Absolute motion analysis

The idea

For a mechanism like a slider-crank, write one equation for the position of the part you care about in terms of the input angle. Then differentiate it with respect to time once for velocity and again for acceleration. It’s called “absolute” because every position is measured from a fixed origin.

The method

1. Geometry: write the position (e.g. the slider’s $x$) in terms of the crank angle $\theta$, using trig or the cosine rule.
2. Differentiate with respect to time → velocity. Every $\theta$ term picks up a $\dot{\theta }=\omega$.
3. Differentiate again (product rule) → acceleration, with $\ddot{\theta }=\alpha$.
4. Substitute the values at the instant asked for, and check the signs make sense physically.

Worked example: a Scotch yoke

A crank of radius $r=0.1$ m drives a slotted yoke, whose position is $x=r\cos \theta$. The crank turns at a constant 20 rad/s. Find the yoke’s velocity and acceleration at $\theta =30°$.

Answer

$x=0.1\cos \theta$ $\dot{x}=-0.1\sin \theta \dot{\theta }$ → at 30°: $-0.1\times 0.5\times 20=$ −1.0 m/s (moving towards −x) $\ddot{x}=-0.1\cos \theta {\dot{\theta }}^2-0.1\sin \theta \ddot{\theta }$. Constant ω means $\ddot{\theta }=0$: $\ddot{x}=-0.1\times 0.866\times 400=$ −34.6 m/s²

The yoke moves in simple harmonic motion, which is why this mechanism is used in some pumps and shakers.

When the geometry is a triangle

For a hydraulic cylinder pushing a link, the cylinder length $s$ and the link angle are tied by the cosine rule: $s^2=a^2+b^2-2ab\cos \theta \Rightarrow 2s\dot{s}=2ab\sin \theta \dot{\theta }$ Given the cylinder’s extension rate $\dot{s}$, solve for $\dot{\theta }$. Differentiate again for $\ddot{\theta }$, remembering that a constant rate means $\ddot{s}=0$, not $\dot{s}=0$.

Common mistakes

• Plugging in numbers before differentiating.
• Losing the second term of the product rule on the second derivative.
• Sign of the velocity: decide which way is positive and check your answer agrees with the picture.

Check yourself

1. Same Scotch yoke, but at θ = 90°. Velocity and acceleration?

$\dot{x}=-0.1\times 1\times 20=-2$ m/s (its fastest), $\ddot{x}=0$ (cos 90° = 0).