Question 6 of 7: Double pendulum with a stopper (impact)

Question 6: Double pendulum with a stopper (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. Two uniform rods (each $m$, $l$); rod 1 pinned to ground at $O$, rod 2 pinned to the end of rod 1. A rigid stopper on the equilibrium vertical contacts rod 1 at $0.9l$ from $O$ when rod 1 returns to vertical ($\theta_1=0$). Restitution $e$, impact duration $\Delta\tau$. Small initial angles $\theta_{1,0},\theta_{2,0}$, released from rest.

Find. The free-vibration response of the system in the interval between the first and second stopper impacts (i.e. the motion "just prior to the second impact").

Approach. Build the linearised 2-DOF equations, get the modes, propagate the free response from the release, apply the impulsive restitution law at $\theta_1=0$ to jump the velocities, and propagate again to the next contact.

Step 1 — Linearised equations of motion

Using absolute angles $\theta_1,\theta_2$, the kinetic and potential energies of two uniform rods expand (small angles) to a mass and stiffness matrix

$$\mathbf M=ml^2\!\begin{bmatrix}\tfrac43 & \tfrac12\\[2pt]\tfrac12 & \tfrac13\end{bmatrix},\qquad \mathbf K=mgl\!\begin{bmatrix}\tfrac32 & 0\\[2pt] 0 & \tfrac12\end{bmatrix},\qquad \mathbf M\ddot{\boldsymbol\theta}+\mathbf K\boldsymbol\theta=\mathbf 0 .$$

Step 2 — Natural frequencies and modes

Setting $\det(\mathbf K-\omega^2\mathbf M)=0$ with $\lambda=\omega^2 l/g$ gives $7\lambda^2-42\lambda+27=0$, so $\lambda_{1,2}=0.732,\ 5.268$ and

$$\boxed{\omega_1=0.856\sqrt{g/l},\qquad \omega_2=2.295\sqrt{g/l}.}$$

The mode-shape ratios $\theta_2/\theta_1$ follow from $(\tfrac32-\tfrac43\lambda)\theta_1=\tfrac12\lambda\,\theta_2$:

$$\boldsymbol\phi_1=\begin{Bmatrix}1\\ 1.43\end{Bmatrix}\text{(in phase)},\qquad \boldsymbol\phi_2=\begin{Bmatrix}1\\ -2.10\end{Bmatrix}\text{(out of phase)}.$$

Step 3 — Free response before the first impact

Released from rest at $\boldsymbol\theta(0)=\{\theta_{1,0},\theta_{2,0}\}$, the modal expansion $\boldsymbol\theta=\eta_1(t)\boldsymbol\phi_1+\eta_2(t)\boldsymbol\phi_2$ with $\dot{\boldsymbol\theta}(0)=0$ gives

$$\boldsymbol\theta(t)=c_1\cos(\omega_1 t)\,\boldsymbol\phi_1+c_2\cos(\omega_2 t)\,\boldsymbol\phi_2,$$

where $c_1,c_2$ solve $c_1\boldsymbol\phi_1+c_2\boldsymbol\phi_2=\{\theta_{1,0},\theta_{2,0}\}$, i.e. $c_1=\dfrac{2.10\,\theta_{1,0}+\theta_{2,0}}{3.53},\ c_2=\dfrac{1.43\,\theta_{1,0}-\theta_{2,0}}{3.53}$. The first impact occurs at the earliest time $t_1$ with $\theta_1(t_1)=0$ (rod 1 back to vertical); the pre-impact velocities are $\dot\theta_1^-=\dot\theta_1(t_1),\ \dot\theta_2^-=\dot\theta_2(t_1)$.

Step 4 — Impulsive restitution at the stopper

At contact the stopper applies a horizontal impulse $\hat P$ to rod 1 at $0.9l$; positions are frozen ($\theta_1=0$) and only velocities jump. In generalised coordinates the impulse appears as $\hat Q_1=0.9l\,\hat P,\ \hat Q_2=0$ (the contact point lies on rod 1, independent of $\theta_2$), so $\mathbf M\,\Delta\dot{\boldsymbol\theta}=\{0.9l\hat P,\,0\}$. The second row gives the internal constraint

$$\tfrac12\Delta\dot\theta_1+\tfrac13\Delta\dot\theta_2=0\;\Rightarrow\;\Delta\dot\theta_2=-\tfrac32\Delta\dot\theta_1 .$$

Restitution at the contact point (horizontal speed of the $0.9l$ point reverses and scales by $e$): $0.9l\,\dot\theta_1^+=-e\,(0.9l\,\dot\theta_1^-)$, hence $\dot\theta_1^+=-e\,\dot\theta_1^-$ and $\Delta\dot\theta_1=-(1+e)\dot\theta_1^-$. Therefore

$$\boxed{\dot\theta_1^+=-e\,\dot\theta_1^-,\qquad \dot\theta_2^+=\dot\theta_2^-+\tfrac32(1+e)\,\dot\theta_1^-.}$$

(The finite duration $\Delta\tau$ enters only through the mean contact force $\bar P=\hat P/\Delta\tau$; it does not affect the impulsive velocity jump.)

Step 5 — Free vibration up to the second impact

Restarting the clock at the first impact ($\tau=t-t_1$), the state is $\theta_1=0,\ \theta_2=\theta_2(t_1)$, with velocities $\dot\theta_1^+,\dot\theta_2^+$ from Step 4. Projecting onto the modes,

$$\boldsymbol\theta(\tau)=\sum_{i=1,2}\Bigl[A_i\cos(\omega_i\tau)+B_i\sin(\omega_i\tau)\Bigr]\boldsymbol\phi_i,$$

with $A_i$ set by the post-impact position $\{0,\theta_2(t_1)\}$ and $B_i=\dfrac{\text{(post-impact modal velocity)}_i}{\omega_i}$ set by $\{\dot\theta_1^+,\dot\theta_2^+\}$ (modal coordinates obtained from the same $\boldsymbol\phi_1,\boldsymbol\phi_2$ inverse used in Step 3). This expression is the free vibration during the interval; the second impact is the next root $\tau_2>0$ of $\theta_1(\tau)=0$, and "just prior to the second impact" means evaluating the boxed response at $\tau\to\tau_2^-$.