff}}$1.691 N·m/rad
$\omega_n$9.196 rad/s
(i) Period $T$ (about upright)0.683 s
(ii) Response$\theta=0.1305\sin(9.196t)$ rad (7.48°)

Check: This system has two stable equilibria — the upright ($\theta=0$, stable because $k_t>mgL$) and the hanging position ($\theta=180^\circ$, always stable, $\omega_n=\sqrt{(k_t+mgL)/I_O}=23.6$ rad/s, $T=0.267$ s). The “inverse pendulum” wording points to the upright branch, which is solved above; the hanging-branch period is noted for completeness.

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