Question 6 of 6: Question 6 (paper Question VI) — Slider-Crank Velocity and Acceleration Analysis (Part B · Dynamics, Parts A and B 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-May. 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. Question 5's impact angle (18°) is on the opposite side of vertical from the 20° release angle, not the same side. Both readings are corroborated below by clean, self-consistent numerical results (e.g. member D–F resolves to an exact 2.5 m, and the opposite-side impact reading alone reproduces the rising-then-falling trajectory the figure draws for sphere B).
Question 6 (paper Question VI) — Slider-Crank Velocity and Acceleration Analysis (Part B · Dynamics, Parts A and B equal value)
Given. Piston pin $a$ slides on the fixed vertical cylinder centreline, 8.5 in directly above the fixed crank pivot $c$; connecting rod $ab$ makes 11° with that vertical centreline at $b$, which is 2 in horizontally from the centreline. Piston velocity at the instant shown: $v_a=180$ in/s, downward.
Given data / reconstructed geometry
Quantity
Value
Vertical distance $a$ to $c$
8.5 in
Horizontal offset of $b$ from centreline
2 in
Angle of rod $ab$ from vertical, at $b$
11°
$v_a$ (downward)
180 in/s
Connecting-rod length $ab$ (computed)
10.482 in
Crank length $bc$ (computed)
2.683 in
Find. A) the angular velocity $\omega_{bc}$ of the crank; B) the angular acceleration $\alpha_{ab}$ of the connecting rod and the acceleration $\mathbf{a}_a$ of the piston pin, given $\omega_{bc}$ constant.
Figure 6 (reconstructed). Piston pin a slides vertically; connecting rod ab links a to crank pin b; crank bc rotates about the fixed pivot c.
Approach. Fix coordinates with $c$ at the origin and $a$ directly above it; recover $b$'s coordinates from the given angle and offset. Apply the rigid-body relative-velocity equation twice (crank about the fixed pivot c, then rod about the moving pin b) to solve $\omega_{bc}$ and $\omega_{ab}$ simultaneously from the known (vertical) direction of $\mathbf{v}_a$; repeat with the relative-acceleration equation, using $\alpha_{bc}=0$, to solve $\alpha_{ab}$ and $\mathbf{a}_a$.
Geometry. With $c=(0,0)$, $a=(0,8.5)$ in: rod length $=2/\sin11^\circ=10.482$ in, so $b=(-2.000,\,-1.789)$ in, and crank length $bc=\sqrt{2.0^2+1.789^2}=2.683$ in.
Part A — velocity analysis. $\mathbf{v}_b=\boldsymbol{\omega}_{bc}\times\mathbf{r}_{b/c}$, then $\mathbf{v}_a=\mathbf{v}_b+\boldsymbol{\omega}_{ab}\times\mathbf{r}_{a/b}$ with $\mathbf{v}_a=(0,-180)$ in/s known. Writing both scalar components: $$1.789\,\omega_{bc}-10.290\,\omega_{ab}=0, \qquad -2.000\,\omega_{bc}+2.000\,\omega_{ab}=-180$$ Solving simultaneously: $$\boxed{\omega_{bc}=108.9\ \text{rad/s (CCW)}}, \qquad \omega_{ab}=18.94\ \text{rad/s (CCW)}$$
Part B — acceleration analysis. With $\alpha_{bc}=0$ (constant $\omega_{bc}$): $\mathbf{a}_b=-\omega_{bc}^2\mathbf{r}_{b/c}=(23,742,\,21,247)\ \text{in/s}^2$. Then $\mathbf{a}_a=\mathbf{a}_b+\boldsymbol{\alpha}_{ab}\times\mathbf{r}_{a/b}-\omega_{ab}^2\mathbf{r}_{a/b}$, with $\mathbf{a}_a=(0,a_{ay})$ having a known (vertical) direction. Solving the horizontal component for $\alpha_{ab}$ and substituting into the vertical component: $$\boxed{\alpha_{ab}=2237\ \text{rad/s}^2\text{ (CCW)}}, \qquad \boxed{\mathbf{a}_a=22{,}017\ \text{in/s}^2\text{ (upward)}}$$ (the piston, moving down at the instant shown, is decelerating — its acceleration points upward, opposite the velocity.)
