Question 6 of 6: Pendulum with spring & damper, perfectly-plastic impact

Question 6: Pendulum with spring & damper, perfectly-plastic impact (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. Uniform bar $m_1=10\ \text{kg}$, length $l=1\ \text{m}$, hinged at top (hanging pendulum). Spring $k=1000\ \text{N/m}$ at arm $l/3$; viscous damper $c=50\ \text{N}\cdot\text{s/m}$ at arm $l$ (lower end). Block $m_2=1\ \text{kg}$ at $v=2.5\ \text{m/s}$ strikes the lower end; impact perfectly plastic (block sticks). $g=9.81\ \text{m/s}^2$.

Find. (a) EOM in $\theta$; (b) $\omega_n$; (c) $\theta(t)$ after impact.

Approach. Take moments about the hinge for the small-angle rotation $\theta$ from vertical: assemble the effective inertia, damping, and stiffness (spring + gravity restoring). Get the post-impact angular velocity from conservation of angular momentum about the hinge, then write the underdamped free response.

1. Effective inertia (after the block sticks at the lower end). Bar about the top hinge $I_{bar}=\tfrac13 m_1 l^2 = 3.333\ \text{kg}\cdot\text{m}^2$; the stuck block adds $m_2 l^2 = 1.0$: $$I = \tfrac13 m_1 l^2 + m_2 l^2 = 4.333\ \text{kg}\cdot\text{m}^2.$$
2. (a) Equation of motion. Moments about the hinge for small $\theta$: the spring (arm $l/3$) contributes $k(l/3)^2\theta$; the damper (arm $l$) contributes $c\,l^2\dot\theta$; gravity on the hanging bar and block is restoring, adding $\big(m_1 g\tfrac{l}{2}+m_2 g\,l\big)\theta$. Hence $$I\ddot\theta + c\,l^2\dot\theta + \Big[k\tfrac{l^2}{9}+m_1 g\tfrac{l}{2}+m_2 g\,l\Big]\theta = 0,$$ i.e. $$\boxed{4.333\,\ddot\theta + 50\,\dot\theta + 169.97\,\theta = 0}$$ with $k_t = 111.11+49.05+9.81 = 169.97\ \text{N}\cdot\text{m/rad}$.
3. (b) Natural frequency and damping. $$\omega_n=\sqrt{k_t/I}=\sqrt{169.97/4.333}=\boxed{6.263\ \text{rad/s}}\ (0.997\,\text{Hz}).$$ Critical damping $c_{cr}=2I\omega_n=54.28$, so $\zeta = c\,l^2/c_{cr} = 50/54.28 = 0.921$ — underdamped. Damped frequency $\omega_d=\omega_n\sqrt{1-\zeta^2}=2.437\ \text{rad/s}$.
4. Post-impact angular velocity (plastic impact). Angular momentum about the hinge is conserved through the instantaneous impact; before, only the block moves ($m_2 v$ at arm $l$); after, the combined body rotates at $\omega_0$: $$m_2 v\,l = I\,\omega_0 \;\Rightarrow\; \omega_0 = \frac{(1)(2.5)(1)}{4.333} = 0.577\ \text{rad/s}.$$
5. (c) Ensuing motion. Initial conditions $\theta(0)=0$, $\dot\theta(0)=\omega_0$. The underdamped free response is $$\theta(t)=e^{-\zeta\omega_n t}\,\frac{\omega_0}{\omega_d}\sin(\omega_d t) = \boxed{0.237\,e^{-5.77\,t}\sin(2.437\,t)\ \text{rad}}$$ The pendulum swings once to about $\theta_{\mathrm{max}}\approx 0.036$ rad ($2.0^\circ$), reached at $t\approx0.164$ s, then decays to rest within roughly one second ($\zeta=0.92$, near-critical).