Question 6 of 6: Rod Released from a Spring-and-Cable Support — Free Motion After Sudden Release

Question 6: Rod Released from a Spring-and-Cable Support — Free Motion After Sudden Release (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 rod $AB$, mass $m=1$ kg, length $L=2$ m, initially horizontal and at rest, supported by a spring (vertical, at $A$) and a cable (vertical, to the ceiling, at $B$). At $t=0$ the cable snaps and (by the stated assumption) the spring simultaneously loses contact with the rod, so for all $t>0$ the rod is a free rigid body subject to gravity alone.

Find. Angular velocity $\omega$ and acceleration of the centre of mass $a_{\text{cm}}$ at $t=1$ s.

Approach. Before $t=0$ the rod is in static equilibrium (spring and cable share the weight equally, since the C.M. sits at the midpoint), so $\omega(0)=0$ and $v_{\text{cm}}(0)=0$. For $t>0$, with both supports gone, the only force acting is gravity, which acts through the C.M. and so exerts zero net torque about it — the angular velocity therefore cannot change from its $t=0$ value, while the C.M. is in ordinary free fall.

1. Confirm the pre-release state. Taking moments about $A$ for the rod in equilibrium (weight $mg$ at the midpoint, 1 m from each end; cable tension $T$ at $B$, spring force $F_s$ at $A$): $T(2)-mg(1)=0\Rightarrow T=mg/2$, and vertically $F_s=mg-T=mg/2$. Both supports share the weight equally, and both $\omega(0)=0$, $v_{\text{cm}}(0)=0$ (nothing was moving before release).
2. Apply Newton–Euler for $t>0$. With the spring assumed detached the instant the cable snaps, the free-body diagram for $t>0$ has only the weight $mg$, acting at the C.M.: $$\sum \vec F = m\vec a_{\text{cm}} \ \Rightarrow\ \vec a_{\text{cm}}=\vec g\ (\text{constant, downward}),\qquad \sum M_{\text{cm}} = I_{\text{cm}}\dot\omega = 0\ \Rightarrow\ \dot\omega=0.$$ Since $\dot\omega\equiv 0$ for all $t>0$ and $\omega(0)=0$, we get $\omega(t)\equiv 0$ for every $t>0$ — the rod never begins to rotate; it simply falls, remaining horizontal.
3. Evaluate at $t=1$ s. $$\boxed{\omega(1\,\text{s}) = 0}, \qquad \boxed{a_{\text{cm}}(1\,\text{s}) = g = 9.81\ \text{m/s}^2\ \text{(downward, unchanged from any other instant)}}.$$ (As a bonus, $v_{\text{cm}}(1\,\text{s})=g\cdot t = 9.81$ m/s downward — numerically equal to $g$ purely because $t=1$ s exactly.)