Rolling-Disk Impact & Damped Free Vibration: 22-Mec-A2 Kinematics and Dynamics of Machines, May 2013, Q5

Question 5: Rolling-Disk Impact & 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. Spring–mass–damper: $m=10\ \text{kg}$, $k=3000\ \text{N/m}$, $c=300\ \text{N}\cdot ext{s/m}$, at rest. Rolling disk: $m_1=5\ \text{kg}$, $R=0.5\ \text{m}$, $\omega_d=10\ \text{rad/s}$; sticks (plastic impact); friction ignored after impact.

Find. $x(t)$ of the combined system after impact.

Fig. 5 — the spring-mass-damper struck by the rolling disk (left), and the resulting damped free response x(t) with its decaying envelope (right).

Approach. Get the disk’s translational speed from the rolling condition, apply linear-momentum conservation for the plastic impact to obtain the post-impact velocity, then solve the standard underdamped free-vibration initial-value problem for the combined mass.

Disk speed (rolling without slip). $v_1=\omega_d R=10(0.5)=5\ \text{m/s}$.
Plastic impact — linear momentum. With no impulsive ground friction, the disk’s spin decouples and only translation couples to the block: $$m_1 v_1=(m+m_1)V_0\ \Rightarrow\ V_0=\frac{5(5)}{15}=1.667\ \text{m/s}.$$ The vibrating mass is $M=m+m_1=15\ \text{kg}$.
Natural frequency and damping. $$\omega_n=\sqrt{k/M}=\sqrt{3000/15}=14.14\ \text{rad/s},\quad \zeta=\frac{c}{2\sqrt{kM}}=\frac{300}{424.3}=0.707.$$
Damped natural frequency. $\omega_d^{*}=\omega_n\sqrt{1-\zeta^2}=14.14\sqrt{0.5}=\boxed{10.0\ \text{rad/s}}$ (underdamped).
Initial conditions & response. $x(0)=0$, $\dot x(0)=V_0$. For $x=e^{-\zeta\omega_n t}(A\cos\omega_d^* t+B\sin\omega_d^* t)$: $A=0$, $B=V_0/\omega_d^*=0.1667\ \text{m}$, and $\zeta\omega_n=10\ \text{s}^{-1}$: $$\boxed{x(t)=0.167\,e^{-10t}\sin(10t)\ \text{m}.}$$

Quantity	Value
Disk approach speed	5 m/s
Post-impact velocity $V_0$	1.667 m/s
$\omega_n$ / $\zeta$ / $\omega_d^{*}$	14.14 rad/s / 0.707 / 10 rad/s
Damped free motion	x(t) = 0.167 e^−10t sin(10t) m

Check: the impact uses linear-momentum conservation, assuming the ground friction impulse during the very short contact is negligible so the disk’s rotational kinetic energy stays in its (now free) spin. If instead the rolling constraint is enforced through the impact, use angular momentum about the contact and the answer for $V_0$ changes.

Topic. Plastic impact into a single-DOF system, then underdamped free vibration.

Key relations. Rolling $v=\omega R$; momentum $m_1v_1=(m+m_1)V_0$; $\omega_n=\sqrt{k/M}$, $\zeta=c/(2\sqrt{kM})$, $\omega_d^*=\omega_n\sqrt{1-\zeta^2}$; response $x=e^{-\zeta\omega_n t}(A\cos\omega_d^*t+B\sin\omega_d^*t)$.

Why this works. The impact is impulsive, so the spring/damper forces (finite) do no work during it and momentum is conserved. The block starts from the equilibrium position ($x_0=0$) with the acquired velocity, giving a pure sine decay. Here $c$ was chosen so $\zeta=1/\sqrt2$, which makes $\omega_d^*$ exactly 10 rad/s.

Common pitfall. Adding the disk’s rotational inertia to the vibrating mass — once friction is ignored the spin is decoupled, so only 5 kg of translational mass is added. Also, do not use energy conservation for a plastic (sticking) impact.

Source. Rao, Mechanical Vibrations, §2.6 (free damped vibration) and §on impulse response / initial conditions.

Question 5 of 6: Rolling-Disk Impact & Damped Free Vibration

Question 5: Rolling-Disk Impact & Damped Free Vibration (20 marks)