Question 3 of 6: Crank-Slider-Slotted Rod Kinematics

Question 3: Crank-Slider-Slotted Rod Kinematics (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Crank $AB=100\text{ mm}$ pinned at fixed point $A$, rotating with $\vec\omega_{AB}=3\vec k\text{ rad/s}$, $\vec\alpha_{AB}=-1\vec k\text{ rad/s}^2$; slider $B$ rides in the slot of rod $CD$, pinned at fixed point $C$, with $CB=300\text{ mm}$ measured along the rod; at the instant shown the crank sits $30^\circ$ above the horizontal and the rod $60^\circ$ above the horizontal (i.e. $30^\circ$ off vertical at $C$), so the crank and rod meet at $B$ with an included angle of $30^\circ$, matching the figure.

Find. The velocity and acceleration of slider $B$ relative to the slotted rod ($v_{rel}$, $a_{rel}$, directed along the slot), together with the rod's own $\omega_{CD}$, $\alpha_{CD}$.

Approach. Get $B$'s absolute velocity and acceleration from the crank alone (pure rotation about the fixed point $A$), then re-express the same point using axes rotating WITH rod $CD$ (origin at the fixed point $C$) via the rotating-reference-frame equations, and solve for the rod's $\omega_{CD},\alpha_{CD}$ and the slider's relative motion $v_{rel},a_{rel}$.

1. Absolute velocity and acceleration of $B$ from the crank. With $\vec r_{B/A}=AB(\cos30^\circ\vec i+\sin30^\circ\vec j) = (0.0866,\,0.0500)\text{ m}$, pure rotation about fixed $A$ gives $$\vec v_B=\vec\omega_{AB}\times\vec r_{B/A},\qquad \vec a_B=\vec\alpha_{AB}\times\vec r_{B/A}-\omega_{AB}^2\vec r_{B/A}$$ Evaluating: $$\boxed{\vec v_B = -0.1500\,\vec i + 0.2598\,\vec j\ \text{m/s}},\qquad \boxed{\vec a_B = -0.7294\,\vec i - 0.5366\,\vec j\ \text{m/s}^2}$$
2. Set up the rotating frame on rod $CD$. Let $\hat e_r=(\cos60^\circ,\sin60^\circ)$ point along the rod from $C$ toward $D$, and $\hat e_t$ perpendicular to it (positive-rotation direction), so $\vec r_{B/C}=CB\,\hat e_r = (0.1500,\,0.2598)\text{ m}$. Since the slot is straight, the slider's motion relative to the rod is purely radial: $\vec v_{rel}=v_{rel}\hat e_r$, $\vec a_{rel}=a_{rel}\hat e_r$ (no transverse relative component). The rotating-axis kinematics equations for the same point $B$ are $$\vec v_B=\vec\omega_{CD}\times\vec r_{B/C}+\vec v_{rel},\qquad \vec a_B=\vec\alpha_{CD}\times\vec r_{B/C}-\omega_{CD}^2\vec r_{B/C}+\vec a_{rel}+2\vec\omega_{CD}\times\vec v_{rel}$$
3. Solve the velocity equation (two scalar equations, two unknowns $\omega_{CD},v_{rel}$) by equating this to Step 1's $\vec v_B$: $$\boxed{\omega_{CD} = \frac{\sqrt3}{2} = 0.8660\text{ rad/s (CCW)}},\qquad \boxed{v_{rel} = 0.1500\text{ m/s}}$$
4. Solve the acceleration equation. With $\omega_{CD}$ and $v_{rel}$ now known, evaluate the Coriolis term $2\vec\omega_{CD}\times\vec v_{rel} = (-0.2250,\,0.1299)\text{ m/s}^2$ and the centripetal term $-\omega_{CD}^2\vec r_{B/C}$, then solve the remaining two scalar equations for $\alpha_{CD},a_{rel}$ against Step 1's $\vec a_B$: $$\boxed{\alpha_{CD} = 0.345\text{ rad/s}^2\text{ (CCW)}},\qquad \boxed{a_{rel} = -0.604\text{ m/s}^2}$$
5. State the requested relative motion. The slider's velocity relative to the slotted rod is $0.150\text{ m/s}$ directed from $C$ toward $D$ (outward along the slot); its acceleration relative to the rod is $0.604\text{ m/s}^2$ directed from $D$ toward $C$ (i.e. the outward slide is momentarily decelerating).