Inverted crank-slider — complete kinematic analysis: 22-Mec-A2 Kinematics and Dynamics of Machines, December 2016, Q2

Question 2: Inverted crank-slider — complete kinematic analysis (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. An inverted slider-crank: crank 2 ($O_2A$) drives coupler 3, which carries an offset arm $AB\perp$ the sliding line and then slides through the pivoted block (follower 4) at $O_4$.

Symbol	Meaning	Value
$r_1=O_2O_4$	frame (ground) length	13.5 cm
$r_2=O_2A$	crank length	5 cm
$r_3=AB$	offset arm ($AB\perp BC$)	4.25 cm
$\theta_2$	crank angle	$110^\circ$
$\omega_2$	crank speed (constant)	200 rpm $=20.944$ rad/s

Find. The angular displacement, velocity and acceleration of the follower 4 and coupler 3, and the linear (sliding) displacement, velocity and acceleration between them, at the stated instant.

Inverted slider-crank, drawn to scale for $\theta_2=110^\circ$. $O_2$ and $O_4$ are fixed pivots; block 4 pivots at $O_4$ and coupler 3 slides through it. The $90^\circ$ corner at $B$ makes $r_3=AB$ the perpendicular offset of pin $A$ from the sliding line.

Positional parameters and loop equations

Approach. Take $\theta_3$ (the sliding-line / coupler angle, which the follower shares, $\theta_4=\theta_3$) and the slide distance $s=B\!\to\!O_4$ as the two unknowns, and close the vector loop $O_2\!\to\!A\!\to\!B\!\to\!O_4\!\to\!O_2$; differentiate once for velocity and once for acceleration.

Vector-loop closure. With $A\!-\!O_2=r_2\angle\theta_2$, the offset $A\!\to\!B=r_3\angle(\theta_3+90^\circ)$, the slide $B\!\to\!O_4=s\angle\theta_3$, and $O_4\!\to\!O_2=r_1\angle180^\circ$: $$r_2\cos\theta_2-r_3\sin\theta_3+s\cos\theta_3-r_1=0,\qquad r_2\sin\theta_2+r_3\cos\theta_3+s\sin\theta_3=0 .$$ Solving these two equations simultaneously for $\theta_3$ and $s$ (Newton–Raphson) gives $$\boxed{\theta_3=\theta_4=-32.65^\circ,\qquad s=15.34\text{ cm}.}$$
Velocity (differentiate the loop, $\dot\theta_2=\omega_2$). The two time-derivatives are linear in $\omega_3$ and $\dot s$: $$\begin{bmatrix}-r_3\cos\theta_3-s\sin\theta_3 & \cos\theta_3\\[2pt] -r_3\sin\theta_3+s\cos\theta_3 & \sin\theta_3\end{bmatrix}\!\begin{bmatrix}\omega_3\\ \dot s\end{bmatrix}=\begin{bmatrix} r_2\omega_2\sin\theta_2\\ -r_2\omega_2\cos\theta_2\end{bmatrix}.$$ With $\omega_2=20.944$ rad/s this yields $$\boxed{\omega_3=\omega_4=5.43\text{ rad/s (CCW)},\qquad \dot s=86.6\text{ cm/s}.}$$
Acceleration (differentiate again, $\dot\omega_2=0$). The same coefficient matrix acts on $(\alpha_3,\ddot s)$, driven by the known centripetal / Coriolis terms; solving, $$\boxed{\alpha_3=\alpha_4=33.6\text{ rad/s}^2,\qquad \ddot s=-11.49\text{ m/s}^2\;(-1149\text{ cm/s}^2).}$$

Because coupler 3 and follower 4 form a sliding pair, they share the same absolute angular motion; their relative motion is the pure translation $s,\dot s,\ddot s$ along the coupler axis. The negative $\ddot s$ shows the slider is decelerating at this instant. A finite-difference check of the closed loop over $\pm\omega_2\,\Delta t$ reproduces every value to four figures.

Quantity	Value at $\theta_2=110^\circ$
Coupler / follower angle $\theta_3=\theta_4$	$-32.65^\circ$
Coupler / follower angular velocity $\omega_3=\omega_4$	5.43 rad/s (CCW)
Coupler / follower angular acceleration $\alpha_3=\alpha_4$	33.6 rad/s$^2$
Slide displacement $s$ ($B\to O_4$)	15.34 cm
Sliding velocity $\dot s$	86.6 cm/s
Sliding acceleration $\ddot s$	$-11.49$ m/s$^2$

Topic: Vector-loop kinematics of an inverted (swinging-block) slider-crank.

Key relations: loop closure $\sum \mathbf r_i=0$; first and second time-derivatives give linear systems in $(\omega_3,\dot s)$ and $(\alpha_3,\ddot s)$ sharing one Jacobian; the sliding pair forces $\theta_4=\theta_3$, $\omega_4=\omega_3$, $\alpha_4=\alpha_3$.

Why this works: Two scalar loop equations exactly match the two configuration unknowns of a single-DOF mechanism, so the position problem is closed. Differentiating a constraint that is identically zero keeps it zero, so each derivative is again a (now linear) constraint; the same geometric Jacobian appears at every order, only the right-hand "known" side changes, which is why velocities feed the acceleration solve.

Common pitfall: forgetting the $90^\circ$ offset arm $r_3$ (treating it as an ordinary slider-crank), or mixing up which member is the coupler vs. follower — here they rotate together and only the slide distinguishes them. Also, the Coriolis term $2\dot s\,\omega_3$ must be retained in the acceleration right-hand side.

Source: Norton, Design of Machinery §4.7 & §7 (inverted slider-crank, analytical acceleration); Wilson & Sadler Ch. 2–3.

Question 2 of 7: Inverted crank-slider — complete kinematic analysis

Question 2: Inverted crank-slider — complete kinematic analysis (20 marks)

Positional parameters and loop equations