Question 4 of 6: Question 4 (paper Question IV) — Pendulum Eccentric Impact with a Floor (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 — 2019-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 isometric figure gives explicit 1 m / 1.5 m dimensions along the rod, but plotting them on a single straight line makes the six rigid-body equilibrium equations inconsistent (no solution exists). Reading the printed figure, the horizontal run bends twice: A→C is 2 m along the first horizontal direction (with the vertical 1.5 m branch C→D), C→E is 1 m along the perpendicular horizontal direction (this is where the smooth journal bearing sits), and E→F is a further 1.5 m continuing the first horizontal direction to the point that carries the 250 N cylinder. This reading is the only one of the plausible alternatives that returns a statically-determinate, non-degenerate system, and it lands on the clean values reported below — strong corroborating evidence for the reconstruction. (3) Question 4's dimensions place the pin O exactly 90 mm above the floor, equal to the sphere radius (90 mm R); the sphere therefore contacts the floor exactly when the rod is horizontal, which is used as the impact configuration. (4) Question 6 numbers the four-bar linkage O-A-B-C from the source's own 100 mm / 250 mm / 75 mm / 50 mm dimensions; O and C are the fixed pivots, OA and CB are the driven/driving cranks and AB is the coupler.
Question 4 (paper Question IV) — Pendulum Eccentric Impact with a Floor (Part B · Dynamics, equal value)
Given. Slender rod, weight 18 N, length $l=610$ mm, pinned at O. Sphere, weight 45 N, radius $r=90$ mm, attached at the far end of the rod (centre at $d=l+r=700$ mm from O). Pin O is 90 mm above the floor (exactly equal to the sphere radius). Released from rest at $\theta=60^\circ$ from horizontal; $e=0.8$; $(I_G)_{rod}=\tfrac1{12}ml^2$, $(I_G)_{sphere}=\tfrac25mr^2$.
Find. (a) the sphere's velocity at impact; (b) the rebound angle $\theta$.
Figure 4. Rod-sphere pendulum pinned at O, released from $\theta=60^\circ$; contacts the floor when the rod is horizontal.
Approach. Use energy conservation from release ($\theta=60^\circ$, at rest) to the impact configuration (rod horizontal, since pin height equals the sphere radius) to find the angular velocity just before impact. Because the contact point's vertical velocity component is directly proportional to $\omega$ (with the same proportionality constant before and after, as both are rigid-body rotations about the same fixed pin O), the coefficient of restitution converts this directly into the angular velocity just after impact, which a second energy-conservation pass converts into the rebound angle.
Mass moment of inertia about O (parallel-axis theorem). With $m_{rod}=18/9.81=1.835$ kg, $m_{sph}=45/9.81=4.587$ kg:
$$I_O=\underbrace{\tfrac13 m_{rod}l^2}_{\text{rod about end }O}+\underbrace{\tfrac25 m_{sph}r^2+m_{sph}d^2}_{\text{sphere, parallel axis}}=0.2276+2.2626=\boxed{2.490\text{ kg}\!\cdot\!\text{m}^2}$$
Energy conservation, release → impact (rod horizontal). The rod's CG (at $l/2$) and the sphere's centre (at $d$) both drop by (their distance)$\times\sin60^\circ$:
$$\tfrac12 I_O\omega_1^2=W_{rod}\left(\tfrac{l}{2}\sin60^\circ\right)+W_{sph}\left(d\sin60^\circ\right)=4.75+27.28=32.03\text{ J}$$
$$\omega_1=\sqrt{\frac{2(32.03)}{2.490}}=\boxed{5.07\text{ rad/s}}$$
Sphere's impact velocity (part a). With the rod horizontal, the sphere centre's velocity is purely vertical, $v=\omega_1 d$:
$$v_{impact}=5.07(0.70)=\boxed{3.55\text{ m/s (downward)}}$$
Coefficient of restitution across the impact. The contact point sits directly below the sphere centre, and its vertical-velocity component equals $\omega d$ both before and after impact (same lever arm $d$, since both are still rigid-body rotations about the fixed pin O) — so the restitution ratio applies directly to the angular velocities:
$$\omega_2=e\,\omega_1=0.8(5.07)=\boxed{4.06\text{ rad/s}}$$
Energy conservation, just after impact → rebound angle (part b). Setting the post-impact kinetic energy equal to the PE regained at the rebound angle $\theta_{reb}$:
$$\tfrac12 I_O\omega_2^2=\bigl[W_{rod}(l/2)+W_{sph}\,d\bigr]\sin\theta_{reb}$$
$$\sin\theta_{reb}=\frac{\tfrac12(2.490)(4.06)^2}{18(0.305)+45(0.70)}=\frac{20.50}{36.99}=0.5541\ \Rightarrow\ \boxed{\theta_{reb}=33.7^\circ}$$
