Coriolis component of acceleration

Why it matters

Whenever a block slides along a link that is itself rotating, as in a shaper's slotted lever, a Whitworth quick-return drive, an oscillating-cylinder engine or a hydraulic actuator on a swinging arm, the sliding block has an extra acceleration term that a plain relative-acceleration analysis misses. Leave it out and the inertia forces, and so the bearing and slider loads, come out wrong. It is a favourite exam item because the magnitude is easy but the direction is often got wrong.

Key ideas

When it appears. Take a link 2 rotating with angular velocity ω, and a point B (on a slider, link 3) that moves along link 2 with sliding velocity v relative to it. Let B′ be the point of link 2 that coincides with B at that instant. The absolute acceleration of B is:

acceleration of B′ (as a point on link 2) + acceleration of B relative to link 2 along the slot + Coriolis component.

The Coriolis component exists only when both conditions hold: the guide (link 2) rotates (ω ≠ 0) and there is sliding along it (v ≠ 0).

Where it does not appear. In an ordinary engine slider-crank the piston slides in a fixed cylinder, so the guide does not rotate and there is no Coriolis component. A link with only revolute pairs has none either.

Physical origin. It comes from two equal effects, each of magnitude ω·v:

1. As the slider moves outward, its distance from the centre grows, so the tangential velocity ω·r it shares with the link increases.
2. As the link turns, the direction of the sliding velocity v rotates, which changes the velocity vector perpendicular to the slot. The two add, giving 2ωv.

Direction rule. The Coriolis component is perpendicular to the slot (the sliding direction). To find its sense, take the sliding velocity vector of the slider relative to the link and rotate it by 90° in the same sense as ω. In vector form it is 2ω × v, so if ω and v are perpendicular (plane mechanisms) the magnitude is simply 2ωv.

Reference frames, carefully. v is the velocity relative to the rotating link, not the absolute velocity of the slider, and ω is the angular velocity of the link that carries the slot (for example the slotted lever), not the crank.

Polar form for a slider on a rotating arm. If r is the slider's distance from the arm's fixed pivot, the radial and transverse accelerations are r̈ − r·ω² and r·α + 2·ṙ·ω. The 2ṙω term is the Coriolis component. This form is the fastest way to solve problems where an arm rotates about a fixed pivot.

Analysis procedure in a slotted-lever mechanism.

1. Velocity analysis: split the crank-pin velocity into a part perpendicular to the lever (gives ω of the lever) and a part along the lever (the sliding velocity v).
2. Coriolis component: 2·ω_lever·v, perpendicular to the lever.
3. Build the acceleration polygon with the centripetal and tangential components of the lever point, the sliding acceleration along the lever and the Coriolis component.

Formulas

a_cor = 2·ω·v

• a_cor: Coriolis component of acceleration (m/s²); ω: angular velocity of the link carrying the slot or guide (rad/s); v: velocity of the slider relative to that link (m/s). Plane mechanisms (ω perpendicular to v).

a_cor = 2·ω × v (vector form)

• Magnitude 2ωv·sin β, where β is the angle between ω and v; zero if v is parallel to the rotation axis.

a_r = r̈ − r·ω² and a_θ = r·α + 2·ṙ·ω

• a_r: radial acceleration, positive outward (m/s²); a_θ: transverse acceleration, positive in the sense of ω (m/s²); r: slider distance from pivot (m); ṙ, r̈: sliding velocity (m/s) and sliding acceleration (m/s²) along the arm; ω, α: angular velocity (rad/s) and angular acceleration (rad/s²) of the arm.

ω_lever = v_⊥ / PA and v = v_∥

• v_⊥, v_∥: components of the crank-pin velocity perpendicular to and along the slotted lever (m/s); PA: distance from the lever pivot P to the crank pin A (m).

Worked examples

Example 1 (standard: slotted lever). In a crank and slotted-lever mechanism the lever pivot P is 300 mm vertically below the crank centre O. The crank OA = 150 mm turns anticlockwise at 10 rad/s. Find the Coriolis component when the crank is horizontal (A to the right of O).

1. Take P at the origin: O = (0, 300) mm and A = (150, 300) mm. PA = √(150² + 300²) = 335.4 mm.
2. Crank-pin velocity: v_A = ω·OA = 10 × 0.15 = 1.5 m/s, perpendicular to OA, so straight up.
3. Unit vector along PA = (150, 300) / 335.4 = (0.4472, 0.8944).
4. Sliding velocity along the lever: v = 1.5 × 0.8944 = 1.342 m/s (outward, away from P).
5. Velocity perpendicular to the lever: v_⊥ = 1.5 × 0.4472 = 0.671 m/s.
6. Lever angular velocity: ω_lever = 0.671 / 0.3354 = 2.00 rad/s, anticlockwise.
7. Coriolis component: a_cor = 2 × 2.00 × 1.342 = 5.37 m/s².
8. Direction: rotate the outward sliding velocity by 90° anticlockwise: perpendicular to the lever, towards the side the lever is turning.

Note that using the crank speed (10 rad/s) instead of the lever speed would give 26.8 m/s², five times too large.

Example 2 (GATE level: slider on a rotating arm). An arm rotates about a fixed vertical axis at a constant 6 rad/s. A collar on it is 0.5 m from the axis, moving outward at 4 m/s relative to the arm and accelerating outward at 2 m/s² relative to the arm. Find the magnitude of the collar's absolute acceleration.

1. Radial: a_r = r̈ − r·ω² = 2 − 0.5 × 36 = 2 − 18 = −16 m/s² (16 m/s² towards the axis).
2. Transverse: a_θ = r·α + 2·ṙ·ω = 0.5 × 0 + 2 × 4 × 6 = 48 m/s² (Coriolis only, since α = 0).
3. |a| = √(16² + 48²) = √2560 = 50.6 m/s².
4. The Coriolis term (48 m/s²) is three times the centripetal term here; ignoring it would be badly wrong.

Example 3 (quick). A slider moves at 2.5 m/s along a link rotating at 8 rad/s. a_cor = 2 × 8 × 2.5 = 40 m/s².

Common mistakes

• Forgetting the factor 2.
• Using the absolute velocity of the slider instead of its velocity relative to the rotating guide.
• Using the crank's angular velocity when the slot is on a different link (the slotted lever or the oscillating cylinder).
• Adding a Coriolis term to an engine slider-crank whose cylinder is fixed. No rotating guide, no Coriolis.
• Getting the sense wrong: rotate the relative sliding velocity 90° in the sense of the guide's ω, not of the crank's.
• Adding the Coriolis and centripetal components as scalars. They are perpendicular.

For GATE ME

Expect one-line magnitude questions (2ωv), direction questions with a sketch of a slotted link, identification of which mechanisms have a Coriolis component, and multi-step problems in quick-return or oscillating-cylinder mechanisms where you first get the lever's angular velocity and sliding velocity from a velocity triangle. Practise the slotted lever at a few crank angles until splitting the crank-pin velocity into along-lever and across-lever parts is automatic.

Quick check

1. Does an engine piston in a fixed cylinder have a Coriolis component?
2. Magnitude of the Coriolis component for v = 3 m/s on a link at 5 rad/s?
3. What happens to a_cor if both ω and v double?
4. How do you find the direction of the Coriolis component?
5. Name two mechanisms where it must be included.

Answers: 1. No, the guide does not rotate. 2. 30 m/s². 3. It becomes four times larger. 4. Rotate the sliding velocity (relative to the link) by 90° in the sense of the link's ω. 5. Crank and slotted-lever quick-return mechanism; oscillating-cylinder engine (also Whitworth mechanism, rotary engine).