23-Mechatronics-A5 Mechanical Design, December 2019, Q9

Question 7 (Part B, 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. Simple pendulum, $L=1\,\text{m}$, $m=4\,\text{kg}$, released from rest at $\theta_0=8^\circ$; a rigid vertical wall sits at the pendulum's equilibrium ($\theta=0$) position, so every downswing ends in an impact there; coefficient of restitution $e=0.9$; each impact lasts $t_i=0.010\,\text{s}$.

Find. The pendulum's motion (peak angle and timing) through the 1st and 2nd full down-swing/impact/up-swing cycle.

Fig. 7 — the pendulum's amplitude decays geometrically each time it strikes the wall at $\theta=0$: $\theta_0=8^\circ\to\theta_1=7.2^\circ\to\theta_2=6.48^\circ$.

Approach. $\theta_0=8^\circ$ is small enough ($\sin\theta_0\approx\theta_0$ to 0.1%) that simple-harmonic pendulum theory applies between impacts; the wall removes energy only in the instant of contact, governed by $e=v_{\text{rebound}}/v_{\text{impact}}$. Because the wall sits exactly at the bottom of the swing, the pendulum can only ever swing on the release side, executing a repeating down–impact–up sequence of shrinking amplitude.

Natural frequency and quarter period. $$\omega_n=\sqrt{g/L}=\sqrt{9.81/1}=3.132\ \text{rad/s},\qquad T=2\pi/\omega_n=2.006\ \text{s},\qquad T/4=0.5015\ \text{s}$$ (a quarter-period is the SHM time from any peak to the bottom, or the bottom to the next peak, independent of amplitude in the linear approximation.)
Angular speed arriving at the wall (energy method, small-angle): $\omega_{\text{impact}}=\theta_0\,\omega_n = 0.1396\times3.132=0.4373\,\text{rad/s}$ (exact energy formula $\sqrt{2g(1-\cos\theta_0)/L}=0.4370\,\text{rad/s}$ agrees to 0.08%, confirming the small-angle approximation is excellent here).
Apply the coefficient of restitution at the wall. $\omega_{\text{rebound}}=e\,\omega_{\text{impact}}$, so the next peak amplitude scales the same way as the SHM amplitude–velocity relation ($\theta_{\text{peak}}=\omega_{\text{peak-approach}}/\omega_n$): $$\theta_{n+1} = e\,\theta_n$$
Cycle 1 (start $t=0$, $\theta=\theta_0=8^\circ$): swings down, reaches the wall ($\theta=0$) at $t=T/4=0.5015\,\text{s}$; impact lasts $0.010\,\text{s}$ (to $t=0.5115\,\text{s}$); rebounds and swings back up, reaching its new peak $$\boxed{\theta_1 = e\,\theta_0 = 0.9\times8^\circ = 7.200^\circ}\quad\text{at } t=0.5115+0.5015=\boxed{1.0130\ \text{s}}$$
Cycle 2: swings back down from $\theta_1$, reaches the wall again at $t=1.0130+0.5015=1.5145\,\text{s}$; impact to $t=1.5245\,\text{s}$; rebounds to $$\boxed{\theta_2 = e\,\theta_1 = e^2\theta_0 = 0.81\times8^\circ = 6.480^\circ}\quad\text{at } t=1.5245+0.5015=\boxed{2.0260\ \text{s}}$$

Event	Time	Angle
Release	$t=0$	$\theta_0=8.00^\circ$
Impact 1 (at wall)	$t=0.5015$ s	$\theta=0$, $\omega=0.437$ rad/s
Peak of cycle 1	$t=1.0130$ s	$\theta_1=7.200^\circ$
Impact 2 (at wall)	$t=1.5145$ s	$\theta=0$
Peak of cycle 2	$t=2.0260$ s	$\theta_2=6.480^\circ$

Check: the wall's position is not stated numerically in the source text; placing it at the pendulum's equilibrium ($\theta=0$) is the natural reading given no other angle is supplied, and matches the figure's description of the mass swinging to strike a vertical wall at the bottom of its arc. Between impacts $\theta(t)$ is a standard cosine/sine SHM segment of amplitude $\theta_n$ centred on each impact instant.

Topic: Free vibration with amplitude decay from repeated impacts (coefficient of restitution).

Key relations: SHM pendulum $\omega_n=\sqrt{g/L}$; impact COR $e=|\vec v_{\text{after}}|/|\vec v_{\text{before}}|$; for amplitude-proportional impact speed, successive peaks form a geometric sequence $\theta_n=e^n\theta_0$.

Why this works: between impacts the system is conservative (undamped SHM), so all the energy loss is concentrated at the wall; because impact speed is proportional to swing amplitude (SHM), the restitution ratio on velocity translates directly into the same ratio on the next peak angle.

Common pitfall: applying $e$ to the angle in energy terms (i.e. writing $\theta_{n+1}=e^2\theta_0$ after one bounce) instead of to the velocity — COR is defined on relative approach/separation velocity, not energy, so the angle ratio per bounce is $e^1$, not $e^2$.

Source: Rao, Mechanical Vibrations, Ch. 2 & 8 (free vibration, impact and coefficient of restitution).

Question 9 of 10

Question 7 (Part B, 20 marks)