Question 4 of 6: Question 4 (paper Question IV) — Drum-and-Wheel Rolling Kinematics (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 — 2017-December. Candidates were required to answer any 2 of 3 questions in Part A (Statics) and any 2 of 3 in 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 — figure reconstruction. Three figures needed a reconstructed reading, each corroborated by an independent numerical check rather than assumed: (1) Question 1's wall cable anchors C and D — the only geometry (C on the +x side, D on the −x side of the pipe) that returns positive (physically valid, tension-only) cable forces for both cables; any same-side reading returns a negative "tension" in one cable, which is impossible. (2) Question 2's overall frame height is not printed directly; it follows from the stated 30° leg angle and the 4 m offset of the lower joints, and the resulting member forces close exactly by symmetry (equal reactions, two exact zero-force members) — strong corroborating evidence the reconstruction is right. (3) Question 5's second suspension cable (block A) is read as the same 6 ft length as sphere B's cable, since the figure's one printed "6 ft" dimension spans a horizontal reference line common to both hanging positions.
Question 4 (paper Question IV) — Drum-and-Wheel Rolling Kinematics (Part B · Dynamics, equal value)
Given. Wheel (outer) radius $R=180\ \text{mm}=0.18\ \text{m}$, rolling without slipping on the ground at C; drum (inner) radius $r=120\ \text{mm}=0.12\ \text{m}$, rope leaving tangentially at F (top of the drum, directly above centre A); D is on the wheel rim level with A. (a) $v_E=0.6\ \text{m/s}$ constant. (b) $v_E=0.6\ \text{m/s}$, $a_E=1.5\ \text{m/s}^2$, at the same instant.
Find. (a) $\mathbf{v}_D$ and $\mathbf{v}_{E/A}$; (b) $\mathbf{a}_A$, $\mathbf{a}_C$, $\mathbf{a}_F$.
Figure 4. Compound wheel/drum, centre A, rolling contact C, rim point D, rope tangent point F.
Approach. The rope pays off the drum without slipping, so the rope speed equals the speed of the material point F on the drum at that instant; combined with rolling-without-slipping at C (instantaneous centre of zero velocity), this fixes the wheel's angular velocity. The same logic with time-derivatives gives the angular acceleration, after which every point's acceleration follows from the rigid-body acceleration equation about the translating centre A.
Angular velocity from the rope constraint. F is at height $R+r$ above the instantaneous centre C, so $v_F=\omega(R+r)$; since the rope does not slip on the drum, $v_E=v_F$: $$\omega=\dfrac{v_E}{R+r}=\dfrac{0.6}{0.18+0.12}=\boxed{2.0\ \text{rad/s}}$$
Velocity of the centre A. A is $R$ above C: $v_A=\omega R=2.0(0.18)=\boxed{0.36\ \text{m/s}}$ (horizontal, same sense as $v_E$).
Part (a) results. D is at perpendicular distance $R\sqrt2$ from C (45° from the vertical through C): $$v_D=\omega R\sqrt2=2.0(0.18)(1.414)=\boxed{0.509\ \text{m/s}}\ \text{(directed 45}^\circ\text{ below horizontal, away from the wheel)}$$ E is a point on the (translating, non-rotating) rope, not on the rigid wheel, so its velocity relative to A is simply the vector difference: $$\mathbf{v}_{E/A}=v_E-v_A=0.6-0.36=\boxed{0.24\ \text{m/s}}$$
Angular acceleration. Differentiating the rope constraint (the geometry $R+r$ is fixed): $$\alpha=\dfrac{a_E}{R+r}=\dfrac{1.5}{0.30}=\boxed{5.0\ \text{rad/s}^2}$$
Acceleration of A, C, F (rigid-body acceleration equation about A). Since A translates on a straight horizontal path, $a_A=\alpha R=5.0(0.18)=\boxed{0.90\ \text{m/s}^2}$ (horizontal). For any point P on the wheel, $\mathbf{a}_P=\mathbf{a}_A+\boldsymbol\alpha\times\mathbf{r}_{P/A}-\omega^2\mathbf{r}_{P/A}$. At C (directly below A, $\mathbf{r}_{C/A}=(0,-R)$): the horizontal ($\alpha R$) terms cancel exactly against $a_A$, leaving pure centripetal acceleration: $$\boxed{a_C=\omega^2R=(2.0)^2(0.18)=0.72\ \text{m/s}^2\ \text{(vertically upward, toward A)}}$$ At F ($\mathbf{r}_{F/A}=(0,r)$): $$a_{Fx}=a_A-\alpha r=0.90-5.0(0.12)=0.30\ \text{m/s}^2,\qquad a_{Fy}=-\omega^2r=-0.48\ \text{m/s}^2$$ $$\boxed{a_F=\sqrt{0.30^2+0.48^2}=0.566\ \text{m/s}^2}$$
