Non-inertial frames, fictitious forces, tides, and constraints

A cart can move backward relative to an accelerating train

Imagine a 2 kg cart on an ideal horizontal track inside a train. Choose the cart as the system. A device applies a constant horizontal force of +4 N to it; rightward is positive. Ignore friction, and let the vertical support and weight cancel. These are exact supplied model values, not measurements.

Assume an inertial laboratory frame: in this Newtonian model, a body with no net physical force has constant velocity there. Velocity describes the rate and direction of position change; acceleration describes the rate of velocity change. For positive constant mass m, the net physical force satisfies F = ma. Thus the cart’s laboratory acceleration is (+4 N)/(2 kg) = +2 m/s². One newton is one kg m/s².

Now let the train’s origin accelerate at A₀ = +3 m/s² relative to that laboratory. Its axes stay parallel to the laboratory axes. This accelerating train frame is non-inertial; rotating axes also produce a non-inertial description. The cart’s relative acceleration is a′ = a − A₀ = −1 m/s². The train gains rightward velocity faster than the cart does. A negative relative acceleration does not mean that the applied force has reversed.

Subtract the moving origin before changing the force equation

Let x be the cart’s laboratory coordinate, s the train origin’s laboratory coordinate, and x′ the cart’s train coordinate. Use the same time t in both descriptions: x′ = x − s. For 0 ≤ t ≤ 2 s, supply x(t) = 1 m + (1 m/s²)t² and s(t) = (1.5 m/s²)t². Subtraction gives x′(t) = 1 m − (0.5 m/s²)t².

The corresponding velocities are v = (2 m/s²)t, v₀ = (3 m/s²)t and v′ = −(1 m/s²)t. Their constant rates of change are +2, +3 and −1 m/s². At t = 0 all three velocities are zero; afterwards the cart moves right in the laboratory and left relative to the train. The position table records coordinates, not distances travelled.

Multiplying a′ = a − A₀ by m gives ma′ = F − mA₀. The added term F_frame = −mA₀ is called a fictitious, inertial or pseudo force. Here it is −6 N, so the effective sum is +4 N − 6 N = −2 N. It accounts for the accelerating coordinates; it is not another physical interaction on the cart. With parallel axes and A₀ = 0, even a uniformly moving origin adds no such term.

Rest in rotating coordinates can require an inward physical force

Consider a separate 2 kg particle moving uniformly around a fixed inertial origin in a horizontal plane, at radius r = 2 m. An ideal taut, massless string connects it to the origin; vertical forces cancel, and there are no other in-plane physical forces. Supply counterclockwise angular velocity ω = +1 rad/s for 0 ≤ t ≤ 2 s, viewed from the positive axis perpendicular to the plane. Choose coordinate axes rotating at that same rate, so this particle has fixed rotating coordinates.

A radian is the angle unit. To use the numerical radian rate in the following formulas, define κ = ω/(1 rad), so here κ = +1 s⁻¹. The inertial tangential speed is κr = 2 m/s. Changing direction around the circle requires inward acceleration of magnitude κ²r = 2 m/s², even though speed is constant. Taking outward as positive, its acceleration component is −2 m/s² and the string’s force component is −mκ²r = −4 N.

In this constant-rate rotating frame, the centrifugal term has outward component +mκ²r = +4 N. It balances the inward string force in the relative equation: −4 N + 4 N = 0. The relative velocity is zero, so the Coriolis term is zero. The inertial physical force is still inward. “Centripetal” describes that inward resultant; it is not an extra force to add to the string force.

Coriolis depends on velocity relative to the rotating axes

Now use a different instantaneous state, without the fixed-radius string constraint. Keep the same fixed origin, positive κ = 1 s⁻¹, mass 2 kg and instantaneous radius 2 m. Let the particle’s relative velocity be 0.5 m/s directly outward, with zero real in-plane force. At this instant let the rotating Cartesian x′ axis point outward and its y′ axis point along the counterclockwise tangent. These are the two directions in which we report relative acceleration components.

For this selected radial relative velocity, the centrifugal x′ component is +mκ²r = +4 N. The Coriolis y′ component is −2mκv′_x = −2 × 2 kg × 1 s⁻¹ × 0.5 m/s = −2 N. It points opposite the positive tangent. Dividing the effective components by m gives (a′_x, a′_y) = (+2, −1) m/s². These are instantaneous components, not a supplied future path or the accelerations of the string-held particle.

Reverse the relative radial velocity and the Coriolis component reverses; the centrifugal component at the same radius and rotation rate does not. Zero relative velocity makes Coriolis zero, but does not generally make centrifugal zero. These formulas use a fixed origin and constant angular velocity. An accelerating origin adds a translation term, and changing the angular velocity vector requires an Euler term; the selected cases do not include those changes.

Tides remain after a common acceleration is subtracted

Use a separate snapshot of an exact prescribed Newtonian central field. On an outward-positive radial line, let g(r) = −μ/r², where μ = 36 m³/s² and r > 0. Compare test particles at 2, 3 and 6 m from the fixed external source. Their influence on that source is ignored. The negative sign means all three gravitational accelerations point toward the source.

Choose a nonrotating, freely falling reference origin at 3 m at this instant. Its acceleration is A₀ = g(3 m) = −4 m/s². The relative acceleration at another location is g(r) − A₀. The table gives −5, 0 and +3 m/s². A positive far-side residual means that location falls inward less rapidly than the reference; gravity there has not become repulsive.

The far-minus-near acceleration is (−1) − (−9) = 8 m/s² in the inertial description, and (+3) − (−5) = 8 m/s² in the falling description. Subtracting the same A₀ cannot change that difference. Such spatial differences are tidal accelerations. These values describe one instant: the radii are not fixed future positions, and the accelerations need not stay constant as the particles move. This model does not predict an ocean height or replace a relativistic treatment of gravity.

Compare force components and tidal accelerations on separate scales

The diagram compares the train cart and the string-held rotating particle above, followed by the tidal snapshot. Solid arrows with filled triangular heads represent physical forces; dashed arrows with open triangular heads represent frame terms. Both force panels use the same newton scale. Their sums are calculations, not additional forces drawn on either body.

The lower panel instead shows relative acceleration, in m/s², with dotted lines and diamond heads. A square marks the zero residual at 3 m. Each row starts at the same component zero; vertical row spacing does not show where bodies are located. Read the signed components from each named axis rather than comparing force and acceleration arrow lengths.

A constraint restricts motion and can supply a physical force

A constraint is a stated restriction on allowed motion, such as the taut string’s fixed radius. It does not by itself specify every aspect of motion: fixing radius does not fix the particle’s angle or angular speed. Our rotating example also supplied uniform angular motion. Newton’s equation then gives the tension needed to maintain that motion: 4 N inward.

Tension is a physical pull along a stretched flexible string, not a fictitious term. A string cannot supply a compressive push. A proposed solution requiring that would contradict the taut-string model and require a different constraint or a slack-string treatment. A track’s normal support is likewise a physical contact force; its value must be consistent with the permitted motion and the other forces.

If the string is removed at an instant in the supplied circular-motion case, and no other real in-plane force acts afterwards, the inertial acceleration becomes zero. The particle then continues along the release tangent with constant speed 2 m/s, while those conditions hold. Its rotating coordinates need not follow a straight line. The radial-moving Coriolis example above already omitted the string; it must not inherit the held particle’s tension.

Declare the frame, the interactions and the constraints together

For a new problem, choose the body or system, name an assumed inertial reference and specify the moving origin and axes. List physical interactions with their sources. Add only the frame terms appropriate to the chosen translation and rotation, and keep the constraint and initial-state assumptions visible. An effective balance in non-inertial coordinates need not mean zero physical force in an inertial frame.

Here mass is positive and constant, the supplied motions are nonrelativistic, and the geometry and fields are idealized Newtonian models. Real bearings, strings, gravity fields and tracks can require additional interactions or more detailed models. The coordinate calculations do not establish those idealizations as measured facts.

The next subject, Constraints and generalized coordinates, develops ways to describe motion with fewer coordinates when restrictions are known. The distinction already matters here: changing coordinates, adding a frame term and imposing a physical constraint are different steps.

Worked examples

Try it yourself

1. Keep the 2 kg cart and +4 N physical force, but choose a parallel train frame with A₀ = +2 m/s². Find a′ and the effective sum. Is the physical force zero?

   Show the answer and reasoning

   a′ = 0 m/s² and the effective sum is 0 N. The physical force remains +4 N.

   The inertial acceleration is 4/2 = 2 m/s². Subtract A₀ = 2 m/s²; the frame term is −4 N and cancels +4 N in the relative equation. The inertial acceleration is still nonzero.

2. In a different stipulated train model, take m = 4 kg, F = +4 N and A₀ = +3 m/s² with parallel axes. Find inertial and relative accelerations, the frame term and the effective sum.

   Show the answer and reasoning

   a = +1 m/s², a′ = −2 m/s², F_frame = −12 N and the effective sum is −8 N.

   First divide the physical force by mass: 4/4 = 1. Then subtract 3 to obtain −2 m/s². The frame force is −4 × 3 = −12 N, so 4 − 12 = −8 N = (4 kg)(−2 m/s²).

3. Supply a separate uniform circular motion with m = 1 kg, r = 3 m and ω = +2 rad/s in the same horizontal, ideal-string setting. In matched co-rotating coordinates, find the required tension magnitude, centrifugal magnitude, inertial inward acceleration magnitude and tangential speed.

   Show the answer and reasoning

   Tension and centrifugal magnitudes are each 12 N; inertial inward acceleration magnitude is 12 m/s² and tangential speed is 6 m/s.

   κ = ω/(1 rad) = 2 s⁻¹. Hence κ²r = 4 × 3 = 12 m/s², mκ²r = 12 N and κr = 6 m/s. The physical tension points inward, while the centrifugal term points outward; relative velocity and Coriolis are zero.

4. In the separate instantaneous rotating state, reverse v′_x from +0.5 to −0.5 m/s, leaving m = 2 kg, r = 2 m and κ = +1 s⁻¹ unchanged. What are the Coriolis y′ component and centrifugal x′ component?

   Show the answer and reasoning

   Coriolis is +2 N along the positive tangent; centrifugal is still +4 N outward.

   The selected-state Coriolis component is −2mκv′_x = −2 × 2 × 1 × (−0.5) = +2 N. The centrifugal component depends on m, κ and r, which did not change. No string-held constraint is imposed.

5. For the same tidal snapshot, use a different nonrotating origin with supplied acceleration A₀ = +1 m/s². Find the relative accelerations at 2 and 6 m and their far-minus-near difference. Is this origin freely falling at 3 m?

   Show the answer and reasoning

   The residuals are −10 and −2 m/s²; their difference is 8 m/s². This origin is not freely falling at 3 m in the supplied field.

   Subtract +1 from −9 and −1. The common subtraction cancels from the difference: −2 − (−10) = 8 m/s². At 3 m the field is −4 m/s², which differs from the supplied +1 m/s² origin acceleration.

6. Remove the string from the supplied 2 kg, r = 2 m circular-motion case at an instant, and assume zero real in-plane force thereafter. State the inertial path, speed and acceleration while those conditions hold. Must the rotating-coordinate path be straight?

   Show the answer and reasoning

   The inertial path is a straight release tangent, speed is 2 m/s and acceleration is zero. The rotating-coordinate path need not be straight.

   Removing the physical constraint removes its inward tension. Zero net real in-plane force preserves the instantaneous inertial velocity. The former tangent speed is κr = 2 m/s. Changing rotating axes can still change the coordinates of that straight inertial path.

What to carry forward

• For parallel translating axes, a′ = a − A₀ and ma′ = F − mA₀.
• A frame term accounts for chosen coordinates; tension and contact forces describe physical interactions.
• Centrifugal and Coriolis terms have different dependencies; rest in rotating coordinates need not be inertial rest.
• A common acceleration subtraction cannot remove spatial acceleration differences.
• Specify the physical constraint, motion conditions, quantity kinds and model domain before interpreting a balance.