Question 6 of 6: Rigid Rod: Angular Acceleration After a Cable Snaps

Question 6: Rigid Rod: Angular Acceleration After a Cable Snaps (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 $AB$, $L=2\text{ m}$, $m=1\text{ kg}$, horizontal; end $A$ rests on a vertical spring (in static equilibrium immediately before the snap); end $B$'s cable has just snapped (tension drops to zero); the rod is at rest at this instant ($\omega=0$, so no centripetal terms appear).

Find. The angular acceleration $\alpha$ of rod $AB$ immediately after the cable snaps.

Approach. Use the PRE-snap static equilibrium to fix the spring force (a spring's force is set by its compression, which cannot change instantaneously), then apply the planar rigid-body equations $\Sigma F=m\vec a_G$, $\Sigma M_G=I_G\alpha$ to the POST-snap free body (spring force plus weight only) to solve directly for $\alpha$.

1. Fix the spring force from the pre-snap equilibrium. Before the snap, $\Sigma F_y=0$ and $\Sigma M_G=0$ (rod at rest, symmetric supports at $\pm L/2$ from the centre $G$) give $$F_A+T_B=mg,\qquad F_A\left(\tfrac L2\right)=T_B\left(\tfrac L2\right)\ \Rightarrow\ F_A=T_B=\frac{mg}{2}$$ Immediately after the cable snaps, $T_B\to0$ but the spring force is UNCHANGED at this instant: $$\boxed{F_A = \frac{mg}{2} = \frac{1(9.81)}{2} = 4.905\text{ N}}$$
2. Apply $\Sigma F=m\vec a_G$ to the post-snap free body (weight $mg$ at $G$; spring force $F_A$ at $A$ only; no horizontal forces): $$a_{Gx}=0,\qquad a_{Gy} = \frac{F_A-mg}{m} = \frac{4.905-9.81}{1} = -4.905\text{ m/s}^2 = -\frac{g}{2}$$
3. Apply $\Sigma M_G=I_G\alpha$. With $\vec r_{A/G}=(-L/2,0)$ and only the spring force $(0,F_A)$ producing a moment about $G$ (the weight passes through $G$): $$\Sigma M_G = -F_A\cdot\tfrac L2,\qquad I_G=\frac{mL^2}{12}=\frac{1(2)^2}{12}=0.3333\text{ kg}\cdot\text{m}^2$$ $$\alpha = \frac{-F_A L/2}{I_G} = \frac{-(4.905)(1)}{0.3333}$$ $$\boxed{\alpha = -14.72\text{ rad/s}^2}$$ (the negative sign, with $A$ on the $-x$ side of $G$, means the rotation is such that end $A$ — over the spring — swings upward while end $B$ drops.)
4. Cross-check via the endpoint accelerations. From $\vec a_A=\vec a_G+\vec\alpha\times\vec r_{A/G}$ and $\vec a_B=\vec a_G+\vec\alpha\times\vec r_{B/G}$: $$a_A = +g = 9.81\text{ m/s}^2\ \text{(upward)},\qquad a_B = -2g = -19.62\text{ m/s}^2\ \text{(downward)}$$ Both are clean multiples of $g$, confirming the algebra.