Question 4 of 6: Question 4 (paper Question IV) — Pendulum with Spring, Work-Energy Method (Part B · Dynamics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-3 / May 2019 — Statics and Dynamics. Closed book; one 8½″×11″ self-prepared note sheet permitted; approved Casio/Sharp calculator. Candidates were instructed to complete 5 of 6 questions (3 of 3 Part A, 2 of 3 Part B) — all 6 are solved below. Reference texts: Hibbeler, Engineering Mechanics: Statics, 14th ed.; Hibbeler, Engineering Mechanics: Dynamics, 14th ed.
Two genuinely under-dimensioned figures required a stated assumption per the exam's own Note 1 ("if doubt exists… submit a clear statement of any assumption made"): Question 1's pin at A is taken as free to rotate about the pipe's own axis A–E (so $M_{Ay}=0$), and cable anchor C is read at the same plan position as A but 1 m higher; Question 2's support at D is read as a roller (horizontal reaction only) rather than a second pin, since two full pins on 7 members (m+r=11 vs 2j=10) is statically indeterminate and unsolvable by first-year statics — the roller reading returns exact, self-consistent reactions. Question 6's rod angle was measured directly from the printed figure (≈31.4° below horizontal) since it is not given numerically.
Question 4 (paper Question IV) — Pendulum with Spring, Work-Energy Method (Part B · Dynamics, equal value)
Given. Pendulum pivoted at O, mass $m=30$ kg, $k_G=0.3$ m (not 0.6 m, which is the separate $OB$ dimension), $OG=0.35$ m. Disk (radius 0.1 m) rigidly fixed to the rod's lower end, centred at A, so $OA=OG+r=0.45$ m. Fixed point $B$ is 0.6 m from O, level with O. Spring $AB$, $k=320$ N/m, unstretched at $\theta=0^\circ$ (A on line OB then). Released from rest at $\theta=0^\circ$.
Given data
Quantity
Value
$m$
30 kg
$k_G$
0.300 m
$OG$
0.350 m
$OA$
0.450 m
$OB$
0.600 m
$k$ (spring)
320 N/m
Find. The angular velocity $\omega$ at $\theta=90^\circ$.
Figure 4. Pendulum O-G-A pivoted at O, spring AB, released from θ=0° to θ=90°.
Approach. Work-energy theorem for the rigid body about the fixed pivot O: gravity does positive work as G falls; the spring stores energy as it stretches; no other forces do work (pin is frictionless).
Mass moment of inertia about O. $I_O=m(k_G^2+OG^2)=30(0.3^2+0.35^2)=30(0.09+0.1225)=\boxed{6.375\text{ kg}\cdot\text{m}^2}$.
Geometry check — spring stretch. At $\theta=0^\circ$: $A$ lies on line $OB$, so unstretched length $L_0=OB-OA=0.6-0.45=0.15$ m. At $\theta=90^\circ$: $OA\perp OB$ (right triangle), so $AB=\sqrt{OA^2+OB^2}=\sqrt{0.45^2+0.6^2}=0.75$ m (a 3-4-5 triangle, $0.45{:}0.6{:}0.75=3{:}4{:}5$). Stretch $x_2=0.75-0.15=\boxed{0.60\text{ m}}$.
Height drop of G. G swings from level with O ($\theta=0^\circ$) to directly below O ($\theta=90^\circ$): $\Delta h=OG=0.35$ m.
Work-energy balance ($T_1+V_1=T_2+V_2$, taking $\theta=90^\circ$ as datum): $mg\,\Delta h=\tfrac12 I_O\omega^2+\tfrac12 k x_2^2$. $30(9.81)(0.35)=103.005$ J $=\tfrac12(6.375)\omega^2+\tfrac12(320)(0.6)^2=3.1875\,\omega^2+57.6$.
Solve for $\omega$. $\omega^2=\dfrac{103.005-57.6}{3.1875}=14.246\Rightarrow \boxed{\omega=3.774\text{ rad/s}}$.
