Question 4 of 6: Question 4 (paper Question IV) — Rod Released from a Compressed Spring (Part B · Dynamics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination December 2016 — 04-BS-3, Statics and Dynamics, 3 hours, closed book (one 8½″×11″ self-prepared note sheet permitted; Casio or Sharp approved calculator only). Candidates complete 2 of 3 questions from Part A (Statics) and 2 of 3 from Part B (Dynamics); all 6 are solved below.
Separately, Question 1's bracket figure (Figure 1) gives only one anchor point's coordinates (E) plus two angle pairs for the other two cables; every printed dimension (50 mm×2, 100 mm×2, 75 mm, the two 45° arcs at C, the 30°/60° pair at D) was used, but the exact 3-D position of points B and C required one reasonable reconstruction, disclosed at first use below. All other figures (2–6) reconstruct cleanly and unambiguously from the printed dimensions, confirmed by internal checks that return clean values (member lengths BD = 5 ft and DE = 10 ft in Question 6) — strong evidence the readings are correct.
Question 4 (paper Question IV) — Rod Released from a Compressed Spring (Part B · Dynamics, equal value)
Given. $m=15\ \text{kg}$, $L=2\ \text{m}$, $OB=0.4\ \text{m}$ (so $OA=1.6\ \text{m}$), spring $k=300\ \text{kN/m}$ compressed $x=25\ \text{mm}$ at release, rod initially horizontal.
Given data
Quantity
Value
$m$
15 kg
$L$ (=AB)
2 m
$OB$
0.4 m
$OA$
1.6 m
$k$
300 kN/m
Spring compression $x$
25 mm
Find. The rod's angular velocity $\omega$, and the pivot reaction components $O_x,O_y$, as the rod passes through the vertical position.
Figure 4 — rod AB pivoted at O; end A rests on a spring compressed 25 mm; centre of mass C is 0.6 m from O on the A side.
Approach. The 25 mm spring stroke subtends a negligible angle at the 1.6 m radius OA (≈0.9°), so the spring's stored energy converts to kinetic energy essentially while the rod is still horizontal; from there, energy conservation (spring energy and the change in gravitational PE of the mass centre) gives $\omega$ at the vertical position, and Newton's second law (with the angular acceleration found from moments about O) gives the pivot reactions.
Mass centre location and mass moment of inertia about O. $C$ is at the rod's midpoint, 1.0 m from A, so $OC=|OA-L/2|=|1.6-1.0|=0.6\ \text{m}$.
$$I_C=\tfrac{1}{12}mL^2=\tfrac{1}{12}(15)(2^2)=5.0\ \text{kg}\cdot\text{m}^2,\qquad I_O=I_C+m\,OC^2=5.0+15(0.6^2)=\boxed{10.4\ \text{kg}\cdot\text{m}^2}$$
Spring energy released. $E_{spring}=\tfrac{1}{2}kx^2=\tfrac12(300{,}000)(0.025^2)=\boxed{93.75\ \text{J}}$.
Height gained by C reaching the vertical position. Releasing the spring pushes A upward, so the rod rotates 90° with the A-side (and C, which is on the A-side of O) swinging up; C rises by $OC\sin90^{\circ}=0.6\ \text{m}$.
$$\Delta(mgh)=mg(OC)=15(9.81)(0.6)=88.29\ \text{J}$$
Angular acceleration at the vertical position. With the rod vertical, C lies directly on the vertical line through O, so gravity's moment arm about O is zero: $\sum M_O=I_O\alpha=0\ \Rightarrow\ \alpha=0$. The mass centre therefore has only a centripetal (radial) acceleration, $a_C=\omega^2(OC)$, directed from C toward O (i.e. straight down, since C is above O).
Pivot reactions (Newton's second law on the rod). Horizontally, $a_{Cx}=0\Rightarrow O_x=0$. Vertically (up positive), $a_{Cy}=-\omega^2(OC)$:
$$O_y-mg=m(-\omega^2\,OC)\ \Rightarrow\ O_y=mg-m\omega^2(OC)=15(9.81)-15(1.025^2)(0.6)=\boxed{137.7\ \text{N}}$$
