Question 5 of 6: Question 5 (paper Question V) — Pendulum Strikes a Wall: Rebound Angle via Coefficient of Restitution (Part B · Dynamics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3, Statics and Dynamics — 2018-Dec. Candidates were required to complete 2 questions from PART A (Statics) and 2 questions from PART B (Dynamics); every question is solved below so this paper serves as a complete study resource (Questions 1-3 = paper Part A, I-III; Questions 4-6 = paper Part B, IV-VI).
Check — source reconstruction notes. (2) Question 1's figure gives explicit coordinates for D, E, F and an explicit height (3 m) for wall anchor C, but not a separate height for wall anchor B; reading the drawing against its own calibrated vertical scale (using the printed "3 m" and "1 m" dimensions as reference) places B at very nearly the same height as C, and only that reading makes the six equilibrium equations for the 5-unknown system (3 reactions at A + 2 cable tensions) consistent AND returns exact, textbook-clean numbers (cable lengths of exactly 3 m each, tensions of exactly 4200 N each) — strong corroborating evidence for the reconstruction. B = (0, -2, 3) m and C = (0, 2, 3) m (both relative to A) are used below. (3) Question 3's figure dimensions the horizontal offset of the top pivots E, F (275 mm) and the vertical drop from E/F to C/D and from C/D to A/B (500 mm each), but not a horizontal offset for C, D; the crate's own 300 mm grip width fixes A, B at ±150 mm, so the jaw arm C-A is read as tapering inward from x = -275 mm (at C, in line with E) to x = -150 mm (at A) over its 500 mm drop — the taper is placed on the lower arm segment (C to A) rather than the upper one (E to C), consistent with the visible bend in the source drawing.
Question 5 (paper Question V) — Pendulum Strikes a Wall: Rebound Angle via Coefficient of Restitution (Part B · Dynamics, equal value)
Given. Slender rod, mass 6 kg, length 2 m, pinned at A (bracket stands 100 mm off the wall). Solid sphere, mass 10 kg, radius 300 mm, rigidly attached at the rod's free end B (sphere centre 2 m from A). Released from rest at $\theta_1=90^\circ$ (horizontal). Coefficient of restitution with the wall $e=0.6$. $I_{rod,end}=\tfrac{1}{3}mL^2$; $I_{sphere,c}=\tfrac{2}{5}mR^2$.
Given data
Item
Value
Rod mass, length
6 kg, 2 m
Sphere mass, radius
10 kg, 300 mm
Pivot standoff from wall
100 mm
Release angle $\theta_1$
90° (horizontal, from rest)
Coefficient of restitution $e$
0.6
Find. The rebound angle $\theta_2$ (from vertical) at which the pendulum momentarily comes to rest.
Figure 5. Pendulum: 6 kg rod (2 m) pivoted at A, 10 kg sphere (radius 300 mm) at the free end B; pivot stands 100 mm off the wall.
Approach. Use energy conservation to find the angular velocity just before the sphere first touches the wall, apply the coefficient of restitution as an angular-impulse–momentum relation about the fixed pivot A to find the angular velocity just after impact, then apply energy conservation a second time to find the momentary-rest angle.
Mass moment of inertia about A. $I_A=\tfrac{1}{3}(6)(2)^2+\left[\tfrac{2}{5}(10)(0.3)^2+10(2)^2\right]=8+40.36=\boxed{48.36\ \text{kg}\cdot\text{m}^2}$.
Impact angle. With the pivot 100 mm off the wall and the sphere (radius 300 mm) approaching from the release side, the sphere's surface reaches the wall when $x_A+L\sin\theta=R\Rightarrow \sin\theta_i=\dfrac{0.3-0.1}{2}=0.1\Rightarrow \boxed{\theta_i=5.74^\circ}$ (still on the release side, just short of vertical).
Angular velocity just before impact (energy conservation, $\theta_1=90^\circ\to\theta_i$). $\tfrac{1}{2}I_A\omega_i^2=g\left[m_{rod}\left(\tfrac{L}{2}\right)+m_{ball}L\right]\cos\theta_i=9.81(26)\cos(5.74^\circ)$, giving $\boxed{\omega_i=3.24\ \text{rad/s}}$.
Impact (coefficient of restitution about A). Since the sphere's velocity is tangential ($v=\omega L$) and the wall's normal (horizontal) component is proportional to $\omega$ by the same geometric factor that the impulsive wall force uses in the angular-momentum equation, restitution reduces directly to $\omega_{after}=-e\,\omega_i$: $\boxed{\omega_{after}=0.6(3.24)=1.94\ \text{rad/s}}$ (reversed direction).
Rebound swing to momentary rest (energy conservation, $\theta_i\to\theta_2$). $\tfrac12 I_A\omega_{after}^2 = g(26)\left[\cos\theta_i-\cos\theta_2\right]$, i.e. $\cos\theta_2=\cos\theta_i(1-e^2)=0.9950(0.64)=0.6368\Rightarrow \boxed{\theta_2=50.4^\circ}$.
