Slider-crank worked example

Why this one

One question that uses every Week 1–3 skill: geometry with Pythagoras, position vectors, cross products, relative velocity, relative acceleration (tangential and normal parts), a sliding constraint, and reading the signs of the answers. It’s the same kind of mechanism as the A2 rig, and it’s set up like a test question. Try each part before opening the answer.

Theory: Relative motion analysis · Absolute motion analysis · Rigid body rotation

The question

A slider-crank has crank OA = 0.2 m pinned at O and connecting rod AB = 0.5 m. Slider B moves along a horizontal guide level with O. At the instant shown, the crank is at 60° above the horizontal and is turning anticlockwise at ω = 8 rad/s, and speeding up at α = 30 rad/s².

Find:

• (a) The position of B, and the position vector ${\mathbf{r}}_{B/A}$
• (b) The velocity of A
• (c) The angular velocity of rod AB and the velocity of slider B
• (d) The acceleration of A
• (e) The angular acceleration of rod AB and the acceleration of slider B
• (f) Is the slider speeding up or slowing down? Describe the motion of each link.

What's different from the lecture example

Here the crank is speeding up, so A has a tangential acceleration as well as a normal one. The slider moves horizontally, so the constraint kills the j parts instead of the i parts. And the answers come out negative, so you have to interpret the signs.

Set-up

Origin at O, i right, j up, k out of the page, anticlockwise positive. Cross products you need (not on the formula sheet): $\mathbf{k}\times \mathbf{i}=\mathbf{j}\mathbf{k}\times \mathbf{j}=-\mathbf{i}\Rightarrow \omega \mathbf{k}\times (x\mathbf{i}+y\mathbf{j})=-\omega y\mathbf{i}+\omega x\mathbf{j}$

(a) Geometry

A: ${\mathbf{r}}_A=0.2\cos 60°\mathbf{i}+0.2\sin 60°\mathbf{j}=0.1\mathbf{i}+0.1732\mathbf{j}$

B is on the line through O, so it’s 0.1732 m below A. Pythagoras on the rod: $\text{horizontal gap}=\sqrt{0.5^2-0.1732^2}=\sqrt{0.22}=0.4690\text{m}$ So $x_B=0.1+0.4690=0.569\mathbf{m}$ from O, and ${\mathbf{r}}_{B/A}=0.4690\mathbf{i}-0.1732\mathbf{j}$ The rod sits at ${\tan }^{-1}(0.1732/0.4690)=20.3°$ below horizontal.

(b) Velocity of A

${\mathbf{v}}_A={\mathbf{\omega }}_{OA}\times {\mathbf{r}}_A=8\mathbf{k}\times (0.1\mathbf{i}+0.1732\mathbf{j})=-1.386\mathbf{i}+0.8\mathbf{j}\text{m/s}$ Size $=8\times 0.2=1.6$ m/s, perpendicular to OA (up and to the left), as it should be for an anticlockwise crank.

(c) Angular velocity of AB, velocity of B

${\mathbf{v}}_B={\mathbf{v}}_A+{\mathbf{\omega }}_{AB}\times {\mathbf{r}}_{B/A}$ B can only move along the guide, so ${\mathbf{v}}_B=v_B\mathbf{i}$ (no j part). ${\omega }_{AB}\mathbf{k}\times (0.4690\mathbf{i}-0.1732\mathbf{j})=0.1732{\omega }_{AB}\mathbf{i}+0.4690{\omega }_{AB}\mathbf{j}$ $v_B\mathbf{i}=(-1.386+0.1732{\omega }_{AB})\mathbf{i}+(0.8+0.4690{\omega }_{AB})\mathbf{j}$ j (one unknown, so do it first): $0=0.8+0.4690{\omega }_{AB}$ → ${\omega }_{AB}=-1.706rad/s$, i.e. 1.71 rad/s clockwise

i: $v_B=-1.386+0.1732(-1.706)=-1.681m/s$, i.e. 1.68 m/s to the left

(d) Acceleration of A

The crank is speeding up, so A has both parts: ${\mathbf{a}}_A={\mathbf{\alpha }}_{OA}\times {\mathbf{r}}_A-{\omega }_{OA}^2{\mathbf{r}}_A$ $=30\mathbf{k}\times (0.1\mathbf{i}+0.1732\mathbf{j})-64(0.1\mathbf{i}+0.1732\mathbf{j})$ $=(-5.196\mathbf{i}+3\mathbf{j})-(6.4\mathbf{i}+11.085\mathbf{j})=-11.596\mathbf{i}-8.085\mathbf{j}{\text{m/s}}^2$ Check: tangential $=30\times 0.2=6$, normal $=64\times 0.2=12.8$, total $=\sqrt{6^2+12.8^2}=14.1$ m/s², and $\sqrt{11.596^2+8.085^2}=14.1$ ✓

(e) Angular acceleration of AB, acceleration of B

${\mathbf{a}}_B={\mathbf{a}}_A+{\mathbf{\alpha }}_{AB}\times {\mathbf{r}}_{B/A}-{\omega }_{AB}^2{\mathbf{r}}_{B/A},{\mathbf{a}}_B=a_B\mathbf{i}$ Tangential: ${\alpha }_{AB}\mathbf{k}\times (0.4690\mathbf{i}-0.1732\mathbf{j})=0.1732{\alpha }_{AB}\mathbf{i}+0.4690{\alpha }_{AB}\mathbf{j}$

Normal: $-(1.706{)}^2(0.4690\mathbf{i}-0.1732\mathbf{j})=-2.909(0.4690\mathbf{i}-0.1732\mathbf{j})=-1.364\mathbf{i}+0.504\mathbf{j}$

j: $0=-8.085+0.4690{\alpha }_{AB}+0.504$ → ${\alpha }_{AB}=16.2rad/{\mathbf{s}}^2$ anticlockwise

i: $a_B=-11.596+0.1732(16.16)-1.364=-10.2m/{\mathbf{s}}^2$, i.e. 10.2 m/s² to the left

Note the square: ${\omega }_{AB}^2$ is positive even though ${\omega }_{AB}$ is negative.

(f) Interpreting the answers

• $v_B$ and $a_B$ are both negative (both to the left), so the slider is speeding up as it moves left. Same sign means speeding up; opposite signs means slowing down.
• ${\omega }_{AB}$ is clockwise but ${\alpha }_{AB}$ is anticlockwise, so the rod’s rotation is slowing down at this instant.
• Crank OA: fixed-axis rotation about O (revolute joint). Rod AB: general plane motion (translation plus rotation). Slider B: rectilinear translation along a prismatic joint. Pins at O, A and B plus a prismatic joint between the slider and its guide make a four-link chain (ground, crank, rod, slider) with one degree of freedom: set the crank angle and everything else is fixed.

Answers

	Answer
$x_B$	0.569 m from O
${\mathbf{v}}_A$	$-1.386\mathbf{i}+0.8\mathbf{j}$ m/s (1.6 m/s)
${\omega }_{AB}$	1.71 rad/s clockwise
$v_B$	1.68 m/s left
${\mathbf{a}}_A$	$-11.60\mathbf{i}-8.09\mathbf{j}$ m/s² (14.1 m/s²)
${\alpha }_{AB}$	16.2 rad/s² anticlockwise
$a_B$	10.2 m/s² left (speeding up)

Common mistakes

1. Forgetting A’s tangential acceleration. It’s only zero when the crank turns at a constant rate.
2. Using the wrong constraint. A horizontal slider has no j velocity or acceleration, so the j equation is the one with a single unknown.
3. Sign of ${\omega }^2$. Square the number, including its minus sign: $(-1.706{)}^2=+2.909$.
4. Getting ${\mathbf{r}}_{B/A}$ the wrong way round. It runs from A to B, so here the j part is negative because B is lower than A.
5. Reporting a negative as a size. Say “1.68 m/s to the left” or “−1.68 i m/s”, not just “−1.68”.

Check yourself

1. At what crank angle would the rod have zero angular velocity for an instant?

When the crank points straight up or down (90° or 270°). Then ${\mathbf{v}}_A$ is horizontal, the same direction as the slider, so the rod just translates and ${\omega }_{AB}=0$.

2. If the crank turned at a constant 8 rad/s instead, which terms in part (d) disappear?

The tangential part, ${\mathbf{\alpha }}_{OA}\times {\mathbf{r}}_A$. Then ${\mathbf{a}}_A=-64{\mathbf{r}}_A=-6.4\mathbf{i}-11.09\mathbf{j}$ m/s², pointing straight at O.